Download Math Resource

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the World
Scholar’s Cup ®
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Daniel Berdichevsky
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EDITOR & A L PAC A- I N - C H I EF
YO
Al geb ra
t h roug h
C a l cul u s
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Craig Chu
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RESOURCE
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MATH
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2010
2011
MATHEMATICS RESOURCE | 1
Mathematics Resource
2010: the great depression
Table of Contents
Preface to the Mathematics Resource .............................................................................. 3
I. General Mathematics.................................................................................................... 4
Fractions, Percents, and Decimals ...................................................................................................... 4
Permutations and Combinations ........................................................................................................ 5
Event Probabilities ............................................................................................................................. 9
II. Algebra ...................................................................................................................... 11
The Function ................................................................................................................................... 11
Function Properties, Composition, And Inversion ........................................................................... 12
Linear Equations and Functions....................................................................................................... 16
Quadratic Equations in One Variable .............................................................................................. 23
Polynomial Equations and Functions ............................................................................................... 32
Things Unequal ............................................................................................................................... 39
Rational Expressions and their Graphs ............................................................................................. 44
Exponents and Logarithms ............................................................................................................... 51
Exponential And Logarithmic Functions, Plus More On Inverses .................................................... 56
Graphical Analysis of Functions—Compositions and Modifications ................................................ 61
Sequences and Series ........................................................................................................................ 65
The Whirled Series .......................................................................................................................... 70
Coordinate Geometry: Circles.......................................................................................................... 71
Coordinate Geometry: Ellipses ......................................................................................................... 73
Coordinate Geometry: Hyperbolas................................................................................................... 74
Coordinate Geometry: Parabolas ...................................................................................................... 75
III. Geometry (Optional) ............................................................................................... 76
Introduction to Lines, Planes, and Angles ........................................................................................ 76
More on Lines and Rays, and a Bit on Planes ................................................................................... 77
Even More on Lines, But First A Bit on Angles ................................................................................ 78
An Introduction to the Numerical Perspective of Triangles.............................................................. 83
A Brief Continuation of the Numerical Perspective of Triangles ...................................................... 87
An Introduction to the Abstract Concepts of Triangles .................................................................... 88
Revisiting the Numerical Perspective on Triangles ........................................................................... 91
A Plethora of Parallelograms, and They Brought Their Friends........................................................ 95
Circular Logic .................................................................................................................................. 96
The Angle-Secant Invasion of Normalcy .......................................................................................... 98
Princess Cirque and her Maidens Chordelia, Secantia and Tangentia............................................. 106
MATHEMATICS RESOURCE | 2
Additional Polygons ....................................................................................................................... 111
The Third Dimension .................................................................................................................... 113
More to the Third Dimension ....................................................................................................... 119
IV. Trigonometry ........................................................................................................ 121
Trigonometry Basics ...................................................................................................................... 121
Trigonometric Functions ............................................................................................................... 123
Trigonometric Functions and Identities ......................................................................................... 128
Radians .......................................................................................................................................... 138
The Laws of Sine and Cosine ......................................................................................................... 139
Appendix A: Completing the Square ........................................................................... 145
Appendix B. More Factoring ....................................................................................... 148
Appendix C. Vectors (Trigonometry) .......................................................................... 151
F. Even and Odd Functions (Algebra) ......................................................................... 158
About the Author ........................................................................................................ 160
by
Craig Chu
Caltech B.S.
For Mr. Maung and the people of Burma.
DemiDec and Scholar’s Cup are registered trademarks of the DemiDec Corporation. Academic Decathlon and USAD are registered trademarks of the United States
Academic Decathlon Association. DemiDec is not officially affiliated with the United States Academic Decathlon Association.
MATHEMATICS RESOURCE | 3
Preface to the
Mathematics Resource
You probably don’t need to be reading this resource. Math,
after all, is universal; most of the topics covered in the World
Scholar’s Cup and Academic Decathlon you’ve probably already
encountered in your classes or could read about in textbooks you already have.
Still, I’m glad you’re here. I enjoy sharing the mathematical experience with students interested in it, and I like to
do it with examples—so most of this resource consists of example problems worked out step by step.
Not every topic covered in this resource will come up on your tests this year—as the Academic Decathlon has
greatly simplified and shortened the curriculum for 2010-2011. But, although I cut the Calculus chapter entirely, I
left the other “extraneous” topics in this guide intact—because learning a little more math won’t hurt you, and can
help you put what you’ll be tested on in its larger context.
Please note, too, that math, unlike the other World Scholar’s Cup subjects, lacks debate topics. Especially at the
most fundamental levels, math is less about judgment and persuasion that it is about precision and preparation.
A couple final updates: the Academic Decathlon now permits programmable calculators to be used without
clearing their memories. And the World Scholar’s Cup replaced math for 2010-2011 with a special topic, the
Modern Metropolis. So if you’re a World Scholar, put this way until at least 2012.
Good luck!
Craig Chu
Price Waterhouse 1
Caltech '04
1
This book actually goes to press on my final day at PriceWaterhouse.
MATHEMATICS RESOURCE | 4
I. General Mathematics
We start not at the very beginning but off to the side
somewhere: with a study of general mathematics
applications. Algebra, geometry, etc. are all very narrowly
defined, and the odd duck topics that fail to quite fit in any of
them are grouped under “general mathematics.” We assume the
reader is well-versed in the arithmetic of fractions, and begin with percentages and decimals.
Fractions, Percents, and Decimals
1 th
A percentage is a portion of a whole, with each 1 percent representing 100
of the whole. It is represented with the
symbol “%.” To work with percentages as numbers, it is usually more convenient to convert them to decimals. For
our purposes, it suffices to know that conversion of a percentage to a decimal for multiplication purposes involves
dividing by 100. That is, each percentage point should be converted to 0.01. Conversely, converting a decimal
portion of a number to a percentage entails multiplying by 100 instead.
Example:
Convert 3%, 29%, 93%, and 161% to decimals.
Solution:
We take each percentage and divide by 100. This requires moving the decimal point two places leftward,
or simply using a calculator if an even more mindless approach is preferred. The equivalent decimals are
0.03, 0.29, 0.93, and 1.61.
The fourth number brings up an interesting point. People often ask, “Can we have percentages over 100%?” If
100% represents a decimal of 1.00, then we realize that every number over 1 represents something over 100%! So
indeed, we can have percentages over 100%; however, it remains on a case-by-case basis to determine whether this
gives a nonsensical answer. For example, to say that Stephen ate 125% of a pie means he ate more than one whole
pie; he ate ¼ of an additional pie! If, however, I am doing a science experiment and find that a procedure produces
110% of the energy that was input, I have made a mathematical error. 2
The most common applications of fractions, decimals, and percents often involve sales and discounts. They will
almost always give part of the whole and then ask for a calculation involving finding the whole or a different part of
the whole. These can usually be accomplished by setting up an algebraic equation and solving. 3
Examples:
1) A $950 computer is on sale for 20% off. What is its new price?
2) A shirt that was discounted 35% now sells for $17.68. What was its original price?
3) A refrigerator is marked down with a 20% discount. After some time, the retailer increases this new
price by 20%. What percentage of its original price is its final new price?
4) A state has a holiday on which consumers will not have to pay sales tax. If sales tax is ordinarily
8.25%, what percentage discount is this equivalent to?
2
Energy must come from somewhere; I cannot receive more than I put in. There is either a mathematical mistake here, or a
Nobel Prize in physics.
3
I understand this is only the beginning of the resource, and we haven’t covered algebra yet. I do, however, expect the reader is
familiar with pre-algebra, and the equations hopefully will have nothing more involved than multiplication and division.
MATHEMATICS RESOURCE | 5
Solutions:
1) If the sale is for 20% off, note that it means it is being sold for 80% of its original price.
(original price)(percentage kept of original price) = (new price)
950 × 0.8 = x
760 = x
$760
2) A discount of 35% means that the price is 65% of the original price. We can set up a simple equation:
(original price)(percentage kept of original price) = (new price)
x ⋅ 0.65 = 17.68
x = 27.2
$27.20
3) For simplicity, we can give the refrigerator a hypothetical original price. $100 is always a convenient
number to choose. This means that after its discount, the new price is $80. Then increasing the new
price by 20% means it is 120% of the new price:
(original price)(percentage kept of original price) = (new price)
80 × 1.2 = 96
Since its original price was $100 and its price now is $96, it is only 96% of its original price. Choosing
any possible hypothetical price will give us the exact same answer.
4) We can again take a hypothetical price of $100 for simplicity. When sales tax is included, the new
price will be (100)(1.0825) = 108.25. Without sales tax, a price of only $100 means that it is only
100
= 0.9238 , or 92.38% of its original price. Since 100% comprises the full price, this is equivalent to
108.25
a discount of 7.62%.
Permutations and Combinations
We now encounter the topic of counting. But we have been counting since kindergarten and preschool. What
more could there be to learn now? Unfortunately, the simple one-by-one counting of cars or apples that we can
already do only scratches the surface of what mathematicians call combinatorics.
For example, if nine women run a race, in how many ways can the gold, silver, and bronze medals be awarded? In a
state with a lottery where 6 numbers are picked out of 54, what is the probability of winning the lottery? With 3
different textbooks for first-semester calculus and 2 textbooks for second-semester calculus, how many ways are
there to teach a full-year calculus course? If I have screen names of 212 buddies and my internet messaging
program can only store 200, how many different buddy lists can I generate?
It’s these questions and more that combinatorics seeks to be able to count, all with fairly unimposing formulas. We
first begin with the Multiplication Principle.
Multiplication Principle
If you wish to pick one each of n types of objects, each with a
number a1, a2, …, an of possible choices, then the total
number of ways in which to do so is
a1 × a2 × … × an
MATHEMATICS RESOURCE | 6
Example:
If a man is putting together an outfit comprising a shirt, a pair of pants, and a pair of boxer shorts, how
many outfits can he make? 4 His closet has 4 shirts, 6 pairs of pants, 3 pairs of briefs, and 4 pairs of boxer
shorts.
Solution:
The man’s shirts, pants, and boxers number 4, 6, and 4, respectively. This means the total number of ways
to dress with these is 4 × 6 × 4 = 96. Note that the number of briefs the man owns is irrelevant if he is not
choosing one.
Example:
Suppose Karl has a collection of 20 CDs. If he wishes to fill a small carrying case with 3 of them, how
many ways are there to fill the mentioned case?
Solution:
This time we are not given a1, a2, and a3 explicitly so we must figure out what they are. We have twenty
choices for the first CD put into the case, so a1 = 20. After one is removed, there are only 19 CDs left with
which to fill the next space, giving us a2 = 19. Likewise, the third space gives us a3 = 18. The total number
of ways to fill the CD wallet is then 20 × 19 × 18 = 6840.
The procedure just outlined is one which describes a permutation. A permutation is an arrangement of a collection
of objects; it is an arrangement in which order matters. We calculate a permutation as we did above in the second
example: ordering 3 objects out of 20 involves multiplying 20, 19, and 18 (the first three numbers counting
downward).
Permutations
To order r objects out of n, we calculate nPr, (sometimes written P(n, r).) The
permutation takes the highest r numbers near n and multiplies them, counting
downwards:
n
Pr =
n!
(n − r)!
The exclamation point in this case is the mathematical symbol for the factorial.
Taking the factorial of n is defined as follows.
n! = n (n-1) (n-2) … (2) (1)
Example:
Work the previous example quickly using the formula for a permutation.
Solution:
We are ordering 3 objects out of 20, and this means we want to find 20P3.
20
4
P3 =
20! 20 ⋅ 19 ⋅ 18 ⋅ 17 ⋅ 16 ⋅ ... ⋅ 1
=
= 20 ⋅ 19 ⋅ 18 = 6840
17!
17 ⋅ 16 ⋅ ... ⋅ 1
…even including fashion atrocities of course…
MATHEMATICS RESOURCE | 7
(Note that all of the factors 17 and below cancel out, leaving the same product as above, 20 × 19 × 18.
Again, these are simply the three highest numbers decreasing from 20; that is often much easier to
memorize than a factorial formula.)
Example:
In how many ways can the letters in “math” be arranged?
Solution:
There are 4 letters total. For the first letter in the arrangement, we have the choice of 4 letters. After that,
there are 3 choices left to choose for the second letter, and so on. This means that there are a total of 4 × 3
× 2 × 1 = 24 arrangements of the letters.
These examples apply when we are picking objects out of a larger selection and placing them in an order. But
there’s another type of counting problem in which order does not matter. For example, what if we had considered
the problem three examples ago, with 20 CDs and a wallet that holds 3, with the caveat that now, we don’t care so
much about “how many ways are there to fill the mentioned case” as we do about “how many sets of 3 CDs can fill
the case?” There is a fine and subtle distinction here that is important to understand: in the second scenario, we
don’t care about the order in which we pick the CDs, only which CDs are chosen. The answer to the second
question will be smaller than the answer to the first question by a factor that accounts for repeated sets of CDs in
different orders. Consider the following example.
Examples:
1) A club of 10 people wishes to elect a president, and then elect a vice president of the 9 remaining.
How many ways are there in which to do this?
2) A club of 10 people wishes to elect two co-presidents. The co-presidents’ jobs are identical. How many
ways are there in which to do this?
Solutions:
1) This is a permutation problem. There are 10P2 ways to do this, or 90.
2) This problem gives us a spin on an old problem. Now, the two people elected have positions that are
identical. This means that of the 90 election possibilities solved for in example (1), half of them will be
duplicates, where the same pair of candidates was elected in a different order. There are thus 45 ways
to elect co-presidents out of 10 people.
The generalization of this formula for these combinations is given below, as well as an as-yet undiscussed formula
for the Arrangement Principle, explained afterwards.
Combinations
n
To choose r objects out of n, we calculate nCr, (sometimes written   or
r 
C(n, r)). The formula for the combination C(n, r) is the formula for the
permutation, divided by r factorial:
n
=
Cr
P
n!
=
r!
r! (n − r)!
n r
MATHEMATICS RESOURCE | 8
Arrangement Principle
If we wish to arrange n things, then the total number of arrangements of them
is n!. However, if r of them are indistinguishable, then we divide by r factorial.
This can be done for any number of sets of indistinguishable things:
n!
r1! ⋅ ... ⋅ rm!
Why are these concepts, then, being listed together? You may wonder how they are related to each other. The
answer lies in the fact that they both involve dividing by a factorial in order to reduce duplicate countings. As often
happens, many mathematical concepts are best explained through example rather than through definition. Look
carefully back at example (2) above. We took a combination. That is, when choosing 2 candidates out of 10, if the
positions are indistinguishable, then we simply needed to calculate C(10,2), which was 45. A flurry of examples
follows that should help solidify the concepts of permutations, combinations, and the generalized multiplication
principle.
Examples:
1) Kim is playing with a standard deck of 52 playing cards. If she deals her friend Cecile a standard
random hand of 5 cards, how many different hands are possible for Cecile to have?
2) An Olympic race has 8 runners. In how many ways can the gold, silver, and bronze medals be given
out?
3) An elementary school race has 8 runners. The top 3 runners will receive certificates that are identical.
In how many ways can the certificates be given out?
4) How many arrangements are there of the letters in “math”? ... in “mathematics”? ... in
“onomatopoeia”?
Solutions:
1) Dealing 5 cards randomly out of a 52 card deck is like choosing 5 objects out of a set of 52 where
order does not matter. The answer will be simply 52C5
 52 
=

