Download MATHEMATICS CONTEST 2003 SOLUTIONS PART I: 30 Minutes

MATHEMATICS CONTEST 2003
SOLUTIONS
PART I: 30 Minutes; NO CALCULATORS
Section A. Each correct answer is worth 1 point.
1. Find the sum of the first six prime numbers.
Solution: 2 + 3 + 5 + 7 + 11 + 13 = 41 (the number 1 is not a prime).
2. Give the official (and more common) name for a regular quadrilateral.
Solution: “Regular” means equal sides and equal interior angles, so it’s a square.
3. Express the repeating decimal 0.17 as a ratio of two positive integers in simplest form.
Solution 1: Let x = 0.17; then 10x = 1.7, and 10x − x = 1.777777 . . . − 0.177777 . . . = 1.6.
16
8
If 9x = 1.6, then x = 1.6
9 = 90 = 45 .
Solution 2:
Recall that
7
9
= 0.7, so we want
7
9
− 35 . (Or
1
10
+
7
90 ,
or
1
6
+
1
90 .)
4. If 2x + 1 = 2003, find the value of 3x − 1000.
Solution: Solve for x: If 2x + 1 = 2003, then 2x = 2002, so x = 1001. Then 3x − 1000 =
3(1001) − 1000 = 2003.
5. A non-isosceles triangle has integral sides of 4, 5, and x. Find all possible values of x.
Solution: The triangle inequality theorem says that the sum of any two sides must be greater than
the third side, so 5 − 4 < x < 4 + 5, or 1 < x < 9. But the sides are integral (they are integers),
so x must belong to the set {2, 3, 4, 5, 6, 7, 8}. Additionally, the triangle is non-isosceles (no
two sides can be equal). Therefore, x is in {2, 3, 6, 7, 8}.
6. Using some or all of the digits 0–9 (no digit more than once), construct the largest possible
six-digit odd number with a 9 in the tens place.
Solution:
9
8
9
87659
876593
9 must be in the tens place.
Biggest possible number.
Biggest possible number (continued).
Biggest possible number (continued), number must be odd.
7. 7 + (−7) = 0. This is an example of what basic property of addition?
Solution: Additive inverse property
or
Opposites property.
Section B. Each correct answer is worth 2 points.
8. If (x 5 + x 4 + x − 5) is divided by (x + 1), find the remainder.
Solution 1 (Long division):
x4
+1
5
4
x +1 x +x +x −5
−(x 5 + x 4 )
x −5
−(x + 1)
−6 ←− the remainder
Solution 2 (Synthetic division): Works when dividing by x − c. Rewrite x + 1 as x − (−1), then
write out the coefficients, multiply and add; the remainder is the last number on the bottom.
−1
1
1
−1
1 0
0
0
0
0
0
0
1 −5
0 −1
1 −6
Solution 3 (Math knowledge): The remainder theorem says you can just plug in −1 to get the
remainder: (−1)5 + (−1)4 + (−1) − 5 = −6
9. Find the 2003rd digit after the decimal point in the decimal representation of 4/7.
Solution: Rational numbers (such as fractions) either stop or repeat. If you do long division far
enough, the pattern is clear:
4 ÷ 7 = 0.57142857142857...
or
0.571428.
So it repeats every six numbers. Now divide 6 into 2003, and observe that the remainder is 5,
so we choose the 5th repeated digit: The answer is 2.
A
10. In the figure on the right, the length of tangent A P is 12,
PD = 8, and chord AB bisects chord CD. If EB = 3, find
the length of AE.
Solution: (A P)(A P) = (P D)(PC), so 122 = (8)(PC),
or PC = 18. Then C D = 10, and because AB bisects
C D, C E = E D = 5. Now (AE)(E B) = (C E)(E D), so
(AE)(3) = (5)(5), and we conclude that AE = 25
3.
D
E
3
C
B
√
2
11. Express in simplest radical form (no radicals in the denominator):
√
4−2 2
Solution: This is sometimes called “rationalizing the denominator”:
√
√
√
√
√
√
8+5 2
12 + 6 2 + 4 2 + 4
16 + 10 2
3+ 2 4+2 2
=
=
√ ·
√ =
√
√
8
4
4−2 2 4+2 2
16 + 8 2 − 8 2 − 8
3+
P
12
8
12. Write the exact numerical value of x if log8 128 = x. Express in simplest form.
Solution 1: Transform to exponential form: 8x = 128. Now rewrite both 8 and 128 as powers of
2: 8 = 23 and 128 = 27 , so 8x = (23 )x = 23x = 27 . Therefore, 3x = 7, or x = 7/3.
Solution 2:
Use the change-of-base formula and properties of logarithms:
x = log8 128 =
ln 128
7 ln 2
ln 27
7
=
=
=
3
ln 8
ln 2
3 ln 2
3
Section C. Each correct answer is worth 3 points.
13. The ellipse x 2 + 2y 2 + 12y − 10x − 57 = 0 has a major axis with two endpoints. Find the
coordinates of the endpoint that lies in quadrant IV. Express in ordered pair form, (x, y).
