Difference between revisions of "2017 AMC 12A Problems/Problem 25"
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Consider <cmath>(t_2x^2+t_6x^6)^8(t_1x^1+t_3x^3+t_5x^5+t_7x^7)^4.</cmath> The expansion will be of the form <math>\sum_i\left(\sum_{\sum a=i} \binom{8}{a_2,a_6}\binom{4}{a_1,a_3,a_5,a_7}{t_1}^{a_1}{t_2}^{a_2}{t_3}^{a_3}{t_5}^{a_5}{t_6}^{a_6}{t_7}^{a_7}x^i \right)</math>. Note that if we reduced the powers of <math>x</math> mod <math>8</math> and fished out the coefficient of <math>x^4</math> and plugged in <math>t_i=1\ \forall\,i</math> (and then multiplied by <math>\binom{12}{4,8}</math>) then we would be done. Since plugging in <math>t_i=1</math> doesn't affect the <math>x</math>'s, we do that right away. The expression then becomes <cmath>x^{20}(1+x^4)^8(1+x^2+x^4+x^6)^4=x^{20}(1+x^4)^{12}(1+x^2)^4=x^4(1+x^4)^{12}(1+x^2)^4,</cmath> where the last equality is true because we are taking the powers of <math>x</math> mod <math>8</math>. Let <math>[x^n]f(x)</math> denote the coefficient of <math>x^n</math> in <math>f(x)</math>. Note <math>[x^4] x^4(1+x^4)^{12}(1+x^2)^4=[x^0](1+x^4)^{12}(1+x^2)^4</math>. We use the roots of unity filter, which states <cmath>\text{terms of }f(x)\text{ that have exponent congruent to }k\text{ mod }n=\frac{1}{n}\sum_{m=1}^n \frac{f(z^mx)}{z^{mk}},</cmath> where <math>z=e^{i\pi/n}</math>. In our case <math>k=0</math>, so we only need to find the average of the <math>f(z^mx)</math>'s. | Consider <cmath>(t_2x^2+t_6x^6)^8(t_1x^1+t_3x^3+t_5x^5+t_7x^7)^4.</cmath> The expansion will be of the form <math>\sum_i\left(\sum_{\sum a=i} \binom{8}{a_2,a_6}\binom{4}{a_1,a_3,a_5,a_7}{t_1}^{a_1}{t_2}^{a_2}{t_3}^{a_3}{t_5}^{a_5}{t_6}^{a_6}{t_7}^{a_7}x^i \right)</math>. Note that if we reduced the powers of <math>x</math> mod <math>8</math> and fished out the coefficient of <math>x^4</math> and plugged in <math>t_i=1\ \forall\,i</math> (and then multiplied by <math>\binom{12}{4,8}</math>) then we would be done. Since plugging in <math>t_i=1</math> doesn't affect the <math>x</math>'s, we do that right away. The expression then becomes <cmath>x^{20}(1+x^4)^8(1+x^2+x^4+x^6)^4=x^{20}(1+x^4)^{12}(1+x^2)^4=x^4(1+x^4)^{12}(1+x^2)^4,</cmath> where the last equality is true because we are taking the powers of <math>x</math> mod <math>8</math>. Let <math>[x^n]f(x)</math> denote the coefficient of <math>x^n</math> in <math>f(x)</math>. Note <math>[x^4] x^4(1+x^4)^{12}(1+x^2)^4=[x^0](1+x^4)^{12}(1+x^2)^4</math>. We use the roots of unity filter, which states <cmath>\text{terms of }f(x)\text{ that have exponent congruent to }k\text{ mod }n=\frac{1}{n}\sum_{m=1}^n \frac{f(z^mx)}{z^{mk}},</cmath> where <math>z=e^{i\pi/n}</math>. In our case <math>k=0</math>, so we only need to find the average of the <math>f(z^mx)</math>'s. | ||
− | \begin{align | + | \begin{align} |
z^0 &\Longrightarrow (1+x^4)^{12}(1+x^2)^4,\\ | z^0 &\Longrightarrow (1+x^4)^{12}(1+x^2)^4,\\ | ||
z^1 &\Longrightarrow (1-x^4)^{12}(1+ix^2)^4,\\ | z^1 &\Longrightarrow (1-x^4)^{12}(1+ix^2)^4,\\ | ||
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z^6 &\Longrightarrow (1+x^4)^{12}(1-x^2)^4,\\ | z^6 &\Longrightarrow (1+x^4)^{12}(1-x^2)^4,\\ | ||
z^7 &\Longrightarrow (1-x^4)^{12}(1-ix^2)^4. | z^7 &\Longrightarrow (1-x^4)^{12}(1-ix^2)^4. | ||
− | \end{align | + | \end{align} |