5
P5 52 ⋅ 51⋅ 50 ⋅ 49 ⋅ 48
=
= 2,598,960
5!
5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
52
2) We wish to find out how many ways in which there are to order 3 racers out of 8. Because we are in a
situation where order matters, we wish to use a permutation:
P = 8 ⋅ 7 ⋅ 6 = 336
8 3
3) Now we hope to only choose 3 runners out of 8. The order in which they finish no longer matters, as
8 8⋅7 ⋅6
all the certificates are identical. The answer will be a combination.   =
=56
 3  3 ⋅ 2 ⋅1
4) For the first part of this question, we are only hoping to order 4 letters, all of which are unique. There
are (4!) arrangements of “math,” or 24. However, arranging the letters in “mathematics” is more
complicated. The letters m, a, and t are all duplicated. We want to divide through by the duplicate
arrangements since these letters are identical. By the arrangement principle, the total number of
11!
arrangements is then 2!2!2!
= 4,989,600 . Arranging the letters “onomatopoeia” obeys the same principle,
except there are FOUR of the letter o and two a. That means its number of arrangements is
12!
.
4!2! = 9,979,200
MATHEMATICS RESOURCE | 9
Event Probabilities
Now that we have learned many of the basics involved in counting, we can proceed to its most frequent
application, that of probability. Loosely defined, the probability of an outcome is the portion of experiments that
will have that outcome true if many many experiments are repeated. 5 There are several ways in order to calculate a
probability. The one that is usually most convenient is presented here.
Probability of Event A
The probability of event A, denoted by P(A), can be calculated as follows:
P(A) =
number of outcomes in which A is true
total possible number of outcomes
Each outcome counted in both the numerator and denominator must be equally likely.
Let’s try to illustrate this formula with an example.
Example:
In rolling a standard six-sided die, what is the probability of rolling a prime number?
Solution:
We know that there are six possible outcomes: 1, 2, 3, 4, 5, and 6, each of which is equally likely. Of these
six outcomes, three of them, 2, 3, and 5, fit the criterion that the number be prime. Thus, the probability
is 36 , or 12 .
Example:
In simultaneously rolling two standard six-sided dice, what is the probability of rolling a total of 9?
Solution:
We cannot just count the outcomes of 2 through 12, as they are not equally likely. In this case, we should
count all of the possibilities that the two dice may cause:
1,1
1,2
1,3
1,4
1,5
1,6
2,1
2,2
2,3
2,4
2,5
2,6
3,1
3,2
3,3
3,4
3,5
3,6
4,1
4,2
4,3
4,4
4,5
4,6
5,1
5,2
5,3
5,4
5,5
5,6
6,1
6,2
6,3
6,4
6,5
6,6
The shaded boxes above indicate the combinations of dice rolls which give a total value of 9. Since there
are 36 equally likely outcomes listed above, the probability of rolling a total of 9 is then 364 = 19 .
5
To truly define “probability” in a rigorous mathematical way would require knowledge often not covered in high school.
Luckily, the loose definitions are pretty intuitive and should serve our purposes.
MATHEMATICS RESOURCE | 10
Example:
What is the probability of a royal flush if dealt a random hand from a standard 52 card deck? What is the
probability of a straight flush (that is not a royal flush)? A straight flush is defined as receiving all 5 cards in
the same suit [a flush], also forming a direct sequence in value [a straight]. A royal flush occurs when a
straight flush has 10, Jack, Queen, King, and Ace for its values.
Solution:
For this problem, we have to exercise our counting skills again. To calculate the probability of an event, we
wish to know how many outcomes exist in which it is true, and we wish to know how many total
outcomes exist.
We already know 52C5 outcomes exist. How many outcomes include is a royal flush? There are exactly 4.
Each suit has only one possible combination that gives rise to a royal flush (10, J, Q, K, A), and there are 4
4
4
1
52 C 5
2598960
649740
suits. The probability of a royal flush is then
= =
.
To calculate the probability of a straight flush, we have to reconsider the numerator in the fraction above.
There are no longer only 4 possibilities for successful outcomes. Each suit has 8 possibilities for a straight
flush, (2, 3, 4, 5, 6) through (9, 10, J, Q, K). We are not counting royal flushes as straight flushes;
otherwise, each suit would instead have 9 possibilities. This means that there are a total of 32 ways to
achieve a straight flush. We divide this by the total number of possible poker hands to find the probability
of a straight flush:
32
32
2
= =
2598960 162435
52 C 5
MATHEMATICS RESOURCE | 11
II. Algebra
Assuming the reader is already acquainted with constants,
variables, equations, and expressions, this resource begins
algebra discussing a very important mathematical construct:
the function. Functions are very important in mathematics and
the sciences, as they often determine the relationship between
parameters in a problem. That is, as one parameter varies, can we
determine how the other one varies? Is the relationship between variables more complicated than
a function relationship? 6
The Function
The standard notation used for functions is that of f(x) notation, read aloud as “f of x.” It should be conceptualized
as follows: f is predetermined by a relationship of x. For any value of x, we can “input” it into the function and
arrive at a single value of f that corresponds to it. Writing x in this fashion, enclosed by parentheses, indicates that
it can change; it is independent and can completely determine the value of the dependent variable. Keep in mind
also that f is a conventionally used letter for functions, but anything is allowable.
Definition
A single-variable function is a mathematical relationship between two
variables such that for every possible value of the independent variable, there
is no more than one possible and correct value of the dependent variable. 7
Examples:
Determine in the following if f, g, h, and y are functions of x.
1) f = x3 – 8x + 1
2) g =
2x + 5
3) 5x + 2y2 – 14 = 0
4) 3x2 – x + 5h = -12
Solutions:
1) We examine to see if this equation fits the criterion to be a function. We want it to be true that for
any one value of x, there is only one correct value of f. Because the equation comes in a form where f is
already explicitly solved for, we can tell this is true. For any value of x, the value of f MUST be x3 – 8x
+ 1, no matter what. There cannot be another value of f for this value of x, and this IS a function.
2) We do the exact same procedure as was done in example (1) just above. We notice, then, that for any
single value of x, the only valid value of g is 2x + 5 . Again, having the dependent variable explicitly
solved for makes this process very easy.
6
7
Or a functional relationship? – Daniel
With any amount of luck, no high school level competition will ever ask its students to consider “multi-variable functions.”
MATHEMATICS RESOURCE | 12
3) Unlike the previous two examples, the dependent variable is no longer solved for. Since it is only
algebra, however, we can simply solve for it ourselves.
5x + 2y2 – 14 = 0
2y2 = -5x + 14
y2 = − 52 x + 7
← collect all the x terms on one side
← begin solving for y
At this point, however, we have to take a minute to think. 8 If we were to have an equation such as “y2
= 4,” we would have two solutions; we would have both y = 2 and y = -2. This means that solving for
y gives us two solutions, y =± − 52 x + 7 . For any value of x, there may be up to two values of y, and
thus y is NOT a function of x.
4) As in (3) above, we again have to solve for the dependent variable to see if the given relation is a
function. We solve for h.
3x2 – x + 5h = -12
5h = -12 +
h =- +
12
5
1
5
x – 3x2
3
5
x- x
2
← isolate h on the left side
← finish solving for h
No matter what value we substitute for x, there can be only one specific value for x. The relation given
here does give h as a function of x.
Function Properties, Composition, And Inversion
What is the purpose of a function? Does of a function have specific properties that make it of special interest to us?
It is fine to simply define it as we have above, but it stands to reason that there might be something more to the big
picture. Indeed there is.
There are many properties of functions that will be expected on the curriculum this year that will likely show up in
some form or another on competition exams. The first major concept arises when we consider one very
fundamental question: what can be placed into a function? For example, suppose we have a function that we have
already decided to designate f(x). Can we place values in for x instead of variables? The answer, as you may already
know, is yes.
We shall choose to note all of the possible combinations of x and f(x) as ordered pairs. The first number (known in
math circles as the abscissa) will be the value c of the independent variable, that of x. In other words, x is a variable,
but we want to examine x = c. The second number (known as the ordinate) will be the function evaluation. This
means that we substitute the constant c for x and evaluate the function accordingly.
Example:
Find the function evaluations below given f(x) = x3 + x .
1) f(3)
2) f(0)
3) f(-1)
8
It’s hard, I know, but it will only be a minute.
MATHEMATICS RESOURCE | 13
Solutions:
As we discussed, we are hoping to find f(c), the function evaluation at c. This means that we simply
substitute c every place we see an x and find the value.
We substitute 3 every place an x occurs:
4) f(3) = 33 +
3 ≈ 28.732
5) f(0) = 03 +
0 =0
6) f(-1) = (-1)3 +
−1
f(-1) is undefined
This means that if we are representing all the combinations of (x, f(x)) possible, we have the two ordered
pairs of (3, 28.732) and (0,0). We can take these two ordered pairs, along with all of the other ones
possible, (1, 2) for another example, along with (2, 9.414), we can create a plot of all of these points. This
graph will have the independent variable values, x = c, along the horizontal axis, and the function
evaluations, f(x) = f(c), along the vertical axis.
Example:
Plot a graph of f(x) = x3 + x . Describe its domain and range, the sets of values possible for its
independent and dependent variables.
Solution:
We already have three of the points desired listed in the example and paragraph above. We can also note
from example (3) that any negative c values will result in not having a real function evaluation.
f(x)
3
2
1
x
-0.5
0
1
2
3
Here we have the appropriate graph plotted. It increases uniformly through its domain, that being the nonnegative
real numbers. We can say the domain is 0 ≤ x < ∞. The same is true of the range; it is the set of nonnegative real
numbers. Similarly, we could write 0 ≤ f(x) < ∞. In what is referred to as set notation 9, these inequalities could also
be written as x ∈ [0, ∞) and f(x) ∈ [0, ∞).
9
These are read as follows: “x is an element of the set closed at 0 and open at infinity. f(x) is an element of the set closed at 0
and open at infinity.” Similarly, x ∈ [-3, 4] would be read aloud as “x is an element of the closed set from -3 to 4.”
MATHEMATICS RESOURCE | 14
Using this notation, a square bracket refers to an endpoint that is included in the inequality, and a parenthesis
refers to an endpoint that is not included. The point of ∞ is always considered to be excluded from the interval,
and is used to mean that a number can increase (or decrease) unendingly. Perhaps a bit more elaboration is
necessary with regards to domain and range. The domain of a function, as was defined above, represents the
collection of all values of the independent variable that can be used in the function. Very often, particularly with
polynomial functions (which will be explained in the next section), a function’s domain is the entire set of real
numbers. A function’s range, then, is the collection of all possible values that the dependent variable will take on.
Now that basic function terminology is mastered, we have one last major question to explore: what can be input
into a function? That is, given f(x), what can be substituted in for x? The answer is, in a nutshell, anything. Any
mathematical expression and combination of variables and constants can be substituted for x. This is even true of
other functions!
Composite Functions
A composite of two functions is what results when one becomes a function of
the other. That is, given functions f(x) and g(x), the composite functions are
the results:
f(g(x)), sometimes denoted (f  g)(x) , and g(f(x)), sometimes denoted (g  f)(x)
Examples:
For (1)-(3) below, take the composite functions given f(x) and g(x) as defined here.
f(x) =
x +8
g(x) = x2
1)
(f  g)(x)
2)
(g  f)(x)
3)
(g  f)(2x)
4)
True or false: Given arbitrary functions h(x) and k(x), it must be true that the domain of h(k(x)) is
smaller or the same size as the domain of h(x).
Solutions:
1) (f  g)(x) = f(g(x))
= f(x2)
← take the function f with the new argument
= x2 + 8
2) (g  f)(x) = g(f(x))
=g
=
(
(
x +8
x +8
)
)
2
← take the function g with the new argument
← simplify by squaring the square root
3) (g  f)(2x) = g(f(2x))
=
(
(
2x + 8
2x + 8
= 2x + 8
← rewrite the composite function
← substitute f(x) in for the argument of g
=x+8
=g
← rewrite the composite function
← substitute g(x) in for the argument of f
)
)
2
← rewrite the composite function
← substitute f(2x) in for the argument of g
← take the function g with the new argument
← simplify by squaring the square root
MATHEMATICS RESOURCE | 15
4) All we know is that we have two functions. In problems like this, it is always good to try to create a
counterexample to the statement, thus proving that it is false. If it seems like it might be impossible
to make such a counterexample, the statement is likely true. To make a counterexample to this
statement about limiting domains, we need to find a function which has a small domain. Let us
examine h(x) = x . As we discussed only a page ago, the domain of this function is the nonnegative
real numbers. Now let’s look at k(x) = |x|, the absolute value function. 10 Taking h(k(x)) gives us
x .
Our original function h(x) required a nonnegative argument; however, this new composite function
can take any negative OR positive number, as the square root will eventually only be taken over a
positive number. The domain of h(x) is 0 ≤ x < ∞. The domain of h(k(x)) is ∞ < x < ∞. The
statement is FALSE.
We have now ascertained that we can take functions of functions; theoretically speaking, we can simply
just proceed to “stack” these functions on top of each other and take more and more composite functions.
One of the questions last remaining in your mind now is likely to be this: since we can take functions of
functions, is there also a way we can undo functions? The answer, as often happens, is yes. Given one
function, we can construct another so that when we take the composite of the two, we arrive back at x.
Inverse Functions
Given a function f(x), its inverse, denoted f-1(x), is a function such that we have
f(f-1(x)) = (f  f −1 )(x) = x.
An inverse relationship, which may or may not fit the definition of a function,
is found by exchanging the independent and dependent variables and then solving
the new relationship for the dependent variable.
Say that we know that f(7) = -1. That means when we consider the inverse, we know that f-1(-1) = 7. What really
happens, though, when we take the inverse of a function? Well, the inverse “relation” may or may not fit the
definition of a function, as stated in the box above. How do we know when it will? The definition of a function
states that no element of the domain can evaluate to more than one element of the range. If we invert this, with the
function’s range becoming the inverse’s domain, then we know that nothing in the inverse’s domain can map to
more than one element in the inverse’s range. In terms of the original function, nothing in the range can come
from more than one place in the domain.
Definitions
A function is known as one-to-one when every range value corresponds to
exactly one domain value. That is, the function doesn’t map to any range value
more than once. Mathematically stated, this means that given a function f(x),
x1 ≠ x2 must mean that f(x1) ≠ f(x2)
Functions that are one-to-one have inverses that are also functions. There will be more on this later, particularly
when we discuss exponential and logarithmic functions. 11
10
The absolute value function will be covered in more detail later in the resource. For now, you should know that the absolute
value of any number is positive, whether that number is positive or negative.
11
Sorry to tease and run like this, but we don’t have the necessary tools to take very many inverses. We haven’t even discussed
types of functions yet!
MATHEMATICS RESOURCE | 16
Lastly before moving on, we must discuss the idea of the zero of a function. A zero refers to a value of the
independent variable that causes the dependent variable to be 0. Sometimes it is also referred to as a root. That is,
since a function is normally written “f(x) = <an expression in x>,” zeroes are the x values that give rise to “0 = <an
expression in x>.” Rather than move on directly to an example, we will work through a few examples in the
following sections.
Linear Equations and Functions
We turn our attention now from functions to more general equations. Assuming that the reader knows how to
solve single-variable equations like “x + 5 = -12,” we can turn our attention to slightly more intricate topics than
that. What if we are dealing with an equation like “-2x + 6y – 4 = x + 2” ? We now have two variables, x and y,
instead of just x. This is a bit like having x and f(x), but we don’t know for sure whether either of the variables is a
function of the other. What is presented here is a linear equation in two variables. This means that there is no
exponent in either of the variables that is higher than 1.
If we try solving the linear equation here for one of the variables, we’ll get an expression containing the other
variable instead of simply a number. Solving for x gives us the equation x = 2y – 2. Similarly, solving for y gives the
equation y = 21 x + 1. Clearly, we need some new ideas. What sorts of numbers can satisfy the equation? Maybe we
can rely on our old friend logic to find a few combinations of x and y that make the equation true. One such
solution is “x = 4, y = 3,” while another is “x = -2, y = 0,” and still another is “x = 0, y = 1.”
We can take these pairs of numbers just like we took (c, f(c)) above in our function evaluations. The three
combinations of x and y above would be written as (4, 3), (-2, 0), and (0, 1)—in each case, we write the x value as
the first of two numbers, hence the term “ordered pair.” Other possible examples of ordered pairs that satisfy this
equation are ( 12 , 54 ) and (-4, -1). If, as mathematicians, 12 we want a way to organize all of the possible solutions to
this linear equation at the same time, we can do the same type of graph as above, with the first number, or the
abscissa, representing the x-coordinate and the second number, the ordinate, representing the y-coordinate. This
two-variable equation has an infinite number of solutions; the five ordered pairs listed above appear below left.
5
4
)
y
y
3
(-2,0)
-4
-3
(-4,-1)
In
-2
2
(0,1)
1
-1
0
-1
3
(4,3)
2
,
1
x
1
2
3
4
-4
-3
-2
-1
x
0
-1
-2
-2
-3
-3
1
2
3
4
the left
graph,
we see the five points on the xy coordinate-plane. The idea of using two numbers to represent a place on a plane is
known as the Cartesian-Coordinate system. The primary thing that we notice about the graph on the left is that
the five points that are all solutions to the equation appear to be lying on a straight line. On the right, we confirm
our guess and show that the five points are indeed on a straight line. Any linear equation that has two variables “x”
and “y” has an infinite number of solutions, and those solutions can be graphed on a plane as a straight line that
extends infinitely in both directions. Just remember that the word “linear” itself has “line” as its first four letters.
Paper isn’t very good at denoting infinite length, but in actuality, even the point (200, 101) exists on the line and is
a solution to the equation.
12
If you’re not a mathematician, then at least pretend you are for the time being. – Craig
MATHEMATICS RESOURCE | 17
The equation itself, “-2x + 6y – 4 = x + 2,” gives us much information. Using a bit of algebraic rearranging, we can
transform this to an equivalent equation, 3x – 6y = -6. An equation with two variables in this ax + by = c form is
said to be in standard form. Some textbooks or mathematicians will instead consider the standard form of a linear
equation to be ax + by + c = 0. These definitions are all relative, and this form is basically the same as the first,
except that the “c” constant will have the opposite sign as it does if you put it in the first form.
Example:
Rewrite the two-variable linear equation 12x – 3y = 9 + 17x – y + 2 in standard form.
Solution:
We add terms to each side of the equation to get to our solution, which can take two different forms.
12x – 3y = 9 + 17x – y + 2
-5x – 3y = - y + 11
-5x – 2y = 11
5x + 2y = -11
5x + 2y + 11 = 0
← subtract 17x from each side, combine 9 and 2
← add y to each side
← multiply by -1 to make the lead coefficient positive
← add 11 to each side for the other standard form
What else can we say about the graphs above? There is a point, called the x-intercept, where the line
intersects the x-axis. That x-intercept is (-2,0). There is another point, called the y-intercept, where the line
intersects the y-axis. That y-intercept is (0,1).
Example:
What is the only point that can be both an x-intercept and a y-intercept for the same line?
Solution:
For a point to be a line’s x-intercept and y-intercept simultaneously, it must be on both axes. The only
such point is the point (0,0), known as the origin.
All graphed lines with have both an x-intercept and a y-intercept, with the exceptions of completely horizontal and
completely vertical lines.
Example:
What are the x-intercept and y-intercept of the standard form line 3x + 7y = 84 ?
Solution:
The x-intercept of a line occurs on the x-axis, when y = 0. Thus, we can find the x-intercept by
substituting y = 0 into the equation.
3x + 7(0) = 84
3x = 84
x = 28
The x-intercept is (28,0).
The y-intercept of the line, then, occurs on the y-axis, when x = 0. The y-intercept can then be found
when we substitute x = 0 into the equation.
3(0) + 7y = 84
7y = 84
y = 12
The y-intercept is (0,12).
There is one other important descriptor of graphed lines: their steepness, or slope. In algebra classes, a line’s slope is
commonly taught as “rise over run.” What that means mathematically is that to find the slope of a line, you take
MATHEMATICS RESOURCE | 18
the vertical change and divide by the horizontal change between any two arbitrary points on the line. For example,
if we revisit the line we graphed earlier, we have five points already labeled on the line. (Remember that the line has
an infinite number of points on it—we just happen to have five conveniently labeled.) If we take any two of these
points and calculate the vertical change divided by the horizontal change (rise divided by run), we can find the
slope.
5
4
)
y
y
3
(-2,0)
-4
-3
-2
2
(0,1)
1
-1
3
(4,3)
0
-1
2
,
1
x
1
2
3
-4
4
-3
-2
-1
-1
-2
-2
-3
-3
1
-2
2
3
4
2
3
-3
1
6
3
-4
x
0
-1
x
0
-1
1
2
3
4
-2
-3
Example:
Find the slope of the line above.
Solution:
We want to find vertical change over horizontal change. This means we want to find change in “y” and
divide by change in “x.” I arbitrarily pick two points: in this case, I’ll choose (-2,0) and (4,3). y goes from 0
to 3 so the change in y is 3. x goes from -2 to 4 so the change in x is 6. The slope is then 36 , or 12 . Note that
we could have taken the points in the reverse order; the final answer would have been the same. If we had
said that y goes from 3 to 0, the change in y would have been -3. x going from 4 to -2 would have given a
change of -6. That means there would have been a slope of −−63 , or 12 .
Frequently, rather than expressing equations in standard form (ax + by = c), mathematicians prefer
expressing two-variable linear equations in slope-intercept form, or y = mx + b form.
Example:
Express the equation 2x – 4y = -12 in slope-intercept form, and find the line’s x-intercept and y-intercept.
Solution:
First, we wish to rewrite the equation in slope-intercept form.
2x – 4y = -12
-4y = -2x – 12
y = - 41 (-2x – 12)
y=
1
2
x+3
MATHEMATICS RESOURCE | 19
To solve for the x-intercept, we substitute y = 0, as discussed above.
0=
1
2
x+3
-3 = x
-6 = x
1
2
The x-intercept is (-6,0).
To solve for the y-intercept, we substitute in x = 0, also as discussed above.
y = 21 (0) + 3
y=3
The y-intercept is (0,3).
Because the substitution of x = 0 allows us to find the y-intercept, we know that in slope-intercept form y
= mx + b, (0,b) must be the y-intercept. In the example problem above, (0,3) was the y-intercept. This
allows us to graph lines very quickly if they are given in slope intercept form. “m” is the slope, and “b” is
the y-intercept.
Example:s
Find the slope and y-intercept of
a) 13x + 12y = -5
b) mx + ny = p
Solutions:
To find the slope and y-intercept of these equations, we only need to rewrite them in slope-intercept form.
After we do, m and b are the slope and y-intercept, respectively.
a) 13x + 12y = -5
12y = -13x – 5
x – 125
y = - 13
12
The slope is - 13
, and the y-intercept is (0, - 125 ).
12
b) mx + ny = p
ny = -mx + p
y = - mn x + pn
← subtract mx from both sides
← divide both sides by n
The slope is - mn , and the y-intercept is (0, pn ).
Examples:
By rearranging into slope-intercept form, quickly graph the lines
a) 12x + 15y = 30
b) 3x – 4y = -12
c) y – 12 = 0
Solution:
a) 12x + 15y = 30
15y = -12x + 30
y = - 54 x + 2
We know that this line must intersect the y-axis at (0,2) and have a slope of - 54 . In the graph below for
(a), there is a rise of -4 (a fall of 4) proportional to a run of 5.
MATHEMATICS RESOURCE | 20
b) 3x – 4y = -12
-4y = -3x – 12
y = 34 x + 3
This line must now have a y-intercept of (0,3) and a slope of 34 . In the graph for (b), there is a rise of 3
proportional to a run of 4.
y
c) y – 12 = 0
y = 0x +
There is a rise of 0, no
matter what the run
3
1
2
y=
1
2
2
1
-4
-3
-2
-1
x
0
-1
1
3
2
4
-2
-3
The last example was written to set the stage for another lesson concerning lines. Equations of the form y =
c or x = c create horizontal and vertical lines, respectively. People often forget which type of equation
creates which line. Remember, though, that the line resulting from an equation is a graph of all the points
that can satisfy the equation. With that in mind, the graph of x = 2 must contain the points (2,0), (2,-3),
(2,5), (2,-10), (2,7), etc. If those points are graphed on a Cartesian Coordinate plane, then they will form a
vertical line. Likewise, a graph of the equation y = -3 contains all of the points (0,-3), (5,-3), (-2,-3), (12,, a horizontal line has a slope
3), etc. and forms a horizontal line. In addition, since slope is defined as rise
run
of 0 (no rise with arbitrary run) while a vertical line has an undefined slope (arbitrary rise divided by zero
run). 13 Sometimes, a vertical line is also said to have “no slope.”
Example:
What are the equations of the x and y axes?
y
a)
y
b)
4
3
2
-4
-4
-3
-2
-2
Solution:
-3
1
x
0
-1
2
3
1
-1
3
1
3
2
5
4
-4
-3
-2
-1
x
0
-1
1
2
3
4
-2
-3
The x-axis is a horizontal line that crosses the y-axis at the point (0,0). The x-axis must have equation y =
0. The y-axis is a vertical line that crosses the x-axis at the point (0,0). The
y-axis must then have x = 0 as its equation.
13
Remember that any division by 0 is always undefined. This complication actually gives rise to many of the most active fields
of research in mathematics these days.
MATHEMATICS RESOURCE | 21
Example:
What is the equation in slope-intercept form of a line that passes through (-2,3) and (3,5)?
Solution:
The first logical thing to do in this case is find the slope. We are already given two points on the line, so all
we must calculate is the change in y and the change in x. The slope must then be 3 −5(−−32) = 25 , and we know
that m = 25 in the equation y = mx + b. We now need a logical way to find b in the equation. This
equation must be true for all of the points along the line, including the two we were already given;
intuitively, if we substitute one of the given points into the equation, we can solve for the missing variable
b. I’ll arbitrarily choose the second point (3,5) and substitute.
y = mx + b
5 = 25 (3) + b
19
=b
5
← substitute the (3, 5) point into the equation to solve for b
We already knew the slope and have now solved for the y-intercept. Thus, the equation in slope-intercept
form is
y=
2
5
x+
19
5
The above example illustrates one way of finding the equation of a line given two points (or one point and the
slope). Substitution into the slope-intercept form is one very intuitive method of finding the equation of a line.
Another method is substitution into the point-slope formula. Given slope m and a point (x1,y1) on a line, we can
solve for the equation of the line using the formula y – y1 = m (x – x1). We can also use the formula in reverse to
quickly graph a line given its point-slope form.
Example:
Use the point-slope formula to find the equation of a line with slope 25 , passing through the point (-2,3).
Solution:
We recognize this as the same line that was found above. We should get the same answer. We take the
given point and substitute as (x1, y1) in the point-slope formula.
y – y1 = m(x – x1)
← substitute
y – 3 = 25 (x – (-2))
y–3=
y=
2
5
2
5
x+
x + 54
← distribute
19
5
y
Example:
3
3
Quickly graph the equation y – 2 = − (x + 3)
5
Solution:
At first, we may be tempted to rearrange this equation into slopeintercept form, but in the point-slope form, it is already ready for
graphing. We see a point on the line is (-3, 2); we also know there
is a slope of − 35 . Those two facts alone are enough to form a graph.
(-3,2)
2
1
-3
-4
-3
-2
5
-1
0
-1
x
1
2
3
4
-2
-3
Remember, there are three different forms for the equation of a line:
standard form (ax + by = c or also ax + by + c = 0), point-slope form (y – y1 = m(x – x1)), and slope-intercept form
(y = mx + b). Each form has different properties, and which form is most appropriate will have to be determined
on a case-by-case basis.
MATHEMATICS RESOURCE | 22
Graphs of Equations in Two Variables
Standard form:
Slope-intercept form:
Point-slope form:
ax + by = c
y = mx + b
y – y1 = m(x – x1)
The slope is - ba , and the yintercept is (0, bc ).
The slope is m,
y-intercept is (0, b).
and
the
The slope is m. (x1, y1) is a point
on the graph.
For competition exams, you should also know that two lines that are perpendicular 14 have slopes that are additive
multiplicative inverses of each other (we more commonly say that one slope is the negative reciprocal of the other).
That is, for two lines to be graphed perpendicularly to each other, one will have slope m and the other will have
slope - m1 . This also means that if we have two slopes such that slope1 × slope2 = -1, their respective lines are
perpendicular. These ideas of perpendicularity and parallelness in graphed lines are listed in the curriculum outline
under the guise of “geometry,” but they make the most logical sense to learn here in “algebra,” when solidifying the
concepts of graphed linear equations.
Example:
Find the equation of a line in slope-intercept form that is perpendicular to y = 3x and is given to have a yintercept of (0, 10).
Solution:
We don’t even have to graph the line we are given to be able to find our answer. The line we are given has
a slope of 3, and therefore a line perpendicular to it will have slope of − 13 . We are also given the line’s yintercept, and thus our line is y = − 13 x + 10.
“But,” you may ask, “wasn’t the title of this section linear equations and functions?” Indeed it was. So where do we
begin considering functions instead of just linear equations? The surprising answer is that we already have: if we
consider y to be y(x) instead, then we have our variable y as a function of x! Recall that the definition of a function
was that a single domain element could not evaluate to more than one range element. In the event of [nonvertical]
lines, that will always be true!
Example:
Make a logical argument to show nonvertical lines are always
3
functions. Use the previous example’s equation, y – 2 = − (x +
5
3), to make your argument. In explicit function form (in this
3
1
case also slope-intercept form), this equation is y(x) = − x + .
5
5
Solution:
y
3
2
1
-4
-3
-2
-1
x
0
-1
1
2
3
4
-2
We have the graph of the equation from before, as it is already
-3
expressed in point-slope form; it appears at left. How can we
examine the effects at a given domain element? In these function cases, x represents our domain elements;
at a given x, we cannot have two range elements. A given x can be represented as a vertical line; recall that x
= c forms a vertical line at c. This means that when no vertical line can intersect the graph more than once,
14
I will define the term “perpendicular” in the geometry section.
MATHEMATICS RESOURCE | 23
no domain element can evaluate to more than one range element. This is known as the vertical line test for
functions. No nonvertical line could double back on itself to violate this property.
Similarly, every line that is not horizontal must cross the x-axis. Just as all nonvertical lines pass the vertical line
test, all nonhorizontal lines must cross the x-axis. As we have discussed, the x-axis represents the idea of y = 0, and
thus if we have a function in this form, every nonhorizontal line must have [exactly] one zero. A horizontal line
(unless it is itself the x-axis) will offer no zeroes. The vertical line test is formalized below.
Vertical Line Test
y
3
2
1
-4
-3
-2
-1
0
-1
x
1
2
3
4
-2
With the dependent variable graphed
on the vertical axis and the
independent variable graphed on the
horizontal axis, a graph represents a
function when a sweeping vertical line
never intersects the graph more than
once at any given instant. At left, the
sweeping vertical line is shown dotted.
-3
Quadratic Equations in One Variable
We now move on from linear equations in two variables to quadratic equations in one variable. As a linear
equation represented an equation with a maximum exponent of 1, a quadratic equation represents the next step: it
is an equation in which there is a maximum exponent of 2 (and an exponent of 2 must exist in the equation).
Moreover, the only variable exponents can be 1 and 2; there can be no fractional or decimal exponents involved.
Example:
Decide if the following one-variable equations are linear, quadratic, or neither.
1) x + 3 = 12
4) 7 – x + 3x2 = 17
2) 19 – x2 = 17
5) 3x = 12 + 2x – 5 x2
3) 12 + x = x3/2
6)
11 x = 11 x
Solutions:
1) This is a clear-cut example of a linear equation.
2) At first, this looks as if it might be neither, as there is a square root; however, the square root
occurs over a constant and not a variable. It’s as if the right side of the equation were 4.12
instead. This is a quadratic equation.
3) There is a non-integral exponent on a variable. This is neither linear nor quadratic.
4) This is a clear-cut example of a quadratic equation.
5) Just like (2), this seems at first as if it might be neither, but the square root could easily be
replaced with a 2.23…, and then it would be more obvious that it is a quadratic equation.
6) There is a square root, equivalent to an exponent of 12 , on a variable. This is neither linear nor
quadratic.
In working with quadratic equations, there are some important concepts to review and be sure we have down
pat before we move on to more complicated things. I have assumed many times that the distributive property
is completely understood in its entirety, but before proceeding, let us review the general distributive property.
MATHEMATICS RESOURCE | 24
General Distributive Property
(a1 + a2 + … + an) (b1 + b2 + … + bm)
=
a1(b1 + b2 + … + bm)
+
a2(b1 + b2 + … + bm)
+
….
+
an(b1 + b2 + … + bm)
The distributive property can be applied to each term in turn.
The general distributive property will help us here. In order to multiply two quadratic or linear expressions
together, we can break apart the first expression and multiply each term with the second expression. More
generally, this means that every term of the first expression must be multiplied by every term of the second
expression before like terms are combined.
Examples:
a) Multiply: (ax + b)(cx + d)
b) Multiply: (x + 2)(-4x – 1)
Solutions:
a)
(ax + b)(cx + d)
= ax(cx + d) + b(cx + d)
= acx2 + adx + bcx + bd
= acx2 + (ad + bc)x + bd
Often, what we have here is called the FOIL algorithm, referring to First, Outside, Inside, Last
terms being multiplied together. If we are dealing with linear binomials, the “Outside” and
“Inside” terms will usually combine as like terms.
b) This example provides us with a specific case to which we can apply the FOIL algorithm. The
first terms multiply to give us –4x2, the outside terms give –x, the inside give –8x, and the last
give -2.
(x + 2)(-4x –1)
= -4x2 – 9x – 2
Solving quadratic equations in one variable often takes some ingenuity. Looking carefully at examples (a) and
(b) above, we see that there is a process to turn the product of two linear expressions into a quadratic
expression. With some examination and a bit of creative thinking, we need to be able to reverse the process.
The reason may not be apparent, but recall we need to solve these quadratic equations that we will be getting.
If we can eventually know that the product of two expressions is 0, then we know that either one or the other
of the expressions must itself be 0. In this way, factoring will directly help us find the solutions of given
quadratic equations, provided we write them as a product equal to 0. As always, a nice solid example will help
explain things much more nicely than a paragraph ever could.
Example:
Solve the equation x2 – 7x = -12 by factoring.
MATHEMATICS RESOURCE | 25
Solution:
We look upwards at the example (a) above, and we find a general form for multiplication of two
linear expressions. It is rewritten below.
acx2 + (ad – bc)x + bd = (ax + c)(bx + d)
What this means is that we want to find a, b, c, and d such that we can rewrite our equation. First,
we add 12 to both sides of our equation, resulting in x2 – 7x + 12 = 0.
Now, we see that the product “ac” must be 1. This makes things a lot easier for us, as they tell us that
a = 1 and that c = 1. What remains is finding b and d in this algorithm. We know that they have to
multiply to be +12. Moreover, they have to add to be -7. We think about it for a bit and realize that
the best choice is that they are -3 and -4. (Order does not matter, as they are going into two things
that are multiplied together.)
x2 – 7x = -12
x2 – 7x + 12 = 0
(x – 3)(x – 4) = 0
x – 3 = 0 or x – 4 = 0
x = 3 or x = 4
← add 12 to each side
← factored as explained in the above paragraph
← a product of 0 means that one factor must be 0
← solved the two equations
So we can see from our example that quadratic equations can have two solutions. The process of
solution by factoring is a useful one, though it may not always be possible. When it is, however, the
procedure becomes a mental procedure of guess and check.
Factoring will often require some ingenuity and such, and copious examples should help illustrate the
concept enough that it will become easier.
Examples:
Factor the following expressions as completely as possible.
a)
b)
c)
d)
e)
f)
x2 + 6x + 5
2x2 + 5x – 12
6x2 – x – 1
x2 – y2
x2 + 2x + 1 – 4y2
x4 – 16
Solutions:
a) This is much like the example above. Here, the proper solution for b and d is that one must
equal 5 while the other equals 1.
x2 + 6x + 5 = (x + 1)(x + 5)
b) For this quadratic polynomial, we no longer have a leading coefficient of 1. This means that for
the variables a and c, one of the two variables must now equal 2. This preliminarily gives us 2x2 +
5x – 12 = (2x + b)(x + d). We still have that bd = -12, but the leading coefficient of 2 means that
2d + b = 5. Through some guessing and checking, we have the solution of b = -3, d = 4.
2x2 + 5x – 12 = (2x – 3)(x + 4)
c) We have the dilemma now that we are unsure of what the values of a and c are. There are rarely
any organized ways of factoring quadratic polynomials. Often, the best course of action is
repeated guessing until a solution seems to work. The possible combinations for a and c are 2, 3
and 1, 6. Also, here we know that b and d must involve 1 and –1. The valid solution is
6x2 – x – 1 = (2x – 1)(3x + 1)
MATHEMATICS RESOURCE | 26
d) Something appears strange in this quadratic polynomial. It has only two terms instead of three.
Many of those reading this already know what’s coming, but let’s try to approach this the
standard way. We have a leading coefficient of 1 on the x2 term. We seem to be missing our
linear x term entirely. The constant term at the end is given to be y2. This means now that b + d
= 0 and bd = -y2. This seems simple enough: we declare one to be y and the other to be –y. The
formula x2 – y2 = (x – y)(x + y) is a formula for the difference of two squares. It can be applied
whenever we have a quadratic binomial that is 15 a difference of two squares.
e) This polynomial seems odd because it now has four terms instead of three. This can be
performed as a two-step factorization. First, we note the quadratic polynomial in x, then the
difference of two squares.
x2 – 2x + 1 – y2 = (x – 1)(x – 1) – y2 = (x – 1 – y)(x – 1 + y) = (x – y – 1)(x + y – 1)
f)
This is our last example of factoring. We can still approach it as a difference of two squares.
Perhaps if we rewrite our problem using the fact that x4 = (x2)2 we might be able to shed some
light on the problem.
x4 – 16 = (x2)2 – 42 = (x2 – 4)(x2 + 4) = (x – 2)(x + 2)(x2 + 4)
Note that after our first factorization, another difference of two squares appears!
Difference of Two Squares
x2 – y2 = (x – y)(x + y)
It should come as no surprise to us, however, that quadratic equations can have two solutions. After
all, if we have an equation such as x2 = 4, do we have two solutions? Absolutely! We have the
solutions x = 2 and x = -2. Both the negative and positive of a given value will square to be the same
thing, and this gives us two solutions. Keep in mind that according to radical notation, the positive
one is represented by
, and the negative one is represented by −
.
Examples:
Solve x2 = 9 by
1) just taking roots of the equation as needed
2) factoring
Solutions:
1) As discussed, this equation has two solutions.
x2 = 9
x=
9 = 3 or
x = − 9 = -3
2) As we discussed in the previous example, we need one side of our equation to be 0 for a factoring
solution method to work. We begin by subtracting 9 from each side of the equation, then we
apply the difference of two squares factoring method:
x2 = 9
x2 – 9 = 0
(x – 3)(x + 3) = 0
x=3
or
15
x = -3
as its name would seem to imply…
MATHEMATICS RESOURCE | 27
When solving a quadratic equation in one variable, factoring is only one of several solution methods we could
use. There are others, including a method known as “completing the square” and then a handy solution
formula simply referred to as the quadratic formula. 16
The Quadratic Formula
Given an equation written in the form ax2 + bx + c = 0, the two solutions
of the equation are given by
x=
−b ± b 2 − 4 ⋅ a ⋅ c
.
2⋅a
The expression under the radical sign, b2 – 4ac, is known as the
discriminant and can determine the nature of the solution(s).
The quadratic formula is one of those things in algebra that should just be memorized for ease of use. You will
likely not have enough time on a given exam to re-derive it when needed, even if you do understand the
derivation in Appendix A. Songs are always a useful mnemonic device: both Pop Goes the Weasel and Row,
Row, Row Your Boat can fit the formula nicely.
“x equals negative b, plus or minus the square root… of b squared minus 4 a c, all over 2 a!”
“x equals negative b, plus or minus the root… [of] b squared minus 4 a c, divided by 2 a!”
Now that we’ve taken that brief musical interlude, I do hope that the formula is memorized, as it is hard to do
the following examples without it.
Examples:
Solve the following equations by the quadratic formula. In each case, state what value the
discriminant has.
a) 6x2 – x – 12 = 0
b) x2 – 6x + 9 = 0
c) x2 – 6x + 12 = 0
Solutions:
a) 6x2 – x – 12 = 0
x=
−( − 1) ± (1) − (4)(6)( − 12)
2⋅6
← apply the quadratic formula outright
x=
(1) ± 289
12
← simplify
=
x
1 + 17 18 3
1 − 17 −16
4
or x =
= =
=
= −
12
12
3
12
12 2
In this case, the discriminant had the value of 289.
16
In reality, the quadratic formula is actually derived from the method of completing the square. I don’t want to take the
space here to detail the derivation of the formula or explain the process of completing the square. However, the
derivation is available in a three-page resource appendix if you’d like to see it done for yourself.
MATHEMATICS RESOURCE | 28
b) x2 – 6x + 9 = 0
6 ± 36 − (4)(1)(9)
2
6± 0
x=
2
6
x=
=3
2
In this case, the discriminant had the value of 0.
x=
c) x2 – 6x + 12 = 0
6 ± 36 − 4(1)(12)
2
6 ± −12
x=
2
We (as of yet) can’t take the square root of a negative number! This quadratic equation has no
real solutions. In this case, the discriminant was negative and had the value of -12.
x=
We have to say that there are no real solutions when we have a negative discriminant. That much is certain, as
no real number when squared can offer a negative number. However, are there “nonreal” solutions? Indeed
there are! This is the most opportune time for yet another digression, this one a very imaginative one.
When mathematicians distinguish between the “real numbers” and the “imaginary numbers,” many people
get the idea somehow that the imaginary numbers are somehow less substantive and corporeal than the real
numbers. Imaginary numbers, however, are simply numbers. Frequently, mathematics arrives at equations
such as x 2 = −4 , and, until recently, mathematicians were forced to write, “no solution,” victims of a number
system we devised for ourselves. No real number can square to give a negative number. But mathematicians
realized it was ludicrous to say there were simply no numbers that squared to be negative. So they defined one.
Since the “new” number was decidedly not real, it was saddled with the unfortunate moniker “imaginary”.
The imaginary unit i
i 2 = -1
i=
Examples:
Simplify the following expressions.
a)
b)
−1
−16
c) − −4
d)
−4 × −12
e) i7
Solutions:
a)
−1 = i
b)
−16 =
−1 ⋅ 16 = 4i
c) − −4 =− −1 ⋅ 4 =−2i
d)
−4 × −12 =2i ⋅ (2 3)i =4 3i 2 =−4 3
( 1)3 ⋅ i =−i
e) i 7 =(i 2 )3 i =−
−1
MATHEMATICS RESOURCE | 29
On the whole, you can see that the square roots of negative numbers tend to obey the same rules as the
ordinary real numbers. Notice in example (d) how the negative sign was square rooted before other
simplification started. Make a special note that the rules for simplifying square roots do not apply when
radicands are negative. Here is an incorrect way to work example (d):
Wrong Solution for (d):
−4 × −12 =
( −4)( −12) =
48 = 4 3
The negatives must be accounted for before they are multiplied away and disappear. Imaginary numbers can
now actually be used for those quadratic equations that we believed had no solution, those pesky quadratic
equations with a negative discriminant. Now that we know how to handle it, the negative square root now
simply gives us an imaginary number, and our final answer will have a dual nature: a real part and an
imaginary part. A complex number is any number of the form a + bi, where a represents the number’s real
part, and bi represents the number’s imaginary part.
It’s important to make a special note that all real numbers and all imaginary numbers are complex numbers.
After all, the real number “4” that we have dealt with as far back as 1st grade can technically be rewritten as
“4+0i” and the imaginary number 2i about which we have just learned could itself be written as “0+2i.”
Arithmetic of complex numbers is somewhat instinctive. If we are to add two complex numbers
a + bi and c + di, then we can simply add the real parts and add the imaginary parts. If we want to subtract
two complex numbers, we subtract the real parts and imaginary parts. Multiplication of complex numbers
results from the FOIL algorithm we have already learned, with like terms simplified together. The term
containing i2 is made into a negative real number. These rules are restated in a table below.
The Arithmetic of Complex Numbers
Addition: (a + bi) + (c + di) = (a + c) + (b + d)i
Subtraction: (a + bi) – (c + di) = (a – c) + (b – d)i
Multiplication: (a + bi) • (c + di) = ac + adi + bci + bdi2 [FOIL] =
ac + adi + bci – bd =
(ac – bd) + (ad + bc)i
Division over the complex numbers, however, is far more complicated. Much like square roots, complex
numbers aren’t considered “simplified” until all imaginary units are out of all denominators; also like square
roots, when complex numbers are found in denominators, multiplication by a conjugate is the easiest way to
remove the imaginary part of the denominator.
Example:
Simplify the following complex fractions, and write your answers in a + bi form.
1
a)
i
2 −i
b)
3 + 4i
Solutions:
a)
1
=
i
1 −i
⋅ =
i −i
← the complex conjugate of 0 + i is 0 – i
MATHEMATICS RESOURCE | 30
−i
=
−i 2
−i
=
−( −1)
−i
b)
2 −i
=
3 + 4i
2 − i 3 − 4i
⋅
=
3 + 4i 3 − 4i
6 − 8i − 3i + 4i 2
=
9 − 16i 2
6 − 8i − 3i − 4
=
9 + 16
2 − 11i
=
25
2 11
− i
25 25
← perform the fraction multiplication
← i2 = -1
← simplify
← the complex conjugate of
is
← perform the FOIL multiplication
← i 2 = −1
← simplify
← put answer into a + bi form
Examples:
Solve the following complex equations for x.
a) 8x = 12ix + 13
b) 3x2 + 4ix = 12
Solutions:
a) This equation is a linear equation so we need only to solve for x.
8x – 12ix = 13
← subtract 12ix from each side
x(8 – 12i) = 13
← factor x to try to isolate the variable
13
x=
← divide by 8 – 12i
8 − 12i
13 8 + 12i
← begin simplifying the complex division
=
x
⋅
8 − 12i 8 + 12i
104 + 156i
← perform the FOIL multiplication
x=
64 − 144i 2
104 + 156i
← i 2 = −1
x=
208
2 + 3i
← simplify the fraction
x=
4
1 3
← put the answer in a + bi form
x=
+ i
2 4
b) This is a quadratic equation, and we will need the quadratic formula to solve.
3x2 + 4ix = 12
← put the equation into ax2 + bx + c = 0 form
3x2 + 4ix – 12 = 0
x=
−4i ± 16i 2 − 4(3)( −12)
2(3)
← apply the quadratic formula
MATHEMATICS RESOURCE | 31
−4i ± −16 − (-144)
6
−4i ± 128
x=
6
−4i ± 8 2
x=
6
−2i ± 4 2
x=
3
4 2 2
4 2 2
=
x
− i or x =
−
− i
3
3
3
3
x=
← simplify with i 2 = −1
← simplify
← simplify the radical
← reduce the fraction
← put into a + bi form
With the use of complex numbers, we can completely finish our discussion of solving quadratic
equations. Recall earlier that we said there was no real solution to the equation x2 – 6x + 12 = 0.
When we tried to solve it, we obtained the square root of a negative number. Now, we can solve it in
its entirety over the complex numbers, and
−12 will pose no problems for us.
Example:
Solve the equation x2 – 6x + 12 = 0 that was presented earlier, this time considering all complex
numbers when creating your solution.
Solution:
x2 – 6x + 12 = 0
6 ± 36 − 4(1)(12)
2
6 ± −12
x=
2
6 ± 2i 3
x=
2
x = 3 + 3 i or x = 3 – 3 i
x=
Example:
Solve the quadratic equation x2 = -10x – 50 over the complex numbers.
Solution:
This problem presents us with an example in which we will have a negative discriminant in the
problem. This will pose no obstacle for us, and we can press on.
−10 ± −100
2
−10 ± 10i
x=
2
x =−5 ± 5i
or
x =−5 + 5i
x=
x =−5 − 5i
The two solutions are not only complex numbers, they are complex conjugates, a pair of complex
numbers in the form a + bi and a – bi. Any real polynomial (defined in the next section) that has a
complex number as a solution will also have the complex conjugate as another root.
MATHEMATICS RESOURCE | 32
The Discriminant
Given a quadratic equation in the form ax2 + bx + c = 0, we can examine the discriminant
b2 – 4ac.
If the discriminant is
positive, the quadratic
equation has 2 unique
real solutions.
If the discriminant is 0, the square
root in the quadratic formula
becomes 0. Accordingly, the ±
operation fails to produce two
solutions. There is 1 unique real
solution.
If the discriminant is
negative, the quadratic
equation has 2 nonreal
solutions that are
complex conjugates.
Polynomial Equations and Functions
Now that we have finished discussing linear and quadratic expressions, we can move on to more general
polynomials. A polynomial in one variable is an expression of the form c n x n + c n −1 x n −1 + ... + c1 x + c 0 . n here is
a nonnegative integer and is called the degree of the polynomial. Each ci is referred to as the coefficient for xi.
Linear expressions are polynomials of degree 1, and quadratic expressions are polynomials of degree 2. We
only wish to generalize some of the properties we have learned into higher degrees, sometimes called higher
orders. Some of these confusing terms are summarized in the table below.
Polynomial Terminology
Polynomials have special
names based on the number of
terms in them.
two terms:
binomial
three terms:
trinomial
Polynomials have special names based on their degree.
degree of 1:
degree of 2:
degree of 3:
degree of 4:
degree of 5:
linear
quadratic
cubic
quartic
quintic
We have learned quite a bit about some of the lower order polynomials; if we set a linear or quadratic
polynomial equal to 0, the maximum number of real solutions is 1 or 2, respectively. We will later see that
this property even generalizes to higher order polynomials. In the meantime, we need to make sure that we
can do arithmetic on polynomials as it arises. I have taken for granted to some extent that their addition and
subtraction is intuitive. We combine the coefficients of the like terms; that is, the terms with the same
exponent can combine to form a single term, as illustrated in one quick review example that follows.
Example:
Perform the following operations:
(3x3 + 6x2 – x + 12) + (14x2 – 2x + 10) – (4x3 – 2)
Solution:
In order to combine like terms, we combine terms with the same x power. The operations are
performed below, with like terms first grouped together before being combined.
(3x3 + 6x2 – x + 12) + (14x2 – 2x + 10) – (4x3 – 2) =
(3x3 – 4x3) + (6x2 + 14x2) + (-x – 2x) + (12 + 10 + 2) =
-x3 + 20x2 – 3x + 24
Note a few things about this example. A subtraction was performed in the last operation. Be sure
never to forget to distribute the subtraction sign through the last polynomial before like terms are
combined. Also, the third polynomial of degree 3 seems to be “missing” a few terms. This is perfectly
MATHEMATICS RESOURCE | 33
acceptable. We simply consider the polynomial to have x2 and x coefficients of 0, much as we did
when we factored a difference of two squares earlier. Also, as we combine like terms, we match up the
terms that have identical exponents. As such, the 3x3 and -4x3 terms were combined, as were the 6x2
and the 14x2 terms, etc. These combine together to give us our final answer.
We applied the General Distributive Property earlier, as we were multiplying linear binomials to form
quadratic trinomials. Recall for the purposes of polynomial multiplication that it also applies when we are
multiplying together more than two or three terms. We could multiply very large polynomials with the same
process. In order to multiply two polynomials together, we always break apart the first and multiply each term
with the second polynomial. More generally, this means that every term of the first polynomial must be
multiplied by every term of the second polynomial before like terms are combined.
Example:
Perform the operations: (-2x3 + 5x – 2)(x2 + 2x – 1)
Solution:
We first apply the General Distributive Property before applying distribution to each term.
(-2x3 + 5x – 2)(x2 + 2x – 1)
= -2x3(x2 + 2x – 1) + 5x(x2 + 2x – 1) – 2(x2 + 2x – 1)
= -2x5 – 4x4 + 2x3 + 5x3 + 10x2 – 5x – 2x2 – 4x + 2
= -2x5 – 4x4 + 7x3 + 8x2 – 9x + 2
← break apart the first polynomial
← distribute each against the second
It is the fourth operation, division, of polynomials that poses the most problem. Believe it or not, it helps to
first review the general procedure for long division in a step-by-step fashion. A diagram of long division
appears at right, with the three primary repeated steps labeled. Notice their sequential progression.
1. How many times does the divisor
divide the current place?
2 4 2 8
4
3. What we find to
be left at the end
is our remainder.
9 7 1 3
−8
1 7
−1 6
1 1
− 8
3 3
− 3 2
1
2. Multiply, then subtract,
before bringing down the
next place. Repeat.
Polynomial division is very much like this long division; what is done to each digit of the divisor becomes
instead done to each term of the polynomial. Each time, we place a factor in the quotient, then multiply by
the divisor and subtract from the appropriate places in the dividend. Since it is complicated, an example
would of course provide the best explanation. Notice the detailed steps labeled in order and explained, and
read through them carefully.
Example:
x4 + x3 + 1
. Then, check your answer by confirming that Quotient
4x + 3
× Divisor + Remainder = Dividend, as it should in any division problem.
Perform the polynomial division:
MATHEMATICS RESOURCE | 34
Solution:
First, we rewrite the dividend with 0’s used as placeholders to represent the “missing” terms. We can
then perform the division analogously as above. Here, the division is labeled with descriptions of the
steps described below. Hold on to your seat; this gets complicated.
1
3
4x + 3
6
4
5
3
7
9
2
x + x + 0x + 0x + 1
4
3
− (x +x )
3
2
x + 0x
3
2
− (x +x )
2
-x + 0 x
2
− (-x −x)
x+ 1
− (x +)
2
4
8
1
4
1.
3
9
1 2
x 3 + 16
x − 64
x + 256
+
229
256
4x + 3
Compare the leading terms. 4x • (what) = x4. That quantity must then be 14 x3 for the term to
multiply correctly. The fraction cancels with the 4, and the variables become x4.
2. Having decided that 4x + 3 goes into x4 a number of 14 x3 times, we now multiply the two and
subtract from the first two terms of the dividend, just as we do in long division. Our chosen
value of 14 x3 causes the first term of x4 to cancel, as it should. We can thus cancel it out and
proceed on to the next term in the dividend.
3. We now want to know how many times 4x + 3 divides 14 x3. Similarly to step 1, we can compare
the leading terms to arrive at a value of 161 x2.
4. By multiplying 161 x2 by 4x + 3 and then subtracting from 14 x3 + 0x2, we cancel the 14 x3 terms
and need to bring down the ever-marvelous 0x term. This step is a repetition of step 2, applied to
a different term in the dividend.
5. − 163 x2 + 0x divided by 4x + 3 yields − 643 x. This is a repetition of step 1.
6. ( − 163 x2 + 0x ) − [- 643 x(4x + 3)] = (- 163 x2 + 0 x) − (- 163 x2− 649 x) = 649 x. Bring down the 1.
9
7. Similarly to steps 1, 3, and 5, I now have 256
added to the quotient.
27
9
229
8. ( 649 x + 1) − 256
(4x + 3) = ( 649 x + 1) − ( 649 x + 256
) = 256
9. We have reached our last step. There are no terms in the polynomial left to bring down, and 4x +
229
229
. This fraction, 256
, is our remainder.
3 cannot divide 256
It remains now to check our answer. We recall the General Distributive Property.
(
=
1
4
9
229
x 3 + 161 x 2 − 643 x + 256
) ( 4x + 3 ) + 256
1
4
9
229
x 3 ( 4x + 3 ) + 161 x 2 ( 4x + 3 ) − 643 x ( 4x + 3 ) + 256
( 4x + 3 ) + 256
27
229
= ( x 4 + 34 x 3 ) + ( 14 x 3 + 163 x 2 ) + ( − 163 x 2 − 649 x ) + ( 649 x + 256
) + 256
= x4 + x3 + 1
MATHEMATICS RESOURCE | 35
This concludes our discussion of polynomial arithmetic. Note that polynomial division is a complicated topic
that requires a lot of practice. Never shy away from it because it seems complicated. There will not be any
more examples of it in this resource, however, as there is a shortcut in most cases.
Whenever dividing by a linear polynomial, there is a faster method that can be used, commonly referred to as
synthetic division (or sometimes synthetic substitution). This is very difficult to illustrate in text; it is near
impossible to give an accurate and understandable description of the process. It is much easier to learn this
process by demonstration from a teammate or teacher if possible. As always, an example would serve best.
This is not set off as an example but will be in the text body of the resource itself. Here, I will compare long
polynomial division to synthetic division in a side-by-side fashion. Also note that there are many ways to write
synthetic division. They are all equivalent and offer identical results, but they may appear very different. If
you have learned it before, don’t be caught off guard if synthetic division as it appears in this resource does
not match what you are used to. Lines and boxes can be written in different places and ways without changing
the division process.
Both of these examples at left and
right
compute
the
quotient 3 2 -1 0 -42
3
2
3
2
6 15 45
x – 3 2x − x + 0x − 42
2x − x − 42
3
2
. x 3−3 Admittedly, the
− (2x – 6x )
x −3
2
2 5 15 3
5x + 0x
synthetic division at right looks much
2
−(5x – 15x)
simpler. In order to divide by the term x – c, place the value c in a
15x − 42
box to the top left of your division problem. The coefficients of
− (15x − 45)
3
the dividend can be placed in a row next to the value c. From
here, we repeatedly “multiply and add.” In this example, the 2 is
first brought down, illustrated by arrows. It is then multiplied by 3 to obtain a 6. The arrows shown illustrate
this. The 6 is then added to the next coefficient to give us 5. 5 × 3 is 15, which is placed below the 0 and then
added. This process is repeated. The bottom row, consisting of [2, 5, 15, 3] represents the quotient and
remainder. If the remainder happens to be 0, then the divisor was a factor of the dividend. As with standard
polynomial division, practice and repetition are the best ways to learn synthetic division. I will work one
standard example and then proceed to rework the earlier long division as a synthetic division problem.
2
2x + 5x + 15 +
Example:
Use synthetic division to find the remainder of
2x 6 − 3x 5 − 30x 4 + 164x 3 + 4x − 2
.
x+4
Solution:
This would be a horrendously long division problem, but with the process of synthetic division, the
6th degree numerator doesn’t pose too much of a problem. It just adds a few extra columns to our
problem. Don’t forget the 0 placeholder for the x2 term.
-4
2 -3 -30 164 0
4
-2
-8 44 -56 -432 1728 -6928
2 -11 14 108 –432 1732 -6930
From here, we can tell immediately that we have
2x 6 − 3x 5 − 30x 4 + 164x 3 + 4x − 2
6930
, and we have
=2x 5 − 11x 4 + 14x 3 + 108x 2 − 432x + 1732 −
x+4
x+4
a remainder of –6930. There is also a very useful Remainder Theorem that states that this value is
also the value if –4 were to be subtituted into the polynomial for x.
MATHEMATICS RESOURCE | 36
Remainder Theorem
If f(x) is a polynomial, then f(c) is the remainder when f(x)
is divided by x – c.
Example:
x4 + x3 + 1
. Recall how horribly tedious it was.
4x + 3
Now perform the same operation using synthetic division.
Recall the earlier long division that was performed:
27
− 256
Solution:
There is one major setback with this problem: the divisor is not of the form x – c. We cannot apply
synthetic division to this problem. Instead of quitting here, let’s try to use some ingenuity to work
through our problem; there has to be some other method of attack…. 17 Eureka! Instead of using 4x +
1 x4 + x3 + 1
. This is a
3 as our divisor, let’s use x + 34 . This means our problem can be rewritten as •
4
x + 34
very clever way to work around our setback. Now we can perform the synthetic division and multiply
1
our final answer by . The synthetic division is shown below.
4
1
1
0
0
1
1
229 256
x4 + x3 + 1 3 1 2 3
. If we take
= x + 4 x − 16 x + 649 +
3
x+4
x + 34
equation, we get the desired result. Notice the remainder is the same.
From the synthetic division, we can see
x4 + x3 + 1
=
4x + 3
1
4
9
x 3 + 161 x 2 − 643 x + 256
+
1
4
of this
229 256
4x + 3
Note that each term in the quotient is multiplied by a factor of 14 ; the remainder, however, was in a fraction
with the divisor as the denominator. In multiplying the denominator by 4, we have already multiplied the
term by 14 , and the numerator doesn’t change. The remainder remains the same! If the remainder is 0,
however, we know a lot more than just the remainder (or lack thereof). If a divisor goes evenly into a
dividend, it means that it was a factor of that dividend. If we are trying to find the zeroes of a polynomial
function, we can do synthetic division until we find c such that the x – c factor gives a remainder of 0. Keep
in mind that if it’s a linear or quadratic function, we can immediately find the roots without dealing with
synthetic division; however, with a degree of 3 or higher, some guesswork may be necessary. What can we
guess? How many zeroes/roots are we looking for? There are actually theorems that can give us these answers.
Proving these theorems can be quite complicated, but you should just accept them at face value while you’re
in high school. The remainder theorem can also be rewritten under another name, the factor theorem, but the
two theorems remain equivalent for all purposes.
17
This “…” represents thinking time. Let’s say it’s about 20-30 seconds of hard thinking.
MATHEMATICS RESOURCE | 37
Factor Theorem
Fundamental Theorem of Arithmetic
Rational Roots Theorem
(x – c) is a factor of
f(x) when f(c) = 0.
All polynomials of degree n, when
considered over the complex numbers, will
have n roots. An odd degree polynomial
must have at least one real root. Any real
polynomial’s roots with imaginary parts
with come as complex conjugate pairs.
Given a polynomial of the form
axn + … + c,
Recall that f(c) is the
remainder when f(x) is
divided by (x – c).
all of the rational real roots will come
in forms like ± db , where b is a factor
of constant c and d is a factor of
coefficient a.
After we have determined a few roots of a polynomial, we can divide out the linear factors (x – c) and lower
the degree of the polynomial. If we have a fractional root of the form ± db , we should instead divide out the
linear factor (dx  b) . This has been a lot of exposition and informational text, without an example, and as
always, a few strong examples should solidify some concepts here.
Example:
Factor the following polynomial functions over the real numbers; give all of the roots (real and
complex) that can be easily determined.
1) f(x) = 2x3 + x2 – 5x + 2
2) g(s) = 3s4 – 25s3 + 64s2 + 12s – 80
3) h(t) = t4 – 8t3 + 18t2 – 8t + 17
Solution:
1) We will have to use some synthetic division with trial and error in order to find factors of this
cubic. By the rational roots theorem, we know that both the numerator and denominator have to
be factors of 2 (if there are rational roots at all), in either positive or negative form. This means
that the factors we have to try are ± 12 , ±1, ±2. This leaves us with 6
factors to try by trial and error. What should we try first? 1 is always
1 2 1 -5 2
a nice easy start. The synthetic division follows.
2 3 -2
What luck! On our very first try, we have found a root of the
2 3 -2 0
polynomial! By the remainder theorem, this means that 1 is a root
of the polynomial; by the factor theorem, this means that (x – 1) is a
factor of the polynomial. From the quotient in the synthetic division, we know that 2x3 + x2 – 5x
+ 2 = (x – 1)(2x2 + 3x – 2)
The quadratic trinomial on the right is a standard quadratic trinomial, and we can just factor it as
normal.
2x3 + x2 – 5x + 2 = (x – 1)(2x2 + 3x – 2) = (x – 1)(2x – 1)(x + 2)
The zeroes of the polynomial function are 1, -2, and 12 .
2) The first thing we have to remember is not to be thrown by the variables. Functions can use
whatever letters they please for both function name and independent variable, and here we have g
and s, respectively. It’s exactly the same concept as f and x. By the rational roots theorem, we
have a lot of possible rational roots to check: ±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20, ±40, ±80,
± 13 , ± 32 , ± 34 , ± 53 , ± 83 , ± 103 , ± 163 , ± 203 , ± 403 , ± 803 . Wow! There are 40 possible roots to
check! Hopefully an easy one will be a zero of the polynomial before we spend hours doing this
problem. 1 is almost always the first and easiest root to check, and its synthetic division is done
below.
MATHEMATICS RESOURCE | 38
1
3
-25
64
12
-80
3
-22
42
54
-22
42
54
-26
3
Blast! It turns out that 1 is not a root of the polynomial! (The remainder fails to be zero.) Sigh….
next on our list, we approach -1. Perhaps we will have more luck from this?
4
3
3 -28 92 -80
4 -32
3 -24 60
80
0
Finally! We have a bit of
luck! -1 is a root of the polynomial,
and the factor theorem, along with the synthetic division above, gives us the equation on the next
line:
3s4 – 25s3 + 64s2 + 12s – 80 = (s + 1)(3s3 – 28s2 + 92s – 80)
We now need to attempt to factor the cubic polynomial on the right. Since it has the same lead
coefficient and constant term, we have to examine the same set of numbers as possible roots! Let’s
take a “random” guess and try 34 .
Success! Our “random” guess was indeed a root! The root comes in a fractional form, and a factor
of 3 will have to be commuted between factors. After doing this, the factor theorem now tells us
(3s3 – 28s2 + 92s – 80) = (s – 34 )(3s2 – 24s + 60) = (3s – 4)(s2 – 8s + 20)
Now that we have broken the last higher-degree polynomial down into a product with a
quadratic polynomial in it, we can use the quadratic formula to find the last two [possibly nonreal] roots. The computations have been left out for conciseness (and because the reader probably
understands the quadratic formula by now), and the last two roots are 4 ± 2i. So, after an entire
page of effort, we have factored and found roots for our quartic polynomial.
g(s) = 3s4 – 25s3 + 64s2 + 12s – 80 = (s + 1)(3s3 – 28s2 + 92s – 80)
= (s + 1)(3s – 4)(s2 – 8s + 20)
Roots are -1,
4
3
, 4 + 2i, and 4 – 2i.
3) This looks terrible! We have another 4th degree polynomial function! There is, however, a
beautiful silver lining here. By the rational roots theorem, there are only four possible rational
zeroes to consider: ±1, ±17. The synthetic division for all four factors follows.
1
1
1
-17
1
-8
18
-8
17
1
-7
11
3
-7
11
3
20
-8
18
-8
-1
1
1
17
17
1
-25 443 -7539 128180
18
-8
17
-1
9
-27
35
-9
27
-35
52
-8
18
17
-17 425 -7531 128163
1
-8
1
9
-8
153 2907
17
49283
171 2899 49300
MATHEMATICS RESOURCE | 39
4) We see here that this quartic polynomial has no rational roots. This means that it cannot be
factored by standard means. Only high-tech computer programs would be able to factor it
adequately. We also know, only doing multiplication and division, such large function
evaluations as h(17) = 49300 and h(-17) = 128180 by the factor theorem.
Things Unequal
Since this resource has had a fairly large number of digressions, it is reaching time for another one. We turn
away from our study of polynomials and functions for a bit to study other things that are similar to equations,
but instead say that things do not equal each other. Any mathematical statement saying that the value of two
expressions is not equal is an inequality. As luck (and maybe the math deities) would have it though, the
terminology and techniques of inequalities are remarkably similar to their counterparts in the world of
equations. The solution of an inequality is the collection of numbers that makes the inequality true, and
equivalent inequalities are inequalities with the same solution sets for the variable(s) involved.
≤
-
≥
-
<
>
≠
-
“less than or equal to”
or “at most”
“greater than or equal to”
or “at least”
“less than”
“greater than”
“not equal to”
In the box above, there are five inequality symbols listed with their most common verbal equivalents. Every
mathematical inequality will use one of the five symbols. Inequalities are solved in much the same way as
equations; the same operations are done to both sides, creating equivalent inequalities, until the variable is
isolated and explicitly stated. The solution of an equation is generally a simpler more explicit equation—x = 4,
for example—so it shouldn’t surprise you in the least to learn that the solution of an inequality is generally a
simpler inequality – something like x > 5 – that gives all possible values of the variable.
Note that the inequality “x < 7” is different from “x ≤ 7” only in that x cannot be 7 exactly—with the
exception of that one detail, the two inequalities x ≤ 7 and x < 7 have the same solution set.
Example:
Solve 4x + 3 < 11 for x.
Solution:
4x + 3 < 11
4x < 8
x<2
← subtract 3 from each side
← divide each side by 4
The solution for x in this inequality is x < 2, meaning that the variable x could take on any value less
than (but not including) 2 and still create a true statement. On this number line, the shaded region
represents the values that x could take.
0
2
The solutions to an inequality are often graphed on a horizontal number line for clearness. Here, the number
line is shaded to the left, meaning that values less than two will satisfy the inequality. Also note the shaded
line ending in the open circle, indicating that all values up to but not including 2 are valid solutions to this
particular inequality. A closed circle would mean that all values less than and including 2 are correct.
MATHEMATICS RESOURCE | 40
It’s been pointed out in words and by example that the procedure for solving inequalities is very similar to
that used to solve equations. There is, however, one significant aspect to solving inequalities that the algebra
enthusiast (or non-enthusiast, for that matter) must be acutely aware of: when both sides of an inequality are
multiplied (or divided) by a negative number, the inequality symbol must “flip”—that is, the ≥ symbol must
become ≤, and the < symbol must become >. At first, this may seem to make no logical sense but if you
remember that “to negate” is just another way to say “take the opposite,” then it might start to make sense. If
we have -x ≤ -2 and we multiply both sides by -1, then we’re looking for “the opposite” inequality. Is that x ≥
2 or x ≤ 2? Look at the number lines…
-2
0
2
-2
0
2
-x:
x:
logical
-2
0
2
x:
uh…no
Remember, the inequality sign must be flipped when a multiplication by a negative is introduced.
Unfortunately, the time will arise when quadratic inequalities need to be solved for x. Quadratic inequalities
are much more complicated than the linear variety we solved just on the previous page. There are several ways
to go about finding their solutions; I have one presented in this resource. The first logical step to take is to
solve as if it were an equation. The solution(s) found, if any, become what are sometimes referred to as
“critical points” or “points of interest.” You should note here that “critical point” takes on a different, more
formal meaning with regards to differential calculus. I use the term here only with regard to quadratic
inequalities. (Also note that it appears in quotation marks and is not bolded.)
These points of interest then form intervals of sorts, most often with alternating intervals. We can then
arbitrarily pick points out of these intervals to see whether or not they satisfy the inequality, and then those
intervals can then be the solution to our inequality.
Example:
Solve x2 + x > 6 for x.
Solution:
We should first solve the inequality as an equation, then form intervals of interest as described above.
Afterwards we can test the intervals of interest to see which are solutions to the inequality.
x2 + x = 6
x2 + x – 6 = 0
(x + 3)(x – 2) = 0
x = 2 or x = -3
MATHEMATICS RESOURCE | 41
We then have three intervals of interest, -∞ < x < -3, -3 < x < 2, and 2 < x < ∞. Our last task is just to
pick arbitrary points within these intervals to see which are solutions to the inequality. In the leftmost
interval, let’s pick -5. The equation is x2 + x > 6. Is it true that
(-5)2 + (-5) > 6? It certainly is! The leftmost interval is a correct solution interval to the equation.
Substituting 0 and arriving at the [incorrect] inequality 0 > 6 means the middle interval is NOT a
correct solution interval. Picking a point in the last interval (say, 3 perhaps) will give us 9 + 3 > 6,
which is true, confirming that the last interval also gives a correct solution to the inequality. A correct
solution to this inequality occurs when we have -∞ < x < -3 or 2 < x < ∞. 18
The solution is pictured below.
-3
0
2
Solutions to quadratic inequalities usually come in one of two forms, though there are many forms possible. If
the quadratic equation associated with the inequality has two solutions, the solution will most commonly be
either two broken intervals that go to -∞ and ∞ or some interval of finite length between two numbers. If the
inequality sign had been the opposite of what it was, then the example above would have had the solution of 3 < x < 2, and the graph would have shaded the dark area in the middle instead. If, instead, the associated
quadratic equation has one or no [real] solutions, then the solution graph could be any of many possibilities.
It could be a single point, and it could be all real numbers excluding a single point. It could also be all real
numbers in their entirety, or even no real numbers. It all depends on how the solutions occur if there are any,
and what the expression’s relation to the inequality is around those solutions.
Though competitors are unlikely to be tested on it, solving an inequality concerning a polynomial of higher
degree would involve much the same process as used here. After real solutions are found, if they can be found
easily, it is just a matter of trying test values within these intervals of interest to see whether each given interval
satisfies the inequality. Those will be the solution intervals, assuming all of the real roots have been found. If
real roots remain unsolved, it may turn out only part of an interval of interest is in the solution to the
inequality.
With inequalities firmly under our belt, we can proceed to the last major concept in this section before
returning to our topic of substance, functions. The last topic remaining is a favorite for many people, that of
absolute value.
Absolute value is a very unique mathematical operator. No matter what value it takes in, it spits out the
corresponding nonnegative value as a result. (There will be a very formal mathematical definition soon, as the
last example immediately below.) On paper, the absolute value of a quantity is represented by a pair of vertical
lines surrounding that quantity.
Examples:
Find the following: |-2|, |4|, |-12.08|, | − 35 |, |12|, and |x|.
18
Note how this is phrased that the solution set occurs when we have one criterion or the other. While it would be
correct to say that -∞ < x < -3 and 2 < x < ∞ are solution intervals, it would not be correct to say that the solution is
-∞ < x < -3 and 2 < x < ∞. In mathematics, use of the word “and” requires that two or more criteria all be met at the
same time. Using “or” instead means that any of the criteria will suffice. There is more explanation of this in the section
on absolute value inequalities.
MATHEMATICS RESOURCE | 42
Solution:
Again, no matter what quantity the absolute value operator takes in, it gives the corresponding
positive-valued result (or 0). This means that the five examples listed above take the values 2, 4,
12.08, 35 , and 12, respectively. The last example, the simplification of |x|, is a bit harder to write. We
cannot simply write the answer as |x| = x because we do not know the value of x. If x were to equal -3,
for example, we would have just asserted |-3| = -3. The formal definition of absolute value is:
|x| = x, if x ≥ 0
|x| = -x, if x < 0
for real x
“Whoa!” you say. “How is it possible that the absolute value of anything can be negative?” The
answer is that it cannot. Look closely at that definition again and think “the opposite” when you see a
negative sign. |x| = -x is only a true equation if x < 0. Try it with a few numbers. Input a positive
number, and you get that positive number back. Input a negative number—you get that number’s
opposite. It works! Now that we have understanding concerning the properties of absolute value, we
are left with the inevitable: solving equations and inequalities with absolute value. Comprehension
comes here most easily with examples.
Example:
Solve |x| = 4 for all possible values of x.
Solution:
We want to know what numbers have an absolute value of 4. This is straightforward; the possibilities
are either x = 4 or x = -4.
Example:
Solve |y| = 11 for all possible values of y.
Solution:
We now want to know what values have an absolute value of 11. This is not difficult, either; the
possibilities are either y = -11 or y = 11.
These previous two examples are probably the easiest that absolute value equations can possibly get. Note that
in both instances there are two possible solutions. This will be the case as long as the quantity within the
absolute value bars equals some quantity greater than 0.
| x | = c means that x = c or x = -c, as
long as c > 0
This takes care of the simplest absolute value equations. What now about the slightly more complicated ones?
Let’s again inspect a few examples.
Example:
Solve |z – 12| = 3 for all possible values of z.
Solution:
This is only marginally more complicated than the previous examples. We know that the quantity
inside the absolute value bars must be either 3 or -3 so we know that we have the equations z – 12 = 3 or z – 12 = 3. From there, we can solve the equations individually to obtain z = 9 or z = 15.
MATHEMATICS RESOURCE | 43
Example:
Solve |3a + π| = 14 for all possible values of a.
Solution:
This absolute value equation now is again becoming more complicated. We know that the quantity
inside the absolute value bars, 3a + π, must be equal to either -14 or 14 so we write the customary
two equations and solve both.
3a + π = -14
3a = -14 – π
or
or
3a + π = 14
3a = 14 – π
−14 − π
3
or
a=
or
a ≈ 3.619
a=
a ≈ -5.714
14 − π
3
These examples illustrate the concept of absolute value. Whatever quantity sits comfortably inside the
absolute value bars must equal either the positive or the negative of the value that it is set equal to.
Sadly, the official decathlon curriculum this year does not expect decathletes to solve equations
concerning absolute value. Instead, it lists “solution of basic inequalities containing absolute value.”
Inequalities containing absolute value are a bit more complicated than equations but are still quite
manageable.
Much like absolute value equations, absolute value inequalities are probably best understood by examples.
Example:
Solve |x| ≤ 2 for x.
Solution:
Possible values of x that can make this a true equation are 1, 0, -1, 1.8, 1.201, -1.99, -0.3, 0.97, 2,
and -1.41. x will be any value between -2 and 2 to make this inequality true. In math language, this
means that both x ≤ 2 and x ≥ -2. We might also just simplify our lives entirely by writing -2 ≤ x ≤ 2.
One thing that is very important to note here is that many textbooks refer to absolute value as the
distance from 0 on a number line. In that sense, the inequality itself says “x is no more than 2 units
away from 0 on a number line.” It must have radius 2 and surround the origin.
-2
0
2
Example:
Solve |y| ≥ 2 for y.
Solution:
It would make sense if the solution of this problem included the numbers that were not in the
solution of the previous example. Possible values of y that can make this a true equation by having
absolute value greater than 2 are 3, 10, 1400, -2.091, -5, -12, and -100. Essentially, the solution
giving all possible y values is all numbers such that either y ≥ 2 or y ≤ -2. Considering our alternate
definition of absolute value, the inequality reads “the distance of y to the origin is greater than or
equal to 2.” No problem.
-2
0
2
MATHEMATICS RESOURCE | 44
These two examples illustrate the general solutions to absolute value inequalities, stated concisely in the box
below.
| u | > c means that
u>c
or
u < -c
| u | < c means that
u < c and
u > -c
(in other words, that -c < u < c provided that c > 0)
There is one very important thing to note in this general formula: the difference between the word “and”
versus the word “or.” If you didn’t read the footnote regarding this distinction given three pages ago, please go
back and do so. Given two inequalities, saying “or” means that either of the two inequalities can be true.
Saying “and” means that both of the given inequalities must be true. Saying x < 2 and x > -2 means that x
must be somewhere between -2 and 2 on the number line. Saying x < 2 or x > -2 means essentially that x
could be any real number. Saying that x > 2 or x < -2 is a way of indicating x could be any real number
outside of the interval from -2 to 2. Saying that x > 2 and x < -2 means that there is no solution. It seems like
a list of facts to memorize, but in reality there is only one fact. “And” means that both conditions must be
true while “or” means that only one is required to be true. We can finish up our work with absolute value
inequalities with one last example.
Example:
Solve | -3q + 5 | > 7 for q.
Solution:
Given the form of the question, we know that we break apart the given expression into two
inequalities joined by an “or.”
-3q + 5 > 7
-3q > 2
q <−
2
or
3
or
or
← break the absolute value inequality apart
← add the additive inverse of 5 to each side
-3q + 5 < -7
-3q < -12
← flip the inequality signs
q>4
-
0
4
2
We
3
have
now finished what will hopefully be the last major digression of the mathematics resource; in particular, it
should be the last digression of the algebra section. The time has come to discuss three major types of
functions.
Rational Expressions and their Graphs
We will begin with the structures that make a logical progression from polynomials. They go by the moniker
of rational expressions.
Definition
A rational expression in one variable is a quotient of
two polynomials, one as the numerator and one as
the denominator.
MATHEMATICS RESOURCE | 45
What are some rational expressions? To answer this question, recall that we want a fraction with both
numerator and denominator as polynomials.
Example:
List some rational expressions.
Solution:
x 2 + 5x − 6
,
x2 − 1
2x 4 − 19x 3 − 27x 2 + 279x + 405
,
2x 4 + 13x 3 − 43x 2 − 297x − 315
x 3 − 4x 2 + 3x − 1, 1
Note in particular the last two rational expressions out of the four listed. Remember that any constant can be
a polynomial, even “1” alone. Thus, any polynomial is itself a rational expression because any polynomial can
be written as polynomial
. 1 is then a rational expression because 11 is a ratio of polynomials. Rational expressions,
1
in a manner of speaking, resemble fractions (formally called rational numbers). Instead of having integers in
the numerator and denominator, rational expressions have polynomials in those positions instead.
Similarly to fractions, rational expressions have a cancellation law whereby we can simplify. Also, addition and
subtraction require that a common denominator be found, and multiplication occurs “straight across.”
Multiplication and division usually prove to be the easier arithmetic topics in rational expressions; finding a
common denominator for addition and subtraction conventionally requires multiplying by a clever form of 1.
This will make more sense after the examples that follow.
Rational Expression Arithmetic
Cancellation Law
ac a
=
b≠0 c≠0
bc b
Addition and Subtraction:
a b a±b
=
±
c≠0
c c
c
Multiplication:
a d ad
=
•
b≠0 c≠0
b c bc
There is one thing that all three of these properties have in common. In every case, we must have the property
that the denominator is not equal to 0. Remember that division by 0 is completely undefined. 19 In graphs,
division by 0 can cause many types of complications, which will be discussed in more detail right after this
arithmetic.
Examples:
Simplify the following.
a)
b)
c)
d)
19
x 2 + 5x − 6
x2 − 1
1
7
−
x − 3 (x − 3)(x + 4)
x +2 x −6 x −6
•
÷
x − 5 2x − 1 x − 5
1
1
+ 2
x − 2 x − 5x + 6
If you prefer dramatic terms, consider it a breakdown in all of the laws of mathematics! Complications of division by 0
are the cause of several current areas of research in mathematics.
MATHEMATICS RESOURCE | 46
Solutions:
a) We should first factor both the numerator and denominator to see if there are common factors
that can cancel out. We practiced factoring polynomials earlier, so these should not faze us in the
least.
x 2 + 5x − 6
=
x2 − 1
(x + 6)(x − 1)
=
(x − 1)(x + 1)
x+6
x +1
, as long as x – 1 ≠ 0 and x + 1 ≠ 0.
Therefore,
x 2 + 5x − 6 x + 6
as long as x ≠ 1 and x ≠ -1.
=
x2 − 1
x +1
b) In order to perform subtraction, we must have a common denominator between the two rational
expressions. The easiest common denominator to reach would be (x – 3)(x + 4). We multiply the
first fraction by a clever form of 1, in this case xx ++ 44 .
1
7
−
x − 3 (x − 3)(x + 4)
x+4
7
=
← find the common denominator
−
(x − 3)(x + 4) (x − 3)(x + 4)
x −3
1
=
for x ≠ 3 and x ≠ -4
=
(x − 3)(x + 4) x + 4
c) There is a mathematical definition of division that we can exploit. Division by c is defined as
multiplication by the reciprocal of c, or multiplication of 1 c . Applying this to the third factor
below, we can arrive at a multiplication problem that we can simplify.
x +2 x −6 x −6
•
÷
x − 5 2x − 1 x − 5
x + 2 x −6 x −5
=
⋅
⋅
x − 5 2x − 1 x − 6
(x + 2)(x − 6)(x − 5) x + 2
=
=
(x − 5)(2x − 1)(x − 6) 2x − 1
← invert the division into multiplication
for x ≠ 5 and x ≠ 6 and x ≠
1
2
d) This addition problem is much like example (b). The easiest common denominator to reach
would be x2 – 5x + 6. To reach it, we multiply the first fraction by xx −−33 , a clever form of 1.
1
1
+
x − 2 x 2 − 5x + 6
1⋅ (x − 3)
1
=
+
(x − 2)(x − 3) (x − 2)(x − 3)
x −3
1
=
+
(x − 2)(x − 3) (x − 2)(x − 3)
x − 3 +1
x−2
1
, as long as x ≠ 2 and x ≠ 3
= = =
(x − 2)(x − 3) (x − 2)(x − 3) x − 3
MATHEMATICS RESOURCE | 47
Rational expressions have the potential to be a difficult topic. The standard rules of arithmetic always apply.
The most difficult aspect of arithmetic will usually be finding a common denominator before adding and
subtracting.
When one of these rational expressions is given as a function, the process of graphing them and
understanding the properties of these graphs becomes paramount. Major aspects of interest always occur when
the denominator equals 0. At locations where this occurs, one of two things may happen. There may be either
a vertical asymptote or a removable discontinuity.
Definitions
An asymptote, either vertical or horizontal,
represents a line that a graph can get
arbitrarily close to without intersecting.
A removable discontinuity is a location
where a graph stops completely and starts
over again in the exact same location.
The only major way to identify these defining characteristics will be to actually stop and form the graph
yourself whenever given a rational expression function. Remember: the places of interest occur where the
function has a division by 0.
Example:
x −2
, giving the locations of any vertical asymptote(s) and
x2 + x − 6
removable discontinuities that occur.
Analyze the graph of f(x) =
Solution:
As we discussed when we first began graphing, the graph of something such as a function is a
representation of different combinations of values for the independent and dependent variables. At
below left there is a chart giving several points on the graph. We focus carefully around the points x =
2 and x = -3. Since the denominator is a quadratic polynomial that factors as (x2 + x – 6) = (x + 3)(x –
2), it is those values that will cause the denominator to be 0.
f(x) =
x
-20
-5
-4
f(x)
-0.0588
-0.5
-1
x
-2.8
-2
-1
-3.2
-5
-3.1
-2.9
-10
10
x −2
x +x −6
f(x)
2
x
5
1
0.5
1.9
2.1
2.2
f(x)
0.2041
0.1961
0.1923
0
0.3333
3
0.1667
1
1.8
0.25
0.2083
4
20
0.1429
0.0435
MATHEMATICS RESOURCE | 48
f(x)
There is a removable
discontinuity at x = 2. The
graph stops there and then
starts again.
x
2
-3
There is a vertical asymptote
at x = -3, as the graph gets
arbitrarily close to it.
There is also a
horizontal asymptote
along the x-axis, at
y = 0.
There are a couple major points to note before moving on. The first is that even though you are not allowed
to use a graphing calculator at competition, if you are studying with one, you likely cannot see the removable
discontinuity. The reason for this is that a graphing calculator evaluates the function at many points, each
represented by a pixel. However, most likely, there is not a pixel representing the exact spot x = 2. Also note
that if we attempt to simplify the rational expression given, a common factor cancels out:
=
f(x)
x −2
x −2
1
, x ≠ 2.
=
=
2
x + x − 6 (x + 3)(x − 2) x + 3
1
. It has an identical graph to the example just worked, except it is
x +3
missing the removable discontinuity at x = 2. That is, when a value makes the rational expression undefined
by division by 0, but a cancellation law could produce a rational expression not undefined there, it is a
removable discontinuity. When even a simplified form has an undefined value there, it means the location in
question has a vertical asymptote. The last property to consider is the horizontal asymptote, if it exists. The
example above has its horizontal asymptote at y = 0. Properties for all these are organized into the table below.
Graph this other rational expression:
Asymptotes and Removable Discontinuities
When given a rational expression to analyze, first apply the cancellation law as much as possible to simplify
the rational expression. Since a rational expression must be a quotient of polynomials, let R(x) represent our
rational expressions, and let P(x) and Q(x) represent our polynomials. Call the original rational expression
p(x)
P(x)
and the simplified rational expression r(x) =
for simplicity.
R(x) =
q(x)
Q(x)
Vertical
asymptotes occur
where both the
simplified and
original rational
expressions have a
division by 0.
Removable discontinuities
(informally called “holes”)
occur when R(x) contains
a division by 0, but
simplifying gives a new
function r(x) that does not
have division by 0.
If the degrees of polynomials P(x) and Q(x) are equal,
then there is a horizontal asymptote at y = c, where c is
the ratio of P and Q’s leading coefficients.
If the degree of P(x) is less than the degree of Q(x), there
is a horizontal asymptote at y = 0.
If the degree of P(x) is greater than the degree of Q(x),
there is no horizontal asymptote.
MATHEMATICS RESOURCE | 49
Examples:
Give the asymptotes and removable discontinuities of the following rational expressions.
x2 − 4
8x 2
1)
2
4)
x +x −6
x 2 − 6x + 5
4
x 2 + 5x − 14
2)
5)
2
x + 3x − 10
x+7
3)
4
x +1
2
Solutions:
1) Our first task is to note where this original expression has a denominator of 0. Since we are now
experts in factoring, we see that the denominator can be factored:
(x2 + x – 6) = (x + 3)(x – 2)
So our locations of interest occur at x = 2 and at x = -3.
The next step is to factor the numerator and simplify the rational expression.
x2 − 4
(x + 2)(x − 2) x + 2
for x ≠ 2
=
=
2
x + x − 6 (x + 3)(x − 2) x + 3
Since x = 2 is undefined in the original rational expression but the x – 2 cancels out of the
denominator, this means that x = 2 represents a removable discontinuity. Also,
x = -3 is an undefined location in both of these rational expressions so x = -3 is a vertical
asymptote. Lastly, the degree of the numerator and denominator is the same, and the horizontal
asymptote occurs at y = 1, as the ratio of the leading coefficients.
2)
3)
4)
4
is a rational expression that is already in its most simplified form. It has no common
x + 3x − 10
factors between its numerator and denominator. The denominator factors as (x2 + 3x – 10) = (x +
5)(x – 2), and our points of interest occur at x = 2 and at x = -5. Since the rational expression
does not simplify, both x = 2 and x = -5 are vertical asymptotes. Also, since the denominator has
degree 2 and the numerator has degree 0, we know there is a horizontal asymptote at y = 0.
2
4
is similar to example (2) above. However, the denominator does not factor. From the
x +1
quadratic formula, we know that the denominator has no real roots. This means that there are no
vertical asymptotes. For the same reasons as #2, though, there is still a horizontal asymptote at y
= 0. There are no removable discontinuities.
2
8x 2
is already fully simplified. There are no common factors between the numerator and
x 2 − 6x + 5
denominator.
8x 2
8x 2
=
x 2 − 6x + 5 (x − 1)(x − 5)
Vertical asymptotes: x = 1 and x = 5
Removable discontinuities: none
Horizontal asymptotes: y = 8
MATHEMATICS RESOURCE | 50
5)
x 2 + 5x − 14 (x + 7)(x − 2)
=
= x − 2 for x ≠ -7
x+7
x+7
Recall that polynomials themselves are rational expressions, and this rational expression is just a
linear polynomial. Despite the confusion, however, the rules still apply as before.
Vertical asymptotes: none
Removable discontinuities: x = -7
Horizontal asymptotes: none
In dealing with the graphs of rational expressions, the domains and ranges can usually be determined from the
method we just completed studying. If the domain comprises all of the values of x that can be successfully
input into the function, then the domain will be all real numbers except for those values of x for which we
have an asymptote or discontinuity. The range is then usually all real numbers except those for which we have
a horizontal asymptote. The range will usually stay on one side of the horizontal asymptote, alternating sides
as the function passes across vertical asymptotes. There are occasional exceptions to this rule, when the
function will stay on the same side of the horizontal asymptote. Substitute arbitrary trial values to determine if
this is the case.
Examples:
Determine the domain and range of the following 5 functions.
x2 − 4
x2 + x − 6
4
2) f(x) = 2
x + 3x − 10
4
3) f(x) = 2
x +1
1) f(x) =
8x 2
x 2 − 6x + 5
x 2 + 5x − 14
5) f(x) =
x+7
4) f(x) =
Solutions:
These 5 examples are the same 5 rational expressions that were just analyzed in the previous example.
1) From there, we know that the domain is all real numbers except for -3 and 2. Since there is a
horizontal asymptote at y = 1, the range is then all real numbers except for 1. (There is a vertical
asymptote so that the function values change to different sides of 1 on either side.)
2) By looking back again at the previous examples, we find that this has two vertical asymptotes at x
= -5 and x = 2. There is also a horizontal asymptote at y = 1. This means that domain is all x such
that -∞ < x < -5 or -5 < x < 2 or 2 < x < ∞. Range is all f such that -∞ < f < 1 or 1 < f < ∞. These
are sometimes written as x ∈ (-∞, -5) ∪ (-5, 2) ∪ (2, ∞) and y ∈ (-∞, 1) ∪ (1, ∞). Alternately,
we could just say x ≠ -5, 2, and y ≠ 1.
3) With no vertical asymptotes, all of the function values will occur on one side of the horizontal
asymptote. Since the denominator must be positive all the time, the function must always be
positive. We have the domain of all real x and the range of all positive f.
4) Looking above, we now have two vertical asymptotes and one horizontal asymptote.
Domain: x ∈ (-∞, 1) ∪ (1, 5) ∪ (5, ∞)
Range: f(x) ∈ (-∞, 8) ∪ (8, ∞)
5) There are no asymptotes in this rational expression. Aside from the removable discontinuity at x
= -7, the function is very well-behaved. The graph will look like a line with a hole. The domain is
x ∈ (-∞, -7) ∪ (-7, ∞). The range is f ∈ (-∞, -9) ∪ (-9, ∞).
MATHEMATICS RESOURCE | 51
Exponents and Logarithms
There are two major forms of functions left to discuss before we can finally leave the realm of algebra. Before
we do so, we need to review (or learn for the first time) the rules and properties of working with exponents
and logarithms. Exponents can be described as those superscripts that come attached to constants and
variables, representing the number of times they have been multiplied by themselves. Hopefully, their
concepts are mostly review here, as they have been used to some small extent already in this resource.
An exponent represents the number of times its base is multiplied by
itself. That is, xn = x ⋅ x ⋅ x ⋅ … ⋅ x ⋅ x, with a total of n x’s appearing.
When n is not an integer, the concept of the exponent is a more subtle one that will have to be abstractly
imagined. I am, however, very eager to move on to the concept of the logarithm, a concept that many high
school students feel underprepared for when studying algebra. First, it helps to define what we mean.
Definition
A logarithm base b,
equals
denoted as a function by
y
f(x) = logb(x), means that
b
bf(x) = x. We must have b
to the
and x > 0. 20 If the base b
is omitted, then log(x) commonly indicates the common logarithm of base 10. The
special transcendental number e = 2.7182… used as a base gives rise to the natural
logarithm, denoted ln(x).
log (x) = y

