Download Solution 1 - JEJAK 1000 PENA

2017, AMC 12 A
Problem 1
At a gathering of people, there are people who all know each other and people who
know no one. People who know each other hug, and people who do not know each other
shake hands. How many handshakes occur?
Solution
Let the group of people who all know each other be , and let the group of people who know
no one be . Handshakes occur between each pair
such that
and
, and
between each pair of members in . Thus, the answer is
Solution - Complementary Counting
The number of handshakes will be equivalent to the difference between the number of total
interactions and the number of hugs, which are
and
, respectively. Thus, the
total amount of handshakes is
Problem 2
Chloé chooses a real number uniformly at random from the interval
.
Independently, Laurent chooses a real number uniformly at random from the interval
. What is the probability that Laurent's number is greater than Chloe's number?
Solution
Suppose Laurent's number is in the interval
. Then, by symmetry, the probability of
Laurent's number being greater is . Next, suppose Laurent's number is in the interval
. Then Laurent's number will be greater with probability . Since each case is
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equally likely, the probability of Laurent's number being greater is
answer is C.
, so the
Alternate Solution: Geometric Probability
Let be the number chosen randomly by Chloé. Because it is given that the number Chloé
choose is interval
,
. Next, let be the number chosen randomly by
Laurent. Because it is given that the number Chloé choose is interval
,
. Since we are looking for when Laurent's number is Chloé's we write the
equation
. When these three inequalities are graphed the area captured by
and
represents all the possibilities, forming a rectangle 2017
in width and 4034 in height. Thus making its area
. The area captured by
,
, and
represents the possibilities of Laurent winning,
forming a trapezoid with a height 2017 in length and bases 4034 and 2017 length, thus
making an area
. The simplified quotient of these two areas is the
probability Laurent's number is larger than Chloé's, which is
.
Problem 3
In the figure below, semicircles with centers at and and with radii 2 and 1, respectively,
are drawn in the interior of, and sharing bases with, a semicircle with diameter
. The two
smaller semicircles are externally tangent to each other and internally tangent to the largest
semicircle. A circle centered at is drawn externally tangent to the two smaller semicircles
and internally tangent to the largest semicircle. What is the radius of the circle centered at ?
Solution 1
Connect the centers of the tangent circles! (call the center of the large circle
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)
Notice that we don't even need the circles anymore; thus, draw triangle
:
with cevian
and use Stewart's Theorem:
From what we learned from the tangent circles, we have
,
,
,
,
, and
, where is the radius of the circle centered
at that we seek.
Thus:
Solution 2
Like the solution above, connecting the centers of the circles results in triangle
with
cevian
. The two triangles
and
share angle , which means we can use
Law of Cosines to set up a system of 2 equations that solve for respectively:
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(notice that the diameter of the largest
semicircle is 6, so its radius is 3 and
is 3 - r)
We can eliminate the extra variable of angle by multiplying the first equation by 3 and
subtracting the second from it. Then, expand to find :
, so =
Solution 3
Let
be the center of the largest semicircle and be the radius of
. We know that
,
,
,
, and
. Notice that
and
are bounded by the same two parallel lines, so these triangles have the
same heights. Because the bases of these two triangles (that have the same heights) differ by
a factor of 2, the area of
must be twice that of
, since the area of a triangle
is
.
Again, we don't need to look at the circle and the semicircles anymore; just focus on the
triangles.
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Let
equal to the area of
and
equal to the area of
states that the area of an triangle with sides
. Heron's Formula
and is
where ,
or the semiperimeter, is
The semiperimeter
Formula to obtain
of
is
Use Heron's
Using Heron's Formula again, find the area of
with sides
, , and
Now,
Solution 4
Let
, the center of the large semicircle, to be at
Therefore
is at
Let the radius of circle
and
is at
, and
to be at
.
be .
Using Distance Formula, we get the following system of three equations:
By simplifying, we get
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.
.
By subtracting the first equation from the second and third equations, we get
which simplifies to
When we add these two equations, we get
So
Problem 4
A square with side length is inscribed in a right triangle with sides of length , , and so
that one vertex of the square coincides with the right-angle vertex of the triangle. A square
with side length is inscribed in another right triangle with sides of length , , and so that
one side of the square lies on the hypotenuse of the triangle. What is
?
Solution 1
Analyze the first right triangle.
Note that
and
. Solving,
are similar, so
. This can be written as
.
Now we analyze the second triangle.
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Similarly,
and
are similar, so
, and
. Solving for
Thus,
. Thus,
, we get
.
.
Solution 2 (Shortcut using answer choices)
As stated in Solution 1, the square in the first triangle
has a side length of
Then we look at the answers. The only answer with a factor of in the denominator is
.
.
Problem 5
How many ordered pairs
such that is a positive real number and is an integer
between and
, inclusive, satisfy the equation
Solution
By the properties of logarithms, we can rearrange the equation to read
. If
, we may divide by it and get
, which implies
. Hence, we have possible values , namely
Since
is equivalent to
, as we can assign
to each
, each possible value yields exactly
. In total, we have
solutions.