Solution:
Put in standard form by completing the squares:
x 2 − 10x
+ 2y 2 + 12y
= 57
x − 10x + 5 + 2(y + 6y + 3 ) = 57 + 52 + 2(32 )
2
(x − 5)2
(x − 5)2
100
2
2
+ 2(y + 3)2
(y + 3)2
+
50
2
= 100
=1
The ellipse is centered at (5, −3), and the major axis is therefore horizontal with length 2a = 20;
the endpoint in quadrant IV is (5 + 10, −3) or (15, −3).
14. Softball player Berni Williams has 120 hits in 300 at-bats for a current batting average of .400.
In today’s game, she will have 5 at-bats. What is the probability that she will get exactly 2 hits?
five at-bats has
Solution: The number of hits among
5 a binomial distribution (more on this
5 her next
2
3
below), so P(exactly 2 hits) = 2 (0.4) (0.6) , where 2 is “5 choose 2” (sometimes written
2 3
as 5 C2 ), which is equal to 10. This evaluates to 10 25 35 = 216
625 = 0.3456.
If you have never heard of the binomial distribution, here are two ways to think about it:
1. Write out every possible sequence of two hits and three outs:
HHooo, HoHoo, HooHo, HoooH, oHHoo, oHoHo, oHooH, ooHHo, ooHoH, oooHH
There are 10 such sequences, and each has probability (0.4)2 (0.6)3 .
2. The binomial distribution gets its name from the terms in the expansion of the binomial
(q + p)n , where p is the probability of success (a hit), q = 1 − p is the probability of failure
(an out), and n is the number of attempts. Thus:
(0.6 + 0.4)5 = 0.65 + 5(0.64 0.41 ) + 10(0.63 0.42 ) + 10(0.62 0.43 ) + 5(0.61 0.44 ) + 0.45 .
Each term in this sum corresponds to the probability of a certain number of failures and successes;
we want the third term.
x + 2
> 1 and graph its solution on the given number line.
15. Solve the inequality 3 x +2
x +2
Solution: Either
> 1 or
< −1, so that x > 1 or x < −5. This can be drawn either
3
3
of these ways:
←−−−−◦
◦−−−−→
←−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→
--7 --6 --5 --4 --3 --2 --1 0 1 2 3
←−−−
−−)
(−−
−−−→
or ←−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→
--7 --6 --5 --4 --3 --2 --1 0 1 2 3
PART II: 30 Minutes; CALCULATORS NEEDED
Section A. Each correct answer is worth 1 point.
1. Find the exact value of the average of the squares of the first six natural numbers.
1 + 4 + 9 + 16 + 25 + 36
91
Solution:
=
= 15 16 = 15.16.
6
6
2. On the first test you scored 80. On the second test you scored 89. Find the percent of increase
to the nearest hundredth of one percent.
amount of change
Solution: The rate of change (increase or decrease) is:
.
original
9
= 0.1125. Then change to a percent (multiply by 100): 11.25%
We have 80
3. Let a = log2 8; let b = log3 8. Find the value of a − b correct to the nearest hundredth.
Solution: log2 8 is equal to 3, but finding log3 8 calls for a calculator and the change-of-base
formula: logb x = log(x)/ log(b), where “log” is either the natural logarithm (ln) or common
log (log10 ). On full-display calculators, you can type something similar to:
log(8)/log(2) - log(8)/log(3)
This gives 1.107210739; to the nearest hundredth, the answer is therefore 1.11.
4. Let sin θ = − 57 , with terminal side in quadrant IV. Write the exact value of tan θ.
52 √
+ x 2 = 72 , so x 2 = 24.
Solution: Use the Pythagorean theorem:
√
Since x is positive, we have x = 24 = 2 6. Finally,
√
y
−5
−5 6
opposite
= = √ =
.
tan θ =
adjacent
x
12
2 6
5. Name the author of the first geometry book, Elements.
Solution: Euclid (Greek, around 300 B.C.).
x
θ
7
–5
6. Otto’s Auto Rental charges $52 per day, and 32 cents per mile for an automobile rental. Arnold
paid $578.40 to rent a car for three days. How many miles did he drive?
Solution: The cost equation is: Cost = 52 × days + 0.32 × miles.
Since he rented the car for three days, we can simplify the equation: C = 156 + 0.32m.
Then put in his cost: 578.40 = 156 + 0.32m.
And solve for m: 422.40 = 0.32m, so m = 1320 miles.
7. Find the domain of the function: f (x) = x 2 − 2.
Solution 1 (algebraic):
This function does not have:
1) square (or other even) roots of a negative number
2) division by zero
3) logarithms.
Consequently, there are no input values that cause any difficulty, so the domain is all real numbers,
or (−∞, ∞).
Solution 2 (graphing calculator): The graph appears to extend in both the positive and negative
x-directions forever. So we would guess that the domain is all real numbers. (This does not
really prove that the domain is all real numbers, but it gives support for that conclusion.)
Solution 3 (math knowledge):
The domain of any polynomial is all real numbers.
Section B. Each problem is worth 2 points.