b =x
The definition given above is intimidating, as it seems to have several parts, and we only want to understand
what the logarithm is and how to do manipulations with it. The second and third parts of the definition are
related to notation issues. Typically in high school, using logarithms (“logs” for short sometimes) is usually
related to the common logarithm, base 10. In the physical sciences, the natural logarithm tends to be used
more, as the number e shows its head in very strange places. To help with memorization, remember that the
base will stay as the base; the logarithm is then the exponent, and the argument of the log is the value that this
power takes. It helps to picture things visually with the spiffy arrows provided in the definition box.
In any case, this resource fully intends to make logarithms understandable even to those who feel completely
overwhelmed by the concept. Logarithms primarily are to be used as a tool, not to be studied too much for
their own properties. They are used as a tool for solving exponential equations, and (unfortunately) will
require some memorization to learn the necessary properties. The key thing to remember in every case is that
the logarithm represents an exponent. Notice that in the above box, the f(x) is the exponent itself. Say it out
loud a few times: a logarithm is an exponent. A logarithm is an exponent. A logarithm is an exponent. Now
that you’ve learned the key concept of logarithms and exponents, take a look at the properties table below. If
you are reading this resource as a file, you may want to print this page for reference until the properties are
committed to memory. If you are reading this resource on paper, you may want to tag this page for now.
20
In reality, the word “must” here is a little misleading. Logarithms in which the base and/or the argument are negative
take imaginary and complex values.
MATHEMATICS RESOURCE | 52
Properties of Exponents
x •x =x
Properties of Logarithms
1
7
logb(xy) = logb(x) + logb(y)
xa
= x a −b
xb
2
8
 
logb  x  = logb(x) – logb(y)
y
(x )
3
9
logb ( x n ) = n logb(x)
4
10
5
11
6
12
a0 = 1 for any a ≠ 0
*
*
loga (1) = 0
b logb (c) = c
γ
γ
logb(bx) = x
a
b
a b
a+b
= xab
(xy)a = xa • ya
a
x
xa
=
 
ya
y
x
a
b
= b xa
log c (x)
for arbitrary c
log c (b)
logb(x) =
1
log b x = log 1  
b  x 
log a b =
1
log b a
The properties are listed here in a table without proof. Proving most of these properties is straightforward,
although a few require some real ingenuity; some of them are proved in the example problem below. These
properties should become second nature to you after working enough practice problems concerning
exponents and logarithms. Some people prefer to memorize properties 1 and 7 as a pair, as well as 2 and 8
also as such. Most of the other properties match up a little bit more loosely, unfortunately. Property 10 for
logarithms, the change of base property, is a very important one. Most calculators come built with only
logarithms base e and 10. Properties * are definitions; that is, the raising of a number to the exponent of 0 is
defined to have a value of 1; properties γ result directly from the definition of a logarithm. That is, for
property γ, logarithm and exponent are inverse functions; they undo one another. Remember, since we
repeated to ourselves that a logarithm is itself an exponent, all we have to do is manipulate some of the
properties of logarithms in order to solve for unknown exponents in equations such as these. Turn to the
DemiDec Mathematics Workbook for more practice after reading these examples. There are also some lessons
regarding simplifying radicals and extraneous solutions that are tucked away within some of the examples.
Examples:
Solve the following equations for the unknown variable. In problem (2), keep your answer in terms of e.
1) 45 = x
2) ln(4x – 7) = 3
3)
3
z
= 529
z −3
7) 9p = 10
( p5/4 )
8) 7 4 + x = 197
9) k = log
7
x
10
4) (w12)(w2/3) = 92
10) 2 log8 x = log8 16
5) 5x = 30
11) log6(x + 3) + log6(2 – x) = 1
6) 74 + x = 197
12) 32x + 3x + 1 – 4 = 0
Solutions:
1) This problem is not hard to do since x has already been isolated. It is just an exercise in
understanding what 45 represents.
MATHEMATICS RESOURCE | 53
x = 45 = 4 ⋅ 4 ⋅ 4 ⋅ 4 ⋅ 4 = 1024
On a scientific calculator, first press 4, then the exponent button. It often looks something like yx.
Then press 5. This will calculate 45.
2) This is our first exercise with logarithms. Before we panic, we should see what we can piece together.
It is a natural logarithm, which means that it has a base of e ≈ 2.7.
ln (4x – 7) = 3
loge (4x – 7) = 3
← rewrite the logarithm with its new base
e3 = 4x – 7
← rewrite the logarithm into exponent form
e3 + 7
=x
4
← solve for x in the standard manner
Note here that e is just a constant, like π. We can’t simplify this final solution without taking an
approximation. As the directions requested, we can leave the final answer in terms of e as shown.
3) This problem is an exercise in working with exponent properties. The equation is worked below with
the properties cited by number as appropriate.
z3
= 529
z −3
z6 = 529
(z6)1/6 = 5291/6
← property 2
← same operation to both sides (raised to 1/6 power)
z = 6 529
← property 3 on the left, property 6 on the right
We could just leave our answer here as it is. Instead, we can work on simplifying the radical. A root
(or radical) is considered simplified when the factors of the quantity inside cannot be lowered
through factoring. See the simplification here.
z = 5291/6
= (232)1/6
1/3
= 23 = 23
≈ 2.844
3
← factor the base
← apply property 3
4) (w12)(w2/3) = 92
38
w 3 = 92
← apply property 1
At this point, the problem becomes the same as example (3). If we want to solve for w here, we raise
both sides to an exponent of 383 .
w = 933/38 ≈ 1.430
You may want to do this problem on your scientific calculator of choice. Be very careful when typing
3
in the numbers that you don’t wind up accidentally taking 9338 ≈ 21167 . Remember to use logic! That
answer would have been completely nonsensical
5) We now have a problem where we will have to apply a logarithm. We know that 52 = 25, and solving
for x here should give us a value very close to 2. We can start by rewriting the problem as a logarithm
instead of an exponent. Then the change of base formula will allow us to find x by calculator.
MATHEMATICS RESOURCE | 54
5x = 30
x = log5 30
log 30
x=
log 5
1.477...
x = 0.698... = 2.113...
← definition of the logarithm
← change to base 10 (property 10)
← use a calculator to evaluate the base 10 logarithms
To take a logarithm on most calculators, you type the argument before pressing the “log” button.
6) 74 + x = 197
4 + x = log7 197 ← definition of the logarithm
log 197
← change to base 10 (property 10)
4+x=
log 7
4 + x ≈ 2.715…
← use a calculator like in the previous example
x ≈ -1.284…
← subtract 4 from each side
7) Here the examples become a little more complicated. What we have here is an equation where we can
no longer just rewrite the given equation in logarithm form and solve. We will have to do some
manipulation to see if we can isolate the variable p. We’ll start by taking the logarithm of each side.
To keep things simple, we can use logarithm base 10 since it is included on the calculator.
( p5/4 )
9p = 10
( p5 4 )  ← take the logarithm base 10 of each side
log (9p) = log 10



p log 9 = p5/4
← property 9 on the left; γ on the right
5/4
p log 9 – p = 0 ← subtract p5/4 from each side, preparing to factor*
← factor p on the left side
p (log 9 – p1/4) = 0
1/4
← if two things multiply to be 0, one must be 0
p = 0 or log 9 = p
← raise the equation to the 4th power
p = 0 or p = (log 9)4
p = 0 or p ≈ 0.8291… ← evaluation by calculator
At the step labeled (*), you should be very careful not to divide both sides by p. If you do, you will
miss one of the two solutions, p = 0. When you divide both sides of an equation by an expression
containing a variable, you might lose a solution if that expression equaling 0 could produce a
solution.
8) This problem looks nearly identical to example (6) above. Unfortunately, now we cannot simply
rewrite the problem in logarithmic form and be done with it. This problem will require taking the
logarithm of each side, and it becomes quite difficult.
7 4 + x = 197
x
(
log ( 7 4 + x ) = log 197
(4 + x) log 7 =
x
)
← take the logarithm base 10 of each side
← apply property 9
x log 197
log 7
x
=
log 197 4 + x
← divide both sides by (4 + x) log 197
[This problem is about to become much harder.]
2
 log 7 
x

 = 2
x + 8x + 16
 log 197 
(
)
log7 2
log197
x2 + 8
(
)
log7 2
log197
x + 16
← square both sides
(
)
log7 2
log197
=
x ← multiply both sides by (x2 + 8x + 16)
MATHEMATICS RESOURCE | 55
(
)
log7 2
log197
x 2 + 8


x + 8 −


2
(
1
log7
log197
(
log7 2
log197
)
)

 x + 16 =
0
2


-1  x + 16

(
)
log7 2
log197
=
0
← subtract x from each side
← divide through by
(
)
log7 2
log197
9) We now have a quadratic equation, which we are fully prepared to solve. The discriminant takes
(
)
2
log197 2
value 8 − log7  − 4 ⋅ 1 ⋅ 16 ≈ −63.605 , and the negative discriminant indicates there are no real