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with
solutions
Problem 6
A set is constructed as follows. To begin,
. Repeatedly, as long as possible, if
is an integer root of some polynomial
for some
, all of whose coefficients are elements of , then is put into . When no more
elements can be added to , how many elements does have?
Solution 1
At first,
.
At this point, no more elements can be added to
. To see this, let
with each in . is a factor of , and is in , so has to be a factor of some element
in . There are no such integers left, so there can be no more additional elements.
has elements
Problem 7
For certain real numbers , , and , the polynomial
three distinct roots, and each root of
is also a root of the polynomial
What is
?
has
Solution
Let
and be the roots of
. Let
Vieta's formulas on the quadratic term of
following:
be the additional root of
and the cubic term of
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. Then from
, we obtain the
Thus
.
Now applying Vieta's formulas on the constant term of
linear term of
, we obtain:
Substituting for
we obtain:
, and the
in the bottom equation and factoring the remainder of the expression,
It follows that
. But
Now we can factor
Then
, the linear term of
in terms of
so
as
and
Hence
.
Solution 2
Since all of the roots of
more than the degree of
for some number
are distinct and are roots of
, we have that
. By comparing
, and the degree of
coefficients, we see that
is one
. Thus,
Expanding and equating coefficients we get that
The third equation yields
, and the first equation yields
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. So we have that
Problem 8
Quadrilateral
is inscribed in circle
, and
and has side lengths
. Let and be points on
such that
and
. Let be the intersection of line
and the line
through parallel to
. Let be the intersection of line
and the line through
parallel to
. Let be the point on circle other than that lies on line
. What is
?
Solution
It is easy to see that
with a ratio of
First we note that
Then
with a ratio of
, so
Now we find the length of
quadrilateral is cyclic, we can simply use the Law of Cosines.
. Because the
By Power of a Point,
. Thus
-solution by FRaelya
Problem 9
Call a positive integer
if it is a one-digit number or its digits, when read from
left to right, form either a strictly increasing or a strictly decreasing sequence. For example,
,
, and
are monotonous, but ,
, and
are not. How many
monotonous positive integers are there?
Solution 1
Case 1: monotonous numbers with digits in ascending order
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There are
ways to choose n digits from the digits 1 to 9. For each of these ways,
we can generate exactly one monotonous number by ordering the chosen digits in ascending
order. Note that 0 is not included since it will always be a leading digit and that is not
allowed. The sum is equivalent to
Case 2: monotonous numbers with digits in descending order
There are
ways to choose n digits from the digits 0 to 9. For each of these ways,
we can generate exactly one monotonous number by ordering the chosen digits in descending
order. Note that 0 is included since we are allowed to end numbers with zeros. The sum is
equivalent to
is not positive. Thus there are
We discard the number 0 since it
here.
Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9.
Thus there are
monotonous numbers.
Solution 2
Like Solution 1, divide the problem into an increasing and decreasing case:
Case 1: Monotonous numbers with digits in ascending order.
Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer
cannot begin with 0.
To get a monotonous number, we can either include or exclude each of the remaining 9
digits, and there are
ways to do this. However, we cannot exclude every digit at
once, so we subtract 1 to get
monotonous numbers for this case.
Case 2: Monotonous numbers with digits in descending order.
This time, we arrange all 10 digits in decreasing order and repeat the process to find
ways to include or exclude each digit. We cannot exclude every digit at once,
and we cannot include only 0, so we subtract 2 to get
monotonous
numbers for this case.
At this point, we have counted all of the single-digit monotonous numbers twice, so we must
subtract 9 from our total.
Thus our final answer is
.
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Problem 10
The number
has over
positive integer
divisors. One of them is chosen at random. What is the probability that it is odd?
Solution
If a factor of
is odd, that means it contains no factors of . We can find the number of
factors of two in
by counting the number multiples of , , , and that are less than or
equal to .After some quick counting we find that this number is
.
If the prime factorization of
has factors of , there are choices for each divisor for
how many factors of should be included ( to inclusive). The probability that a randomly
chosen factor is odd is the same as if the number of factors of is which is
.
Solution by: vedadehhc
Solution 2
We can write
as its prime factorization:
Each exponent of these prime numbers are one less than the number of factors at play here.
This makes sense;
is going to have factors:
, and the other exponents
will behave identically.
In other words,
has
factors.
We are looking for the probability that a randomly chosen factor of
that do not contain multiples of as factors.
will be odd--numbers
From our earlier observation, the only factors of
that are even are ones with at least one
multiplier of , so our probability of finding an odd factor becomes the following:
Solution submitted by David Kim
Problem 11
A coin is biased in such a way that on each toss the probability of heads is
probability of tails is
and the
. The outcomes of the tosses are independent. A player has the choice
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of playing Game A or Game B. In Game A she tosses the coin three times and wins if all
three outcomes are the same. In Game B she tosses the coin four times and wins if both the
outcomes of the first and second tosses are the same and the outcomes of the third and fourth
tosses are the same. How do the chances of winning Game A compare to the chances of
winning Game B?
