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2017 AMC 12A Problems/Problem 25

Problem

The vertices $V$ of a centrally symmetric hexagon in the complex plane are given by \[V=\left\{ \sqrt{2}i,-\sqrt{2}i, \frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i) \right\}.\] For each $j$, $1\leq j\leq 12$, an element $z_j$ is chosen from $V$ at random, independently of the other choices. Let $P={\prod}_{j=1}^{12}z_j$ be the product of the $12$ numbers selected. What is the probability that $P=-1$?

$\textbf{(A) } \dfrac{5\cdot11}{3^{10}} \qquad \textbf{(B) } \dfrac{5^2\cdot11}{2\cdot3^{10}} \qquad \textbf{(C) } \dfrac{5\cdot11}{3^{9}} \qquad \textbf{(D) } \dfrac{5\cdot7\cdot11}{2\cdot3^{10}} \qquad \textbf{(E) } \dfrac{2^2\cdot5\cdot11}{3^{10}}$

Solution 1

It is possible to solve this problem using elementary counting methods. This solution proceeds by a cleaner generating function.

We note that $\pm \sqrt{2}i$ both lie on the imaginary axis and each of the $\frac{1}{\sqrt{8}}(\pm 1\pm i)$ have length $\frac{1}{2}$ and angle of odd multiples of $\pi/4$, i.e. $\pi/4,3\pi/4,5\pi/4,7\pi/4$. 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 $6^{12}$. 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 $n$ of the numbers $\pm \sqrt{2}i$; then we must have $\left(\sqrt{2}\right)^n\cdot\left(\frac{1}{2}\right)^{12-n}=1 \Longrightarrow n=8$. Having $n=8$ will take care of the length of the product; now we need to deal with the angle.

We require $\sum\theta\equiv\pi \mod 2\pi$. Letting $z$ be $e^{i\pi/4}$, we see that the angles we have available are $\{z^1,z^2,z^3,z^5,z^6,z^7\}$, where we must choose exactly 8 angles from the set $\{z^2,z^6\}$ and exactly 4 from the set $\{z^1,z^3,z^5,z^7\}$. If we found a good combination where we had $a_i$ of each angle $z^i$, then the amount this would contribute to our count would be $\binom{12}{4,8}\cdot\binom{8}{a_2,a_6}\cdot\binom{4}{a_1,a_3,a_5,a_7}$. We want to add these all up. We proceed by generating functions.

Consider \[(t_2x^2+t_6x^6)^8(t_1x^1+t_3x^3+t_5x^5+t_7x^7)^4.\] The expansion will be of the form $\sum_i\left(\sum_{\sum a=i} \binom{8}{a_2,a_6}\binom{4}{a_1,a_3,a_5,a_7}{t_1}^{a_1}{t_2}^{a_2}{t_3}^{a_3}{t_5}^{a_5}{t_6}^{a_6}{t_7}^{a_7}x^i \right)$. Note that if we reduced the powers of $x$ mod $8$ and fished out the coefficient of $x^4$ and plugged in $t_i=1\ \forall\,i$ (and then multiplied by $\binom{12}{4,8}$) then we would be done. Since plugging in $t_i=1$ doesn't affect the $x$'s, we do that right away. The expression then becomes \[x^{20}(1+x^4)^8(1+x^2+x^4+x^6)^4=x^{20}(1+x^4)^{12}(1+x^2)^4=x^4(1+x^4)^{12}(1+x^2)^4,\] where the last equality is true because we are taking the powers of $x$ mod $8$. Let $[x^n]f(x)$ denote the coefficient of $x^n$ in $f(x)$. Note $[x^4] x^4(1+x^4)^{12}(1+x^2)^4=[x^0](1+x^4)^{12}(1+x^2)^4$. We use the roots of unity filter, which states \[\text{terms of }f(x)\text{ that have exponent congruent to }k\text{ mod }n=\frac{1}{n}\sum_{m=1}^n \frac{f(z^mx)}{z^{mk}},\] where $z=e^{i\pi/n}$. In our case $k=0$, so we only need to find the average of the $f(z^mx)$'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 $x=1$ and take the average to find the sum of all coefficients of $x^{\text{multiple of 8}}$. Plugging in $x=1$ makes all of the above zero except for $z^0$ and $z^4$. Averaging, we get $2^{14}$. Now the answer is simply \[\frac{\binom{12}{4,8}}{6^{12}}\cdot 2^{14}=\boxed{\textbf{(E)} ~\frac{2^2\cdot 5\cdot 11}{3^{10}}}.\]