solutions. Luckily, the variable is already isolated, and there is no need to do any difficult equation
solving. We can use logarithm properties to find k with a scientific calculator.
k = log
7
10
12
log 10
12
log 7
1
log 10
k = 21
2 log 7
k=
k=
log 10
log 7
← change of base formula and property 6
← property 9
← ½ cancels from the numerator and denominator
k = 1.183…
← evaluation by calculator
10) This problem is a bit different than the others because there are two separate logarithms that must be
accounted for. Several of the properties will be used in ways opposite the way we have been using
them.
2 log8 x = log8 16
log8 x2 = log8 16
log8 x 2
8
=8
2
x = 16
x = 4 or x = -4
log8 16
← property 9, used in reverse
← exponentiate both sides with base 8**
← property γ
← solution of the new equation
There are two things to note here. The first is that x = -4 is an extraneous solution. That is, it satisfies
the transformed equation but not the original one, and so it should not be considered a solution to
the equation. When we defined logarithms, we said that both the base and the argument had to be
positive. x = -4 violates this rule. The only solution to this equation is x = 4. We got this as a solution
when we changed the argument to x2, which must always be positive, but it cannot work when the
argument is really supposed to be x.
The other thing to note is that the line labeled (**) is usually skipped. This is fine. If you have two
logarithms equal to each other and they have the same base, then it must be that the arguments are
equal. This makes logical sense, but the reason it is true is shown in line (**)
11) log6(x + 3) + log6(2 – x) = 1
log6 ((x + 3)(2 − x)) = 1
6 = (x + 3)(2 – x)
6 = -x2 – x + 6
x2 + x = 0
x(x + 1) = 0
x = 0 or x = -1
← property 7, used in reverse
← definition of the logarithm
←*
In this case, there are two solutions, but neither of them is extraneous. Both, when substituted into
the original equation, give positive arguments and satisfy the equation.
MATHEMATICS RESOURCE | 56
12) This problem could be viewed as a challenge problem. It is like nothing we have worked so far, and it
will require some ingenuity to solve. The trick here is to find a substitution that will transform this
into an equation that we can solve. Observe:
32x + 3x + 1 – 4 = 0
(3x)2 + (3x)(3) – 4 = 0
u2 + 3u – 4 = 0
(u + 4)(u – 1) = 0
u = 1 or u = -4
3x = 1 or 3x = -4
x = 0 or [no solution]
x = 0 is the only solution
← properties 1 and 3, used in reverse
← Let’s make the substitution u = 3x
← factoring the quadratic equation
← solve the quadratic equation
← make the substitution back
← property * for the first equation; for the second
equation, note that no exponentiation with a
positive base gives a negative result
That about does it for logarithms for now. Hopefully, with twelve example problems, you will be able to find
one that answers and explains any questions you may have. All of the properties of exponents and logarithms
are exploited in some form in one question or another. Carefully follow your way through these examples to
make sure that you could solve each of them if they were to arise on a competition exam. With the
manipulations of exponents and logarithms mastered, there is only a bit left to cover before we can conclude
our study of algebra.
Exponential And Logarithmic Functions, Plus More On Inverses
Knowing now what we do about exponents and logarithms, it’s come time to investigate what happens if we
have exponential or logarithmic functions. That is, given y = 10x or y = log x, what do we know about the
graph of y? Can we discern its domain and/or range? Concerning the topic of inverses that we visited such a
long time ago, can we find the inverse of a given exponential or logarithmic function? The answer to all of
these questions is yes. 21
We proceed as always when graphing a function. We first find many function values, and then we proceed to
plot them on a pair of coordinate axes. Recall that when working the example problems above, we noted that
taking an exponent and taking a logarithm are functions that undo each other. That is, they are inverses.
Review: Inverse Functions
Given a function f(x), its inverse, denoted f-1(x), is the function such that we
have
f(f-1(x)) = (f  f −1 )(x) = x.
An inverse relationship, which may or may not fit the definition of a function,
is found by exchanging the independent and dependent variables and then
solving the new relationship for the dependent variable.
Below, the graphs of 10x and log(x) are placed side by side.
21
Well… technically, the answer to the first question is “a lot.”
MATHEMATICS RESOURCE | 57
f(x) = 10x
f(x) = log(x)
(0,1)
(1,0)
The first thing we notice about the two graphs is how similar they look. There is one point labeled on each
graph. Among our table of properties, we noted that a0 = 1 and that logb(1) = 0, for any positive choice of
variables a and b. The two graphs above have the choices a = b = 10; in turns out that no matter what we
choose for our bases, the point (0,1) will always be on an exponential function’s graph, and (1,0) will be on
the logarithmic function’s graph. It is only required that the bases be positive. What other things do we notice
about the two graphs? Will those properties still hold true if we have exponential or logarithmic functions
with different bases? The answer to this second question is yes. All of the major properties of exponential and
logarithmic functions will hold true no matter what we pick as our base of choice.
What properties are there to note? The major one is the asymptotes. Exponential functions increase from 0
without stopping. To the left of the vertical axis, the exponential function decreases to 0, and it does so
asymptotically; the horizontal axis is an asymptote. Likewise, for the logarithmic function, the vertical axis is
an asymptote. Since the argument of the logarithm must always be positive, the logarithm only extends in the
right-half plane, with a domain of (0,∞). However, its range is the domain of the exponential function. The
logarithmic function has all real numbers as its range.
There is one other major thing to note, and it requires examining both graphs simultaneously. Both of the
graphs are plotted below on the same set of axes. Recall that the functions are inverses; they form a reflection
across the dotted line! It turns out this is true of all function that are inverses!! All of these properties are
summarized in the table below the graph.
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Exponential Functions
Logarithmic Functions
Given positive base b, the exponential function
bu has a horizontal asymptote along the
horizontal axis. Its domain is all real numbers
(-∞,∞) and its range is all positive numbers
(0,∞). It will contain the point (0,1).
Given positive base b, the logarithmic
function logb(u) has a vertical asymptote
along the vertical axis. Its domain is all
positive numbers (0, ∞), and its range is all
real numbers (-∞,∞). It will contain the
point (1,0).
Last Box About Inverse Functions
Given a function f(x), its inverse, denoted f-1(x), is the function such that we have
f(f-1(x)) = (f  f −1 )(x) = x.
Two graphs that are inverses are reflections across a 45° line, the line y = x. An inverse
relationship, which may or may not fit the definition of a function, is found by exchanging
the independent and dependent variables and then solving the new relationship for the
dependent variable.
Examples:
In (1)-(4), find the domain and range of the function. In (5) and (6), find the inverse of the function
given and both the domain and the range of the function and its inverse.
1) f(x) = log41(3 – x)
2) F(s) = 7-s
3) g(t) = e3t – 41•3t
4) f(x) = log(x – 5) – log(11 – 2x)
5) y = e4x – 1
6) y = log
(
2x + 3
)
Solutions:
1) f(x) = log41(3 – x)
We have a logarithmic function. As its argument u goes over 0 < u < ∞, its range is all real numbers.
It is only defined when its argument is positive. We declare u = 3 – x. This gives us:
u=3–x>0
-x > -3
x<3
As x goes from 3 back down to -∞, then u still goes from 0 to ∞.
Domain: x ∈ (-∞,3)
← all real numbers less than 3
Range: f(x) ∈ (-∞,∞)
← all real numbers
2) F(s) = 7-s
Now we have an exponential function. Exponential functions have all real numbers as their domain.
The independent variable is negated before it is exponentiated. That makes no difference to us, as all
of the real numbers are accounted for. Since there is a negated s, the part that normally occurs at ∞
occurs now at –∞, and vice versa. The graph is mirrored across the vertical axis, but it still has the
same domain and range.
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Domain: x ∈ (-∞,∞)
Range: f(x) ∈ (0,∞)
3) g(t) = e3t – 41•3t
Now we have a difference of two composite functions. Things get a little tricky here. Can we still say
that all real numbers make up our domain? We can indeed. There is no value of t that can make this
function value undefined. Any value of t that is put into the function will be defined in both
exponential pieces. The domain is still all real numbers. Discovering the range is an entirely different
story. Both e3t and -3t have their own respective ranges (we consider -3t because it is as if we are
adding that to the first function). The first has a range of all positive numbers. The second has a
range of all negative numbers. What happens when we add them together? It depends on which
function is more important, or “bigger.” For a long time, the second, negative function will be more
important. Since there is a coefficient of 41 and the base is larger (3 instead of ~2.7), the second
function will, for a long time, cause the range to go extremely negative. Once the value of t becomes
very large, we have a different story. The first function, with its exponent of 3t, begins to dominate.
It then happens that the function begins to get really positive. This happens as t increases further
without stopping. That is, the function looks a bit like -3t for a long time, then past a point begins to
look like e3t instead. The range is all positive numbers plus some portion of the negative numbers.
After we have learned some calculus, we will be able to figure out exactly where that point is. (For
now, take my word that it turns out to be -124.31.)
Domain: x ∈ (-∞, ∞)
Range: f(x) ∈ [-124.31, ∞)
4) f(x) = log(x – 5) – log(11 – 2x)
Now we have a difference of two logarithmic functions. We know for a fact that our domain will not
be all real numbers. Our domain can only be the portion of numbers that give both arguments as
positive. That is, the domain will occur when both x – 5 > 0 and 11 – 2x > 0.
Solving both inequalities simultaneously gives x > 5 and x < 5.5. Our domain appears to be very
small! That is, our domain is only 0.5 wide!
When we restrict the domain to be such a small interval, we have shrunk down and compressed the
asymptotes. Since the logarithmic function has an asymptote when we try to go towards log(0),
plugging the value of x = 5 into our function gives us
f(5) = log (0) – log(1) = log(0).
Similarly, trying to take the independent variable as 5.5, we get
f(5.5) = log(0.5) – log (0) ≈ -0.301 – log (0).
The first of these undefined quantities goes towards –∞, and the second one (since it has a negative
sign in front) goes towards +∞. This means that our range is still all real numbers! This is particularly
impressive when you consider that our domain has been shrunk until it is only 0.5 units wide.
Domain: (5, 5.5)
Range: (-∞, ∞)
← note that the function is undefined at the boundaries b/c of log(0)
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5) Many pages ago, I touched on the concept of function inverses; I said we would return to it again
later. This is now a good time to begin exploring such concepts more thoroughly. I had hoped to
wait until exponential and logarithmic functions were learned so that we could easily take the inverses
of functions given.
In this case, we are asked to find the inverse of the function
f(x) = y = e4x – 1
From the concept box above, we know that to find a function’s inverse, we must switch the
independent and dependent variables, then solve for the new dependent variable. In this case, we will
have x and y trade places, then solve for the new y variable. Since we have been dealing with
exponential and logarithmic functions for a while, I will skip some steps when I work through this
problem.
← function with variables switched
← take the natural logarithm (logarithm base e) of each side
x = e4y – 1
ln(x) = 4y – 1
ln(x) + 1 = 4y
ln(x) + 1
y=
4
So we have an inverse function now that is a sum of a logarithm and a constant. Using the techniques
of examples (1) – (4), we can analyze the domain and range of both the original function and its
inverse. Let’s decide here to change our notation a bit, just to avoid confusion. Instead of having two
different y functions, let’s say
f(x) = e4x – 1
ln(x) + 1
f-1(x) =
4
This notation is the same as what I discussed in the concept box above. If it looks confusing, the
superscript of -1 tends to just be mathematical jargon for “inverse.” If you want, you can just think of
f-1 as another letter, like g or h. At its core, we just have two separate names for two functions. In any
case, let’s analyze their domains and ranges. For the first, f(x), we have to recall that exponential
functions can have all real numbers for their domains and give only positive numbers in their ranges.
That is, the new is domain will occur wherever 4x – 1 is a real number. If all real numbers are covered
in this new domain, which means that the range will still be all real numbers. This is indeed the case,
as 4x – 1 is a real number for all real numbers. The inverse function , f-1(x), is a logarithmic function
with a coefficient of ¼, then added to ¼. In this case, the logarithm still has the standard domain and
range we learned (domain of positive numbers and range of all real numbers), and multiplying by a
constant and adding a constant will not affect that.
ln(x) + 1
f-1(x) =
f(x) = e4x – 1
4
Domain: x ∈ (-∞, ∞)
Domain: x ∈ (0, ∞)
Range: f(x) ∈ (0, ∞)
Range: f(x) ∈ (-∞, ∞)
6) y = log
(
2x + 3
)
Now we have another function of which we are seeking the inverse. The same method applies.
x = log
(
2y + 3
ex = 2y + 3
)
← switch variable places and then solve for the new y
← rewrite the logarithm as an exponent instead
MATHEMATICS RESOURCE | 61
← square both sides
e2x = 2y + 3
e2x – 3 = 2y
e 2x − 3
y=
2
e 2x − 3
. We have yet to analyze the domains and
2
ranges of these functions. We know that the argument of a logarithm must be positive for the values
given to be in the domain. A square root is always positive already, so the domain occurs in the
region where 2x + 3 > 0. The domain is x > − 32 . Is the range still the standard range for logarithmic
So we now have f(x) = log
(
)
2x + 3 and f-1(x) =
functions? As x gets close to − 32 from the right on the number line, the argument of the logarithm
gets smaller and smaller, approaching zero. This means the logarithm does what we are used to; it has
an asymptote at x = − 32 , where the function goes really fast towards –∞.
e 2x − 3
, we have yet to analyze domain and range. Is the
2
domain all real numbers like we are used to? Indeed it is. There will not be a case where we can place
a real number into the function and receive a non-real result. The exponent of 2x can take any real
number and give a real number back. The range, however, has changed. Let’s rewrite the inverse
function as f-1(x) = 12 e 2x − 32 . Here we have a sum of two things (a sum, even though one is positive
and one is negative). The first one is an exponential function, with a modified exponent and a
coefficient that is not 1. Those two complications give the first term a domain and range that is
identical to normal exponential functions. The second term then takes whatever value comes from
the first term and subtracts 32 . That means the original range of (0, ∞) becomes (- 32 , ∞).
Now, for the inverse function f-1(x) =
f(x) = log
(
2x + 3
Domain: ( − 32 , ∞ )
Range: (-∞, ∞)
)
e 2x − 3
2
Domain: (-∞, ∞)
f-1(x) =
Range: ( − 32 , ∞ )
There we are. Having worked so many examples of exponential and logarithmic functions, I hope you can
find whatever you need within them. Note in the last two we saw a specific pattern of domains and ranges. In
general, the domain of a function should match the range of its inverse.
Graphical Analysis of Functions-----Compositions and Modifications
Now that we have worked with exponential and logarithmic functions, we have experienced most of the nontrigonometric functions with which we will be working. A decathlete may have to answer a test question or
two regarding what the graphs of particular functions might look like. That is, given a graph in a question,
the equation from which it arose should be selectable. Some of the basic functions which have been covered
are listed and drawn below.
MATHEMATICS RESOURCE | 62
Six Basic Functions
f(x) = x
f(x) = x2
f(x) = ex
f(x) = ln (x)
f(x) =
f(x) =
( 12 )
1
x
x
= 2 -x
Hopefully there has not been too much a surprise in these graphs. We have the exponential and logarithmic
graphs just discussed, along with an exponential graph with negative exponent, along the bottom row. Along
the top row, we have the most basic linear and quadratic functions, in addition to the reciprocal function.
Now, as an application of composite functions, we want to produce a graph of f(x – c), given a graph of f(x).
The topic will come up in detail in reference to graphing trigonometric functions later, which we will see after
this lesson and the section on geometry. For now, suppose we wish to graph f(x) = (x – 1)2 instead of f(x) = x2.
By filling in tables of function values, we can produce the graphs below. Be sure to look very carefully at both
the tables and the graphs. This is a pattern that is intuitive and very important to understand.
Example:
Fill in the tables below and plot the points on graphs.
f(x) = x2
f(x) = (x – 1)2
x
f(x)
x
f(x)
x
f(x)
x
f(x)
-2
4
1
1
-2
9
1
0
-1
1
2
4
-1
4
2
1
0
0
3
9
0
1
3
4
MATHEMATICS RESOURCE | 63
Unlike many drawings in this resource, these drawings are to scale. The primary point of this exercise
was this: the two graphs have exactly the same shape; the only difference is that the second one has
been right-shifted by 1 unit. The reasoning is as follows: because the function is the same except for
“x” being replaced with “x – 1,” all the same function values will happen. In the new case, though,
what used to happen at 5 will now occur at 6, and so on. Whatever graph f(x) has, the composite
function f(x – c) will have the same graph, shifted rightward by c units. The same goes for negative c
values: f(x + c) will have the same graph, shifted leftward by c units. These horizontal shifts are not
the only simple shift possible for arbitrary graphs. Vertical shifts are possible too.
For example, consider the tables below and plot the points on graphs. Compare your results to those
on the next page.
f(x) = x2
f(x) = x2 + 3
x
f(x)
x
f(x)
x
f(x)
x
f(x)
-2
4
1
1
-2
7
1
4
-1
1
2
4
-1
4
2
7
0
0
3
9
0
3
3
12
Again, the drawings given are to scale. This time, the only change is that the second parabola has been shifted
upwards by 3 units. This should make intuitive sense. The same function is given, except that 3 has been
added to it. Therefore, all function values will be 3 greater than they were before. Whatever graph f(x) has, the
composite function g(x) = f(x) + k will have the same graph, with the one exception that it has been shifted
upwards by k units. As before, this also works for negative values of k. That is, the function f(x) – k will have
the same graph as f(x), only shifted downwards by k units.
MATHEMATICS RESOURCE | 64
These vertical and horizontal translations are the easiest types of modifications when working with graphs.
The other modifications possible might best be described as stretching (as opposed to the first modifications,
which one might describe as shifting). The two shifting modifications could be written as (1) f(x – c) in the
case of horizontal shifting and (2) c + f(x) in the case of vertical shifting. In a related way, the two stretching
modifications can be written as (1) f(cx) for horizontal stretching and (2) c f(x) for vertical stretching.
After seeing graphs shift using tables of function values and graphs, l try to understand these stretching
modifications in words without resorting to more tables and graphs. The statement above says that f(cx) refers
to a horizontal shift. We will consider f(2x) for an example. If we have a graph for f(x) in which we had the
point (2, -7), the new graph for f(2x) will have the point (1, -7). Because we are taking f(2x), the value of 1
will be doubled, and the function value becomes f(2) instead of f(1). That is, the point we had was moved
inwards towards the origin by a factor of 2. The x-coordinate became only half of what it was previously. This
will happen for every point on the graph, and the graph will squish inwards horizontally by a factor of 2.
Now let us say we want to examine the graph of 2f(x), given the graph of f(x). If the original graph of f(x)
contained the point (2, -7) now, then the new graph for 2f(x) will contain the point (2, -14). Now in this
case, every point will have its y-coordinate doubled from what it was, and the points will move away from the
origin by a factor of 2. Again, this happens for every point on the graph. The graph will become stretched
outwards vertically by a factor of 2. All of the results we have found are summarized below. In each row entry,
the graph of g(x) is represented by a dotted line for f(x) represented by a solid line.
Graphs of Function Compositions
Composition
Description
g(x) = f(x – c)
shift rightward by c
g(x) = f(x) + c
shift upward by c
g(x) = c • f(x)
stretch vertically by a factor of c
g(x) = f(cx)
squish horizontally by a factor of c
Illustration
MATHEMATICS RESOURCE | 65
Remember that figuring out these results from logic may be a lot easier for you than memorizing the chart
above. In each case, you want to consider a point on the original graph, then think about where that point
would become located on the new graph. That process will happen to every point; whether that process moves
or stretches the graph, and in what direction, should be discernable. One last example will combine two of
these modifications.
Example:
Draw the graph of=
g(x) 2( 13 ) − 1 .
x
Solution:
We want to first produce the graph of f(x) = ( 13 ) . Then, we will stretch the graph vertically by a
x
factor of 2, and proceed to shift it downward by 1 unit.
The first things to note about the graph of f(x) are that is has a horizontal asymptote along the x-axis,
and the graph itself passes through the point (0,1), as any plain exponential function must. Because a
fraction is raised to the exponent of x, the graph will look very similar to the 6th graph in the table at
the start of this section. For consistency’s sake, we can observe another arbitrary point to help in
graphing: the easiest to obtain mentally is (1, 13 ) . Below, three graphs are shown. The first is the graph
of f(x) just deduced here. The second is the graph of 2f(x), which is similar to the first graph except
for a vertical stretch by a factor of 2. Lastly, g(x) = 2 f(x) – 1 is shown, which is identical to the
previous graph except for a downward shift of 1 unit.
(0,0)
Sequences and Series
Sequences and series are odd. Very few schools can make up their mind what “area” of math they belong to!
Many schools teach them in algebra class, although it’s not unheard of for a school to wait until precalculus or
even calculus to discuss these mathematical orphans. For our purposes, we will consider them in the algebra
curriculum and be sure that we understand the intricacies we need. We start with a basic building block, the
definition of a sequence.
Definition
A sequence is a function. Its domain is the counting
numbers, and its range is specified.
Here we have an official mathematical definition of a sequence. It seems complicated, but said more simply, a
sequence is any specified progression. If we were to look at the progression of £, €, £, €,£, €,£, €,£, €,£, €, we
could consider this a sequence of symbols. There are two values in the range, the symbols € and £, and we
could tell right away that any odd-numbered term would have value of £ and any even-numbered term in the
sequence would be €. The range could be symbols, things, or numbers. Most often, we will be concerned with
sequences when the terms of the sequence are numbers.
MATHEMATICS RESOURCE | 66
Example:
1) Determine the 6th term of the sequence: 6, 7, 8, 9, …
2) Determine the 10th term of the following sequence, and also find a formula in terms of n that can
express the nth term: 3, 6, 12, 24, 48, …
3) Find a formula in terms of n that can express the nth term of the following sequence, and use the
formula to find the 1000th term: 3, 5, 7, 9, 11, …
Solution:
1) Examples rarely get more straight-forward than this one. We have a sequence which just counts
numbers, beginning at 6. We can continue this sequence and easily find the 6th term: 6, 7, 8, 9,
10, 11.
2) Here we have a sequence of numbers that doubles at each term. We can continue the sequence to
find the 10th term: 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536. Since we have a sequence that is
doubling, we could choose instead to write it as 3, 3 ● 2, 3 ● 22… The number 2 is a common
ratio between terms, and we can write the nth term as 3, times 2 raised to the n – 1 power, or 3 ●
2n – 1. The most common notations for this would be f(n) = 3 ● 2n – 1, or tn = 3 ● 2n – 1.
3) This sequence is similar to the example above, except that instead of each term being two times
the previous term, each term is two more than the previous term. That is, instead of having the
sequence progress as 3, 3 ● 2, 3 ● 22, 3 ● 23,… , the sequence progresses as 3, 3 + 2, 3 + 2 + 2 (or
3 + 2(2)), 3 + 2 + 2 + 2 (or 3 + 2(3)),… In the earlier example, we had tn = 3 ● 2n – 1. Now,
though, we will instead have tn = 3 + 2(n – 1). Finding the 1000th term is a matter of substituting
1000 for n. So the 1000th term is t1000 = 3 + 2(999) = 2001.
Examples 2 and 3 showed special types of sequences. In Example 2, each term increased by a common ratio,
2. That is, each term was twice the term before it. In Example 3, each term increased by a common
difference, 2. Each term was two more than the term before it. Sequences of these types are special and have
special definitions.
Definitions
An arithmetic sequence is a sequence where each term is equal to a
constant added to the preceding term. This constant is known as the
common difference. A geometric sequence is a sequence where each term
is equal to a constant multiplied by the preceding term. This constant is
known as the common ratio.
Anytime we want to find the term in an arithmetic or geometric sequence, we can apply formulas based on
the pattern we just discovered in examples (2) and (3). Let’s change the notation a bit and use function
notation: t(n) means the nth term of a sequence.
MATHEMATICS RESOURCE | 67
Formulas
Given an arithmetic sequence where each term is some difference d
greater than the term before it and with the first term t(1), the nth term
will be
t(n) = t(1) + [d(n – 1)]
Given a geometric sequence where each term is some ratio r of the term
before it (greater or less) and with first term t(1), the nth term will be
t(n) = t(1) ● rn – 1
Now, based on the title of this section, you may have guessed that we are going to move on to another
concept in addition to sequences. Let’s say that we have a sequence, and it’s a large sequence of 5000 terms.
What would we do if we needed to add up all the terms? Alternately, what if we needed to know the sum of
the first 1000 terms only? These are concepts that are easy to gawk at, and we might think that they are
difficult. However, it is easy to develop methods to find these sums. We could even write these methods down
as formulas, although memorizing formulas is often harder than just remembering how to do the calculations
manually. Observe the following examples.
Example:
1) Take the simplest arithmetic sequence available: 1, 2, 3, 4, 5, 6, etc. That is, the nth term is t(n) =
n. Find the sum of the first 99 terms. Come up with a clever way of doing this quickly.
2) Take the following geometric sequence: 3, 9, 27, 81, 243, etc. The nth term of this sequence is
t(n) = 3n. Find the sum of the first 12 terms. Come up with a clever way of doing this quickly.
Solution:
1) We know that the 99th term will be 99. That leads us to the sum 1 + 2 + … + 99. Now we have
to devise a clever way to take this sum. Let’s try rewriting it a different way. I am going to reorder the terms of the sum, alternating between the front and the back of the numbers.
1 + 2 + 3 + 4 + 5 + … + 97 + 98 + 99
= (1 + 99) + (2 + 98) + (3 + 97) + (4 + 96) + … + (49 + 51) + 50
← reorder
= 100 + 100 + 100 + 100 + … + 100 + 50
← combine
= 4950
← we have 49 pairs, plus the 50 in the middle
By pairing the first and last numbers, we created lots of pairs that added to 100. Since we had 99
terms in total, we have 49 pairs of 100, plus the lone 50 in the middle, so the total sum is 4950.
That was quick! So when we want to take the sum of this arithmetic series, we have (1 + 99) ●
(99 / 2). This pattern will hold for arithmetic series in general. If we pair up the first and last
terms, we will multiply by half of the number of terms we have.
2) Now things are a little more complicated. Taking the sum of a geometric series is more
complicated than the sum of an arithmetic series. In fact, the cleverness here is something that
can often be really challenging. However, after you see it once, you can usually remember the
trick. Let’s try a little bit of mathematical wizardry.
x = 3 + 9 + 27 + … + 59,049 + 177,147 + 531,441
← rewrite the problem and call it x
3x = 9 + 27 + … + 59,049 + 177,147 + 531,441 + 1,594,323
←write out 3x
3x – x = 1,594,323 – 3
← write-out 3x – x, and cancel terms
12
← simplify
x (3 – 1) = 3 – 3
MATHEMATICS RESOURCE | 68
313 − 3
312 − 1
x= =
3•
=
797,160
3 −1
3 −1
← solve for x
This took a little bit of trick math. Who would have thought of calling the sum x, then
manipulating it and solving for x? In the final line where we solve for x, the pattern emerges. We
can take the first term, then multiply it by one less than the ratio raised to the number of terms,
then divide that by one less than the ratio.
What we have done in these two examples is to take the corresponding series sums of each sequence. Now
that we have seen it in an intuitive manner, we can talk about the definitions of series, and some of the
formulas used to calculate them.
Definitions
A series is the sum of the terms of a sequence. The series with n terms is
sometimes called the nth partial sum of the series. An arithmetic series is a
series based on an arithmetic sequence. A geometric series is a series
based on a geometric sequence.
Formulas
Given an arithmetic series where each term is some difference d greater
than the term before it with the firm term t(1), the sum S of the first n
terms will be
S(n) = [n/2][t(1) + t(n)]
Note that if we do not know the nth term, it can be solved for using the
formula established earlier.
t(n) = t(1) + [d(n – 1)]
Given a geometric series where each term is some ratio r of the term
before it (greater or less) and with first term t(1), the sum S of the first n
terms will be
S (=
n ) t (1) •
rn −1
r −1
There is a risk inherent in stating formulas. Sometimes, students will memorize the formulas and not learn
the concepts behind them. I have personally felt that remembering simple examples like the ones above is
often easier than remembering complicated formulas. If sums come up on a competition exam, finding them
using the methods above might be easier than cramming two more formulas into your already-cramped mind.
Let’s try a few more examples and see where it leads us.
Example:
1) Find the sum of the first 65 terms of the following sequence: 7, 11, 15, 19, …
2) Ira Saives invests $1000.00 into the stock market at the age of 25. By some miracle, it earns
exactly 8.0% every year for the next 40 years. He also invests $1000.00 into the stock market at
the age of 26, and then at the age of 27, each year for 39 and 38 years. This goes on until he
reaches age 65. How much does he have in savings at the age of 65? His brother Roth was late to
start investing and did not start investing until the age of 45. Roth figured that he could catch up
if he deposited $2000.00 every year instead of $1000.00. How much does Roth have at the age
of 65? Roth put $42,000 into the market while Ira put in $41,000; did Roth catch up to Ira?
MATHEMATICS RESOURCE | 69
3) Find the sum of the first 5 terms of the following sequence: 8, 4, 2, 1, 0.5, 0.25, … . Find the
sum of the first 15 terms. Find the sum of the first 50 terms. What do you notice? Explain what
is happening.
Solutions:
1) Since the first term is 7 and each term is 4 greater than the term before it, we know that the 65th
term will be 7 + 4(64) = 263. Using the method in the earlier example, we can pair off the terms
into pairs of 270. The number of pairs will be half the number of terms, and the sum is [65 /
2][7 + 263] = 8,775.
2) We will have to use our formulas for geometric series to solve this problem. Investments
compound upon themselves, so Ira’s first deposit after one year is worth $1080. The next year it
is worth $1160.40, since the $1080 earned 8.0% interest and 1080(1.08) = 1160.40. The next
year it is worth $1259.71. After 40 years, Ira’s first contribution is worth 1000(1.08)40. His
second contribution has one fewer year of interest on it, and his second contribution is worth
1000(1.08)39 when he reaches age 65. So the total of Ira’s savings after 40 years will form a
geometric series with 41 terms as follows:
1000(1.08)40 + 1000(1.08)39 + … + 1000(1.08) + 1000
← write the 41 terms
1000 + 1000(1.08) + … + 1000(1.08)39 + 1000(1.08)40
← reverse the order
Now we have a first term of 1000 and a common ratio of 1.08 between terms. We could use the
method we used earlier of setting this sum equal to x, then finding 1.08x and subtracting
between the two. In this case, let’s shortcut and go straight to the formula:
SIra(41) = t(1) ● [(rn – 1)/(r – 1)] = 1000 ● (1.0841 – 1)/0.08 = 280,781.04.
Wow! Ira has a lot of money! Let’s see how his brother Roth does with 21 contributions instead
of 41.
2000(1.08)20 + 2000(1.08)19 + … + 2000(1.08) + 2000
← write the 21 terms
SRoth(21) = t(1) ● [(rn – 1)/(r – 1)] = 2000 ● (1.0821 – 1)/0.08 = 100,845.84.
This is crazy. Roth put in twice as much money as his brother for half as long, but his total
savings at the age of 65 are FAR less than his brother.
3) We have another geometric series here, this one with common ratio of ½. The first term is 8, and
we can apply our formula for geometric series to find the sum of the first 5 terms. Let’s use
decimals here instead of fractions so that we can evaluate our sums on our calculator. It is a quick
adjustment to find the 15th and 50th partial sums as well.
S(5) = t(1) ● [(rn – 1)/(r – 1)] = 8 ● (0.55 – 1) / (0.5 – 1) = 15.5.
S(15) = t(1) ● [(rn – 1)/(r – 1)] = 8 ● (0.515 – 1) / (0.5 – 1) = 15.9995 (and more digits)
S(50) = t(1) ● [(rn – 1)/(r – 1)] = 8 ● (0.550 – 1) / (0.5 – 1) = 15.999999 (lots of 9’s)
These sums are getting closer and closer to 16. If you try grabbing your calculator and taking the
500th, 1000th, or 1000000th sums, you will find that the sums continue to get closer and closer to
16.
What is happening in the 3rd example here? If you took the various pieces of the formula separately on your
calculator, then you probably realized that it is the power of 0.5 that is causing this series to stop growing. As
we raise higher and higher powers of 0.5, the piece of the formula with 0.5n will get closer and closer to 0.
This will happen any time the ratio between terms is less than 1.
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Special Case
An infinite geometric series is a series that has an infinite number of terms.
The sum of an infinite geometric series with ratio less than 1 will approach a
finite number.
Formula
Given a geometric series where each term is some ratio r of the term before it
(greater or less) and with first term t(1), the sum S of the first n terms will be
rn −1
r −1
With r less than 1, this infinite series sum will become
S (=
n ) t (1) •
=
S t (1) •
0 −1
r −1
The Whirled Series
Infinite geometric series with ratio less than 1 approach a finite number. However, there are other special
series that continue forever and also approach finite numbers. They are often not as easy to describe as typical
geometric series, as the terms change by amounts that are not simply ratios.
There are several famous series that mathematicians generally know. It would be worthwhile to memorize
these, as well as one sequence which we will get to in a minute.
One series that many students wonder about is the one where denominators increase by 1. That is, the series
1
1
1
1
is an infinite series. However, it is very important that this series does not reach a finite
2 + 3 + 4 + 5 + ...
number. It will continue to grow forever without stopping.
Three of the most major series are shown below.
Three Special Series
1+
1 1 1
1
1
π2
+ + +
+
+ ... =
4 9 16 25 36
6
1−
1 1 1 1 1
+ − + − + ... =
ln(2)
2 3 4 5 6
1−
1 1 1 1 1
π
+ − + − + ... =
3 5 7 9 11
4
There may be one thing exceedingly odd about the sums above. Even though they have nothing to do with
circles or exponents, these series add up to sums that can be expressed in terms of π and natural logarithms.
This is really amazing when you think about it; these numbers can crop up just about anywhere! Even more
amazing is the fact that these series show that irrational numbers can be found by adding an infinite number
of rational numbers! I usually prefer to learn logic and proofs than just memorize formulas, but proving these
special sums is a real challenge. This might be a rare instance where it helps to just memorize the formulas.
Lastly, we have one very special sequence to discuss. Some readers may already know it.
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The Fibbonacci Sequence
The Fibbonacci Sequence is the sequence of numbers
starting with 1, where each term is equal to the sum of the
two preceding it:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
There is one very special note about the Fibbonacci sequence. By examining it, it is plain to see that the terms
will grow without stopping. In fact, they will grow faster and faster. However, the ratio between terms will
indeed stop growing. As the series grows larger, the ratio between terms will approach the Golden Ratio.
The Golden Ratio
The ratio between consecutive terms of the Fibbonacci sequence will
approach a finite number. That number is denoted ϕ (spelled phi and
pronounced as either “fie” or “fee”) and has a value as follows:
ϕ=
1+ 5
2
Moreover, the nth term of the Fibbonacci sequence can be found by the
following formula:
t (n ) =
ϕ n − (1 − ϕ )n
5
The Golden Ratio probably seems a bit esoteric, and certainly not worth its name, but there is much more to
it than would seem on the surface. A few Google searches will show exactly how frequently the Golden Ratio
crops up—including in The Da Vinci Code—and with remarkable accuracy—except, maybe, in The Da Vinci
Code. It appears often in geometric constructions, whether square pyramids, pentagons, or right triangles.
With that in mind, maybe it’s time to start learning our geometry—and we’ll begin with a little bit of
coordinate geometry, which the Academic Decathlon believes to be algebra. Of course, none of these fields are
really that discrete… but I digress.
Coordinate Geometry: Circles
Someone once suggested to me that circles should
actually be called “rounds.” After all, squares are
called squares and triangles triangles. I thought
about it for a while and then argued that ellipses
were round too. One wouldn’t want to confuse
ellipses and circles.
r
(h,k)
But the take-home message is clear: circles are
round. Perfectly round. The distance from the
middle point—the center—of the circle to any point
along the edge is the same. This distance is called
the radius, and is labeled r in the above diagram.
If you put the circle on what is called the coordinate plane (that is, the graph of x and y) then we can describe
the center as a point on that plane, with x and y values. The center might be (4,0)—4 units to the right of the
origin, and zero units up.
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All we need to know about a circle in order to draw it is the center and the radius—so the equation for a circle
tells us exactly these things:
( x − h) 2 + ( y − k ) 2 = r 2
The x value of the center is h, the y value is k, and the radius is r.
The Opposite X
Careful. The center of a circle is (h,k)—which means that the center of a
circle with equation
( x + 2) 2 + ( y + 4) 2 = r 2
is NOT (2,4), but (2,4)
Remember, this is equivalent to
( x − (−2)) 2 + ( y − (4)) 2 = r 2
Examples:
Try using the formula for circles to solve the following three equations on your own before looking at their
solutions below.
1) Find the equation of a circle with a radius of 6 centered at the origin.
2) What are the center and radius of a circle with equation x 2 + 12 x + y 2 − 4 x = 0
3) Find the equation of a circle centered at (2,0) with a radius the same length as the same perimeter
of a square with area 36.
Solutions:
1) This is about as straightforward as problems involving circles can be. Insert what you know into
the circle equation:
radius = 6
center at (0,0)
( x − h) 2 + ( y − k ) 2 = r 2
( x − 0) 2 + ( y − 0) 2 = (6) 2
x 2 + y 2 = 36
2) We want to transform this equation— x 2 + 12 x + y 2 − 4 x = 0 —into the more familiar form of
the equation for circles:
( x − h) 2 + ( y − k ) 2 = r 2
Look at the x terms first. We will need to complete the square—so that x 2 + 12 x becomes
x 2 + 12 x + 36 , or ( x − 6) 2 . This requires adding 36 to both sides of the equation:
( x + 6) 2 + y 2 − 4 x = 36
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Now let’s complete the square for y 2 − 4 y , turning it into y 2 − 4 y + 4 , or ( y − 2) 2 . This
requires that we add 4 to both sides of the equation:
( x + 6) 2 + ( y − 2) 2 = 40
This is now a standard equation for a circle, with radius
40 and center (-6,2).
3) A square of area 36 has sides of length 6 and therefore a perimeter of 24. The radius of the circle
is the same length as this perimeter, or 24. The center is (2,0). This means we can determine the
equation of the circle to be:
( x − (2)) 2 + ( y − (0)) 2 = 24 2
or
x 2 − 4 x + 4 + y 2 = 576
Coordinate Geometry: Ellipses
Ellipses are all over the place. For example, watermelons are threedimensional ellipses. Even comets tend to have elliptical orbits,
bringing them much closer to the sun on one end of their journey
around the solar system than on the other.
Like a circle, an ellipse has a center. Unlike a circle, however, it does
not have a single radius. The distance from the center to the edge of the
circle depends on where you are along the edge.
It does, however, have two quasi-radii, each called an axis. The semi-major axis is the longer of the two. The
semi-minor axis is the shorter. Each axis leads from the center of the circle to the top or to the side. In the
above diagram, a is the semi-major axis, and b the semi-minor axis.
( x − h) 2 ( y − k ) 2
+
=1
a2
b2
If the semi-major axis were vertical, parallel to the y-axis, then the equation would read:
( x − h) 2 ( y − k ) 2
+
=1
b2
a2
Notice that this is not very different from the equation for the circle, except that the completed squares for x
and y are themselves divided by the squares of a and b—where a and b are the lengths of the axes.
If we’re given that the center of an ellipse is (5,-4), the length of the semi-major axis is 5 and the length of the
semi-minor axis is 4, where the semi-major axis is horizontal, the equation for this ellipse would be:
( x − 5) 2 ( y + 4) 2
+
=1
52
42
x 2 − 10 x + 25 y 2 +8 x + 16
+
=1
25
16
Now we can multiply through by 25 and 16 in order to eliminate the fractions:
16 x 2 − 160 x + 400 + 25 y 2 +200 x + 400 = 400
MATHEMATICS RESOURCE | 74
Given an equation like this one, we would work in reverse to find the center and the lengths of the axes of the
ellipse.
ASYMPTOTES
Coordinate Geometry: Hyperbolas
Circles and ellipses are nice and round. You start at
any one point on the edge, follow it long enough, and
eventually you come back to where you started.
VERTEX
VERTEX
Hyperbolas are not round. They open up and
disappear endlessly into the distance on either side of
the coordinate plane. They are symmetrical, and they
are infinite.
The equation of a hyperbola is that of an ellipse
twisted inside out, with one term subtracted from the
other instead of added. Witness:
( x − h) 2 ( y − k ) 2
−
=1
a2
b2
or
( y − k ) 2 ( x − h) 2
−
=1
a2
b2
Once again, (h,k) is the center of the hyperbola—but the curves, as shown above, don’t actually pass through
the center point. Rather, they “start” at the vertices, which are a units from the center on either side of it—to
the left and right if the x-term is positive, and up and down if the y-term is positive. In the above diagram, the
vertices are a units to the left and right.
Notice that the curves approach but never intersect their asymptotes. Later, we will discuss asymptotes in
more detail, but for the moment, think of them as barrier lines, like black holes, which you can come closer
and closer to without ever reaching.
If we know what the asymptote lines are, we can trace the curves of the hyperbola more accurately.
Fortunately, the lines are easy to find. One is
x−h y−k
−
=0
a
b
And the other is
x−h y−k
+
=0
a
b
For the most part, we solve these problems just as we do those for ellipses—by completing squares, dividing
by the proper a and b to make sure the coefficients of x and y are both 1, and then examining our original
formula and—in this case—that of the asymptotes, to shape the hyperbola correctly.
MATHEMATICS RESOURCE | 75
Coordinate Geometry: Parabolas
Parabolas. When the Enterprise swoops around the
sun to travel through time in Star Trek IV, it does so
along a parabolic curve. Think of a parabola as
something not unlike a fingernail, except expanding
and going on forever on both sides.
FOCAL
POINT
VERTEX
A parabola has a vertex, where the curve “begins” to
move outward. It also has a focus point around which
the curve takes shape.
The section of the curve closest to the vertex is known
as the latus rectum. Think of it as the butt of the
parabola. It covers the whole region of the curve around the vertex and before the focus—as shown below.
The distance from the focus to the end point of the
latus rectum is twice the distance from the focus to
the vertex.
The most familiar parabola of all is
y = x2
This parabola has a vertex at the origin and rises
gently toward infinity.
2p
2p
LATUS
The more complete equation for a parabola opening
upward or downward is
( x − h) 2 = 4 p ( y − k )
And opening left or right,
( x − h) 2 = 4 p ( y − k )
In these equations, (h,k) represents the vertex, and p is the distance from the vertex to the focus. Consider the
sample parabola y + 4 x 2 + 8 x + 4 = 2 . We can separate out the expressions as follows:
y + 4 x 2 + 8x + 4 = 2
4 x 2 + 8x + 4 = 2 − y
4( x 2 + 2 x + 1) = 2 − y
( x + 1) 2 =
−( y − 2 )
4
( x + 1) 2 = − 14 ( y − 2)
The equation has a vertex at (-1,2) and a focus – 161 units from the vertex.
MATHEMATICS RESOURCE | 76
III. Geometry (Optional)
Geometry has fascinated man for centuries. Egyptian
hieroglyphics, Greek sculpture, and the Roman arch all made
use of specific shapes and their properties. Geometry is
important not only as a tool in construction and as a
component of the natural world, but also as the branch of
mathematics commonly studied after algebra, requiring us to
make logical constructs to deduce what we know. USAD has removed geometry—mostly,
except for coordinate geometry—from this year’s curriculum, but reviewing the basics of this
field will help you understand trigonometry, practice algebra, and understand what
coordinate geometry is all about in the first place.
Introduction to Lines, Planes, and Angles
In the study of geometry, most terms are very rigorously defined and used to define our
deductions. A few terms, however, exist primarily as concepts with no strict mathematical
definition. The point is the first of these ideas. A point is represented on paper as a dot. It
has no actual size; it simply represents a unique place. To the right are shown three points,
named C, H, and U (capital letters are conventionally used to label and represent points
in text).
C
U
m H
C
H
U
The next geometric idea is that of the line. A line is a one-dimensional object that
extends infinitely in both directions. It contains points and is represented as a line on
paper with arrows at both ends to indicate that it does indeed extend indefinitely.