The probability of winning Game A is
less than the probability of winning Game B.
The probability of winning Game A is
less than the probability of winning Game B.
The probabilities are the same.
The probability of winning Game A is
greater than the probability of winning Game
The probability of winning Game A is
greater than the probability of winning Game
B.
B.
Solution
The probability of winning Game A is the sum of the probabilities of getting three tails and
getting three heads which is
. The probability
of winning Game B is the sum of the probabilities of getting two heads and getting two tails
squared. This gives us
Game A is
and the probability of winning Game B is
. The probability of winning
, so the answer is
Solution by: vedadehhc
Problem 12
The diameter
of a circle of radius is extended to a point outside the circle so that
. Point is chosen so that
and line
is perpendicular to line
.
Segment
intersects the circle at a point between and . What is the area of
?
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Solution 1
Let
be the center of the circle. Note that
. However, by
Power of a Point,
,
so
. Now
. Since
.
Solution 2: Similar triangles with Pythagorean
is the diameter of the circle, so
.
is a right angle, and therefore by AA similarity,
Because of this,
Likewise,
, so
, so
Thus the area of
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.
.
.
Solution 3: Similar triangles without Pythagorean
Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal
calculation:
Draw
with
on
.
.
.
.(
ratio applied twice)
.
Problem 13
Let
the integers from to
divided by ?
be the -digit number that is formed by writing
in order, one after the other. What is the remainder when is
Solution
We will consider this number
that the number is
and
. By looking at the last digit, it is obvious
. To calculate the number
, note that
so it is equivalent to
Thus it is
and
, so it is
Problem 14
Real numbers and are chosen independently and uniformly at random from the interval
. What is the probability that
, where
denotes the greatest
integer less than or equal to the real number ?
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Solution
First let us take the case that
. In this case, both
the interval
. Similarly, in the case that
, and the probability is
. The probability of this is
, and lie in the interval
and
. It is easy to see that the probabilities for
lie in
for
are the infinite geometric series that starts at and with common ratio .
Using the formula for the sum of an infinite geometric series, we get that the probability is
.
Solution by: vedadehhc
Problem 15
The graph of
, where
is a polynomial of degree , contains points
, and
. Lines
,
, and
intersect the graph again at points
and , respectively, and the sum of the -coordinates of , , and is 24. What is
,
,
,
?
Solution
First, we can define
, which contains points ,
, and . Now we find that lines
,
, and
are defined by the equations
,
, and
respectively. Since we want to find the coordinates of the intersections of these lines and
, we set each of them to
, and
synthetically divide by the solutions we already know exist (eg. if we were looking at line
, we would synthetically divide by the solutions
and
, because we already
know
intersects the graph at and , which have -coordinates of and ). After
completing this process on all three lines, we get that the -coordinates of , , and are
,
which gives us
, and
respectively. Adding these together, we get
. Substituting this back into the original equation, we get
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, and
Solution by: vedadehhc and tdeng
Problem 16
Quadrilateral
There is a point
is
has right angles at
in the interior of
times the area of
and ,
such that
. What is
, and
.
and the area of
?
Solution 1
Let
Theorem,
,
, and
. Since
. Note that
. By the Pythagorean
, the ratios of side
lengths must be equal. Since
,
a point on
such that
is an altitude of triangle
and
. Note that
. Let F be
. Therefore,
and
. Since
and
form altitudes of triangles
and
, respectively, the areas of these
triangles can be calculated. Additionally, the area of triangle
can be calculated, as it is
a right triangle. Solving for each of these yields:
Therefore, the answer is
Solution 2
Draw line
through
,
, with
on
and
on
,
. By weighted average
. WLOG let
.
Meanwhile,
.
. We obtain
, namely
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.
,
The rest is the same as Solution 1.
Problem 17
A set of people participate in an online video basketball tournament. Each person may be a
member of any number of -player teams, but no two teams may have exactly the same
members. The site statistics show a curious fact: The average, over all subsets of size of the
set of participants, of the number of complete teams whose members are among those
people is equal to the reciprocal of the average, over all subsets of size of the set of
participants, of the number of complete teams whose members are among those people.
How many values ,
, can be the number of participants?
Solution
Solution by Pieater314159
Let there be teams. For each team, there are
different subsets of players
including that full team, so the total number of team-(group of 9) pairs is
Thus, the expected value of the number of full teams in a random set of players is
Similarly, the expected value of the number of full teams in a random set of players is
The condition is thus equivalent to the existence of a positive integer
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such that
Note that this is always less than
have that this is equivalent to
, so as long as
is integral,
It is obvious that divides the RHS, and that does iff
divides it iff
. One can also bash out that
possible residues
.
is a possibility. Thus, we
divides it in
. Also,
out of the
Using all numbers from to
, inclusive, it is clear that each possible residue
is reached an equal number of times, so the total number of working
range is
. However, we must subtract the number of "working"
, which is . Thus, the answer is
.
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in that