8. Write the equation of the line through (−3, 2) and perpendicular to the line 3x − y = 7. Express
your answer in slope-intercept form: y = mx + b.
Solution: There are many approaches to this problem. Here is one:
Put 3x − y = 7 in slope-intercept form: y = 3x − 7. Its slope is 3; a line perpendicular to it has
a slope of − 13 . So our equation has the form y = − 13 x + b. Temporarily substitute the point
(−3, 2): 2 = − 13 (−3) + b, or 2 = 1 + b. Therefore, b = 1, and the equation is y = − 13 x + 1.
9. The positive integers a, b, c are said to form a Pythagorean triple if a 2 + b2 = c2 . Write two
distinct Pythagorean triples in the form a, b, c (with a < b < c) such that c = 65.
and you get b2 = c2 − a 2 . Then
Solution√1 (a little algebra and trial & error): Solve for b2 √
b√ = 652 − a 2 . Now try different integers in place of a: 652 − 1 is not an integer, nor is
652 − 22 , etc. The first triples you get this way are 16, 63, 65 and 25, 60, 65. (There are two
other answers as well: 33, 56, 65 and 39, 52, 65.) The trial and error process can be shortened
by using the table or list features on a graphing calculator.
Solution 2 (math knowledge): Notice that 65 = (5)(13), and note that these are the hypotenuse
lengths for two familiar right triangles: 3-4-5 and 5-12-13. We can therefore rescale the first of
these by a factor of 13, and get one triple: 39, 52, 65. The second triple comes from scaling a
5-12-13 triangle by a factor of 5: 25, 60, 65.
10. n! =
15
k. Write the value of n.
k=1
Solution:
15
k = 120. You can do this quickly in a calculator; however, there is a formula:
k=1
n
=
k=1
n(n + 1)
15(16)
=
.
2
2
A little trial and error reveals the answer: 120 = 5!, so n = 5.
11. Find three numbers a, b, and c so that 2, a, b, c, 5 is an arithmetic sequence.
Solution 1: The formula for the nth term of an arithmetic sequence is an = a1 + d(n − 1). Since
a1 = 2 and a5 = 5, we can find d (the difference):
5 = 2 + d(5 − 1)
⇒
5 = 2 + 4d
⇒
3 = 4d
⇒
d=
3
4
Then a = 2.75, b = 3.5, c = 4.25.
Solution 2: The average of the 1st and 5th terms is b: b = 2+5
2 = 3.5. Likewise, a is the average
= 2.75 and
of 2 and b (the first and third terms), and c is the average of b and 5: a = 2+3.5
2
3.5+5
b = 2 = 4.25.
12. In a certain class of 60 students:
20 study German;
25 study Spanish;
18 study French;
5 study German and French;
9 study German and Spanish;
7 study Spanish and French;
4 study all three languages.
How many students are studying no language?
Solution: Use a Venn diagram. For example, observe that there are 4 students taking all three
languages, and 5 studying French and German. Therefore, 1 student is taking French and German
(but not Spanish). This leads to the numbers in the gray shaded area in the diagram on the left.
Continuing in this way, we fill in the diagram and eventually arrive at the figure on the right.
French
10
1
1
10
4
4
French
(18)
3
5
German
Spanish
German
(20)
13
Spanish
(25)
That’s a total of 10 + 10 + 13 + 1 + 5 + 3 + 4 = 46 students taking at least one language. Thus
14 are not taking a language.
Section C. Each problem is worth 3 points.
13. From the top of a 200-ft-tall square water tower, SpongeBob observes a car moving toward
the tower. If the angle of depression of the car changes from 25◦ to 50◦ during the time of
observation, how far does the car travel (to the nearest foot)?
Solution: Angle of elevation always equals
the angle of depression, so we will work
with angle of elevation (it’s easier). In the
figure,
.
tan 25◦ = 200
⇒ x = 428.9
x
tan 50◦ =
200
y
200 ft
50o
y
.
y = 167.8
⇒
25o
x
The distance traveled is x − y; to the nearest foot, the car travels 261 feet.
14. Given matrices A = 3
2
3
x
−1 and B =
5
row, first column of matrix AB is 2003. Find x.
9
4
−7
2
3 . The value of the entry in the second
1
Solution: The second row, first column comes from multiplying row 2 of A and column 1 of B:
9 2
3 3 −1
⇒ (2)(9) + (x)(4) + (5)(−7) = 4x − 17
4 3
2 x 5
−7 1
Set this equal to the given entry value and solve for x: 4x − 17 = 2003, so x = 505.
15. A right circular cone has the same volume as a sphere. If the base of the cone has the same
radius as the sphere, find the height of the cone in terms of that radius r .
Solution: The volume of any cone is V = 13 πr 2 h. The volume of a sphere is V = 43 πr 3 . Since
they have the same volume, set the equations equal to each other (if the radii were not the same,
we would need to use two different variables for r —perhaps r1 and r2 .):
1
2
3 πr h
= 43 πr 3 ,
or
πr 2 h = 4πr 3
Now work on getting h by itself. There are many approaches; the most direct is to divide both
sides by πr 2 , which leaves h = 4r .