Lines are named either with two points that lie on the line or with a script letter. CH

, HC , and m all refer to the same line in the diagram at left.
Similar to the line is the ray. Definitions for rays vary; in general, a ray can be thought of
as being similar to a line, but extending infinitely in only ONE direction. It has an
endpoint and then extends infinitely in any one direction away from that endpoint. You
might want to think of this endpoint as a “beginning-point.” Rays are named similarly to U


lines, but the endpoint must be listed first. Because of this, CH and HC do not refer to


the same ray. CH is shown at right. HC would point the other direction and is not shown.
C
H
Last in our introduction to basic geometry is the angle. The definition for an angle
B
remains pretty consistent from textbook to textbook; it is the figure formed by two
A
5
rays with a common endpoint, known as the angle’s vertex. An angle is named in
6
one of three ways: (1) An angle can be named with the vertex point if it is the only
C
angle with that vertex. (2) An angle can be named with a number that is written
D
inside the angle. (3) Most commonly, an angle is named with three points, the
center point representing the vertex. In the drawing at left, ∠1, ∠C, ∠UCH, and
C
∠HCU all refer to the same angle. In the drawing at right, ∠5 and ∠BAC refer to
1
the same angle while ∠6 and ∠CAD refer to the same angle. ∠BAD then refers to
H
U
an entirely different angle. Note that ∠A cannot be used to refer to an angle in the
diagram at right because there are several angles that have vertex A. There would be
no way of knowing which you meant.
MATHEMATICS RESOURCE | 77
More on Lines and Rays, and a Bit on Planes
Now that we have defined and discussed lines, rays, angles, points, and vertices, we can set up a framework for
geometry. Between any two points, there must be a positive distance. Even better, between any two points A

and B on AB , we can write the distance as AB or BA. The Ruler Postulate then states that we can set up a
one-to-one correspondence between positive numbers and line distances. This means that all distances can
have numbers assigned to them, and we’ll never run out of numbers in case we have a new distance that is soand-so times as long, or only 75% as long, etc. Think of the longest distance you can. Maybe a million
million miles? Well, now add one. A million million and one miles is indeed longer than a million million
miles. You can keep doing this forever. That’s the assumption that the Ruler Postulate seeks to say; it seems
pretty intuitive I hope.
Next, we say that a point B is between two others A and C if all three points are collinear (lying on the same

line) and AB + BC =
AC . For example, given MN as shown, (remember that we could also call it
  
MR , MT , NT , or a whole slew of other names),
M
R
N
T
We might be able to assign distances such that MR = 3, RT = 15, and TN = 13. Intuitively, RN would then
equal 28 (because 15 + 13 = 28), and MT would equal 18 (because 3 + 15 = 18). (Can you find the distance
MN?) Also note that R is between M and T, T is between R and N, T is between M and N, and R is between
M and N. There are many true statements we could make concerning the “betweenness” properties on this
line. In addition, M, R, T, and N are four collinear points.
Example:
Four points A, B, C, and D are collinear and lie on the line in that order. If B is the midpoint of AD
and C is the midpoint of BD , what is AD in terms of CD?
Solution:
Don’t just try to think it through; draw it out. The drawing is somewhat similar to the one above
with points M, R, T, and N. Since C is the midpoint of BD , BD= 2 ⋅ CD and since B is the
midpoint of AD , AD =⋅
2 BD =⋅
4 CD .
2x
A
B
x
x
D
C
Earlier, we came to a consensus concerning what exactly lines and rays are. Now we explicitly define a new
concept: the line segment. A line segment consists of two points on a line, along with all points between
them. (Note that geometry is a very logical and tiered branch of mathematics; we had to define what it meant
to be between before we could use that word in a definition.) The notation for line segments is similar to that
for lines, but there are now no arrows over the ends of the overhead bar. We can also say that the midpoint B
of a line segment AC is the point that divides AC in half; in other words, the point where AB = BC. It
should make sense that every line segment has a unique midpoint dividing the original segment into two
smaller congruent segments. Two things in geometry are congruent when they are exactly the same size: two
line segments are congruent if their lengths are equal, two angles are congruent if they have the same angle
measure (more on that later), two shapes are congruent if all their sides and angles are equal, etc. We write
MATHEMATICS RESOURCE | 78
congruence almost like the “=” sign, but with a squiggly line on top. To say AB is congruent to CD , we write
AB ≅ CD .
The final concept in this little section is the idea of the plane. Most people in geometry learn that a plane is a
surface extending infinitely in two dimensions. The paper you are reading is a piece of a plane. The geometric
definition of a plane that many books offer is “a surface for which containment of two points A and B also

implies containment of all points between A and B along AB .” Essentially, this is a convoluted but very

precise way of saying that a plane must be completely flat and infinitely extended; otherwise, the line AB
would extend beyond its edge or float above or below it. 22
We have now discussed several basic terms, but there remain several key concepts concerning lines and planes
that deserve emphasis. First is the idea that two distinct points determine a unique line. Think about this: if

you take two points A and B anywhere in space, the one and only line through both A and B is AB (which

we could also call BA , of course). Second is the idea that two distinct lines intersect in at most one point. This
is easy to conceptualize: to say that two lines are distinct is to say that they are not the same line, and two
different lines must intersect each other once or not at all. Lastly, consider the idea that three non-collinear
points determine a unique plane; related to that, also consider the concept that if two distinct lines intersect,
there is exactly one plane containing both lines. Take three non-collinear points anywhere in space, and try to
conceive a plane containing all three of them; there should only be one possible. (Do you understand why the
points must be non-collinear? There are an infinite number of planes containing any given single line.)
Similarly, if two distinct lines intersect, take the point of intersection, along with a distinct point from each
line—this gives three non-collinear points, and a unique plane is still determined!
Any plane can be named with either a script letter or three
non-collinear points. Here, this plane could be called
B
plane n, plane AEB, or even plane CEA. We could not,
however, call it plane BEC because B, E, and C are
E
collinear points and do not uniquely determine one plane.
C
A


Note also now that BC and AE intersect in only one
n
point, E. In this drawing, we say that A, B, C, and E are
coplanar points because they all reside on the same plane.
Does it make sense to you that any three points must be coplanar, but four points are only sometimes
coplanar? The fourth point may be “above” or “below” the plane created by the other three 23.
Even More on Lines, But First A Bit on Angles
Just two pages ago, we discussed the Ruler Postulate, which says, in a nutshell, that arbitrary lengths can be
assigned to line segments as long as they maintain correspondence with the positive numbers. Now, we
discuss the Protractor Postulate, which states in a similar fashion that arbitrary measures can be assigned to
angles, with 180° representing a straight angle and 360° equaling a full rotation. 24 The degree measure
assigned to an angle is called the angle measure, and the measure of ∠1 is written “m∠1.” While a straight
22
Many formal definitions in mathematics seem very convoluted upon first (or even tenth) glance, but it’s the fine points
that make these definitions useful to mathematicians.
23
Why do you think that stools with four legs sometimes wobble while stools with three legs never do? A stool with three
legs can always have the ends of its legs match the plane of the floor. If four legs aren’t all the same length, they won’t be
able to all stay coplanar on the floor at the same time.
24
As a student, I often wondered why 360° comprised a rotation. The most accepted theory is that it was arbitrarily
defined at some point in mathematical history. This may have been because 360 is divisible by so many numbers.
MATHEMATICS RESOURCE | 79
angle is 180°, a right angle is an angle whose measure is 90°. Keep in mind that on diagrams, straight angles
can be assumed when we see straight lines, but right angles conventionally cannot be assumed unless we see a
little box in the angle 25. An acute angle is an angle whose measure is less than 90°, and an obtuse angle is an
angle whose measure is between 90° and 180°. Two angles whose measures add to 90° are then said to be
complementary angles, and two angles whose measures add to 180° are said to be supplementary angles.
Adjacent angles are angles that share both a vertex and a ray. An angle bisector is a ray that divides an angle
into two smaller congruent angles. This laundry list of terms is summed up in the picture below.
•
∠ABC is a straight angle. Notice it is flat.
•
∠ABD is a right angle. (We cannot assume a right
angle on drawings, but the little box is a symbol
indicating that ∠1 is a right angle. Whenever you
see a box, feel confident you are dealing with a right
angle.)
E
2
1
A
B
3
C
•
∠ABE is an obtuse angle—its measure is greater
than 90°.
•
∠2 and ∠3 are acute angles—they measure less than 90°.
•
∠CBD is a right angle. (Because ∠1 is a right angle, ∠ABC is a straight angle, and
180-90=90.)
•
∠2 and ∠3 are complementary. Their measures add to 90° since ∠CBD is a right angle.
•
∠ABD and ∠CBD are supplementary. They combine to form straight angle ∠ABC, which measures
180°.
•
∠ABE and ∠3 are supplementary. They, too, combine to form straight angle ∠ABC, which
measures 180°.
•
∠1 and ∠2 are adjacent because they share both a vertex and a ray, as are ∠ABE and ∠3, ∠1 and
∠DBC, and ∠2 and ∠3.
•
∠1 and ∠3 are not adjacent, however; they share vertex B but no common ray.

BD is the angle bisector of ∠ABC because it divides the straight angle which measures 180° into two
smaller congruent angles, each of which measures 90°.

If m∠2 and m∠3 each happened to measure 45°, then BE would be the angle bisector of ∠DBC—
splitting ∠DBC in half.
•
•
Now that the giant list of angle terms has been
sorted through, there are just a couple more before
we can finally move on. Suppose you started at a
point and drew a ray in one direction, then drew
another ray pointed in exactly the opposite
direction. Two collinear rays such as these, with
the same endpoint, are defined to be opposite rays,
25
D
This is especially important to remember on the SAT.
N
Z
I
M
T
MATHEMATICS RESOURCE | 80
and two angles whose side rays are pairs of opposite rays are called vertical angles. For the drawing shown, we




could say that IT and IZ are opposite rays, as are IN and IM . There are then two pairs of vertical angles:
the first is ∠NIZ and ∠MIT, and the second is ∠TIN and ∠MIZ. A critical theorem concerning vertical
angles states any pair of vertical angles is congruent. 26 Therefore, ∠NIT ≅ ∠MIZ and ∠NIZ ≅ ∠MIT.
Two lines that intersect to form right angles are said to be perpendicular lines. Therefore, in the diagram on

 

the previous page, CB and BD are perpendicular. In mathematical shorthand, we say CB ⊥ BD . Two lines
that do not ever intersect are called either parallel if they are coplanar, or skew if they are not coplanar. To say
that two lines a and b are parallel, we write a || b. It is one of the most primary and fundamental tenets in
geometry that given any line and any point not on that line, there exists exactly one line through the point,
parallel to the previously given line. Analogously, it is also true that given the same circumstances, there exists
exactly one line through the point, perpendicular to the given line. 27
m
c
5
1
6
2
r
7
3 4
8
i
g
a
In the diagram here—representing three dimensions—line i is skew to both line c and line r. Line i also
intersects lines a and g and is perpendicular to both. A line is said to be perpendicular to a plane if it is
perpendicular to all lines in that plane that intersect it through its foot (its foot being the point that intersects
the plane). Line i is in this case perpendicular to plane m. Lines c and r appear to be parallel. If the distance
between them remains constant no matter how far we travel along them, then they never intersect. Both lie in
plane m, and thus they would be parallel lines. Line a, intersecting lines c and r, is called a transversal. A
transversal is a line that intersects two coplanar lines in two points; thus, line a is also a transversal for lines c
and g. There are two very useful theorems that would apply now if c and r are indeed parallel. The first is that
when a transversal intersects two parallel lines, corresponding angles are congruent. The second is that when a
transversal intersects two parallel lines, alternate interior angles are congruent. Rather than get bogged down
by definitions, consider these examples. In this diagram, there are four pairs of corresponding angles: ∠1 and
∠3, ∠2 and ∠4, ∠5 and ∠7, and ∠6 and ∠8. There are also two pairs of alternate interior angles: ∠2 and
∠7, along with ∠3 and ∠6. (∠5 and ∠4, along with the pair of ∠1 and ∠8, would be called alternate
exterior angles.) The words “interior” and “exterior” refer to the angle placement in relation to the parallel
lines, and the words “corresponding” and “alternate” refer to the angle placement in relation to the
transversal.
26
In mathematics, an axiom or postulate is something considered true without any proof. It lays the foundation. A
theorem is something proven true (either with axioms/postulates or with other theorems) to establish more mathematics.
Can you prove the theorem “Vertical angles are congruent” ? Hint: it involves comparing supplementary angles.
27
Actually, in non-Euclidean geometry, these two statements are not necessarily true.
MATHEMATICS RESOURCE | 81
r
c
1 2
5 6
3 4
7 8
a
Examples:
Assume c || r and answer the following questions.
a) If m∠4 = 70°, then what is m∠7?
b) If m∠6 = 100°, then what is m∠8?
c) If m∠2 = 80°, then what is m∠7?
d) If m∠1 = 95°, then what is m∠4?
e) If m∠5 = x°, then what is m∠3?
f)
If a⊥r, is a⊥c?
g) Are lines i and r skew? Are lines i and c? Are lines i and g?
Solutions:
a) ∠4 ≅ ∠7 because they are vertical angles. Therefore, m∠7=70°
b) ∠6 ≅ ∠8 because they are corresponding angles. Therefore, m∠8=100°
c) ∠2 ≅ ∠7 because they are alternate interior angles. Therefore, m∠7=80°
d) ∠1 ≅ ∠3 because they are corresponding angles. Then ∠1 is supplementary to ∠4 because ∠3 is
supplementary to ∠4. m∠4 = 180° - 95° = 85°
e) ∠5 ≅ ∠7 because they are corresponding angles. Then ∠7 is supplementary to ∠3 so m∠3 =
180-x.
f)
If a⊥r, it means that ∠3, ∠4, ∠7, and ∠8 are all right angles. Thus, ∠1, ∠2, ∠5, and ∠6 are all
right angles (corresponding and alternate interior angles) so a⊥c.
g) i and r are skew; i and c are skew. i and g, however, are not because they intersect. Lines are skew
only if they are non-coplanar and do not intersect.
Angles have another method of measure as well. Here, we have been concerned with angle measures in
degrees. When angles have some non-integer number of degrees, however, the fractional part of the whole can
MATHEMATICS RESOURCE | 82
be measured in minutes and seconds. Just as within any hour of a day, the term minute refers to 1/60th the
length of the hour, each minute in this case refers to 1/60th of a degree. Likewise, each second becomes 1/60th
the length of a minute. For notation, minutes are represented with a single hash mark, and seconds are
represented with a double hash mark.
Degrees, Minutes, and Seconds
An angle measure can be represented as d° m’ s’’, read aloud as “d degrees, m minutes, s
seconds, where each minute is 1/60th of one degree, and each second is 1/60th of one
minute. This is occasionally referred to as dms notation.
Examples:
1) Convert 32.5° into dms notation.
2) Convert 60.75° into dms notation.
3) Convert 71° 28’ into degrees.
4) Convert 20° 49’ 17’’ into degrees.
5) Convert 112.69917° into dms notation.
Solutions:
1) In the same way that half an hour is 30 minutes, we know that half a degree is 30 minutes. Without
even needing to haul out our calculators, we make the conversion that 32.5° = 32° 30’
2) Here is another easy example we can do without a calculator. Three-fourths of an hour is 45 minutes
when we refer to time, and as in the previous example, the same applies when we work with angles.
60.75° = 60° 45’
3) To make a conversion the other way, recall that each minute is 1/60th of a degree. Using a calculator
that can work with fractions, we have
28
71+ 60
≈ 71.467
71° 28’ ≈ 71.467°
4) Now we have seconds thrown into the mix. Each second is 1/60th of a minute, and because
1 1
1
, each second is then 1/3600th of a degree. Similarly to before, we can make a conversion
60 60 = 3600
by working with fractions:
17
49
20 + 60
+ 3600
≈ 20.8214
20° 49’ 17’’ ≈ 20.8214°
5) Once again, we want to convert ° into dms. However, unlike examples (1) and (2), this is not a
decimal that we are going to be able to convert mentally. This means that we will need to use our
calculators a bit. It is obvious that the degree measurement is going to be 112. To find the minute
measurement, we wish to know how many whole minutes are within a decimal of 0.69917.
Multiplying the decimal by 60 gives 41.9502. That means that there will be 41 whole minutes in the
angle. Notice that even though the number is close to 42, we do not round up; we are interested in
how many whole minutes we can have. This means we have 0.9502 minutes that need to be
converted into seconds. Multiplying by 60 gives approximately 57. The final conversion is
112.69917° = 112° 41’ 57’’
Our findings are summarized below.
MATHEMATICS RESOURCE | 83
Converting Angle Measurements
In order to convert an angle measured in dms into an angle measured only in
s
m
degrees, treat d° m’ s’’ as d + 60
+ 3600
In order to convert an angle in degrees into dms, multiply the fractional part by 60
to find how many whole minutes are in the angle. Multiple the minutes’ fractional
part by 60 again to find how many seconds are in the angle.
An Introduction to the Numerical Perspective of Triangles
60°
13
12
7
5
7
60°
60°
7
7
11
103.6°
7
There are three triangles shown above. 28 What can we say in describing them? The leftmost contains a right
angle and two acute angles, with no two sides congruent. The triangle in the center contains three congruent
acute angles and three congruent sides. The rightmost triangle contains an obtuse angle and two acute angles,
along with two congruent sides. There are specific mathematical terms for each of these properties. The first
set of three terms pertains to the largest angle within a triangle. If the largest angle of a triangle is 90°, the
triangle is a right triangle. If the largest angle of a triangle is acute, we call it an acute triangle, and if the
largest angle of a triangle is obtuse, we call it (logically) an obtuse triangle. The other set of terms concerns a
triangle’s sides. If a triangle has three sides with different lengths, the triangle is said to be scalene. If exactly
two sides are congruent, we have an isosceles triangle, and if all three sides are congruent, the triangle is
equilateral. 29 Lastly, the term equiangular applies to any polygon in which all angles are congruent. Above, it
looks like equilateral and equiangular refer to the same property, but that’s true only for triangles; there are
other shapes for which they are completely different. For instance, a rectangle is equiangular but not always
equilateral, and a star is equilateral but not equiangular.
Example:
Describe the three triangles above as specifically as possible.
Solution:
The triangle on the left is a right scalene triangle. The center triangle is an equilateral / equiangular
and acute triangle. The triangle on the right is an obtuse isosceles triangle.
28
I know that I should build from the ground up in mathematics (especially geometry), but I don’t feel like I need to
define “triangle” before using the word; some previous knowledge is expected.
29
The term equilateral applies not only to triangles, but to other polygons as well. (A polygon is any closed plane figure
with many sides.) Any polygon can be said to be equilateral if all its sides are congruent.
MATHEMATICS RESOURCE | 84
What else can be said about the three triangles
above? Believe it or not, there are still other facts
concerning the triangles that we have not yet
8
uncovered. One, which many students learn
very early, is known as the Triangle Inequality.
3
The triangle inequality is a theorem stating that
4
any two side lengths of a triangle combined
must be greater than the third side length. Some people understand the logic behind this theorem; if you’d
like to, find three sticks (maybe toothpicks if you are willing to work with small objects), and cut them into
lengths of the ratio 3, 4, and 8. Now try to construct a triangle with the sticks. You will find the best you can
do will be somewhat similar to what is pictured on the previous page. No matter how hard you try, the sticks
with lengths 3 and 4 are not long enough to form two sides of a triangle. The
side of 8 is just too long. This is the logic behind the triangle inequality. Two
sides must be able to reach the ends of the third side.
c
Another item that is not completely evident in the triangle pictures but is still
nevertheless important is the fact that the measures of the three angles in a
triangle always sum to 180°. (This is a fact tested in great detail on
standardized tests and one you may have learned years ago.)
Lastly, concerning the three triangles previous, we can describe the triangles
(or at least the right triangle) with the Pythagorean Theorem. Most students are
familiar with the Pythagorean Theorem from previous math courses; even
many algebra courses cover it. The Pythagorean Theorem states that in any
right triangle, the sum of the squares of the legs 30 equals the square of the
hypotenuse. If the legs are “a” and “b” and the hypotenuse is “c,” then a2 + b2
= c2. We can even check the Pythagorean Theorem with the right triangle on b
the previous page. 25 + 144 = 169 is a true equation, and thus the sides form
a right triangle. 31 Most students are familiar with the Pythagorean Theorem
but much less known is the converse of the Pythagorean Theorem: if a2 + b2 >
c2, then the triangle is acute, and if a2 + b2 < c2, then the triangle is obtuse,
where c is the longest side. (If a2 + b2 = c2, then the triangle is right.)
b
a
2
a + b =cc2
b
2
a
2
a + b2 > c2
c
a
2
a + b2 < c2
Examples:
a) Find the hypotenuse of a right triangle with legs 44 and 117.
b) Find the unknown leg of a right triangle with hypotenuse 17 and one leg 8.
c) Find the altitude of an isosceles triangle with congruent sides 41 and base 18. Also find its area.
d) Find the area of an equilateral triangle with side “s”.
Solutions:
2
a) We set a = 44 and b = 117 and use c=
a 2 + b 2 to obtain:
=
c 2 44 2 + 117 2
c 2 = 15625
c = 125
The hypotenuse has length 125.
30
Strictly speaking, we are dealing with the squares of the lengths of these various legs.
Integer possibilities for right triangles are known as Pythagorean Triples. The most common Pythagorean Triples are
3, 4, 5 and 5, 12, 13 and 7, 24, 25. Lesser known Pythagorean Triples include 9, 40, 41 and 12, 35, 37.
31
MATHEMATICS RESOURCE | 85
2
b) We set c = 17 and b = 8 and use a=
c 2 − b 2 to obtain:
=
a 2 17 2 − 82
a 2 = 225
a = 15
The missing leg has length 15.
41
41
c) The altitude of a triangle forms a right angle with respect to its base. If
the base is 18, the drawing looks like the one here.
We now realize that we are looking for the unknown leg of a right
triangle with hypotenuse 41 and leg 9. Using the same procedure as in
example (b), we find the other leg to be 40 units, which is the altitude of
the isosceles triangle in question. 12 s To find the area of this isosceles triangle, we
off all the previous math we’ve had and recall that the formula for the area of a
triangle is A = 12 bh , where b and h represent the base and height of a triangle,
respectively. This gives us:
A = 12 bh
9
18
dust
s
h
s
A =12 ⋅ 18 ⋅ 40 =360
The area is 360 square units.
d) This example is very similar to example (c). We use the Pythagorean Theorem after drawing the
triangle in question and the missing height. We find the height:
( 12 s )
2
+ h2 =
s2
h 2 =s 2 − 14 s 2 =34 s 2
=
h
3 2
=
4s
s 3
.
2
The area of a triangle, as was reviewed above, is 12 bh , so we can find the area: A = 12 bh =
1
2
⋅s⋅
s 3 s2 3
.
=
2
4
s2 3
square units. This is a fact that often
4
rears its ugly head on tests, and it may be worth memorizing. If you find it difficult to memorize, then try to
conceptualize this example to remember where the formula came from.
If we have an equilateral triangle with side of s units, the area is
The Pythagorean Theorem is indeed a theorem, meaning that it has been proven mathematically (and many,
many different proofs of it exist). Here, I will include a brief proof of the Pythagorean Theorem only because
I find it fascinating and not for any competitive purpose. Feel free to skip ahead if you are not interested.
In the drawing at right, four congruent right triangles have been laid,
corner to corner. This forms an outer square with sides of length a+b
and an inner square with sides of length c. To find the area of the inner
square, we can take c2, or we can take the area of the outer square and
subtract the area of the four right triangles. Let’s set these two
possibilities equal to each other.
a
b
b
c
( a + b )2 − 4 ⋅ ( 12 ab ) =
c2
(a
2
+ 2 ab + b
2
c
) − 2ab =
2
c
a
a +b =
c
2
2
a
c
c
b
2
The Pythagorean Theorem is proven. Pretty nifty, huh?
b
a
MATHEMATICS RESOURCE | 86
The last application arises when we consider the coordinate
plane instead of just an arbitrary one. If we have two points in
the Cartesian coordinate plane, we can actually use the
Pythagorean Theorem to find the distance between them. That
is, suppose we have two points that are diagonal. We have their
coordinates. Let’s use the Pythagorean Theorem in order to find
the distance between them. Observe the picture here.
(x2, y2)
d
(x1, y1)
b
a
Suppose we have two points. We have arbitrary points (x1, y1)
and (x2, y2). Then we want to find d, the distance between
them. From the Pythagorean Theorem, we know that a2 + b2 =
d2. Do we know how to find the values of a and b? Indeed we do. They are the differences between the points’
coordinates. That is, a is the value of x2 – x1, as it is the horizontal movement from point 1 to point 2.
Similarly, b is then y2 – y1. There’s another summary box below explaining the discovery we have just made.
Distance Formula
(as derived from the Pythagorean Theorem)
Given points 1 and 2, with coordinates (x1, y1) and (x2, y2), respectively.
The distance d between them satisfies the equation below.
d2 = a2 + b2
d2 = (x2 – x1)2 + (y2 – y1)2
d=
(x 2 − x1 )2 + (y 2 − y1 )2
Example:
Find the distance between (-2, 4) and (4, 12). Then find their midpoint and show the distance
between the midpoint and each of them is half the original distance.
Solution:
To find the distance between the points given, the first thing to notice is that we have a negative
coordinate in one of our points. That makes no difference at all! Since we will be finding the distance
between the x-coordinates, we want to find the distance between -2 and 4. This distance is 4 + 2, or
equivalently 4 – (-2). Also, it makes no difference which point we choose to be our first and which to
be the second. Suppose we chose them in the reverse order. That would just give us negative
distances; before we take the square root, however, both directions’ distances are squared. Negatives
and order make no difference to the distance formula. Here, we apply the formula outright.
d=
(4 − ( − 2))2 + (12 − 4)2=
6 2 + 82= 10
The distance between (-2, 4) and (4, 12) is 10. Now we wish to find their midpoint. I have not had a
chance to detail what many call the “midpoint formula” yet in the resource. I have faith, though, that
you can pick this up quickly: if we have two points’ coordinates and we wish to find their midpoint,
we can take the average of both coordinates. This procedure will give us new coordinates exactly in
the middle of the two points originally given. Let’s call (-2, 4) point A. Similarly, point B will be (4,
12). The midpoint of AB will then be
2
then (1 − ( − 2))2 + (8 − 4)=
( -22+ 4 , 4+212 ) = (1, 8). The distance from A to the midpoint is
2
32 + 4=
5 . The distance between the midpoint and point B is then
MATHEMATICS RESOURCE | 87
(1 − 4)2 + (8 − 12)2 =
( − 3)2 + ( − 4)2 = 5 . The midpoint of AB is then 5 away from the endpoints
of AB , which has length 10.
A Brief Continuation of the Numerical Perspective of Triangles
Anytime you encounter a right triangle, the Pythagorean Theorem will
45°
apply; however, there are two “special right triangles” that have
additional properties as well. One is the
isosceles right triangle, or the 45-45-90 triangle,
30°
and the other is a right triangle in which the
45°
hypotenuse is twice the length of one of the
legs, the 30-60-90 triangle. Just as the names
imply, a 45-45-90 triangle has two congruent
45° angles in addition to its right angle, and a 30-60-90 triangle has angles of
60°
30° and 60° in addition to its right angle. What makes these triangles
“special” is that the relationships between the sides are known and (relatively)
easy to commit to memory. x
Examples:
a) Prove the relationships of the 45-45-90 triangle and the 30-60-90 triangle.
Find x in the diagram, given that ABDC is a square.
A
Solutions:
In an isosceles right triangle, we have two legs that are
congruent. That means that for the Pythagorean
Theorem, a and b are equal. If we set a= b= x , then we
can
solve
for
the
hypotenuse.
2
2
2
c= a + b
c 2 = x 2 + x 2 = 2x 2
=
c
x
C
30°
10
Q
D
B
=
2x 2 x 2
Now for the 30-60-90 triangle, we turn our attention to a triangle with angles of 30°, 60°, and 90°.
The best place to find one is hidden inside the equilateral triangle. As in example (d) on the previous
page, an altitude in an equilateral triangle creates two congruent 30-60-90 triangles. Since these two
smaller triangles are the same size, we know the hypotenuse has twice the length of the shorter leg. If
we say that the hypotenuse is c = 2 x , and the shorter leg is a = x , then we can solve for b.
a2 + b2 =
c2
2
b=
c 2 − a2
b 2 = (2 x )2 − x 2 = 4 x 2 − x 2 = 3 x 2
=
b
=
3x 2 x 3
This resource has not had a chance yet to detail the specifics of special quadrilaterals. From a young age,
though, most people seem to know that a square has four congruent sides and four right angles. The diagonal
of a square bisects the angles so that it divides the square into two congruent 45-45-90 triangles; the diagonal
= AC= CD
= BD
=
of the square is then the hypotenuse of those triangles. Thus, AB
now the value of AB, so the value of QA = AB ⋅
3
2
= 5 2⋅
3
2
=
5 6
2
.
= 5 2 . We know
10
2
MATHEMATICS RESOURCE | 88
An Introduction to the Abstract Concepts of Triangles
Very early on, geometric congruence was defined similarly to algebraic equivalence. If two line segments are
congruent, their lengths are equal; if two angles are congruent, their angle measures are equal. What does it mean,
then, to say that two triangles are congruent? It means, in short, exactly what one
D
would intuitively expect: that all the corresponding parts of the triangles are
N
congruent. To say that ∆DAN ≅
U
∆CHU means that ∠D ≅ ∠C, ∠A ≅
∠H,
and
∠N
≅ ∠U. It also means that DA ≅ CH ,
AN ≅ HU , and DN ≅ CU . Would
correct in this case to say that
A
∆DNA≅∆HCU? The answer is no.
When congruence between two polygons is written, it is written in an
so that corresponding parts of the two polygons are congruent. If we
it be
H
C
order
were
to write ∆DNA ≅ ∆HCU, that would imply AN ≅ UC , which is not one of the congruencies listed earlier.
One of the core concepts and drills practiced in every high school geometry class is the proof. Usually written in a
two-column form, the geometric proof is a series of logical statements proceeding from a list of givens to a desired
conclusion, where each assertion is justified by a mathematical reason (a theorem, postulate, definition, or
property in almost all cases). While some students enjoy the proof as a fun exercise in logic, most loathe it for its
apparent pointlessness. 32 The formal two-column proof will not be tested. The logic behind it, however, is still
quite necessary in order to be successful.
Being able to prove the congruence of triangles is a
difficult skill to master and takes up a majority of the
year in many geometry courses. Just how many pieces
of two triangles must be congruent before we can
know for sure that the entire triangles are congruent?
For example, in the two triangles here, if we wanted to
prove ∆RUD ≅ ∆OCK and we knew only that
C
U
O
K
D
RU ≅ OC and UD ≅ CK , would we be able to R
conclude that the two triangles were congruent? If we
could indeed conclude the triangle congruence, we would automatically know then that all of the corresponding
angles were congruent, and that the third sides were congruent. Unfortunately, in this case, we would not be able
to conclude the triangle congruence. We have only two congruent sides, and we would need also to know that
either the included angles were congruent or the third sides were congruent. The conditions sufficient for proving
triangle congruence are listed below; the logic behind these theorems is a bit too complicated to detail in this
resource.
•
•
32
SSS theorem – if two triangles have all three pairs of their corresponding sides congruent, then those two
triangles are congruent
SAS theorem – if two triangles have two pairs of corresponding sides congruent, along with the
corresponding angle between those sides congruent, then those two triangles are congruent.
It is true that the practice of proof develops logical reasoning skills, but, quite frankly, no one in “the real world” will ask
you to prove a building is a rectangular prism. Even my teacher admitted it once. Just keep it quiet. – Craig
MATHEMATICS RESOURCE | 89
•
•
•
ASA theorem – if two triangles have two pairs of corresponding angles congruent, along with the
corresponding side between those angles congruent, then those two triangles are congruent.
AAS theorem – if two triangles have two pairs of corresponding angles congruent, along with a
corresponding, non-included side congruent, then those two triangles are congruent.
SSA/ASS ambiguous case – if two triangles have two pairs of corresponding sides congruent, along with a
corresponding angle congruent that is not between those two sides, we cannot conclude that those
triangles are congruent.
This will probably seem like old hash to anyone who has had a geometry course but overwhelming for anyone
new to geometry. If you fall in the latter category, take a breather to commit these to memory; you may also wish
to spend a day with a geometry book to practice some proof before continuing. Otherwise, these next examples
might overwhelm you. For those of you who have had a geometry course before, let’s look at a few brief examples,
the first of which explains the ambiguous case. If these proofs do not seem to resemble anything you are used to,
remember that the two-column proof is not the only valid type of proof. Here, I will simply offer the “paragraph
proof.” 33 In addition, you will do well to remember that in testing situations, figures are rarely drawn to scale.
Example:
E
Attempt to prove ∆DEM ≅ ∆DEI given only that EM ≅ EI .
Solution:
We know trivially that ∠D ≅ ∠D and that DE ≅ DE by what
is called the reflexive property. We are also given that EM ≅ EI .
This means we now have two congruent corresponding sides (
M
I
DE & DE ) and ( EM & EI ) along with a congruent D
corresponding angle that is not between the sides (∠D). If it
were possible, we could assert a triangle congruence by the SSA theorem, but the triangles are very
obviously not congruent. In this case SSA is indeed ambiguous in that two obviously different triangles
can be formed with ∠D, side DE , and a third side equal in length to EM.
Example:
 
Prove that in this “shape” ∆PET ≅ ∆RTE, given that PE || TR
 
and that PT || ER . 34
Solution:

 
If we consider that PE || TR and that ET is a transversal
intersecting those parallel lines, we can note the alternate
E
P
T
 
interior angle congruence ∠PET ≅ ∠RTE. If we then consider the other parallel lines PT || ER with the
same transversal, we note another alternate interior angle congruence ∠PTE ≅ ∠RET. Now we use the
reflexive property to say that TE ≅ ET (trivially, because they are obviously the same line segment!). This
gives us the congruence of two corresponding angles and the included side, so we can conclude that
∆PET ≅ ∆RTE by the ASA theorem.
33
… frankly, because I think two-column proofs are too constrictive and sometimes restrict logic instead of letting it flow.
To maintain the logical structure of geometry, I avoid using the term “parallelogram” here because this resource has not yet
defined the term. If you have had geometry before, then you may realize this is a parallelogram; we are proving the diagonal
of a parallelogram divides it into two congruent triangles.
34
R
MATHEMATICS RESOURCE | 90
Example:
Prove HI ≅ HW given that CI ≅ TW ,
T
W
I
C
HC ≅ HT , and ∠T ≅ ∠C.
Solution:
The three given congruencies in this
problem make proving a triangle
congruence somewhat easier (even
though triangle congruence is not our
final goal in this proof). Because
H
M
CI ≅ TW , HC ≅ HT , and ∠T ≅ ∠C, I can say that ∆HTW ≅ ∆HCI
by citing the SAS theorem. One of the key characteristics, then, of
congruent shapes is that the parts of congruent shapes are congruent. In
other words, all corresponding parts of congruent triangles are congruent.
(Many books abbreviate this reason in two-column proofs as “CPCTC.”)
A
R
At any rate, we know that ∆HTW ≅ ∆HCI so we know that HI ≅ HW
because all corresponding parts of congruent triangles are congruent.
Example:
In the drawing shown, MT ≅ MN , MA ≅ MR , ∠MAR ≅ ∠MRA, and
 
AR || TN . Prove that AN ≅ RT .
T
N
Solution:
There are probably several equally valid ways of going about this proof. Here is one possibility. We are


 
given that AR || TN . We then have two transversals that intersect the parallel lines ( MN and MT ) so we
know that pairs of corresponding angles are congruent. That is, we know ∠MAR ≅ ∠MTN and ∠MRA
≅ ∠MNT. We are told that ∠MAR ≅ ∠MRA, so the transitive property tells us that ∠MTN ≅
∠MNT. 35 We will need this particular angle congruency a bit later. Now we examine the other given
information and see what else we can deduce. MT ≅ MN and MA ≅ MR so, subtracting the second pair
of congruent sides from the first (a procedure guaranteed by the subtraction property of equality to
produce two congruent line segments), we can get AT ≅ RN . Furthermore, TN ≅ NT by the reflexive
property, so we can prove ∆ATN ≅ ∆RNT by citing the SAS theorem ( AT ≅ RN , ∠MTN ≅ ∠MNT,
and TN ≅ NT ). We can then say that AN ≅ RT because CPCTC. (The acronym was introduced in the
previous example.)
There are two things to note as we end these examples on triangle congruencies. The first is that none of these
examples cited the SSS or AAS congruency theorems; the procedure for those, however, is essentially the same: the
needed congruencies must first be established, and then the theorem can be cited. The second thing to note is that
the last example was quite challenging. Avoid panicking if you found it difficult to follow; geometric proofs can
become very complicated. Only with practice will you find yourself more comfortable with them.
35
The transitive property as it pertains to geometry is very similar to the transitive property in algebra. In algebra, if x = y and
y = z, then the transitive property tells us that x = z. For a geometric application, simply replace the = with ≅ and you have it.
In this particular geometry problem, we’ve deduced a ≅ b and c ≅ d, and we were given a ≅ c, so our conclusion is b ≅ d.
MATHEMATICS RESOURCE | 91
Revisiting the Numerical Perspective on Triangles
Now that we have discussed the geometry of triangle congruence in detail, you may think it is time to move on to
a new topic. In fact, it seems odd that we covered triangle congruencies at all, as they are not explicitly listed on
the curriculum outline for this year. However, it is useful surprisingly often to know that two triangles are
congruent. Sometimes, only then will you be able to find unknown measures of lengths and angles. Amazingly,
and perhaps unfortunately, there is still more. One theorem that might have made some of the preceding proofs
easier is the Angle-Side Theorem. 36
The Angle-Side Theorem: If two sides of a triangle are congruent, then the angles opposite those sides are
congruent. If two angles of a triangle are congruent, then the sides opposite those angles are congruent. If two
sides are not congruent, then the angles opposite those sides are not congruent, and the larger angle is opposite the
longer side. If two angles are not congruent, then the sides opposite those angles are not congruent, and the longer
side is opposite the larger angle. In the drawing at left, the identical tick marks signify congruence between the
appropriate parts. In the drawing at right, the side and angle with double tick marks are greater in length and
measure than the side and angle with single tick marks.
H
The Angle-Side Theorem is quite versatile and can
be used in a wide variety of proofs and
calculations. Here are two examples.
Example:
Prove HA ≅ HR given that KA ≅ LR and ∠K ≅
∠L.
K
36
A
R
L
Most books treat the parts of this “Angle-Side Theorem” as four separate theorems. I lump them together for two reasons.
First, I think you are bright enough to understand all of it at once. Second, I want this resource to be as concise as possible;
since two-column proofs are absent from the curriculum, the concepts are more important than theorem distinctions.
MATHEMATICS RESOURCE | 92
Solution:
This proof is identical to an example given earlier except that there are now only two given statements
and not three. Perhaps the Angle-Side Theorem can shed some light on the “missing” given. If we know
now that ∠K ≅ ∠L, we can cite the Angle-Side Theorem and know that HK ≅ HL . With this
congruence and the two congruencies given, we can assert ∆HKA ≅ ∆HLR by the SAS theorem. Then
HA ≅ HR because CPCTC.
Example:
Make the most restrictive inequality possible for the length of this triangle’s unknown side. 37
Solution:
Because of the Triangle Inequality (stated two sections
x
ago), we know that x + 3 > 7 and 3 + 7 > x. Therefore,
3
x must be greater than 4 and smaller than 10. Wait,
though, there’s more… let’s apply the Angle-Side
50°
Theorem. The unlabeled angle must be 70° (because
60°
7
the three angles together must add to 180°), and the
Angle-Side Theorem tells us that the comparative
lengths of sides opposite non-congruent angles correspond to the sizes of those angles. So then, since 50°
< 60° < 70°, the sides must satisfy the relationship 3 < x < 7. Thus, the most restrictive statement we can
make about x is 4 < x < 7.
The last triangle topic we need to address now
is the concept of similar triangles. What does it
mean to say that two things are similar? In
literature and English classes, it means that
those two items share certain characteristics or
traits. In geometry, the term “similar” takes on
a very specific definition.
Similar Polygons: Two shapes are similar if all of
their corresponding angles are congruent and
the ratios between corresponding sides are
constant. We write triangle ABC similar to
triangle DEF as ∆ABC ∼ ∆DEF
L
12
8
B
9
m
R
A
6
X
n
U
Example:
Find the unknown sides m and n given that ∆LAX ∼ ∆BUR.
Solution:
We know that the ratios between corresponding sides are equal, and we need only to set up a proportion
between the two triangles’ side lengths.
37
With trigonometry, we could find the value of this missing side. With only geometric methods, though, our capabilities are
more limited. This example concerns the information available from geometry—sorry to all you knowledgeable trig experts
out there. By the end of this resource, we’ll all be knowledgeable trig experts, provided I do my job right.
MATHEMATICS RESOURCE | 93
LA XL
=
BU RB
12 8
=
9 m
We also set up the other proportion necessary to find n:
LA XA
=
BU RU
12 6
=
9 n
With cross-multiplication in proportions, we get the final two equations 12m = 72 and 12n = 52 and the final
m 6;=
n 4.5 .
answers are=
This is all very interesting, of course, but is there anything more to it? I’m afraid so. If you examined the term list
at the beginning of this section, you saw some terms that seemed unsettlingly close to the theorems used to prove
triangles congruent. The list is exactly what your intuition tells you. Earlier, the SSS, SAS, ASA, and AAS
theorems were used to prove two triangles congruent, each theorem means that two triangles are identical in
enough respects to declare the triangles wholly congruent. Proving triangle similarity is done in much the same
manner. Triangles may exist with 0, 1, 2, or 3 angles congruent, and they may exist with 0, 1, 2, or 3 side lengths
in proportion. How many corresponding angles must be congruent, and how many side lengths must be in
proportion before we may assert that two triangles are necessarily similar? The triangle similarity theorems are
listed below. Note if you are comparing them to the triangle congruence theorems that ASA and AAS have jointly
been replaced with a single AA similarity theorem.
•
•
•
SSS similarity theorem – if two triangles exist such that all three pairs of corresponding side lengths form a
constant ratio, then the two triangles must be similar
SAS similarity theorem – if two triangles exist such that two pairs of corresponding side lengths form a
constant ratio and the angles included between those sides are congruent, then the two triangles must be
similar
AA similarity theorem – if two triangles exist such that two pairs of corresponding angles are congruent,
then the triangles must be similar 38
Example:
 