Solution 2

By changing $z_1$ to $-z_1$, we can give a bijection between cases where $P=-1$ and cases where $P=1$, so we'll just find the probability that $P=\pm 1$ and divide by $2$ in the end. Multiplying the hexagon's vertices by $i$ doesn't change $P$, and switching any $z_j$ with $-z_j$ doesn't change the property $P=\pm 1$, so the probability that $P=\pm1$ remains the same if we only select our $z_j$'s at random from \[\left\{a= \sqrt 2,\quad b=\frac1{\sqrt{8}}(1+i),\quad c=\frac1{\sqrt{8}}(1-i)\right\}.\] Since $|a|=\sqrt2$ and $|b|=|c|=\frac12$, we must choose $a$ exactly $8$ times to make $|P|=1$. To ensure $P$ is real, we must either choose $b$ $4$ times, $c$ $4$ times, or both $b$ and $c$ $2$ times. This gives us a total of \[2\binom{12}{4}+\binom{12}{2}\binom{10}{2}=(12\cdot 11\cdot 10\cdot 9)\left(\frac1{12}+\frac14\right)=(2^3\cdot 3^3\cdot 5\cdot 11)\frac13\] good sequences $z_1,\dots,z_{12}$, and hence the final result is \[\frac12\cdot \frac{2^3\cdot 3^2\cdot 5\cdot 11}{3^{12}}=\boxed{\textbf{(E)} ~\frac{2^2\cdot 5\cdot 11}{3^{10}}}.\]

Solution 3

We use generating functions and a roots of unity filter. Notice that all values in $V$ are eighth roots of unity multiplied by a constant. Let $x$ be a primitive eighth root of unity ($e^{\frac{i\pi}{4}}$). The numbers in $V$ are then $\left \{ \frac{x}{2},x^2\sqrt2,\frac{x^3}{2},\frac{x^5}{2},x^6\sqrt2,\frac{x^7}{2} \right\}$. To have $P=-1$, we must have that $|P|=1$, so eight of the $(z_i)$ must belong to \[\left \{ x^2\sqrt2,x^6\sqrt2 \right\}\] and the other four must belong to \[\left \{ \frac{x}{2},\frac{x^3}{2},\frac{x^5}{2},\frac{x^7}{2} \right\}\] So, we write the generating function \[\left (x^2 +x^6 \right )^8 \left (x+x^3+x^5+x^7 \right )^4\] to describe the product. Note that this assumes that the $(z_i)$ that belong to $\left \{ x^2\sqrt2,x^6\sqrt2 \right\}$ come first, so we will need to multiply by $\binom{12}{4}$ at the end. We now apply a roots of unity filter to find the sum of the coefficients of the exponents that are $4\pmod{8}$, or equivalently the coefficients of the powers that are multiples of $8$ of the following function: \[P(x)=\left (x^2 +x^6 \right )^8 \left (x+x^3+x^5+x^7 \right )^4x^4\] Let $\zeta=e^{\frac{i\pi}{4}}$. We are looking for $\frac{P(1)+P(\zeta)+P(\zeta^2)+P(\zeta^3)+P(\zeta^4)+P(\zeta^5)+P(\zeta^6)+P(\zeta^7)}{8}$. $P(1)=P(-1)=2^{16}$, and all of the rest are equal to $0$. So, we get an answer of $\frac{2^{16}+2^{16}}{8\cdot 6^{12}}=\frac{2^{14}}{6^{12}}=\frac{2^2}{3^{10}}$. But wait! We need to multiply by $\binom{12}{4} =\frac{12\cdot11\cdot10\cdot9}{4\cdot3\cdot3\cdot1}$$=5\cdot9\cdot11$. So, the answer is $\boxed{\text{\textbf{(E) }} \frac{2^2\cdot5\cdot11}{3^{10}} }$

Remark

Here is a really good post about the Roots of Unity Filter: https://artofproblemsolving.com/community/c1340h1003741_roots_of_unity_filter