We plug in <math>x=1</math> and take the average to find the sum of all coefficients of <math>x^{\text{multiple of 8}}</math>. Plugging in <math>x=1</math> makes all of the above zero except for <math>z^0</math> and <math>z^4</math>. Averaging, we get <math>2^{14}</math>. Now the answer is simply <cmath>\frac{\binom{12}{4,8}}{6^{12}}\cdot 2^{14}=\boxed{\frac{2^2\cdot 5\cdot 11}{3^{10}}}.</cmath> | We plug in <math>x=1</math> and take the average to find the sum of all coefficients of <math>x^{\text{multiple of 8}}</math>. Plugging in <math>x=1</math> makes all of the above zero except for <math>z^0</math> and <math>z^4</math>. Averaging, we get <math>2^{14}</math>. Now the answer is simply <cmath>\frac{\binom{12}{4,8}}{6^{12}}\cdot 2^{14}=\boxed{\frac{2^2\cdot 5\cdot 11}{3^{10}}}.</cmath> | ||
Revision as of 18:03, 8 February 2017
Problem
The vertices of a centrally symmetric hexagon in the complex plane are given by
For each
,
, an element
is chosen from
at random, independently of the other choices. Let
be the product of the
numbers selected. What is the probability that
?
Solution
It is possible to solve this problem using elementary counting methods. This solution proceeds by a cleaner generating function.
We note that both lie on the imaginary axis and each of the
have length
and angle of odd multiples of
, i.e.
. When we draw these 6 complex numbers out on the complex plane, we get a crystal-looking thing. Note that the total number of ways to choose 12 complex numbers is
. Now we count the number of good combinations.
We first consider the lengths. When we multiply 12 complex numbers together, their magnitudes multiply. Suppose we have of the numbers
; then we must have
. Having
will take care of the length of the product; now we need to deal with the angle.
We require . Letting
be
, we see that the angles we have available are
, where we must choose exactly 8 angles from the set
and exactly 4 from the set
. If we found a good combination where we had
of each angle
, then the amount this would contribute to our count would be
. We want to add these all up. We proceed by generating functions.
Consider The expansion will be of the form
. Note that if we reduced the powers of
mod
and fished out the coefficient of
and plugged in
(and then multiplied by
) then we would be done. Since plugging in
doesn't affect the
's, we do that right away. The expression then becomes
where the last equality is true because we are taking the powers of
mod
. Let
denote the coefficient of
in
. Note
. We use the roots of unity filter, which states
where
. In our case
, so we only need to find the average of the
's.
\begin{align}
z^0 &\Longrightarrow (1+x^4)^{12}(1+x^2)^4,\\
z^1 &\Longrightarrow (1-x^4)^{12}(1+ix^2)^4,\\
z^2 &\Longrightarrow (1+x^4)^{12}(1-x^2)^4,\\
z^3 &\Longrightarrow (1-x^4)^{12}(1-ix^2)^4,\\
z^4 &\Longrightarrow (1+x^4)^{12}(1+x^2)^4,\\
z^5 &\Longrightarrow (1-x^4)^{12}(1+ix^2)^4,\\
z^6 &\Longrightarrow (1+x^4)^{12}(1-x^2)^4,\\
z^7 &\Longrightarrow (1-x^4)^{12}(1-ix^2)^4.
\end{align}
We plug in
and take the average to find the sum of all coefficients of
. Plugging in
makes all of the above zero except for
and
. Averaging, we get
. Now the answer is simply
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Problem ?? |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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