Given FW || LA , find x.
F
6
Solution:
First we should try to prove triangle similarity. We are
 
given that FW || LA , and we observe that the parallel


lines have two transversals, LW and AF . With the
two transversals, we then have two pairs of congruent
alternate interior angles: ∠F ≅ ∠A and ∠W ≅ ∠L.
38
L
x+1.5
D
x
A
9
W
Remember that theorems can be proven from axioms, postulates, and other theorems. Given that AAA is a postulate (If all
pairs of corresponding angles between two triangles are congruent, the triangles are similar.), could you prove the AA
similarity theorem? Hint: It involves the number 180.
MATHEMATICS RESOURCE | 94
Those two pairs of congruent angles are enough to cite the AA ∼ theorem and say that ∆DFW ∼ ∆DAL.
We then set up a proportion between corresponding sides.
DF DW
=
DA DL
6
9
=
x x + 1.5
9=
x 6( x + 1.5)
x =3
Example:
H
Prove the following statements given the
diagram at right.
HB2 = OB ⋅ BU
HU2 = OU ⋅ BU
HO2 = OB ⋅ OU
B
O
U
Solution:
We know that ∠BHU ≅ ∠BOH because they are both right angles. We also know that ∠HBU ≅
∠OBH by the reflexive property. (It may be named differently, but it is still the same angle, and it must
be congruent to itself.) With those two angle congruencies, we cite AA∼ and say that ∆UHB ∼ ∆HOB.
HB OB
. In any proportion,
We set up a proportion between corresponding sides and say that
=
BU BH
however, we can cross-multiply, and this gives us the sought equation. HB ⋅ BH =OB ⋅ BU , or
2
HB=
OB ⋅ BU .
We know that ∠HOU ≅ ∠BHU because they are both right angles. We also know that ∠OUH ≅
∠HUB by the reflexive property. With those two angle congruencies, we cite AA∼ and say that ∆HOU ∼
HU BU
. We then cross∆BHU. We set up a proportion between corresponding sides and say that
=
OU HU
2
multiply within our proportion: HU ⋅ HU =OU ⋅ BU , or HU
=
OU ⋅ BU .
In part (a), we successfully proved that ∆UHB ∼ ∆HOB. If we rearrange the order of the lettering, we can
equivalently say that ∆BHU ∼ ∆BOH. In part (b), we successfully proved that ∆HOU ∼ ∆BHU. We
know that a pair of similar triangles must have all their angles congruent. Therefore, if two different
triangles are similar to the same triangle, they must be similar to each other since all the corresponding
angles are still congruent. Thus, saying both ∆BHU ∼ ∆BOH and ∆HOU ∼ ∆BHU means that ∆BOH
∼∆HOU. 39 [The completion of this example is left as an exercise for the reader.]
[These three theorems are known as the altitude-to-hypotenuse theorems and may be worth memorizing.]
39
I suppose one might call this the transitive property of similarity if he or she were so inclined.
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A Plethora of Parallelograms, and They Brought Their Friends
Only two examples so far in this resource have dealt with special quadrilaterals. (A quadrilateral is any polygon
with four sides.) In one, we proved the diagonal of a parallelogram creates two congruent triangles. In the other,
we used the properties of 45-45-90 triangles on the two identical triangles created when a square is cut by its
diagonal. The parallelogram and square are just two of a group of special quadrilaterals. The special quadrilaterals
of concern to us, along with their definitions and a list of the major properties of each, are given below.
Shape
Mathematical
Definition
Pertinent Information & Useful Properties
Trapezoid
a quadrilateral with
exactly one pair of
parallel sides
The two parallel sides are known as the bases. A trapezoid may or may
not be an isosceles trapezoid, in which the two non-base sides (called
legs) are congruent. If a trapezoid is isosceles, the pairs of base angles
are congruent, as are the diagonals. In addition, sometimes a trapezoid
with one right base angle is called a right trapezoid.
Parallelogram
a quadrilateral with
two pairs of parallel
sides
Not only are both pairs of parallel sides parallel, they are also
congruent. In addition, opposite angles are congruent, and consecutive
angles are supplementary. Also, the parallelogram’s diagonals bisect
each other.
Rectangle
a parallelogram
containing at least
one right angle
Not only is at least one angle a right angle, all four angles are right
angles. All properties of parallelograms apply, and the diagonals are
congruent in addition to bisecting each other.
Rhombus
a parallelogram
containing at least
one pair of
congruent adjacent
sides
Not only is one pair of adjacent sides congruent, all four sides are
congruent. All properties of parallelograms apply, and the diagonals are
perpendicular bisectors of each other. In addition, the diagonals bisect
the angles and form four congruent right triangles.
Square
a parallelogram that
is both a rectangle
and a rhombus
All four angles are right angles, and all four sides are congruent. All the
properties of both rectangles and rhombuses apply, and in addition the
diagonals now form four congruent 45-45-90 right triangles (a.k.a.
right-isosceles triangles).
Example:
A rhombus has sides of length 9 and angles of measure 60°, 60°, 120°, and 120°. Find the lengths of its
diagonals.
Solution:
When a picture is not given, it is usually a good idea to draw one labeled with what we
know and what we need. We need to remember also that it is a property of rhombuses
that the diagonals bisect the angles. So then, each of the four congruent right triangles is
a 30-60-90 triangle with a hypotenuse of 9. A 30-60-90 triangle with a hypotenuse of 9
has a shorter leg of
9
2
and a longer leg of 9 2 3 . Each of those legs makes up half of a
diagonal. The diagonals then have lengths 9 and 9 3 .
9
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Examples:
Classify each of the following statements as true or false. Justify your answer.
a) All squares are rectangles.
b) Some rhombuses are squares.
c) Some rectangles are rhombuses.
d) No trapezoids are parallelograms.
Solutions:
a) TRUE – Squares have four right angles and any parallelogram with at least one right angle qualifies as
a rectangle.
b) TRUE – Squares are parallelograms with four congruent sides and any parallelogram with at least one
pair of congruent adjacent sides qualifies as a rhombus. All squares are rhombuses; some rhombuses
are squares.
c) TRUE – To be a rhombus, a parallelogram must have four congruent sides. It is possible for a
rectangle to have four congruent sides (after all, some rectangles are squares) so it is true that some
rectangles are rhombuses.
d) TRUE – A trapezoid is defined to have exactly one pair of parallel sides. A parallelogram has both
pairs of sides parallel. It is obviously impossible for a quadrilateral to have both one pair of parallel
sides and two pairs of parallel sides. Being a trapezoid and being a parallelogram are mutually
exclusive conditions; no trapezoids are parallelograms.
Circular Logic
We now turn our attention from things polygonal to something far more fun: circles. Circles appear everywhere in
day-to-day life 40, from the tires on cars, to hula hoops, to traffic roundabouts, to the shape of the earth. Besides,
standing and “spinning in triangles” or “spinning in parallelograms” causes much less of a blissful, dizzy blur, and
it makes you look even more ridiculous than spinning in circles would.
It’s obvious what circles are: everyone has dealt with them since
childhood. Unfortunately, the fact that circles are round is not a
mathematical definition. The mathematical definition of “circle”
is the set of all coplanar points that are the same distance from a
fixed point in the plane. That fixed point is called the center, and
the uniform distance is called the circle’s radius. The word
“radius” also sometimes refers to a line segment that connects the
center to a point on the circle (instead of referring to the distance
itself). Circles are named by their center point. In the picture
here, point O is the circle’s center, and the circle has radius 7.
We could also say that AO , RO , and NO are radii of circle O.
(Remember, the radius can refer either to a segment connecting
the center to a point on the circle or to the length of such a
segment.)
40
Those who are pretentious might use the word “ubiquitous.” I just did.
T
R
7
I
O
A
M
N
MATHEMATICS RESOURCE | 97
Example:
What are AO and ON in the
diagram of the circle?
N
Solution:
In a circle, all radii are congruent
since the distance from the center of
the circle to any point on the circle is
constant. Thus, AO = NO = 7.
and NT . There is a special term for such
segments: a chord is any segment that
connects two points on the circle. One of the
three chords shown passes through the center
T
Q
What else do we have in our picture of a
circle here? There are three line segments that
connect two points on the circle: AR , RN ,
I
m
Z
(chord RN ) and any chord that passes through the center is, as most people probably know, called a diameter—
chord RN , then, is a diameter. Like “radius,” the word “diameter” can refer to either the line segment itself, or
the length of such a segment (so in this case the statements “the diameter is 14” and “ RN is a diameter” are
equally correct). In addition to the chords and radii, two other points are shown in the picture, points M and I.
Point M rests comfortably inside the circle while point I is sadly neglected on the outside of the circle. 41 Point M
is considered to be in the interior of the circle—the distance from M to the center is less than the radius. Point I,
logically then, is considered to be in the exterior of the circle—its distance from the center exceeds the radius.
While it may seem that point I is receiving the short end of the stick and does not get to enjoy being in the inner
circle, external points are far from insignificant. Any line passing through an external point of a circle can do one
of three things. (1) not intersect the circle, (2) intersect the circle at exactly one point, or (3) intersect the circle at
two distinct points. There has already been a veritable deluge of mathematical terms thrown at you regarding
circles, but there are still a few more.
A line fitting case (2) is a tangent line, and a line fitting case (3) is a secant line. In the figure showing circle Q,



lines m and TZ are secant lines and NI is tangent to the circle. Point N is the point of tangency for NI .
Example:
Given circle O and the congruence HU ≅ HB ,
∠U ≅ ∠B without citing the Angle-Side Theorem.
H
prove
Solution:
Remember that in a circle, the radius has constant length, and thus
all radii are congruent. That means OB ≅ OU . Also, OH ≅ OH
because of the reflexive property. We can combine these two
statements with the given ( HU ≅ HB ) to assert ∆HOB ≅ ∆HOU
by the SSS theorem. Then ∠U ≅ ∠B because CPCTC.
41
poor point I
B
O
U
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Example:
Any polygon whose vertices all touch a given circle is said to be inscribed in the circle; any polygon whose
sides are all tangent to a circle is said to circumscribe the circle. If equi-lateral triangles are both inscribed
in and circumscribed about a circle, then what is the ratio of the sides between the two triangles?
Solution:
This problem is extremely difficult and will combine many of the concepts thus far. Our diagram will
contain two equilateral triangles with a circle between them. One important fact that will help us is that if
all of the altitudes are drawn in an equilateral triangle, then six 30-60-90 triangles are formed. In the
inscribed triangle, the hypotenuse of these smaller triangles will be equal to the radius. In the
circumscribed triangle, the shorter leg of these triangles is equal to the radius.
O
30°
M
30°
R
r
W
0.5r
60°
D
Before we continue, be sure you can prove on your own that these 30-60-90 triangles do indeed form
when the appropriate lines are drawn. We can see that an angle similarity forms, giving us ∆ODR ~
∆MDW. We also know that all radii in a circle are congruent, so MD = DR. We can assign an arbitrary
length to this radius so I’ll call it r. Because of the 30-60-90 triangles, DW = 0.5r. The proportion
between corresponding sides in the large triangles will be the same as the proportion between any
corresponding parts, so the ratio we are looking for is the same as the ratio of DW to DR. The sides are in
ratio 1 to 2.
The Angle-Secant Invasion of Normalcy
The chords, tangents, and secants that have been covered as they pertain to circles
are unfortunately not a means to an ends. While these line segments exist and sit
comfortably through or next to their circles, they are blissfully unaware that they
too (much like triangles and quadrilaterals) are subject to geometry’s frustratingly
extensive description by properties. Our ultimate goal, at least with regard to
circles, is a thorough understanding of all the properties between the lengths of
tangents, the lengths of secants, the lengths of chords, and the angles they all form.
N
M
100°
R
O
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We’ll begin our investigation with a look at the angles associated with circles, but before we even get to any
specific angles, there is one key concept that must be stated: One full rotation comprises 360°. Look at the circle
to the right and imagine that N is not fixed but instead moves freely around the edge of the circle. Can you see
that if N traveled all the way around the circle, the radius ON would pass through or “sweep out” a full rotation?
It’s that circular rotation that is described in the geo-numeric statement “A circle contains 360°.”
Now look again at the diagram, and notice the darkened section. That particular shape is called an arc. An arc is
mathematically defined as two points on a circle along with all the points connecting them along the circle; arc
NR is written similarly to line segment NR , except that the line above the letters is curved to make a small arc. An
arc is a fraction of the circle’s circumference; its measure is linear. More important than the arc’s length, though, is
the central angle that intercepts the arc. A central angle in a circle is an angle whose vertex is the center of the
circle, and the measure of the central angle is equal to the measure of the intercepted arc. In the circle here, the
measure of arc NR is clearly 100°. More accurately, though, I should say that the measure of minor arc NR is
100°. Notice that the points N and R actually create two distinct arcs: the smaller, darkened piece that we’ve been
calling arc NR and the longer, undarkened, as-yet-unnamed piece. The official designation of that large arc is
major arc NMR—when naming a major arc (any arc covering more than half a rotation is “major”), we include a
third point to indicate the endpoints of the arc, along with a point through which the arc passes. In this particular
case, it should be reasonably clear that major arc NMR has measure 260°, because minor arc NR and major arc
NMR must together form a complete circle of 360°.
Example:
Find the measure of the darkened arcs in the given circle.
A
Solution:
We know that the measure of arc MI is the same as m∠MOI,
which is given to be 25°. To find the highlighted major arc, we
have to find the measure of ∠AON and subtract from 360.
m∠AON= 22° + 25° + 15°= 62° . Thus the measure of arc ACN
is 360 – 62, or 298°.
M
22° 25° 15°
C
I
N
O
In addition to the central angle, there are several other angles related to
circles. If perceptive, you saw the list of vocabulary terms at the beginning of this section. On the next page are six
pictures of circle-related angles. See if you can label them on your own before turning the page again to find out
what they are. Use the list of vocabulary terms from above: inscribed angle, chord-tangent angle, secant-secant
angle, secant-tangent angle, tangent-tangent angle, and chord-chord angle.
MATHEMATICS RESOURCE | 100
I
H
J
G
∠I is a ____________________ angle.
m∠I =
L
A
B
K
M
N
Q
P
∠LNM is a _____________________ angle.
m∠LNM = m∠PNO =
S
R
O
C
U
V
D
T
∠Q is a ___________________ angle.
m∠Q =
X
W
∠U is an __________________ angle.
m∠U =
Z
ℵ
β
E
Y
F
α
∠X is a __________________ angle.
m∠X =
δ
∠αβδ is a _________________ angle.
m∠αβδ =
MATHEMATICS RESOURCE | 101
How many of the angles were you able to identify? The second line of each description was probably a bit
hard to complete without the needed information first. The first lines, though, should have all been fairly
intuitive based on the definitions of secants, tangents, and chords. At any rate, you should have managed
to write down most of the angle
classifications. 12 ∠I is a secant-secant angle.
m∠I =
∠Q is a secant-tangent angle.
m∠Q =
∠X is a tangent-tangent angle.
m∠X =
∠LNM is a chord-chord angle.
m∠LNM = m∠PNO =
∠U is an inscribed angle.
m∠U =
∠αβδ is a chord-tangent angle.
m∠αβδ = mβδ
If those are not the labels that you selected, then go
back and write them in correctly. Now, for our grand anti-climax, the mathematical definition of each type of
angle follows.
•
A secant-secant angle is an angle with vertex exterior to a circle and sides determined by secants to the
circle.
•
A secant-tangent angle is an angle with a vertex exterior to a circle and sides determined by a secant and a
tangent to the circle.
•
A tangent-tangent angle is an angle with a vertex exterior to a circle and sides determined by tangents to
the circle.
•
A chord-chord angle is an angle formed by two intersecting chords.
•
An inscribed angle is an angle with vertex on the circle and sides determined by two chords.
•
A chord-tangent angle is an angle with vertex on the circle and sides determined by a tangent and a
chord.
Now, we must move on to the measurement of these types of angles. There are three general theorems that deal
with the measurements of circle-related angles. The proof supporting the theorems is complicated, but the
theorems themselves are not.
If the vertex of a special circle angle lies in the exterior of the circle, the measure of the angle is half the difference of
the intercepted arcs. [This applies to the secant-secant angle, the secant-tangent angle, and the tangent-tangent
angle.]
If the vertex of a special circle angle lies in the interior of the circle, the measure of the angle is half the sum of the
intercepted arcs. [Note that this only applies to the chord-chord angle.]
If the vertex of a special circle angle lies on the circle, the measure of the angle is half the intercepted arc. [This
applies to the inscribed angle and the chord-tangent angle.] This is like a special case of the first two properties;
the “second arc” now has measure of 0°.
Now you can complete the empty page. Be extra careful to note the alternate equation to find a tangent-tangent
angle.
MATHEMATICS RESOURCE | 102
Example:
Find w, x, y, and z in the two diagrams given.
T
120°
Y
z is m∠YLR,
not m∠YLO
16.5°
x°
160°
80°
w°
z° L
O
y°
R
165°
Solution:
In the first circle, we see that w and x are two of the arc measures in a circle. They are also the intercepted
arcs of a 16.5° secant-secant angle. The remaining arcs are given as 120° and 165°. This gives us the twovariable system below.
w−x =
33
2w = 78
(w − x ) =
16.5
⇒
⇒
⇒ x =6
w+x=
45
w = 39
w + x + 120 + 165 =
360
1
2
In the first problem, w = 39° and x = 6°.
In the second problem, we observe that m∠TOL = 80°. This tells us that the measure of arc TL is also
80°. We hardly have to do any calculations to find y. y is the inscribed angle that intercepts arc TL, and
thus y = 40° because an inscribed angle has half the degree measure of the intercepted arc. Now to find z,
we note that z is the angle measure of a chord-tangent angle, intercepting arc LTR. The measure of arc
LTR is 240°, and the value of z must then be 120°.
In the second problem, y = 40° and z = 120°.
J
M
Example:
Given the drawing of circle O, find m∠MJK.
Solution:
65°
O
K
We see that m∠MOJ = 65°; that means that m∠KOJ = 65° also
because ∆MOJ ≅ ∆KOJ. 42 ∠MJK is a tangent-tangent angle; if
we remember the alternate formula for a tangent-tangent angle,
we know that it is supplementary to minor arc MK. We now
know that the minor arc SL has measure of 130°, so m∠MJK =
50°.
42
Can you prove this triangle congruence on your own? Here it will involve the SSS theorem.
MATHEMATICS RESOURCE | 103
Now that we have finally learned a bit about circles, we can summarize some major concepts of area that we will
need to know. A figure’s area is the size of the space it encompasses. The rectangle and square are not on the table;
I would like to assume you can find those areas without my help. The table lists the areas of circles, sectors of
circles, trapezoids, and parallelograms. One formula for area of a triangle is approached from a direction you may
not have seen before as part of the formula for parallelograms.
Circles
Parallelograms
Area = bh
h
Area = πr2
b
A parallelogram has two pairs of parallel sides. When
one side is considered the base, the perpendicular
distance to the top is the height. The area is then the
product of the base and height lengths.
Sectors of Circles
Triangles: a First Glance
bh
2
Area=
h
b
The formula for the area of a triangle is half that of
the parallelogram. Remember that the height must be
a perpendicular height to the chosen base. This
formula has already been used in this resource prior
to this point, but hopefully this will serve as a visual
reminder.
A sector of a circle is the region bounded by two radii
and an arc. Since a circle contains 360° in a full
rotation, we just wish to take a fraction of the circle’s
area. Thus, we take the area of the circle, then
multiply by the fractional part of 360 contained in
the arc measure.
MATHEMATICS RESOURCE | 104
Trapezoidsi
Area =
More on Triangle Areas
b1 + b2
bh1
2
b1
b2
b2
=s=
a+b+c
2
b
a
h
semiperimeter
c
If we have all of the side lengths in a triangle, we can
find its area. We may not have a height for a given
base, but Heron’s formula can give us the area of the
triangle anyway. (In some textbooks, this is instead
referred to as Hero’s formula.)
Trapezoids are a shape with two parallel sides, called
the bases, and two other sides, called the legs. A lot of
people conventionally have difficulty remembering
the formula for the area of a trapezoid; what I have
discovered as the best way is as follows.
Area = s(s − a)(s − b)(s − c)
(a + b + c)( − a + b + c)(a − b + c)(a + b − c)
Think of the trapezoid as a slightly disabled
=
4
parallelogram. We can still use the height
measurement, but the bases are now unequal lengths The proof and derivation of the formula is fairly
so we should just use their average instead. The area convoluted, but it can be achieved through the Law
is the “height times the average of the bases.”
of Cosines (see the Trigonometry section).
Example:
Find the shaded area at left, given that the measure of arc AB is 90
degrees and the radius of the circle is 7 units.
Solution:
The area shaded is known as a segment of the circle. A segment is a
region bounded by a chord and its corresponding arc. In order to find
this segment’s area, we must draw radii OA and OB . Then, we will
take the area of the appropriate sector and subtract the area of triangle
AOB. Since we are told that the measure of arc AB is 90, we know
that triangle AOB must be a right triangle, with base and
height each equal to 7. The area of the sector is then
90 π49
. The area of the corresponding
Area = π49 ⋅
=
360
4
h
1
1
49
right triangle is Area = bh = ⋅ 7 ⋅ 7 =
.
2
2
2
Thus, the area of the shaded segment is
49π 49 49π − 98
Area =
−
=
≈ 14.0 square units.
4
2
4
A
b
MATHEMATICS RESOURCE | 105
Example:
In the diagram given, the height of the triangle is 1.5 times the height of the parallelogram. Find the area
of the region marked A if the mutual base has length b and the total height is h.
Solution:
2
2bh
We know that the height of the parallelogram is h . Thus, the area of the parallelogram is
. The area
3
3
2
of the triangle we wish to subtract off will be half that. Its height is still h , but the formula for the area
3
1
bh
of a triangle contains an extra factor of . Thus, the area of the triangle is simply . Upon subtracting,
3
2
bh
the area of region A is also .
3
Example:
Find the shaded area if the 7 small circles are congruent and each has a radius of 1. All
of the circles are tangent to each other and are internally tangent to the large circle.
Solution:
Since all of the circles are tangent, we can draw a horizontal line across the center of
the circle to convince ourselves that the radius of the large circle must be 3. Thus, the area we want is the
area of the large circle minus 7 times the area of a small circle.
Areaπ=3 ⋅ 27− π (1 ⋅
2
)2π=
.
Example:
Find the area of the
figure in terms of x,
given that all angles
that appear to be right
angles are indeed right
angles.
5x
6x
4x
Solution:
We first need to use the Pythagorean Theorem to obtain the height of the figure. With a right triangle
having hypotenuse 5x and leg 4x, we know that the remaining leg (and height) must be 3x. Then, we
must add the areas of the rectangle and triangle.
1
1
Area =
br h + bt h =
(6x)(3x) + (4x)(3x) =
24x 2
2
2
Note that I used subscripts to differentiate between the different bases. Subscript r denotes the base of the
rectangle, while subscript t denotes the base of the triangle.
MATHEMATICS RESOURCE | 106
Example:
Two concentric [having the same center] semicircles are given.
The inner semicircle has radius 4, and the outer semicircle has
radius 9. Find the shaded area. The triangle shown is isosceles.
Solution:
We will have to perform some repeated subtractions and additions in order to find this area. The area of
the outer semicircle minus the area of the inner semicircle will give us the area of the semicircular donut
region, labeled in the first picture below. Then, taking the area of the inner semicircle and subtracting the
area of the isosceles triangle will give us the area depicted in the second picture below. [Note that it will
have to be multiplied by ½ to get the desired area.]
1
2
+•
The area in the first picture is the area of the outer semicircle minus the area of the inner semicircle, or
π 92 π 42 65π
−
= . The area of the second picture becomes the area of the inner semicircle, minus the
2
2
2
π 42 1
area of a triangle with base 8 and height 4; its area is thus
− ⋅ 8 ⋅ 4= 8π − 16 . Taking into account
2
2
65π 1
73π − 16
the factor of ½ that is shown in the picture, we have a final area of
+ (8π −=
16)
≈ 106.7 .
2
2
2
Princess Cirque and her Maidens Chordelia, Secantia and Tangentia 43
When the special angle properties of circles are entirely said and done, there is surprisingly little rote
memorization; there are just three possibilities for the placement of the angle’s vertex. When we concern ourselves
with the lengths of chords, secants, and tangents, however, simplicity is a bit more elusive.
A
5
O
B
We have already introduced and defined the chord at the beginning of the section
concerning circles. Unfortunately, we discussed little about it other than its
definition: a chord is a line segment connecting two points on a circle. The chord
itself has yet another catalogue of properties that pertain to it. In any circle, the
distance to a chord from the center is defined as the length of the perpendicular line
segment from the center to the chord. For example, in the diagram here, the
distance from the center to chord AB is 5. The following is a list of theorems
involving chords and their distances from the center.

If two chords in a circle are the same distance from the center, then they are congruent.

If two chords in a circle are congruent, then they are the same distance from the center.

If two chords in a circle are congruent, then their subtended arcs are congruent.

If two chords are not congruent, then the longer chord will be closer to the center.

If two chords are different distances from the center, the one closer to the center will be longer.

A circle’s longest chord is the diameter (it is a distance of 0 from the center).
43
I suppose this title is only creative in one of those sad, twisted ways. Luckily for me, I’m sad and twisted. - Craig
MATHEMATICS RESOURCE | 107
Chord-Chord Power Theorem: Two intersecting chords form four line segments such that the product of one
chord’s line segments equals the product of the other chord’s line segments.
The last of these theorems may seem a bit wordy and hard to understand, but
once you get the words matched with a visual it’s usually pretty clear. For the
diagram to the right, to say that “the product of one chord’s line segments
equals the product of the other chord’s line segments” means only that
DO × OE =IO × OB in this diagram.
B
D
O
E
The Chord-Chord Power Theorem is used quite frequently. Problems in
which chords intersect with specific ratios are fairly common and are not hard
I
to solve. Just remember that the product of one chord’s segments equals the
product of the other chord’s segments. Because you know that secants and tangents also occur with circles, you
probably won’t be surprised now to learn that there are theorems about those segments, too, all of which deserve
memorization. Luckily, the lists of theorems for secants and tangents are shorter and more concise than the list of
theorems for chords. There are only four theorems in total for secants and tangents of circles.
Suppose point D is exterior to circle O. The two tangent lines are drawn that pass through D; the points of
tangency are L and Y.
•
DL ≅ DY - two tangents to a circle from a common point are congruent
•
OL ⊥ LD and OY ⊥ YD - the radius drawn from the circle’s center to a point of tangency is
perpendicular to that tangent
Now suppose that a circle has a secant and a tangent from a common exterior point.
Secant-Tangent Power Theorem: The product of the lengths of the secant and its external part is equal to
the square of the length of the tangent.
Lastly, suppose that a circle has two secants from a common exterior point.
Secant-Secant Power Theorem: The product of the lengths of one secant and its external part is equal to
the product of the lengths of the other secant and its external part.
These four theorems concerning secants and tangents are reiterated graphically below.
MATHEMATICS RESOURCE | 108
L

D
O

Two tangents from a common
exterior point are congruent – in
this case,
Any radius drawn to a point of
tangency is perpendicular to the
tangent – in this case, and
Y
H
If a tangent and a secant are drawn from
a common exterior point, the product of
the secant’s length and the length of its
external part equals the square of the
length of the tangent:
in this case,
[Secant-Tangent Power Theorem]

If two secants are drawn from a common
point, the product of the first secant’s
length and the length of its external part
equals the product of the second secant’s
length and the length of its external part:
in this case,
[Secant-Secant Power Theorem]
E
S
O

U
C
U
S
H
O
N
SC × SU = SN × SH Example:
Find the measure of arc PI given that m∠P = 20°.
Solution:
P
O
All radii in a circle are congruent, and thus OP ≅ OI . By citing the Angle-Side
Theorem, we know that ∠P ≅ ∠I so m∠I = 20°. The interior angles of any triangle
must add to 180° so m∠POI = 140°. Thus, the measure of arc PI is 140°.
I
Example:
Find the radius of a circle in which a chord of length 8 is a distance 3 from the
center.
3
8
4
MATHEMATICS RESOURCE | 109
Solution:
The distance to a chord from the center is given to be the perpendicular distance. We then have the
drawing here; the radius of the circle is the hypotenuse of a right triangle. It is a recognizable 3, 4, 5 right
triangle so the radius of the circle is 5.
Example:
Three circles are externally tangent to each other. A triangle joining the centers of the three circles has
sides 9, 11, and 15. What are the radii of the three circles?
Solution:
Without a picture given, we must draw one.
We stop to draw three circles that are
R
externally tangent to each other. The two
A
largest circles contain the triangle side of
T
length 15. In the drawing we have, RI = 9,
MR = 11, and MI = 15. Let’s give RA the
value of x and see what we can do. If RA is x,
I
N
then MA = 11–x since MR = 11. All radii in
M
a circle are congruent so MN = 11–x also.
Since MI = 15, we know that NI = 15 – (11–
x) = 4 + x. Again, all radii in a circle are
congruent, so TI = 4 + x also. Knowing that
RI = 9 lastly gives us that RT = 9 – (x + 4) = 5–x. RA = x and RT = 5–x, but they are radii of the same
circle so we solve the equation x = 5–x to get the solution x = 2.5. We travel around the circles doing our
algebra substitution to get that the radii are 2.5, 8.5, and 6.5 (traveling counter-clockwise).
Example:
In this diagram, find the radius of the circle given that the
chord has length 16. Solve the problem (a) by using right
triangles and (b) by using the Chord-Chord Power Theorem.
N
T
Solution:
M
Z
To solve for the radius using right triangles, we draw OM
and label it with length r. 44 ∆OTM then is a right triangle with hypotenuse OM .
OT2 + TM2 = OM2
(r – 4)2 + 64 = r2
r2 – 8r + 16 + 64 = r2
-8r = -80
r = 10
44
Try labeling the little circles provided.
I
O
We are given that the chord has length 16, so we can begin
labeling line segment lengths. We know that NT = MT = 8.
We are also given that TI = 4. OI is a radius of the circle so
OI = r. We then know that OT = r – 4.
4
MATHEMATICS RESOURCE | 110
To use the Chord-Chord Power Theorem, we must draw radius OZ to
create diameter ZI .
ZO = r and OT = r – 4
Thus, we know that
ZT = 2r – 4.
By the Chord-Chord Power Theorem, we can calculate the radius.
ZT × TI = NT × TM
(2r – 4) × 4 = 8 × 8
8r – 16 = 64
8r = 80
r = 10
Example:
Find x and y in the given diagram.
Solution:
The problem requires us to employ the Secant-Secant Power
Theorem and the Secant-Tangent Power Theorem in the same
problem. First, we should use exterior point R because we will
only have to solve for one missing variable instead of two.
We use the Secant-Tangent Power Theorem.
RF × RP =
RL2
(x + 3) × 3 =
5
3x + 9 =
25
3x = 16
16
x=
3
2
5
R
L
3
P
y
x
O
F
5
D
4
B
After we have x, we can find y using the Secant-Secant Power Theorem and exterior point B.
We use the Secant-Secant Power Theorem.
BP × BF = BL × BD
(5 + 163 ) × 5 = (4 + y) × 4
( 153 + 163 ) × 5 = 16 + 4y
31
3
× 5 = 16 + 4y
=
155
3
48
3
+ 4y
= 4y
107
=y
12
107
3
MATHEMATICS RESOURCE | 111
Additional Polygons
Thus far, we have covered all the major topics of triangles and quadrilaterals. By far the most important aspect of
these is their definition: triangles and quadrilaterals have 3 and 4 sides, respectively. We can even name shapes
purely by the numbers of their sides. A polygon is any closed two-dimensional shape with many sides. Any
polygon with n sides can be called an n-gon, although there are some specific names that refer to certain n-gons. A
table is below with these terms.
Polygon Names
A polygon is any closed two-dimensional shape with
many sides.
number of sides n
polygon name
3
triangle
4
quadrilateral
5
pentagon
6
hexagon
8
octagon
10
decagon
A polygon is referred to as regular when all of its sides have identical length and all of its angles have identical
measure. Very rarely will any exam question concern itself with polygons that are not regular. For simplicity, we
will work with regular polygons for the rest of this section; try to study by pondering which concepts apply to all
polygons and which apply only to regular ones. We know that the interior angles of any triangle must have
measures that sum to 180°. Look at the diagrams below.
The first picture at left is a quadrilateral [not regular], while we have a regular pentagon and a regular hexagon
afterwards. In each one, I took the lower left vertex and drew diagonals to the unconnected vertices. This fills the
interior of each polygon with triangles, and each triangle’s angle measures comprise the total interior angle
measures of each polygon. The polygons with n sides have (n – 2) interior triangles in this manner, which means
that we can prove that the interior angle measures of polygons sum to (180°)(n – 2), where n is the number of
sides.
A diagonal here is defined to be a line segment connecting two unconnected vertices, drawn through the polygon.
It is a little difficult to prove the statement, but any polygon where all of the interior angles are under 180° can
have n(n2−3) diagonals drawn.
Examples:
What polygons have their total number of diagonals greater than the total interior angle measures?
MATHEMATICS RESOURCE | 112
Let’s call an exterior angle the angle formed outside the polygon when a polygon side is extended past the
polygon. (It is the angle adjacent to the polygon, the smaller of the two formed.) What can we say about
the exterior angle measure for a regular pentagon? What about for a regular hexagon? An n-gon?
Solutions:
We know from the paragraphs above that the total number of diagonals in any n-gon is n(n2−3) , and the
sum of the interior angle measures is (180°)(n – 2). We are looking for the polygons where
n(n − 3)
Multiply
both
sides
by
2
and
then
proceed
to
solve:
> 180(n − 2) .
2
n(n − 3) > 360(n − 2)
n2 – 3n > 360n – 720
n2 – 363n + 720 > 0
363 ±
( −363 )2 − 4(1)(720)
, or approximately 1.994 or 361.006. We need to examine which
Equality occurs at
2
interval will be the correct one in this case (between the two numbers or outside of it). The test value of 0
does satisfy the inequality, and this means that the solution can be written as (-∞, 1.994) ∪ (361.006, ∞).
No polygon can have fewer than 3 sides (otherwise you can’t even make a closed shape), so the first
solution interval is useless to us. Our inequality is thus only usefully satisfied for n > 361.006. Since n is
the number of sides for a polygon, this means that any polygon with at least 362 sides will satisfy the
relationship we want.
Just to clarify a bit, an external angle is drawn for both a pentagon and a hexagon below.
108°
120°
In both cases, we know from our earlier discussion that the sum of the interior angle measures is (180°)(n – 2).
That means that we can divide by n to find the measure of each interior angle. In the case of a pentagon, this is
108°, and for the hexagon, this is 120°. These angles are labeled in the pictures below.
Since straight angles must be 180°, we know that these external angles must be the difference between these angle
measures and 180. That is, for a regular pentagon, an external angle must measure 72°, and for a regular hexagon,
an external angle must measure 60°. For a regular n-gon, this means an external angle must measure
180 – 180(nn − 2)
= 180 – 180n
+ 360
n
n
=
360
n
For ANY n-gon, the measure of an external angle is just the quotient when n is divided into 360!
MATHEMATICS RESOURCE | 113
The Third Dimension
We have now finished all of our analysis of geometry in the plane. We move now into three-dimensional
concepts. Now, instead of the perimeter and area of shapes, we will find the surface area and volume of them.
When we have a shape in three dimensions, the surface area refers to the combined area of its external surface;
volume refers to the collective space that the shapes take up.
There are several types of these three-dimensional shapes to study. The first class is prisms.
A prism consists of two parallel and congruent bases, which are two-dimensional, and all the space in three
dimensions that it takes to travel from one base to the other. 45 The bases then determine the name of the prism.
In the special case that the base is a circle, the prism in question is instead called a cylinder. The distance between
the two bases is known as the prism’s height. Several different prisms are drawn here.
Technically, the third object here is a cylinder and isn’t formally called a prism. I find, however, that it helps to
think of it as a “circular prism” of sorts. All four of the prisms have the same orientation, height drawn vertically.
In reality, all prisms can be drawn in any direction. The height is always the perpendicular distance between the
parallel bases. The prisms are named after their bases. The first, second, and fourth are called rectangular,
triangular, and hexagonal prisms, respectively. Each shape’s surface area is calculated by adding several areas
together. The polygons on each prism that are not bases are called lateral faces. The total surface area is then the
areas of the lateral faces, added to the areas of the bases. The pictures below illustrate how to find the surface area
of a triangular prism. We will eventually have to know every dimension of the triangular prism in order to find
the area of all the faces at right.
Surface area of:
×2
+
45
+
To be mathematically precise, we say that corresponding vertices are joined by sets of parallel rays.
MATHEMATICS RESOURCE | 114
To find the surface area desired, we have to count all of the faces in question as separate polygons, then add all of
their areas together. In the case of the cylinder, there is only one lateral face. It is a rectangle, with one dimension
equal to the circumference of the circle. Perhaps the picture below will help.
surface area
of:
×2
r
r
h
+
h
2πr
Notice for the rectangle, the longer dimension is just the circumference of the circle. It might help to think of this
as unrolling the label off a soup can. 46 Since we know that the area of a circle is πr2, this means that the total
surface area of a cylinder can be completely determined only knowing the radius and height. The circles’ total area
is thus 2πr2, and the rectangle has area 2πrh.
Surface Area of Prisms and Cylinders
The surface area of a prism consists of twice the base area, plus the area of the lateral
faces, which will be rectangles.
The surface area of a cylinder follows the same pattern, with the lateral surface area
equal to 2πrh, where r is the radius and h is the cylinder’s height. The total surface
area of a cylinder is equal to the sum 2πr2 + 2πrh.
In addition to finding these figures’ surface areas, we are interested in finding their volumes. A three-dimensional
figure’s volume represents the amount of space that is taken up. When I defined what it meant to be a prism, I
said that it consists of two congruent bases, with all of the space needed to travel from one to the other. Now we
want to measure the amount of space taken up there. To do so, we just take the base area and then multiply it by
that height. This accounts for all of the space. The same even applies for cylinders. A prism’s volume is often said
to be equal to simply Bh, where B is the base area and h is the height.
Prism and Cylinder Volume
B
h
Any prism or cylinder has volume equal to Bh, where
B is the base area and h is the height.
(Variable h represents the distance extended between
the bases, and thus this formula can be thought of as
finding the space taken up by the prism.)
46
I seem to recall my geometry class having an exercise where we actually did this. It was called “exploration of geometry” or
something to that effect. I would have preferred to call it “busy work” or “play with your food day.”
MATHEMATICS RESOURCE | 115
This formula then means every prism and cylinder has a volume that is easily determined given the appropriate
dimensions.
Examples:
a) Given a rectangular prism with dimensions 3 x 2 x 7, find the surface area and volume.
b) There is a triangular prism where the bases are isosceles triangles. These bases have congruent sides of
length 13 and other side of length 10. The prism’s height is 7. Find the total surface area and volume.
c) Given a cylinder with radius r and height h, find the ratio of the surface area to the volume. 47
d) Given a cylinder where the height is twice the radius r, find the ratio of the surface area to the volume.
Solutions:
a) We have a rectangular prism here with given dimensions. Any rectangular prism will have six faces, and
those faces will come in congruent pairs. Two faces will have an area of 3 × 2 = 6. Another two faces will
have area of 2 × 7 = 14. The last pair of faces will have an area of 3 × 7 = 21. On the diagram below, there
is a hidden congruent face that corresponds to each face showing. For any rectangular prism’s volume
calculation, it does not matter which faces we choose to be our bases. The remaining dimension will then
represent the prism’s height. All three numbers will simply be multiplied.
surface area: 2(14) + 2(6) + 2(21) = 82
7
volume: 2 × 3 × 7 = 42
14
21
3
6
2
b) Now we have a triangular prism to analyze. The most important step
will be finding the area of the base, which is an isosceles triangle. Recall that any
isosceles triangle can have an altitude drawn, creating right triangles. This will let
us find its own height (not to be confused with the height of the prism.
surface
area of:
13
13
2×
5
+
10
7
10
+2×
To calculate this surface area and volume, we need the area of the triangle, the one with dimensions of710
and 13. With the help of the Pythagorean Theorem, we can draw an altitude of this triangle, and we find
that the triangle’s height is 132 − 52 =
12 ,as shown in the picture above. This means that we can
calculate as below:
47
Here we ignore units. Comparing cubic units and square units in a ratio is inconsistent.
13
MATHEMATICS RESOURCE | 116
surface area = 2 [ 12 • 10 • 12 ] + 7 • 10 + 2 • 7 • 13 =
372
volume = Bh =
420
( 12 • 10 • 12 ) • 7 =
c) We now have a cylinder with radius r and height h. We have a circular base, and it has area πr2. This
means the total volume is πr2h. The surface area then consists of two circles’ areas, along with a
rectangle’s. The rectangle’s area is 2πrh [see the diagram two pages ago], and the circles’ combined area
must be 2πr2. The surface area to volume ratio is
2πr 2 + 2πrh πr(2r + 2h) 2r + 2h
.
=
=
πr 2 h
πr 2 h
rh
A common factor of πr cancels from both the numerator and denominator along the way.
d) We now have exactly the same problem, with one small twist. Rather than deal with the variables r and h,
we have a situation where h = 2r. In working the same problem, we can substitute this new situation into
the answer to the previous problem.
2r + 2h 2r + 2(2r) 2r + 4r 3
=
= =
rh
r(2r)
2r 2
r
We can find the surface areas and volumes of prisms and cylinders; is there more to this? That depends on the
question. In dealing with the decathlon’s curriculum for this year, there is not more to learn than surface area
and volume when we concern ourselves with three-dimensional shapes. There are, however, shapes besides
prisms and cylinders. In a somewhat analogous fashion, there are pyramids and cones. Pyramids consist of
only one base, with the other edges no longer parallel to each other, but instead meeting above the pyramid in
a point, which we can call the pyramid’s vertex. The base is a polygon. Cones then are the same concept,
extended to the strange case when we have a circle for the base.
Since pyramids must have polygons for their bases, we know that all of the sides of the pyramids must be triangles.
To find the total surface area, we have to add the base area to these triangle areas. Finding the surface area of a
cone requires the memorization of a formula. The base has area of πr2, but the lateral area in this case is no longer
a rectangle.
Much like with prisms, the height (sometimes called altitude) is the perpendicular distance from the vertex to the
base. That is, from the vertex, it is the shortest distance possible to reach the base. Regular pyramids and cones
have a second height, known as the slant height. The slant height is the distance from the vertex to the center of a
base edge. In the case of the cone, in which there are no base edges, it is the distance from the vertex to any point
on the circular base. The diagrams below should help clarify all of these new terms.
MATHEMATICS RESOURCE | 117
slant height
height / altitude
height / altitude
Notice, among other things, that the height and slant height can combine with a portion of the base to form a
right triangle. Exam questions involving pyramids and cones will almost always find a way to test this relationship
by citing the Pythagorean Theorem as some middle step in the question-solving process.
When we are finding the surface area of these shapes, we must add the surface portions together piecemeal. For
pyramids, this will involve a polygonal base and several triangles. For a cylinder, this surface area will contain a
circle and another shape. We cannot unroll this shape like the label of a soup can; it cannot be properly described
without some moderately-advanced calculus. Study this formula directly: the lateral surface area of a cone is
πr r 2 + h 2 , where r and h are the radius and height of the cone, respectively. 48 This only goes for regular cones,
where the vertex is above the center of the circle.
When we deal with volume, we find that the volumes of pyramids and cones are closely related to their matching
prisms and cylinders. Since we know that a pyramid or cone will fit inside a prism or cylinder with the same base
and height, each must have volume smaller than Bh. It is smaller by a factor that can is difficult to determine
without more advanced mathematics. Provided we have a vertex as drawn instead of a parallel base, giving us a
pyramid or cone instead of a prism or cylinder, the volume formula is V = 13 Bh instead of V = Bh. That is, “when
the top of your shape could skewer a bird, you find volume with an extra factor of one-third.”
Surface Area and Volume of Cones and Pyramids
Pyramids
Surface Area
SA = [base area] + [area of triangles]
Cones
SA = [base area] + [lateral area]
SA = πr2 + π r r 2 + h 2
SA = πr(r + r 2 + h 2 )
Volume
V = 13 Bh
V = 13 Bh =
πr 2 h
3
Examples:
a) A regular square pyramid has lateral edges of length 4 and an altitude of 3. Find its surface area and
volume.
b) A cone of unknown height has its top chopped off. It now has two parallel circles and some lateral
area also. The circles’ radii are 4 and 8, and the slant height of the remaining shape, called a frustum,
is 5. Find the surface area and volume of this shape.
This formula is a bit hard to memorize, but I think of it as the formula for the area of a circle, πr2, with one radius factor
replaced with the slant height. The slant height is then by the Pythagorean Theorem.
48
MATHEMATICS RESOURCE | 118
Solutions:
a) We have a square base, and the lateral edge length and altitude length. We have to find the
dimensions of the square base before we can do anything useful; however, we will have to use some
triangle relationships to be able to do this. The diagram at left shows us what we have known. There
is a right triangle related in the information given, and it has missing leg length of 4 2 − 32 =7 .
The second diagram below explains how this length is related to the square. It is half the diagonal,
and the diagonal is then 2 7 .
3
4
We have a square, comprised of 45-45-90 right triangles when a diagonal is drawn. This means that a side
of the square will have length of
1
2
the diagonal’s length.
7
1
2
2=
7
1
2
2
2 7=
2
=
2 14
2
14 is the
length of the square’s side.
Since we have the altitude, we can now find the pyramid’s volume.
V = 13 Bh
=
(
)
2
=
14 ( 3 ) 14
1
3
Calculating the pyramid’s surface area now requires one more intermediate task.
There are four congruent isosceles triangles. We know that each one has base of
14 , but we do not know their heights. Luckily, drawing altitudes in isosceles
triangles allows us to use the Pythagorean Theorem to find missing lengths. Here,
we use the Pythagorean Theorem to find a missing height, given a “hypotenuse”
of 4. The drawing at right should elaborate on this. 14 height =
42 −
(
1
2
14
)
2
=
area of triangle =
16 − 144 =
( 12 ) (
14
64
4
− 144 =
) ( =)
50
2
700
=
4
50
2
4
=
100 7
4
5 7
2
total surface area = (square area) + 4(triangle area)
total surface area =
(
14
total surface area = 14 +
)
2
+ 4 ( 12 )
(
14
)( )
50
2
700 = 14 + 10 7
b) The first order of business with a problem like this is to draw the situation we have been given. There
is a cone frustum pictured below left, with the given dimensions drawn in.
Finding the surface area and volume of this frustum will take quite a few steps. When finding these
quantities, we will have to consider two cones: there will be the large one, before the top was
removed, and there will be the cone that was removed. Quantities will be figured from both cones
and then subtracted. The first step will be to find the height of the cones.
MATHEMATICS RESOURCE | 119
A
4
4
B
5
8
cone 2
cone 1
5
3
3
C
5
3
4
D
F
4
E
With the diagram of triangles at right, all the missing dimensions have been figured. If we drop a line
directly down from point C, we can figure out the height of the triangles. We already knew that BC
and DE are parallel. Since we dropped CF to be perpendicular, this makes quadrilateral BCFD a
rectangle, and ∆CEF a right triangle. By the Pythagorean Theorem, we find that CF = 3. Because of
BC
, have dimensions
the rectangle, we know that BD = 3 also. Similar triangles ∆ABC ~ ∆ADE, from DE
in the ratio 1:2, and the dimensions of ∆ABC are written in as well. This means cone 1 has height of
6 and radius of 8. Cone 2, then, has height 3 and radius of 4. Now we can do our calculations.
To find the volume of the frustum, we want to find the volume of cone 1 and then subtract the volume
of cone 2.
Volume = 13 Bcone1h cone1 − 13 Bcone2 h cone2 =
1
3
( π ⋅ 8 ) (6) − ( π ⋅ 4 )(3) = 112π
2
1
3
2
To find the surface area, we will have to add the surface areas of the two circles together first. Afterwards,
we will add the lateral surface area. To calculate the lateral area, we take the lateral surface area of cone 1
and then subtract the area of the lateral surface area of cone 2.
Surface Area = [area of circles] + {lateral surface area}
Surface Area = [area of radius 8 circle] + [area of radius 4 circle] +
{(lateral surface area of cone 1) – (lateral surface area of cone 2)}
Surface Area = [π 82] + [π 42] + {(π ⋅ 8 ⋅ 10) – (π ⋅ 4 ⋅ 5)}
Surface Area = 64π + 16π + {80π – 20π}
Surface Area = 80π + 60π = 140π
More to the Third Dimension
We can find the volumes and surface areas prisms, pyramids, cylinders, and cones; we don’t have to beat around
the bush to realize what must come next. We are going to deal now with spheres. A sphere is the last threedimensional shape we are going to study this competition year. It is defined as all the points in space within a
certain distance (known as the radius) from a certain location (known as the center). More commonly, we think
of spheres as balls, or perfectly round shapes. Everyone has seen them; to some extent, even the earth is a sphere.
The most important concepts here to learn are the formulas required to calculate the volumes and surface areas of
given spheres. All that is required to do so is the radius of the sphere.
MATHEMATICS RESOURCE | 120
Spheres
Volumes
Surface Areas
The volume of a sphere with The surface area of a sphere
with radius r is given by 4πr2.
4
radius r is given by π r 3 .
3
r
Commit these formulas to memory. They can be proved using
integral calculus (or even some advanced geometry methods),
but. for now, you will have to just take them at face value.
Example:
Find the volume and surface area of a sphere with radius 5.
Solution:
We substitute 5 for r in our formulas.
V = 34 π r 3 =
4
3
π(5)3 =
500
3
π
A = 4πr2 = 4π(5)2 = 100π
Example:
Suppose a cube is “inscribed” in a sphere; that is, all eight of its vertices just touch the sphere. What is the
ratio of the cube’s volume to the sphere’s volume?
Solution:
This is a difficult problem to solve. We should say that the radius of the sphere is r. This means that the
diagonal-to-diagonal distance across the cube is 2r. See the diagram below.
2 As the diagram at left shows, the cube’s diagonal is 2r. We can choose
to declare that the side of the cube is s. Because of the isosceles right
2r
s
s
triangles in the square on bottom, we know that the face’s diagonal is s 2
. We need to find s now, the side of the cube. By the Pythagorean
Theorem, we use the dotted triangle.
(
s2 + s 2
After doing a bit of algebra on this equation, we find that s =
2r
3
)
2
= (2r)2
. Since the volume of a cube is simply s3,
we can find the volume of the cube to be s3 = 38r 3 . We know that the volume of the sphere outside must
3
be π r . The ratio of the cube’s volume to the sphere’s volume is then
8r 3
3 3
3
4
3
4
3
πr 3
. The factor of the cube of
the radius cancels out, and we can simplify this ratio.
8r 3
3 3
4
3
πr
3
=
8
3 3
4
3
π
=
8
3 3
•
3
2
=
≈ 0.37
4π π 3
This means if a cube is inscribed in a sphere, the cube’s volume is only about 37% of the sphere’s volume.
MATHEMATICS RESOURCE | 121
IV. Trigonometry
Trigonometry has a reputation for being a particularly
difficult branch of mathematics. The Greek word Trigon means
“triangle,” while the word Metron means “measure.” Perhaps a
more accurate description of trigonometry would be as the study
of angles and their relations to each other. We defined the term
angle in the Geometry section, and we have been working with
angles ever since. Perhaps it is now time to discuss the possible things we can derive from their
relations.
Trigonometry Basics
When we wish to consider an angle in a plane, we place it in standard position. That is, its first ray goes along the
positive x-axis. Its second ray sweeps out from that position. Below are four angles, one pictured in each quadrant
of the plane. The symbol θ is a Greek letter pronounced “theta” or “thay-tuh.” It is commonly used to represent
degree measures of angles.
As we can see, angles sizes are not limited to the acute, right, or even obtuse classifications. Angles of measure
greater than 180° also exist. What’s more, angles of measure greater than 360° exist! Since 360° comprises a full
rotation, an angle larger than a full rotation can have all the properties of an angle smaller than a full rotation.
Here is an illustrated example.
y
y
0° < θ < 90°
θ
θ
x
x
90° < θ < 180°
y
y
θ
180° < θ < 270°
θ
x
x
270° < θ < 360°
MATHEMATICS RESOURCE | 122
y
y
θ = 55°
θ = 415°
θ
θ
x
x
The angles of 55° and 415° are known as coterminal angles, angles whose ending ray is at the same location. Does it
make sense to you that I could also say 55°, 415°, and 775° are all coterminal angles? What about negative angles?
Is an angle of -305° also coterminal with this list? Two more diagrams might shed some light on the subject.
y
y
θ = -305°
θ = 775°
θ
x
θ
x
Negative angles just indicate that the terminal ray rotates in the opposite direction from the positive x-axis as
usual. In this way, we can say that any real number can define an angle. We have four coterminal angles listed
here; have we found all of them, or are there more? What about -665° and 1135°? Are these two angles also
coterminal with this list? The answer is yes. For any angle θ, there are an infinite number of coterminal angles
given by the formula θ + 360°n. The variable n here can be any integer, whether it be positive, negative, or zero.
Since these angles are the same, the properties possessed by one are possessed by the others.
Example:
This resource has not yet defined what the trigonometric functions are. Nevertheless, using the concepts of
coterminal angles, we can answer some questions about them. Suppose we have an equation y = sin(θ). We
have not defined what “sin” means, but suppose someone were to tell you that sin(30°)= 12 . Find an infinite
list of angles φ where sin(φ)= 12 . φ is another Greek letter often used to represent degree measures of angles.
It is spelled “phi,” and can be pronounced “fie” or “fee.”
Solution:
We know the coterminal angles of 30° also have the properties of the angle. They are, in effect, the same
angle with different names. Since sin(30°)= 12 , we know any sin(30° + 360°n) = 12 also. We can say that φ =
30° + 360°n gives us the list we want. If you already know some trigonometry, note this is not all angles
such that sin(φ)= 12 . For the complete list, we also need to include 150° + 360°n, but this will be covered in
more detail as the trigonometric functions are explained.
We have said any real number can make a valid angle. However, as we will soon be working with trigonometric
functions, we may need to simplify the way in which we work with these angles. Any given angle, no matter how
MATHEMATICS RESOURCE | 123
large, will have a coterminal angle in the domain 0° ≤ θ < 360°. In addition, every angle, even within the interval
of 0° ≤ θ < 360°, also has what is known as a reference angle. A reference angle is the acute angle formed by the
second ray of an angle and the x-axis. For example, all four of the angles shown graphed below have reference angles
of approximately 30°, though each appears in a different quadrant of the plane.
y
y
θ
θ
x
x
y
y
θ
θ
x
x
In each case, the 30° reference angle is shown by a vertical line. The purpose of reference angles may not be
readily apparent; indeed, it seems like another term has been thrown into the works for no reason. Another page
or two of this resource will soon set the record straight. On rare occasions, reference angles even have their own
notation. For angle θ or φ, the reference angle is sometimes denoted by a small apostrophe, θ’ or φ’. These are
read aloud as “theta prime” and “phi prime,” respectively. This notation is rarely used, as these “primes” have a
completely different meaning in calculus. Nevertheless, some people do indeed use it.
Example:
Find the reference angle of a 210° angle.
Solution:
A 210° angle will look almost identical to the bottom left picture in the above set of four. Since we know
that 180° comprises a straight angle, the small portion left over (angle measure denoted by the vertical
line) must be 30°. 30° is the reference angle.
Trigonometric Functions
This section will contain the meat of the concepts of trigonometry. Trigonometry revolves around six functions,
the so-called trigonometric functions. Given any angle in the plane, we can take a point along the angle, and name
MATHEMATICS RESOURCE | 124
it with the ordered pair (u, v) since we are in a plane. This isn’t any different from (x, y), besides a different
lettering choice. Also, we’ll name the direct distance from the point to the origin as r. This gives us a right triangle
with legs of length u and v, along with a hypotenuse of length r. There is a picture here that can clear things up.
For simplicity, the picture has the given angle θ in the first
quadrant. The graph has u, v, and r labeled, with descriptions of u,
v, and r in other quadrants. Notice that r is always positive, as it is
simply the distance to the origin. We also have the Pythagorean
Theorem from geometry.
u2 + v 2 =
r2
Trigonometry describes this situation. For any given angle θ, we
always have a consistent ratio between the u, v, and r values. There
are six possible ratios, and it is these ratios that define the six
trigonometric functions.
y
Quadrant II:
u negative
v positive
r positive
r
Quadrant I:
u, v, r all positive
(u, v)
θ
Quadrant III:
u negative
v negative
r positive
u
Quadrant IV:
u positive
v negative
r positive
These definitions must be memorized. Perhaps making a mnemonic device might help.
Trigonometric Functions
Given an angle θ, we have a consistent ratio between u, horizontal coordinates; v,
vertical coordinates; and r, the distance to the origin. Those ratios are given as follows.
y
r
θ
v
u
The sine of θ, written sin(θ), is given by sin(θ) = .
The cosine of θ, written cos(θ), is given by cos(θ) = .
The tangent of θ, written tan(θ), is given by tan(θ) = .
The cotangent of θ, written cot(θ), is given by cot(θ) = .
The secant of θ, written sec(θ), is given by
sec(θ) = .
The cosecant of θ, written csc(θ), is given by csc(θ) = .
x
v
x
MATHEMATICS RESOURCE | 125
Example:
Using your knowledge from geometry, find the following exactly.
(1) sin(30°)
(4) sec(120°)
(2) cos(30°)
(5) cot(210°)
(3) tan(45°)
(6) cos(315°)
Solution:
1) We make a brief sketch here. From our knowledge of the 30-6090 triangle, we know that the ratio of v to r will be 1 to 2. That
is, sin(30°) = 12 .
2) We already have a diagram drawn for example (a) above. We
again must dust off the cobwebs for our fifteen-second old
knowledge of the 30-60-90 triangle. This time, the ratio of u to r
will give us that cos(30°) =
3
2
.
1), 2)
3)
3) We need a new sketch now. This one has to remind us of the
properties of isosceles right triangles, the 45-45-90 kind. We
know that the legs are congruent, and the ratio of v to u will
always be 1 to 1. Thus, tan(45°) = 1.
4) To take a trigonometric function of 120°, we have to draw a
120° angle from the positive x-axis. From there, we drop a
vertical segment to create a triangle. To take the secant of our
120° angle, we wish to take the ratio of r to u. Because of special
30-60-90 triangle, we know that the ratio of r to u is 2 to 1.
Moreover, we know that in the second quadrant, u is negative.
This gives us
sec(120°) = −21 = -2.
5) We have another 30-60-90 triangle, this time with u and v both
negative. This time, it’s u to v that we want. We get
cot(210°=
)
3.
− 3
−1
=
6) For our last problem, we have a Quadrant IV angle. To take the
cosine of this angle, we want the u to r ratio. This gives us:
cos(315°)=
1
2
=
2
2
4)
120°
5)
210°
6)
315°
MATHEMATICS RESOURCE | 126
For every angle that has a reference angle that is 30°, 45°, or 60°, we can draw the special triangle and calculate
the values of the trigonometric functions. We can also take trigonometric functions of 0°, 90°, 180°, and 270°. In
order to do so, we need to know that v is zero for 0° or 180° angles; the “triangle” is a horizontal line, and u = r.
This means that sin(0°) = 0, cos(0°) = 1, and tan(0°) = 0. For 90° or 270° angles, we have u = 0. sin(90°) = 1,
cos(90°) = 0, and tan(90°) is undefined. For cot, sec, and csc values, we can just take the reciprocals of the values
here. Below is a table of trigonometric functions for special angle values. Many people commit it to memory; just
remember that if you can work the examples above with no help, you can derive a value when you need it with no
extra memorization needed!
sin
cos
tan
cot
sec
csc
0°
0
1
0
undefined
1
undefined
30°
1
2
3
2
2
2
1
2
3
3
3
2 3
3
2
1
1
2
2
3
3
3
2
2 3
3
0
undefined
0
undefined
1
-2
2 3
3
2
2
3
2
1
45°
60°
90°
3
2
2
2
1
2
120°
135°
150°
0
180°
210°
225°
240°
270°
300°
315°
330°
360°
−
1
2
2
2
3
−
2
-1
−
3
2
2
−
2
1
−
2
−
0
−
1
2
2
2
3
−
2
-1
− 3
−
-1
−
−
3
3
3
3
-1
− 3
− 2
2
2 3
3
2
-1
undefined
2 3
3
-2
−
0
undefined
3
2
2
−
2
1
−
2
3
3
3
1
1
− 2
3
3
3
-2
0
undefined
0
undefined
1
2
− 3
−
2
2
3
2
1
-1
−
3
3
0
−
3
3
-1
−
2
2
− 2
−
2 3
3
-1
−
2 3
3
− 2
− 3
2 3
3
-2
undefined
1
undefined
There are some things to pay attention to in this table. First, the bottom row could have been omitted. Since 360°
and 0° are coterminal, the bottom row is the same as the top row. I decided it helps to include it because it makes
MATHEMATICS RESOURCE | 127
patterns more visible. Second, the tangent, cotangent, secant, and cosecant functions all have places where they are
undefined. The sine and cosine functions are smooth everywhere. Also, this table gives all of the trigonometric
functions for special angle values; in order to find the trigonometric value of an angle that does not generate a
special triangle, a calculator will be necessary. For example, cos(82°) ≈ .1392 by a scientific calculator. 49 In order
to find the cotangent, secant, or cosecant values of angles on a calculator, you will have to note that each is the
reciprocal of tangent, cosine, and sine, respectively. Using the calculator and the x-1 or 1/x buttons, you can find
these trigonometric functions. Also, there are inverse trigonometric functions given on your calculator, indicated
by sin-1 or cos-1. If we were told that tan(θ) = -4, we could type “-4” before pressing “tan-1” to find that θ ≈ 75.9638°. Sometimes in text, there are referred to as arcfunctions; that is, arcsin, arccos, arctan, etc.
Do you remember at the end of the last section, when we defined the reference angle, but there was no apparent
use for the concept? We can use the concept now. In looking at the trig table, we can perceive patterns in the
values. Most importantly, every angle’s trig value has the same absolute value as its reference angle’s trig value. For
example, the absolute values of sin(60°), sin(120°), sin (240°), and sin(300°) are all identical. This reference angle
uniformity is true for all six trig functions, in all four quadrants. As I noted earlier, the cotangent, secant, and
cosecant functions are simply reciprocals of the tangent, cosine, and sine functions; this means that if we
memorize the first three functions in the first quadrant, we can reproduce the rest of the table at will. [The
functions’ values at 0°, 90°, 180°, and 270° can be memorized also if we like.]
We now need some guidelines to remember which functions are positive in each quadrant. As we already learned,
this depends on whether u and/or v is positive or negative. A summary is below in table form, with a mnemonic
device listed afterwards. Remember all of the trigonometric definitions. They are listed again below for emphasis.
sin(θ) =
csc(θ) =
cos(θ) =
sec(θ) =
tan(θ) =
cot(θ) =
49
v
r
r
v
u
r
r
u
v
u
u
v
As many veteran teachers will tell you, having to work with a scientific calculator beats having to use a trigonometric table
of values or a slide rule. I think I saw a slide rule once. It was pretty nifty-looking, but I haven’t a clue how to use one.
MATHEMATICS RESOURCE | 128
u
v
r
positive functions
Quadrant I
+
+
+
all 6 of them
Quadrant II
-
+
+
sin, csc
Quadrant III
-
-
+
tan, cot
Quadrant IV
+
-
+
cos, sec
The most common mnemonic devices for this are “All students try cheating” and “All students take calculus,”
standing for all, sine, tangent, cosine. (The csc, cot, and sec can be remembered as reciprocals of the sine, tangent,
and cosine.)
Example:
Find the following “mentally.”
1) cos(315°)
2) sin(210°)
3) sec(225°)
Solutions:
1) We can make a mental picture of 315° to note that the reference angle is 45°, in Quadrant IV. We
recall that cos(45°) =
cos(315°) =
2
2
2
2
. In Quadrant IV, the cos and sec functions are positive. This gives us
.
2) One quick mental picture later, we see the reference angle of 210° is 30°. sin(30°) = 12 , but we are in
Quadrant III, where only tan & cot are positive. We have sin(210°) = - 12 .
3) We have a reference angle of 45°, located in Quadrant III. In Quadrant III, only tan & cot are
positive. Since cos(45°) =
2
2
, we have sec(45°) = 2 . sec(225°) = − 2
Trigonometric Functions and Identities
Many of the trigonometric functions have a long list of properties. They involve changing between types of
function and changing between arguments. The first properties we have already been using and taking advantage
of. When we were working with special angle values and the trigonometric functions of them, we noticed that
cotangent, secant, and cosecant were simply the reciprocals of tangent, cosine, and sine functions, respectively.
This is true because of the definition of the functions, reiterated below for emphasis. In addition, because of the
properties of fractions, we can rewrite tangent and cotangent as a quotient of sines and cosines or of secants and
cosecants.
MATHEMATICS RESOURCE | 129
v
r
r
v
u
r
r
u
v
u
u
v
sin(θ) =
csc(θ) =
cos(θ) =
sec(θ) =
tan(θ) =
cot(θ) =
r
v
θ
u
Reciprocal Properties
1
csc x
1
x=
sec x
1
tan x =
cot x
x=
Quotient Properties
csc x =
sec x =
cot x =
1
sin x
1
cos x
1
tan x
sin x
sec x
=
cos x csc x
cos x csc x
=
cot x =
sec x
sin x
tan x =
Our next properties come as a set of three; they are derived from the Pythagorean Theorem. Anticlimactically,
they are often called the Pythagorean Properties. In the diagram above, we have the right triangle drawn with the
length of the legs as u and v, and r is the length of the hypotenuse. Depending on the quadrant, the variables u
and v are sometimes negative, but in terms of the Pythagorean Theorem, that tidbit is meaningless. All negative
signs are quickly squared. Thus, we have u2 + v2 = r2. We can divide this equation through by three different
possible things: u2, v2, or r2. In each respective case, we get the following.
1+
v2 r2
=
u2 u2
2
u2
r2
+
1
=
v2
v2
2
v r
1+   =
 
u u
2
u
r
  +1 =
 
v
v
u2 v 2
+
=
1
r2 r 2
2
2
2
u v
1
  +  =
r r
From here, we can cite the definitions of the trigonometric functions to have the three Pythagorean Properties
below.
Pythagorean Properties
1 + tan2 (x) = sec2 (x)
cot2 (x) + 1 = csc2 (x)
sin2 (x) + cos2 (x) = 1
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Examples: 50
1) If cos(θ) =
3
, then find sin(θ).
4
2) If tan(θ) = -10, then find sec(θ).
3) cot2(3θ) – csc2(3θ) = __________
4) tan(cos-1(x)) = __________ [an algebraic expression in terms of x]
Solution:
1) There are two ways to solve this problem. Here is the first: we can
use the trigonometric identity: sin2(θ) + cos2(θ) = 1. This means
that sin2(θ) + 169 = 1. This gives us sin2(θ) = 167 . There are two
possible values of sin(θ). Either sin(θ) =
7
4
or sin(θ) = −
7
4
.
3
, we can draw
4
a triangle with r = 4 and u = 3. The cosine function is positive in both Quadrant I and Quadrant IV.
Here is the second: upon being given that cos(θ) =
By the Pythagorean Theorem, we can solve for v. We have the solution v =
on the quadrant of θ. Therefore, either sin(θ) =
7
4
or sin(θ) = −
7
4
7 or − 7 , depending
.
2) This problem is nearly identical to example (1) above. Negative tangent values may occur in
quadrants II and IV. In one case, sec(θ) has a positive value; in the other, sec(θ) is negative. When we
substitute into the Pythagorean Property, we get sec(θ) = ± 101 .
3) The second Pythagorean Property listed can be rewritten as cot2(x) – csc2(x) = -1. When we look at
this problem, it’s tempting to say that cot2(3θ) – csc2(3θ) = -3, but would that be correct? The answer
is that no, it would not. We can declare a “substitution” of sorts; let’s say that x = 3θ. If we do so, we
can clearly see that cot2(3θ) – csc2(3θ) = cot2(x) – csc2(x) = -1.
4) This example problem is much like examples (1) and (2) again, except that it is phrased differently. If
we wished to rewrite the question, we could rewrite it as “If cos(θ) = x, then what is tan(θ)?” This is
what the expression cos-1(x) refers to: it is an angle whose cosine is x.
1
x
We don’t have a Pythagorean Property relating cosine to tangent. In order to do this problem, we
will have to draw pictures. If cos(θ) = x, the triangle is drawn above. We are going to ignore the
complications of possibly having θ in another quadrant.
We can tell from this drawing that if cos(θ) = x, then we know the v-coordinate is =
v
1 − x 2 by the
Pythagorean Theorem. If we add this v coordinate into our drawing, we can tell tan(θ) =
50
1 − x2
.
x
The notation of cosn(x) represents [cos(x)]n. This helps to avoid confusion with cos(x)n, which sometimes means cos(xn).
MATHEMATICS RESOURCE | 131
With the reciprocal and Pythagorean properties out of the way, we can begin a discussion of other properties. It
has been my habit throughout this entire resource to derive all of the formulas and concepts presented; it seems to
be the best way to promote conceptual understanding instead of rote memorization. The following derivation is a
bit complicated; it’s hard to conceptualize something so convoluted. I found it easier in high school to simply
memorize the composite argument properties. If you do not feel the need to see the derivation, then you can (and
should) skip directly to the box on the next page that states the properties.
Let’s suppose we have a unit circle. On this unit circle, we have
two angles, α and β. We are going to assume that α > β and
that α is less than a full rotation (α < 360°). The derivation
and property are actually valid for all real numbers, but these
assumptions will help make this proof understandable. Next
we draw our angles α and β and label the points where they
intersect the unit circle. This is shown in Figure 1. As
indicated, if one has an angle α and the other has an angle β,
then the angle between them is α – β. Since we are dealing
with a unit circle with r = 1, we should take note that the
trigonometric sines and cosines of these angles are as follows.
sin(α) = v2
cos(α) = u2
sin(β) = v1
cos(β) = u1
(u1, v1)
(u2, v2)
α-β
α
β
Fig.
1
(u3, v3)
Notice that we don’t need to deal with potential negative signs;
if one of these functions is negative, it is already taken care of
because we are using the coordinates of each point, which can
be negative. Each then has a denominator of 1 since r = 1. We
will need these functions later.
(1, 0)
Fig.
2
To prove our property, we will need to rotate this α – β angle so
that one point lies along the positive x-axis at
(1, 0). The other point will have new coordinates (u3, v3). Since
we still have a unit circle and one ray is now placed on the positive x-axis, we can also take the trigonometric
functions of α – β. This rotation is shown in Figure 2.
cos(α – β) = u3
sin(α – β) = v3
Because these two diagrams involve a congruent angle α – β, we know the distance from (u1, v1) to (u2, v2) is
identical to the distance from (u3, v3) to (1, 0). Now, we want to apply the distance formula to say that these
distances are equal.
( u 2 − u1 )
2
+ ( v 2 − v1 ) =
2
( u 3 − 1)
2
+ ( v3 − 0)
2
u 2 2 − 2u1u 2 + u12 + v 2 2 − 2v1 v 2 + v12= u 3 2 − 2u 3 + 1 + v 3 2
It appears like we can’t do a lot more simplification than that, but remember that all three of the points listed are
on the unit circle. By the Pythagorean Theorem51, we know that u12 + v12 = u 2 2 + v 2 2 = u 3 2 + v 3 2 = 1 . By
51
Isn’t it amazing how many things are based on the Pythagorean Theorem?
MATHEMATICS RESOURCE | 132
subtracting 2 from each side, we can simplify our lives greatly. Afterwards, we can multiply both sides of our
equation by − 12 .
−2u1u 2 − 2v1 v 2 =
−2u 3
=
u 3 u1u 2 + v1 v 2
Earlier, we noted the values of the trigonometric functions of our α and β angles. We can substitute those in here
to finally prove our property that cos(α – β) = cos α cos β + sin α sin β. This long convoluted process can be
sin( 
− )
done over in different fashions to arrive at four properties. Also, if we recall that tan(
− )=
cos( 
− ) , we can
derive a composite argument property for the tangent function.
Composite Argument Properties
cos(α − β) =
cos(α)cos(β) + sin(α)sin(β)
cos(α + β) =
cos(α)cos(β) – sin(α)sin(β)
sin(α − β) =
sin(α)cos(β) – cos(α)sin(β)
sin(α + β) =
sin(α)cos(β) + cos(α)sin(β)
tan(α + β) =
tan(α – β) =
tan 
+tan
1 − tan 
tan
tan 
−tan
1 + tan 
tan
Since the derivation is near impossible to conceptualize and recall at a whim, these properties are easier to
memorize than derive. Here is how I remember them: the arguments of the functions will always be α, β, α, β.
The cosine becomes cos-cos-sin-sin and changes the sign. The sine becomes sin-cos-cos-sin and keeps the sign.
Tangent “distributes” the function across the argument and creates a denominator of 1, opposite sign, and a
product.
Example:
Find exact values (in terms of square roots) of the following.
sin(15°)
cos(255°)
Solution:
In order to find these values, we can use the composite argument properties to our advantage. This
requires a bit of ingenuity on our part, but we just need to create an argument of 15°.
sin(15°) = sin(45° – 30°)
= sin(45°)cos(30°) – cos(45°)sin(30°)
=
2
2
⋅
3
2
−
2
2
⋅ 12
6− 2
4
cos(255°) = cos(210° + 45°)
= cos(210°)cos(45°) – sin(210°)sin(45°)
=
MATHEMATICS RESOURCE | 133
( ) ⋅ ( ) − (− ) ⋅ ( )
= −
=
3
2
2
2
1
2
2
2
− 6+ 2
4
With the Composite Argument Properties defined, we can begin using them to find even more 52 new
properties. If we apply the composite argument properties with the arguments all “θ + θ,” we can derive
the double-angle formulas.
sin(θ + θ)
cos(θ + θ)
tan(θ + θ)
= sin(θ)cos(θ) + cos(θ)sin(θ)
= cos(θ)cos(θ) – sin(θ)sin(θ)
=
tan ( 
) + tan ( )
1 − tan ( 
) tan ( )
= 2 sin(θ)cos(θ)
= cos2(θ) – sin2(θ)
=
2tan (  )
1 − tan 2 (  )
Also, remember the Pythagorean Property that sin2(x) + cos2(x) = 1. If we substitute this into the
property just derived for cos(2θ), it gives us two alternate forms. Our new formulas are summarized here:
Double Angle Properties
sin(2θ) =
2 sin(θ) cos(θ)
cos(2θ) =
cos2(θ) – sin2(θ)
=
1 – 2 sin2(θ)
=
2 cos2(θ) – 1
2tan (  )
1 − tan 2 (  )
tan(2θ) =
Example:
If cos(θ) = 34 , find cos(2θ) and sin(2θ) without a calculator. Assume that θ is in Quadrant I.
Solution:
In a previous example, we calculated that if cos(θ) =
sin(θ) =
cos(2θ)
= 2 sin(θ) cos(θ)
= cos2(θ) – sin2(θ)
= 2⋅
=
( 34 )
=
1
8
7
4
⋅ 34
and θ is in Quadrant I, then
. Using this and the double argument properties, we can find the desired values.
sin(2θ)
= 3 87
52
7
4
3
4
2
−
( )
7
4
2
Yes, I’m afraid there are still more properties – quite a few of them in fact. They keep going and going and going.
Remember that trigonometry is usually taken over the course of a semester, between 15-20 weeks of school. Covering the
entire subject concisely means packing in a lot of information.
MATHEMATICS RESOURCE | 134
Next after double angle properties, we wish to find some half angle properties. In order to do so, we want
to start with the double angle property for cosine.
cos(2θ) = 1 – 2 sin2(θ)
cos(2θ) = 2 cos2(θ) – 1
cos(2θ) – 1 = -2 sin2(θ)
cos(2θ) + 1 = 2 cos2(θ)
sin2(θ) =
1 − cos ( 2
2
)
cos2(θ) =
1 + cos ( 2
2
)
Now, we need a renaming of variables. We can say that 2θ = ϕ. That means that θ = 2 . The formula for tangent
takes a bit more manipulation but also appears when a different form of its double angle property is used. The
appropriate square root remains to be taken. Whether it is positive or negative depends on the quadrant in which
φ lies.
Half Angle Properties
 
sin2   =
2
1 − cos ( 
2
)
 
cos2   =
2
1 + cos ( 
2
)
 ϕ  1 − cos ( 
tan2   =
 2  1 + cos ( 
)
)
Example:
6− 2
. Use the half-angle properties to calculate sin(15°) in a
4
different way. Does it yield the same result? [It better; it’s the same value!]
Earlier, we found that sin(15°) =
Solution:
To use the half angle property, we declare φ = 30°.
sin2(15°) =
1 − cos(30°)
2
sin2(15°) =
1 − 23 2 − 3
=
2
4
Now, we know what sin2(15°) is, but in order to find sin(15°), we have to know whether we are taking
the positive or negative square root. Since 15° is in Quadrant I, we take the positive root, and we obtain
sin(15°) =
2− 3
.
2
It looks as if this answer is completely different from the one obtained earlier. If we take a calculator,
6− 2
however, we find that =
4
2− 3
≈ 0.2588 . The two values are exactly the same, as we desired.
2
MATHEMATICS RESOURCE | 135
We are nearing the end of our trigonometric properties. There are two sets of properties left to contend with.
They can be derived from the Composite Argument Properties. Most likely, these properties will not be tested, as
they are exceedingly complex and difficult to memorize. Nevertheless, no trigonometry resource can be complete
without them. My personal recommendation is not to memorize them unless you somehow find yourself with
unexplained copious amounts of free time.
Sum to Product Properties
+ 
 
 - 
sin α + sin β = 2 sin 
 cos 

2


 2 
− 
 
 + 
sin α – sin β = 2 sin 
 cos 

2


 2 
+ 
 
 − 
cos α + cos β = 2 cos 
 cos 

 2 
 2 
+ 
 
 − 
cos α – cos β = −2 sin 
 sin 

 2 
 2 
Product to Sum Properties
sin α sin β
=
1
2
[cos(α − β) – cos(α + β)]
cos α cos β
=
1
2
[cos(α − β) + cos(α + β)]
sin α cos β
=
1
2
[sin(α − β) + sin(α + β)]
These properties allow for the conversion of trigonometric expressions between sum and product forms. The sumto-product properties are often used to assist in the solving of trigonometric equations; the product to sum
properties are sometimes used in calculus to make integration possible.
And ultimately, it sometimes becomes necessary to convert a linear combination of a sine and cosine into a single
function. Said differently, it becomes necessary to change A sin x + B cos x into the form C sin (x – D). In order
to do so, we can apply the composite argument property to the right hand side and analyze the pieces that result.
A sin x + B cos x = C sin (x – D)
A sin x + B cos x = C sin x cos D – C cos x sin D
A sin x + B cos x = (C cos D) sin x – (C sin D) cos x
D is a constant, and thus sin(D) and cos(D) are also. This means we can break apart this last equation to find:
A = C cos D
B = -C sin D
A2 = C2 cos2 D
B2 = C2 sin2 D
A2 + B2 = C2 cos2 D + C2 sin2 D = C2 (cos2 D + sin2 D) = C2
Thus, C = A 2 + B2 , cos D =
A
B
, sin D = − .
C
C
After C is known, there will only be one unique angle D that satisfies the other two criteria. Using a similar
derivation, we could also combine these two functions into a cosine. The two cases are presented below.
MATHEMATICS RESOURCE | 136
Linear Combination of Sine and Cosine
We can combine A sin x + B cos x into a single trigonometric function.
A sin x + B cos x = C sin (x – D),
where C =
cos D =
A
C
A 2 + B2
sin D = −
B
C
A sin x + B cos x = E cos (x – F)
where E =
A 2 + B2
A
C
sin F =
cos F =
B
C
The ability to take a linear combination of sine and cosine proves most useful when solving trigonometric
equations. Examples of its application will be delayed until their need arises later. It will be soon though.
This concludes our discussion of trigonometric properties. While it’s true that there is a massive list of properties,
We turn our attention now to a practical application of trigonometry. As is often surprising, math does indeed
have practical applications! Trigonometry especially seems to have these weird, wacky, and wonderful applications
to real life. Thus far, we have always been working with angles in standard position. That is, one ray lies along the
positive x-axis, and the position of the other ray determines the angle measure. Unfortunately, working directly
with angles has quite a limited usefulness. We can turn our attention specifically to right triangles in order to
begin finding the practical applications we want. Defined as a subset of the standard trigonometric functions, we
have triangles where one right angle is already included. Then for another angle θ, we have an opposite leg and an
adjacent leg. The opposite leg can serve the purpose of v, and the adjacent leg can serve the purpose of u. The
length of the hypotenuse h serves the purpose of r.
Trigonometry Inside a Right Triangle
sin θ
=
cos θ
=
n
θ
=
opposite
o
=
hypotenuse h
adjacent
a
=
hypotenuse h
opposite o
=
adjacent a
csc θ
=
sec θ
=
cot θ
=
hypotenuse h
=
opposite
o
hypotenuse h
=
adjacent
a
adjacent = a
θ
opposite = o
hypotenuse = h
adjacent a
=
opposite o
There are many different mnemonic devices to remember these ratios. The most common ones are “soh cah toa”
[pronounced so-kuh-toe-uh] and “Oscar had a heap of apples.” In both cases, the sine, cosine, and tangent are
memorized in that order with the numerator and denominator remembered as letters. In the first case, “soh cah
toa,” we see the abbreviation for “sine: opposite-hypotenuse, cosine: adjacent-hypotenuse, tangent: oppositeadjacent.” In the second case, “Oscar had a heap of apples,” we find “opposite-hypotenuse, adjacent-hypotenuse,
opposite-adjacent” in the letters. As usual, the cosecant, secant, and cotangent can be memorized as reciprocals of
the three first ratios; often, they are completely unneeded as scientific calculators already have sin, cos, and tan
buttons standard.
MATHEMATICS RESOURCE | 137
With the use of these trigonometric ratios, we can solve for any right triangle in which one of two conditions
holds. 53
We have the lengths of two sides of the triangle.
We have the measure of one angle (aside from the right angle) and the length of one side.
Methods will usually differ depending on what information we are given. However, the general strategy is fairly
consistent. We will analyze the information we have and use a trigonometric ratio where we have only one
unknown. We can solve for that unknown and then find it using our handy-dandy calculator.
Example:
Find the angles in the following right triangle.
Solution:
11
10
Let’s call the top angle θ and the angle on the bottom right φ. We could solve for the missing leg using
the Pythagorean Theorem if we wished to, but there is no need since it is not requested.
In reference to the top left angle θ, we have the lengths of the opposite leg and the hypotenuse. The sine
.
function uses these values. Thus, we know that sin(θ) = 10
11
θ = sin-1 ( 11 )
10
θ ≈ 65.38°
Now, for the bottom right angle φ, we instead have the lengths of the adjacent leg and the hypotenuse.
This means we have the necessary information for the cosine ratio.
cos(φ) =
10
11
φ = cos-1 ( 11 )
10
φ ≈ 24.62°
Note that after we obtained θ, we could have used the fact that every triangle’s internal angle measures
sum to 180° to find φ. There are always multiple ways to solve every problem.
Example:
Use the cosine and tangent ratios to find the missing lengths in the triangle given here.
y
Solution:
We don’t even have to figure out the correct ratios to use; that mystery is solved for us
in the problem.
cos(40°) =
53
5
y
tan(40°) =
40°
x
5
x
5
y cos(40°) = 5
x = 5 tan(40°)
5
y = cos(40
°)
y = 6.53
x = 4.20
When I say we can “solve” for a right triangle, I mean we will be able to solve for every side length and angle measure
within it. No measurement is safe; we can find them all.
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There may be some slight variations from these two examples, but the underlying principles remain the same. We
first examine which measurements we have. After we decide which trigonometric ratios we have enough
information for, we solve for the missing part or parts. All in all, it is a consistent and predictable process. Look to
the appendix in order to learn the Law of Cosines and Sines to learn more.
Radians
In upper-level mathematics, it grew necessary to take the trigonometric functions of real, imaginary, and complex
numbers (instead of simply angles). Mathematicians began needing a way to relate a given angle based in degrees
to the real numbers. This didn’t prove too daunting a task: as we discussed earlier in the geometry section, the
value of 360° in a full rotation was rather arbitrarily defined a long time ago. It has no intrinsic mathematical
meaning, and it is therefore possible to choose a different arbitrary value. If you were a mathematician in such a
predicament, what arbitrary value would you choose?
We can consider a circle of radius 1, the so-called unit circle. In such a circle, we have a
circumference of 2π. The most convenient thing we can do now is to declare a quantity of
2π radians in a circle. That’s right, if we have to choose an arbitrary number, we might as
well make it the same as the circumference of a circle of radius 1. That means if we have an
arc length of 3, then the central angle of that arc has a measure of 3 radians, provided we are
dealing with the unit circle. Each radian is slightly less than 1/6 of a rotation. As always,
there is a diagram. In the diagram, we have an arc length of s subtended by an angle of s
radians.
arc length s
s
Notice I did not write “s° radians.” There is no superscript label for the radian. It is simply written as s. Let’s look
now at how to convert between radian angle measures and degrees. We know that there are 2π radians in a full
rotation, as there are also 360 degrees. Since any angle measure is just a fraction of a rotation, we can set up a
proportion.
r
d
=
2π 360
360r
= 2π ⋅ d
r=
180
π
r
d or d =
π
180
Some common angle measures are compared below.
30°
= π6 radians
45°
= π4 radians
60°
= π3 radians
120°
= 2π3 radians
135°
= 3π4 radians
150°
= 5π6 radians
For example, tan( π6 ) =
3
3
. The angle name may have changed, but no other properties have changed significantly.
MATHEMATICS RESOURCE | 139
The Laws of Sine and Cosine
We have gained an amazing ability using trigonometry: using a limited
amount of information, we can now solve for all the measurements of any
right triangle, provided we are given enough initial information.
Unfortunately, this does nothing for us when we are given a triangle that is
not a right triangle. What can we do then? Do we sit idly by as the exam
leprechauns mock us to a bitter end? Of course not. Just because a triangle has
no right angle is no reason to balk and stare. Let’s start with a (non-right)
triangle and see what sorts of things we can deduce.
B
c
a
h
A
C
b
It is common convention in geometry to label side lengths with lowercase letters; the angle opposite a side is then
labeled with the corresponding capital letter. In the diagram below, the triangle is named ABC, and the angles will
also be called, ∠A, ∠B, ∠C. An altitude of length h has been drawn in. By penciling in this altitude, we have
created right triangles. As we discussed in the last section, we can now use right triangle trigonometry on these
individual triangles. When we do, we will find that sin(A) = hc and sin(C) = ha . Solving these for h, we find two
equations.
h = c sin(A)
h = a sin(C)
Since these are both equal to the height, we can set them equal to each other.
c sin(A) = a sin(C)
With this equation, we can divide both sides by the product of sines to arrive at a proportion.
a
c
=
sin(A) sin(C)
Had we drawn the altitude h in another direction, we would have instead found the same proportion with one of
b
the sides replaced by
. This leads us to the Law of Sines. The Law of Sines says that, in a triangle, the ratio
sin(B)
between the length of a side and the sine of its opposite angle is constant.
Law of Sines
If triangle ABC has legs of lengths a, b, and c, each opposite its
corresponding angle, then
a
b
c
= =
sin(A) sin(B) sin(C)
sin(A) sin(B) sin(C)
= =
a
b
c
The Law of Sines is a general law that applies to all triangles, not just right ones. Using it, we can conditionally
solve for the majority of unknowns in a triangle as long as an angle opposite a side is known.
Example:
Solve for the unknowns in the pictured triangle.
c
32°
a
35°
7
C
MATHEMATICS RESOURCE | 140
Solution:
We know that the angles of a triangle must have measures that add to 180°. From this, we immediately
know that m∠C = 113°. Since we know a side of length 7, we can use the Law of Sines.
sin(32°) sin(35°)
=
7
a
a=
7sin(35°)
sin(32°)
sin(32°) sin(113°)
=
7
c
c=
a ≈ 7.58
7sin(113°)
sin(32°)
c ≈ 12.16
On its surface, the Law of Sines appears to be a wonderful cure-all for any triangle where a side and its opposite
angle both have their measurements known. As you might expect, there is a small catch. If we have an SSA case,
that is, a case where we have the lengths of two sides next to each other and the measure of an angle across from
the first side, then we may not be able to determine a unique solution. There may be two correct solutions or even
none at all! The case of the SSA triangle is known as The Ambiguous Case. In order to understand the ambiguous
case, we have to go back to our original understanding of the sine ratio. Recall that in Quadrants I and II, sin(θ) is
positive. Since angles in Quadrant II can have the same reference angles as angles in Quadrant I, we see that it is
true that sin(15°) = sin(165°) and sin(40°) = sin(140°). More generally, it will always be true that sin(θ) =
sin(180° - θ). When using the Law of Sines, if we arrive at a statement such as sin(θ) = 12 , we have two possible
solutions! θ = 30° or θ = 150°; both are valid solutions. This generates the ambiguous case. Examples tend to be
the best explanation. The first example below illustrates the ambiguous case when there are two correct solutions.
Afterwards, there is an example where there is no solution at all!
Example:
Solve for the triangle where a = 4, b = 7, and m∠A = 28°.
Solution:
Without even having to draw a picture, we can set up the desired proportions with the Law of Sines.
sin(A) sin(B)
=
a
b
=
sin(B)
b sin(A) 7sin(28°)
=
≈ .821575
a
4
This is the point where the ambiguous case becomes a problem. If sin(B) ≈ .821575, then we don’t know
if either m∠B = 55.24° or m∠B = 124.76°. A calculator gives us that B = sin-1(.821575) = 55.24°, but it
could also be that B = (180 – 55.24)° = 124.76° as we discussed above.
Since the angles of a triangle always have measures that add to 180°, knowing m∠B means that we can
determine m∠C.
If m∠B = 55.24°, then m∠C = 96.76°. The Law of Sines gives us c = 8.46.
If m∠B = 124.76°, then m∠C = 27.24°. The Law of Sines gives us c = 3.90.
Both of these are valid solutions! Their respective pictures follow.
MATHEMATICS RESOURCE | 141
96.76°
27.24°
7
7
4
28°
124.76°
28°
4
55.24°
8.46
3.90
As this example just showed, there are two possible triangles with consecutive sides of lengths 4 and 7,
and a 28° angle opposite the side of length 4. Let’s look at another example.
Example:
Solve for the triangle where a = 3, c = 1, and m∠C = 40°.
Solution:
First, we use the Law of Sines to try to find m∠A.
sin(40°) sin(A)
=
1
3
sin(A) = 1.928
There is no solution for m∠A. There is no angle with a sine
greater than 1. The sine function only varies between –1 and +1.
A quick picture can show us why. The side of length 1 isn’t long
enough to reach the other side at the bottom.
3
1
40°
The conclusion that we have to reach is that in order to use the Law of Sines with any amount of reliability, we
have to have two angles. Only then we can know the shape of the triangle.
What, then, do we do if we only have the measurement of one inconveniently placed angle? Worse yet, what do
we do if we don’t have the measurements of any angles at all? There is another law that can help us out in this
case. To see its derivation, we want to place a triangle in the coordinate plane. We will place one vertex at the
origin and one along the positive x-axis.
y
B
c
a
C
b
A
x
When we analyze this picture, we notice that since point A was placed along the positive x-axis, its coordinates are
(b, 0). The coordinates of B are a bit harder to figure, but since point C is at the origin, the distance “a” can serve
as “r” in our trigonometric ratios. The coordinates of B are (a cos(C), a sin(C)).
To arrive at the law we desire, we want to use the distance formula between points A and B and set the answer
equal to c.
MATHEMATICS RESOURCE | 142
c=
(b − a cos(C))2 + (0 − a sin(C))2
c2 = (b – a cos(C))2 + (-a sin(C))2
c2 = b2 – 2ab cos(C) + a2 cos2(C) + a2 sin2(C)
c2 = b2 – 2ab cos(C) + a2 (cos2(C) + sin2(C))
c2 = a2 + b2 – 2ab cos(C)
Law of Cosines
In any triangle, if ∠A, ∠B, and ∠C are opposite sides
a, b, and c, respectively, then
c2 = a2 + b2 – 2ab cos(C)
Unlike the Law of Sines, the Law of Cosines has no ambiguous case. If we have the necessary information for the
Law of Cosines to solve for c or ∠C, we will be able to solve for that unknown with complete certainty. This is
because we know that if cos(θ) is positive, then θ must be in Quadrant I, under 90°. If cos(θ) is negative, then θ
must be in Quadrant II, between 90° and 180°. No triangle can have an angle with measure greater than 180°;
we do not need to concern ourselves for solutions of θ in Quadrants III and IV.
Since ∠C is opposite side c, if we are solving for an angle measurement, we should intentionally declare “c” to be
its opposite side. If we are solving for a side length, either make ∠C its opposite angle or make c and ∠C to be
two known measurements. If we are ever solving for “a” or “b,” we may still have an ambiguous case. A quadratic
equation may result.
Example:
5
Solve for the unknowns in the triangle pictured.
50°
Solution:
Since we have the angle of 50°, we need to declare its measurement to be
m∠C. In that way, we can first solve for c.
8
c2 = a2 + b2 – 2ab cos(C)
c2 = 52 + 82 – 2(5)(8) cos(50°)
c2 = 25 + 64 – 51.423
c2 = 37.577
c = 6.130
After we have solved for c, we wish to solve for our missing angles. In choosing to solve for the large one at the
top, we rename “c = 8” and wish to solve for m∠C.
c2 = a2 + b2 – 2ab cos(C)
82 = 52 + 6.1302 – 2(5)(6.130) cos(C)
82 − 52 − 6.1302
cos(C) =
= −0.023215
−2(5)(6.130)
C = cos-1(-0.023215) = 91.33°
5
50°
91.33°
8
6.130
38.67°
MATHEMATICS RESOURCE | 143
The last angle is solved for much more easily. Recalling that the three angle measures of a triangle always
sum to 180°, we have the last angle measure equal to 180° – 50° – 61.33°, or 38.67°. The solved triangle
is pictured at right.
Example:
Solve for the triangle where a = 4, b = 7, and m∠A = 28°.
Solution:
Since only one angle measurement is known, we should use it and its opposite side as the ∠C, c pair in
our formula for the Law of Cosines. A completely equivalent restatement of the Law of Cosines can be
used.
a2 = b2 + c2 – 2bc cos(A)
42 = 72 + c2 – 2(7)c cos(28°)
0 = c2 – 12.361c + 33
c=
12.361 ± 12.3612 − 4(1)(33)
2
c = 3.90 or 8.46
You may recognize this as the ambiguous case that was solved when we discussed the Law of Sines. Just as before,
there are two completely correct answers. They are redrawn below.
7
7
4
4
28°
28°
3.90
8.46
The Law of Cosines will explicitly allow us to solve for the two cases when they arise. We do not need to be
vigilant enough to catch the ambiguous case ourselves.
The Law of Sines and the Law of Cosines are fantastic ways for us to find the measures and lengths of all of the
angles and sides within a triangle. We can solve for any triangle given to us, provided that three adequate
measurements are given. What, however, of the triangle’s area? We may have its sides and angles, but the amount
of substance inside is still unknown to us.
MATHEMATICS RESOURCE | 144
If we have a triangle with its sides labeled a, b, c
and each side’s opposite angle labeled ∠A, ∠B,
and ∠C, respectively, then we should be able to
find a simple expression for the triangle’s area.
We already know that a triangle’s area is 12 bh,
where b is a base, and h is the height with
respect to that base. If we decide to make side b
the base (just to keep the letters consistent), we
can solve for h.
In the diagram, we have declared b to be the
base of the triangle. The height h is then the
quantity c sin A. This means that our formula
for height, 12 bh, becomes 12 bc sin A.
B
c
A
a
h
C
b
Example:
Find the area of the triangle pictured.
5
Solution: We have the lengths of two sides and the measurement of their
included angle. We can apply the formula outright.
Area =
1
2
⋅5⋅8⋅sin(50°) = 15.321
Area Law
The area of a triangle is given by
Area =
1
bc sin A
2
Here, ∠A is the angle opposite side a. This means that
a triangle’s area is the product of two side lengths and
the sine of their included angle.
50°
8
MATHEMATICS RESOURCE | 145
Appendix A: Completing the Square
First, let us recall something we have already discussed: when we want to solve an equation such as the quadratic
equation x2 = 121, what do we find for our solutions? In this case, gut instinct leads us to x = 11 and x = -11.
General Formula for a Squared Variable
If we are given that x2 = c, where c > 0, then we solve for x by saying that x = ± c .
Example:
Use the general formula for a squared variable to solve (x + 2)2 = 50.
Solution:
(x + 2)2 = 50
x + 2 = ± 50
← general formula for a squared variable
x + 2 = ±5 2
← simplify the square root
x = -2 ± 5 2
← subtract 2 from each side
x = -2 + 5 2 or x = -2 – 5 2
We can use this idea to our advantage in an algebraic technique known as completing the square. Before we can
learn how to complete the square, we have to review one concept concerning the multiplication of two binomials.
Square of a Binomial
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
These formulas come from the FOIL method of multiplying binomials. If any binomial (a ± b) is multiplied by
itself, then the multiplication of the binomials will result in
(a ± b)(a ± b) =
a2 ± ab ± ab + b2 =
a2 ± 2ab + b2
← FOIL binomial multiplication
← combining like terms
If the square of a binomial can be recognized immediately and quickly, then completing the square is a good
technique with which to solve a quadratic equation. When we complete the square to solve a quadratic equation,
we add a common constant to both sides so that the side containing the variable will become the square of a
binomial. For example, if the variables and constants are separated to opposite sides of the equation, and if the
“variable side” of the equation were x2 – 10x, then only the lack of a constant 25 keeps that side from being the
square of a binomial—adding 25 to both sides of the equation would transform the “variable side” to be x2 – 10x
+ 25, which could afterwards be factored into (x – 5)2. The general formula for a squared variable is then valid. As
is almost always true, a well-explained example will resolve much confusion.
Example:
Solve x2 – 10x = 11 by completing the square.
Solution:
Details of this problem were discussed in the paragraph above. We start by adding 25 to both sides and
continue afterwards by factoring the left side into the square of a binomial.
x2 – 10x = 11
MATHEMATICS RESOURCE | 146
← add 25 to both sides to create an equivalent equation
← factor the left expression; simplify the right expression
← general formula for a squared variable
← subtract 5 from each side
x2 – 10x + 25 = 11 + 25
(x – 5)2 = 36
x–5=±6
x=5±6
x = 11 or x = -1
More complications arise if the coefficient of the squared term is not 1. Nevertheless, the procedure of completing
the square remains completely valid. If the squared-term coefficient is not 1, first divide both sides of the equation
by the coefficient of the squared term, then proceed as before. Remember that the constant to be added to both
sides must equal the square of half the linear coefficient. Then, the variable expression can be factored into the
square of a binomial and the general formula for a squared variable can be applied.
Procedure for Completing the Square
Step 1: Arrange the equation so that the variable is entirely on one side and a
constant is on the other.
Step 2: If the coefficient of the squared term is not 1, then divide both sides
of the equation by that coefficient.
Step 3: Add a constant to both sides of the equation. That constant should
be the square of half the linear term’s coefficient.
Step 4: Factor the side containing the variable into the square of a binomial.
Step 5: Apply the general formula for a squared variable.
Step 6: Finish solving for the variable in question.
Example:
Solve 6x2 – x – 12 = 0 by completing the square.
Solution:
6x2 – x – 12 = 0
6x2 – x = 12
x2 – 16 x = 2
← Step 1
← Step 2
x2 – 16 x +
1
144
=2+
x2 – 16 x +
1
144
=
( x − 121 )
289
= 144
2
x – 121 = ±
← Step 3 (half the linear term’s coefficient is 121 )
1
144
← express the constant side as a single fraction
← Step 4
289
144
← Step 5
x – 121 = ± 17
12
← simplify the expression on the right
x = 121 ± 17
12
x=
18
12
=
3
2
← Step 6
or
x=
−16
12
= − 34
Note that when we were first learning the quadratic formula, this is the first equation we solved. In this case, we
have used the method of completing the square instead of the quadratic formula.
Completing the square is an alternative to the quadratic formula that at first seems hardly to be worth the effort it
requires. With practice, though, the technique of completing the square can be quite fast and useful; sometimes, it
is even faster than the quadratic formula.
MATHEMATICS RESOURCE | 147
Example:
Derive the quadratic formula. That is, solve the equation ax2 + bx + c = 0 by completing the square.
Solution:
ax2 + bx + c = 0
ax2 + bx = -c
x2 + ba = - ac
← Step 1
← Step 2
x2 +
b
a
x+
b2
4a 2
= - ac +
x2 +
b
a
x+
b2
4a 2
b
= - ac⋅⋅4a
4a + 4a 2
2
x +
b
a
x +
( x + 2ab )
2
← Step 3 (half the linear term’s coefficient is 2ab )
2
2
b
4a 2
2
= b 4a− 4ac
2
b − 4ac
=
4a 2
x + 2ab =
±
b2
4a 2
2
b2 − 4ac
4a 2
← find a common denom. among the rational expressions
← simplify the expression on the right
← Step 4
← Step 5
x + 2ab =
±
b2 − 4ac
x + 2ab =
±
b2 − 4ac
2a
← simplify the square root
x=
− 2ab ±
2
← Step 6
x=
4a 2
b − 4ac
2a
− b ± b2 − 4ac
2a
← rewrite the square root
← simplify because of the common denominator
Thus, the solutions to the equation ax2 + bx + c = 0 are x =
−b + b 2 − 4ac
−b − b2 − 4ac
and x =
.
2a
2a
MATHEMATICS RESOURCE | 148
Appendix B. More Factoring
When we studied algebra, we learned several methods of factoring with which to simplify our expressions. If we
have a common factor multiplied into an expression, we have been able to factor it in front, as a sort of
distributive property undone. That is, we can say the following.
Linear Factoring
ax + ay = a(x + y)
In addition, we have learned how to factor quadratic polynomials. There are consistent patterns to factoring
quadratic polynomials, and it is a pattern which we have practiced several times.
Quadratic Factoring
acx2 + (ad + bc)x + bd = (ax + b)(cx + d)
Special Case: Difference of Two Squares
a2x2 – b2 = (ax – b)(ax + b)
There is a special case when we have quadratic polynomials: if there is no linear term, then it turns out that we
may be able to factor the quadratic polynomial as a difference of two squares. These patterns are ones that we have
studied extensively as we have studied algebra in detail. However, there are still more factoring patterns that may
arise. They will come in useful to study when we need to do advanced simplification. To learn how these patterns
may arise, we should begin by looking at an example of polynomial multiplication.
Example:
Perform the polynomial multiplication given here.
(x – y)(x2 + xy + y2)
(x + y)(x2 – xy + y2)
Solution:
We apply the general distributive property to the polynomial multiplication.
(x – y)(x2 + xy + y2)
= x(x2 + xy + y2) – y(x2 + xy + y2)
= x3 + x2y + xy2 – x2y – xy2 – y3
= x3 – y3
← split the first binomial into two
← do each distribution
← cancel all the + and – terms
We follow the same procedure as before, with slightly modified signs in our polynomials.
(x + y)(x2 – xy + y2)
= x(x2 – xy + y2) + y(x2 – xy + y2)
= x3 – x2y + xy2 + x2y – xy2 + y3
= x3 + y3
← split the first binomial into two
← do each distribution
← cancel all the + and – terms
We see from these two examples that there are two situations in which we can multiply complicated polynomials
to achieve fairly simple expressions. Using these patterns in reverse gives us the ability to factor differences (and
sums!) of two cubes.
MATHEMATICS RESOURCE | 149
Sums and Differences of Two Cubes
u3 + v3 = (u + v)(u2 – uv + v2)
u3 – v3 = (u – v)(u2 + uv + v2)
This factoring pattern may seem a bit frivolous to memorize; as it turns out, it will even extend to higher order
polynomials than just cubics! A somewhat similar factoring pattern can even apply to sums and differences of 5th
or 7th degree variables. Factoring these higher orders rarely arises in problems, but that does not mean that it is not
possible. Examine the patterns here and notice how they are similar to the pattern for cubes.
Sums and Differences of Like Odd Powers
u5 + v5 = (u + v)(u4 – u3v + u2v2 – uv3 + v4)
u5 – v5 = (u – v)(u4 + u3v + u2v2 + uv3 + v4)
u p ± v p = ( u ± v ) ( u p −1  u p −2 v + u p −3 v 2 ...  uv p −2 + v p −1 ) ,
for an odd integer p
The general pattern is, understandably, very hard to read. Read the pattern as follows. A difference of like odd
powers factors into two factors. The first factor is a linear difference; the second factor is a sum of all the possible
combinations of remaining powers. Then, a sum of like odd powers also factors into two factors also. The first is a
linear sum; the second factor has all the remaining possible combinations of powers, with the operations
alternating inside.
These factorization patterns are useful for simplifying complicated expressions, sometimes including
trigonometric identities.
Example:
Simplify the following expressions.
x3 − y3
, for x ≠ y
x−y
m6 – n6
cos3 x + sec 3 x
cos x + sec x
Solution:
We first factor the numerator of this expression as a difference of two cubes. Then we can cancel the
common factor between the numerator and denominator. Since we are told that x ≠ y, we can cancel
these factors with no worries.
x3 − y3
x−y
(x − y)(x 2 + xy + y 2 )
x−y
2
= x + xy + y2
=
← factor the numerator
← cancel the common factors
MATHEMATICS RESOURCE | 150
Since we have a difference of two squares, we can first factor the binomial that way. Then it will turn into
both a sum and a difference of two cubes.
m6 – n6
= (m3 – n3)(m3 + n3)
= (m – n)(m2 + mn + n2)(m + n)(m2 – mn + n2)
← m6 – n6 is a difference squares of cubes
← factor both the sum and difference of the cubes
This example looks very similar to example (1), except that we are now dealing with trigonometric
functions instead of just variables. We carry on the simplification as in (1), and then we apply
trigonometric properties that we learned in our study of trigonometry. (If reading this appendix before
having read the trigonometry section, this example can be returned to momentarily.)
cos3 x + sec 3 x
cos x + sec x
( cos x + sec x ) ( cos2 x − (cos x)(sec x) + sec 2 x )
← sum of cubes factorization
=
cos x + sec x
= cos 2 x − (cos x)(sec x) + sec 2 x
← cancel the common factor
2
2
1
← use a reciprocal property
= cos x – (cos x) ( cos x ) + sec x
= cos2 x – 1 + sec2 x
= cos2 x – (sin2 x + cos2 x) + sec2 x
= sec2 x – sin2 x
← multiply the reciprocals
← use a Pythagorean property
← cancel the cosine terms
MATHEMATICS RESOURCE | 151
Appendix C. Vectors (Trigonometry)
In day-to-day life, we have always spent our time dealing with scalars. Simply put, a scalar is a number. The
temperature of an object is a scalar; for example, the room I am typing in may be around 92° F. In physics, the
energy of an object is a scalar. An object placed on a table may have 10 Joules of potential energy. There are many
quantities that cannot be represented as numbers, however.
For example, consider directions. In physics, displacements are formally measured in a unit of meters, abbreviated
m; they are a measure of the “movement” that any object or body undergoes. Imagine I placed a large block on
the ground and proceeded to ask, “If I pushed the block 10 meters, where would it then be?” You would most
likely look at me quizzically before asking, “10 meters in what direction?” Things such as displacements, forces,
and electric fields must be represented as vectors. A vector is a numerical abstract, similarly to a scalar. It has a
magnitude, its “size,” so to speak, but it also has a direction. In typeset, we generally distinguish vectors using
boldface. When handwritten on paper, vectors are conventionally denoted with an arrow above their variable
name. Official practice exams seem to indicate vectors using italicized font.
To draw a representation of a vector, we draw an arrow on paper; the length of the arrow represents its
magnitude. For example, the graph below shows a vector v pointed to the “northeast” direction. The magnitude
of v, written ||v||, is then simply its length.
N
v
E
v seems to be directed approximately 30° north of east.
Equivalently, we could say v seems to point approximately
60° east of north.
To stick with the intuitive and logical progression, I will continue to discuss displacements for now. Suppose we
have a giant coordinate plane and place a small object at the origin, (0, 0). In terms of vectors, let’s say I push the

object 6 meters north, displacement a (or a when handwritten). Afterwards, I push the object 8 meters east,

displacement b (or b when handwritten). Where is its final location? As you’ve just been introduced to vectors,
you probably don’t realize that what was just requested was vector addition. To perform the vector addition x + y,
we place the terminal point or head of x where it coincides with the initial point or tail of y. The initial point, or
tail, of a vector is the point where it begins. In the diagram above, v has its tail at the origin. The terminal point,
or head, of a vector is its ending point. In the diagram above, v has its head at some indeterminate location in
Quadrant I. Perhaps a diagram would make more sense than a laundry list of terms. At right, vectors a and b are
labeled, as is their sum, a + b. The vector of a + b points from the tail of a to the head of b. The sum is known as
the resultant vector.
N
b+a
b
a
E
We know a = 6m north. Also, b =
8m east. By the Pythagorean
Theorem, we know ||a + b|| = 10m.
Using some of the amazing
trigonometry skills that we’ve
acquired, we know that the direction
of a + b is 36.87° north of east. [The
tangent is used to solve for the
angle.]
N
b
a
a+b
E
MATHEMATICS RESOURCE | 152
By observing this diagram, we can note that vector addition is commutative like scalar addition. If we had instead
added b + a, we would have the diagram at left. In either case, the vector sum is identical. It has magnitude of 10
and points in the same direction, 36.87° north of east.
What would we do, however, if we had a more difficult vector addition? Forces add vectorally exactly as
displacements do. If two force vectors are applied to an object, their vector sum is the net force the object
experiences. It makes no difference if the forces are applied sequentially or simultaneously. Forces are measured in
units of Newtons, abbreviated N.
Example:
An object at the origin has two forces acting on it. One is a force of 6N at 195°. The other is a force of
4N at 310°. Describe the net force on the object. Here, the two angles given are in standard position on a
coordinate plane.
Solution:
First, we wish to draw the two force vectors given to us, which have been named c and d.
We then wish to place them head-to-tail so that we can add them. This process is shown here.
c
c
d
d
c+d
The final diagram here shows a reproduction of the final triangle, with the appropriate angles labeled. We
can solve for the magnitude of c + d by using the Law of Cosines. Afterwards, we can use the Law of
Cosines again in order to solve for the angle of the c + d vector.
6
50°
40°
θ
15°
4
By the Law of Cosines,
||c + d||2 = 42 + 62 – 2(4)(6) cos(65°)
42 = 62 + 5.6322 – 2(6)(5.632) cos θ
||c + d||2 = 31.714
cos θ =
||c + d|| = 5.632
θ = cos-1(0.7653) = 40.07°
4 2 − 6 2 − 5.632 2
= 0.7653
−2 ⋅ 6 ⋅ 5.632
The θ we have solved for is given in the diagram. To find the true position of the vector c + d, we note
that it is in Quadrant III and add the 195° needed to find that its position is 235.07° from standard
position. Our summary can be stated very concisely. c is 6N at 195°. d is 4N at 310°. c + d is then
5.632N at 235.07°.
MATHEMATICS RESOURCE | 153
From the previous two examples, it becomes clear there is a visual property useful for vectors. The two examples’
diagrams are reproduced below for easy viewing. In these new diagrams, all three of the vectors of interest have
been placed with their initial points at the origin.
a
b+a
b
c
c+d
d
We have from this the Parallelogram Law. The Parallelogram Law states that the sum of two vectors placed at the
origin is visually the diagonal of a parallelogram made with those two vectors. A picture of this is available here
with the same drawings as above and the parallelograms shown. The Parallelogram Law is provable as you move
one vector to have its tail at the other’s head. What results from this is a parallelogram.
a
b+a
b
c
c+d
d
The term “Parallelogram Law” is a bit of a misnomer, as no new undeniable mathematical truth is stated here,
and the law is not used as a problem solving technique. More than anything, it is simply a rule of thumb to be
able to visualize the vector sum of two vectors placed at the origin.
Given two vectors, then, we have a general method for their addition. We first reposition them so they are placed
head-to-tail before using the Law of Cosines in order to find the magnitude and direction of the resultant vector.
We have yet to discuss scalar multiplication and subtraction of vectors, two operations that are also allowed. If we
have a vector g that has magnitude of 10 and direction of 10°, would we be able to find a vector 3g? The answer is
yes. The vector 3g would have a magnitude of 30 and direction of 10°. In a manner of speaking, 3g represents the
vector sum g + g + g.
The vector αv has magnitude of α||v|| and the same
direction as v for α > 0.
There is a central algebraic concept involved now in the subtraction of two vectors. To subtract two vectors and
take the difference a – b, we wish to take a and add the additive inverse of b.
The additive inverse of vector v, written –v, has the same
magnitude as v, but is directed in the opposite direction,
180° from v.
MATHEMATICS RESOURCE | 154
In the first of the three graphs below, vectors x and y are shown. In the second drawing, the Parallelogram Law is
used to give a representation of x + y. In the last drawing, the Parallelogram Law is again used, this time to give a
representation of x – y.
x
y
x+y
x x–y
x
y
-y
Using the Law of Cosines, we could solve for the magnitude and direction of the x + y vector and of the x – y
vector if we wished to, but that’s not necessary right now for me to illustrate the concept of vector subtraction. To
take the difference x – y, we simply added x and –y, where –y was a vector pointed oppositely of y.
Example:
I have invented a “push-pull” machine that can push or pull with any requested force in any direction.
Initially, I set one to push an object with 27N of force, directed eastward. I set a second machine to push
an object with 17N of force, directed at 315° from standard position. Into what quadrant does the net
force point with the current setup? If I changed the setting on the second machine to be “pull” instead of
push, into what quadrant would the net force then point? Which of the two is the stronger net force?
Solution:
This example sets up a problem with two vectors. We will say that u is 27N at 0°. The second force
vector is v, 17N at 315°. The problem then proceeds to ask into which quadrants u + v and u – v point.
The three drawings below show u and v, u + v, and u – v, respectively.
u−v
u
v
-v
u
v
u+v
u
From these diagrams, we can tell that the vector sum u + v, the first setup, points into Quadrant IV. The
difference u – v, the next setup, points into Quadrant I. The stronger net force is the force vector with the larger
magnitude, or the longer vector. This means that the first setup has a stronger net force than the second setup.
Now that the concepts of vector addition and subtraction have been sufficiently explained, we can begin looking
for mathematical shortcuts to the tedious process. While the Law of Cosines certainly allows us to solve for both
the magnitude and the direction of the resultant vector, perhaps we can find an easier method.
This “easier method” is the resolution of a vector into components. When dealing with vectors in two
dimensions, rather than deal with a magnitude in a specific direction, we could choose to decompose our vector
into a horizontal component and a vertical component. Then, instead of having to draw a triangle and involve the
Law of Cosines, we could simply say that the horizontal component of m + n is the sum of the horizontal
components of m and n. Likewise, the vertical component of m + n would just be a simple sum as well.
If we consider a vector of magnitude r with angle θ, our trigonometry has given us a way to consistently solve for
its horizontal and vertical components.
MATHEMATICS RESOURCE | 155
Vector Components
Vector r can be decomposed as ai + bj. Vector i points a
distance 1 in the positive x direction and vector j points a
distance 1 in the positive y direction.
bj
θ
ai
Since ||r|| is the magnitude of the vector r, trigonometry
tells us that
a = ||r|| cos θ and
b = ||r|| sin θ.
We can write this as r = a i + b j or as r = a x̂ + b ŷ
[pronounced “x hat” and “y hat”] or as r = <a, b> (or
even with other notations.)
There are a few things to note. The first is the plethora of notations available for vector components. Three are
given in the box above, and I know several more. They are all equivalent; they first give a number of units in the x
direction, then a number in the y direction. Three-dimensional vectors can also use k and ẑ in the first two
notations; for vectors of higher dimension, mathematicians and physicists tend to list just the components. 54 The
other thing to note is that sometimes θ will not lie in Quadrant I. In these other cases, it makes no difference!
To decompose a vector into components, the cosine and sine functions do our work for us. If θ is in Quadrant II
or Quadrant III, then the cosine function will give us a horizontal component that is negative. If θ is in Quadrant
III or Quadrant IV, then the sine function will give us a vertical component that is negative. Since i and j are
given to point in the positive x and y directions, respectively, a negative “a” or “b” value will cause the component
to point leftward or downward, appropriately. Many times, we can simply leave our final answer in terms of
components. If we wish to use these components to solve for magnitude and direction, we can solve for
b
magnitude using the Pythagorean Theorem. Direction is given by the expression θ = tan-1 ( a ) though we may
have to add 180° if the angle isn’t in Quadrants I or IV.
Example:
w = 5 at 53.13°
x = 13 at 67.38°
Calculate w + x and w – x using the method of components. First give your answers in terms of
components. After drawing the situations, solve for their magnitudes and directions.
Solution:
5
3
The diagrams show the
decomposition of w and x.
4
vector
w = <5 cos 53.13°, 5 sin 53.13°>
w = <3, 4>
x = <13 cos 67.38°, 13 sin 67.38°>
x = <5, 12>
Calculating w + x becomes a matter of adding <3, 4> and <5, 12> to
arrive at <8, 16>. w – x is then <3, 4> – <5, 12> = <-2, -8>.
54
I think they just ran out of letters and became lazy, personally. – Craig
13
5
12
MATHEMATICS RESOURCE | 156
12
3
4
4
3
12
5
5
Now, we wish to find the magnitudes and directions. By the Pythagorean Theorem,
||w + x|| = 82 + 16 2 =
17.889
||w – x|| =
22 + 82 =
8.246
We use a tangent in order to find the directions.
θw + x = tan-1 ( 8 ) = 63.435°
16
θw − x = tan-1 ( -2 ) = 75.964°
-8
However, the difference is in Quadrant III so we have to add 180° to the value given to us by the arctangent.
In summary, w + x has magnitude 17.889 at 63.435°, and w – x has magnitude 8.246 at 255.964°.
At first glance, the process of decomposing vectors into components doesn’t seem to be much speedier or easier at
all! This example took copious space and numerous diagrams. In reality, though, many of these steps could have
been easily consolidated into fewer lines, and with some mental visualization, the diagrams could have been
omitted also. We can rework this example, omitting the extraneous writing, in order to see the simplicity of the
method of components.
Example:
Given w = 5 at 53.13° and x = 13 at 67.38°, calculate the following.
a) w + x
b) w – x
c) 10w − 4x
d) -9w + 5x
Express your answers both in component form and in magnitude and direction form.
Solution:
a) w + x = <5 cos 53.13°, 5 sin 53.13°>+ <13 cos 67.38°, 13 sin 67.38°>
= <3, 4> + <5, 12>
= <8, 16>
||w + x|| = 82 + 16 2 = 17.889
16
θ = tan-1 ( 8 ) = 63.435°
b) w – x = <5 cos 53.13°, 5 sin 53.13°>– <13 cos 67.38°, 13 sin 67.38°>
= <3, 4> – <5, 12>
= <-2, -8>
||w – x|| =
22 + 82 = 8.246
MATHEMATICS RESOURCE | 157
( −8 )
θ = tan-1 −2 = 75.964°
Since it is in Quadrant III, we actually have θ = 75.964° + 180° = 255.964°
c) 10w − 4x = 10 <5 cos 53.13°, 5 sin 53.13°> − 4 <13 cos 67.38°, 13 sin 67.38°>
= <30, 40> – <20, 48>
= <10, -8>
||10w − 4x|| =
( −8 )
102 + 82 = 12.806
θ = tan-1 10 = -38.660°
We can choose to leave θ in this form, or we can choose to give the coterminal value of 321.340°.
d) -9w + 5x = -9 <5 cos 53.13°, 5 sin 53.13°>+ 5 <13 cos 67.38°, 13 sin 67.38°>
= <-27, -36> + <25, 60>
= <-2, 24>
||-9w + 5x|| =
( 24 )
2 2 + 242 = 24.083
θ = tan-1 -2 = -85.236°.
The resultant vector is actually in quadrant II, so we need to take θ + 180°, or 94.764°.
MATHEMATICS RESOURCE | 158
Appendix D. Even and Odd Functions
There is a set of properties that applies not only to trigonometric functions, but also to other algebraic functions.
In trigonometry, we noticed we can take cos(-x) = cos(x). In addition, the sine function has the property that sin(x) = -sin(x). When these properties occur for general functions, they have special names.
Definitions
An even function is a function f(x) for which it is true that f(-x) = f(x). Some
examples of even functions include cosine and secant, as well as polynomials that
have only even powers.
An odd function is a function f(x) for which it is true that f(-x) = -f(x). Some
examples of odd functions include sine, cosecant, tangent, and cotangent, and
polynomials that have only odd powers.
We have used these properties a bit in our study of trigonometric identities. The definitions box states that even
things as simple as polynomials might be odd or even. We have to be very careful that we understand the ideas of
exponents very thoroughly; there is a brief review in the box:
Review
(-x)p = xp, when p is any even integer
(-x)p = -xp, when p is any odd integer
When we square (or raise to the 4th power, etc.) a negative sign, it makes the negative sign go away and become
positive. If we raise the negative sign to an odd power, it remains negative. We can apply that knowledge here; to
show a function is even, we place (-x) into the function where (x) was. If we can simplify the expression back into
f(x), it means we have an even function. If, instead, we get -f(x), we have an odd function. If neither is possible,
then the function is neither even nor odd.
Examples:
Determine if the following functions are even, odd, or neither.
1) f(x) = cos(x) – x2
2) h(x) = x – x3
3) f(x) = x4 + x2 + 1
4) g(x) = csc(x) – sin(x) + x
5) f(x) = x4 – x3
Solutions:
1) We substitute –x into the function every place x occurs. Then we attempt to simplify.
f(-x) = cos(-x) – (-x)2
f(-x) = cos(x) – (-x)2
f(-x) = cos(x) – x2
f(-x) = f(x)
← substitute (-x) for (x)
← we make the simplification cos(-x) = cos(x)
← we make the simplification (-x)2 = x2
← the right side here is exactly our function definition
MATHEMATICS RESOURCE | 159
Because we have found that f(-x) = f(x), we know that f(x) is an even function.
2) Again, we substitute –x into the function every place x occurs.
h(-x) = (-x) – (-x)3
h(-x) = -x – (-x3)
h(-x) = -x + x3
h(-x) = -h(x)
← substitute (-x) for (x)
← make the simplification (-x)3 = -x3
← simplify
← notice that the function is the negative of h(x)
Because we have this relation that h(-x) = -h(x), we know that h(x) is an odd function.
3) We follow the procedure that we have practiced twice in (1) and (2). We substitute (-x) for (x) every
time it appears.
f(-x) = (-x)4 + (-x)2 + 1
f(-x) = x4 + x2 + 1
f(-x) = f(x)
← substitute (-x) for (x)
← we simplify because (-x)4 = x4 and (-x)2 = x2
We have an even function.
4) Much like the sine function, the cosecant function is itself odd. We will be able to use the
simplification csc(-x) = -csc(x) partway through the problem.
g(x) = csc(x) – sin(x) + x
g(-x) = csc(-x) – sin(-x) + (-x)
g(-x) = -csc(x) + sin(x) – x
g(-x) = -g(x)
← substitute (-x) into the given function
← simplify the expressions in each case
We have an odd function.
5) Again, we do the substitution of (-x) for (x) in the function.
f(x) = x4 – x3
f(-x) = (-x)4 – (-x)3
f(-x) = x4 + x3
← substitute (-x) for (x)
← simplify the powers
This is equal neither to –f(x) nor f(x), and the function is neither even nor odd.
We have seen many examples now of identifying even and odd functions from their function definitions,
but we have yet to see any good applications of these properties. We have used an occasional application
when studying trigonometric identities; we simplified sine and cosine functions in order to work more
easily with them. Below is a more intricate use of even and odd functions in algebra.
Example:
Given f(x) = ax4 – bx2 + 35, we know that f(4) = 10. Find f(-4).
Solution:
If we substitute (-x) for (x) in this polynomial, we will find that f(-x) = f(x). That is, we have an even
function. Since f(-x) = f(x), this means more explicitly that f(-4) = f(4). We do not even have to know the
values of the unknown constants a and b so as to know that f(-4) = 10.
MATHEMATICS RESOURCE | 160
About the Author
A proud Caltech graduate, Craig Chu now participates in
DemiDec from New York City, where he is an actuarial
programmer.
While at Caltech, Craig participated enthusiastically in as many
singing groups as possible.
Craig’s future plans include helping elect at least one DemiDec
team member to Congress—possibly himself. He is also a
member of the Board of Directors of the World Scholar’s Cup.
You can email Craig at [email protected] or find him on
Facebook at www.facebook.com/craig.chu.