Solution 4

We can write the points in polar form $(r, \theta)$ as \[V = \left\{ \left(\sqrt{2}, \frac{\pi}{2}\right), \left(\sqrt{2}, \frac{3\pi}{2}\right), \left(\frac{1}{2}, \frac{\pi}{4}\right), \left(\frac{1}{2}, \frac{3\pi}{4}\right), \left(\frac{1}{2}, \frac{5\pi}{4}\right), \left(\frac{1}{2}, \frac{7\pi}{4}\right)\right\}.\]Note that when multiplying complex numbers, the $r$'s multiply and the $\theta$'s add, and since $-1 = (1, \pi),$ we need $8$ complex numbers with $r = \sqrt{2} = 2^{0.5}$ and $4$ with $r = \frac{1}{2} = 2^{-1}.$ By binomial distribution, the probability of this occurring is $\dbinom{12}{4} \left(\frac{1}{3}\right)^8 \left(\frac{2}{3}\right)^4.$ For the $\theta$ part, note that $\frac{4\theta}{\pi}$ must be congruent to $4 \mod 8,$ and by using simple symmetry, the probability of the aforementioned occurring is $\frac{1}{4}.$ This is since $2a + 6b$ is even for all $a + b =8,$ and the number of ordered quadruples $(a_1, a_2, a_3, a_4)$ such that $a_i \in \{1, 3, 5, 7\}$ for all $1 \leq i \leq 4$ and $a_1 + a_2 + a_3 + a_4 \equiv 2k \mod 8$ is the same for all $1 \leq k \leq 4,$ again by using symmetry. Thus, our probability is \[\left(\frac{1}{4}\right) \left(\frac{1}{3}\right)^8 \left(\frac{2}{3}\right)^4 \dbinom{12}{4} = \frac{2^2}{3^{12}} \dbinom{12}{4} = \boxed{\textbf{(E)} ~\frac{2^2\cdot 5\cdot 11}{3^{10}}}\]

-fidgetboss_4000

Solution 5 (Simple)

The absolute value of the first two complex numbers is $\sqrt{2}$ while the absolute value of the latter four is $\frac12$. For the absolute value of the product to be $1$, we need $8$ elements with absolute value $\sqrt{2}$ and $4$ elements with absolute value of $\frac12$.

We pick $8$ elements from $A = \left\{ \sqrt{2}i,-\sqrt{2}i\right\}$ and $4$ elements from $B = \left\{\frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i) \right\}$. We also need to choose which $8$ of $z_j$ will be chosen from $A$ which gives us ${12 \choose 8} \cdot 2^8 \cdot 4^4$ cases. However, suppose we have chosen $11$ elements and we need to choose one more element from $B$. The product of these $11$ elements can have any of these values: $\left\{ 2\mathrm{cis}(\frac{\pi}{4}), 2\mathrm{cis}(\frac{3\pi}{4}), 2\mathrm{cis}(\frac{5\pi}{4}), 2\mathrm{cis}(\frac{7\pi}{4}), \right\}$. For either of these values, there is just one value of $z_{12} \in B$ such that $P = -1$ so we must divide by $4$ which gives ${12 \choose 8} \cdot 2^8 \cdot 4^3$ cases.

Because there are $6^{12}$ ways to pick any $12$ elements, our probability is simply $\frac{{12 \choose 8} \cdot 2^8 \cdot 4^3}{6^{12}} = \frac{2^2}{3^{12}} {12 \choose 8} = \boxed{\text{\textbf{(E) }} \frac{2^2\cdot5\cdot11}{3^{10}} }$

~Zeric

Solution 6 (Casework)

Let $a = \sqrt{2}i$, $b = \frac{1}{\sqrt{8}}(1+i)$, $c = \frac{1}{\sqrt{8}}(-1+i)$. $\bar{a} = -\sqrt{2}i$, $\bar{b} = \frac{1}{\sqrt{8}}(1-i)$, $\bar{c} = \frac{1}{\sqrt{8}}(-1-i)$

The magnitude of $\left\{ a, \bar{a} \right\}$ is $\sqrt{2}$, while the magnitude of $\left\{ b, \bar{b}, c, \bar{c} \right\}$ is $\frac12$. For the product's magnitude to be $1$, we need $8$ elements from $\left\{ a, \bar{a} \right\}$, and $4$ elements from $\left\{ b, \bar{b}, c, \bar{c} \right\}$.

The product of $8$ elements from $\left\{ a, \bar{a} \right\}$ will be either $16$ or $-16$. The product of $4$ elements from $\left\{ b, \bar{b}, c, \bar{c} \right\}$ will be either $\frac{1}{16}$ or $-\frac{1}{16}$.

\[\text{Combinations of } \left\{ a, \bar{a} \right\} \text{ multiplied to 16 is} \left\{ a^8, \bar{a}^8, a^2\bar{a}^6, a^6\bar{a}^2, a^4\bar{a}^4 \right\}\]

\[\text{Combinations of } \left\{ a, \bar{a} \right\} \text{ multiplied to -16 is} \left\{ a\bar{a}^7, a^7\bar{a}, a^5\bar{a}^3, a^3\bar{a}^5 \right\}\]

\[\text{Combinations of } \left\{ b, \bar{b}, c, \bar{c} \right\} \text{ multiplied to } \frac{1}{16} \text{ is} \left\{ b^2\bar{b}^2, c^2\bar{c}^2, b^2c^2, \bar{b}^2\bar{c}^2, b\bar{b}c\bar{c}, b^3\bar{c}, b\bar{c}^3, \bar{b}^3c, \bar{b}c^3 \right\}\]

\[\text{Combinations of } \left\{ b, \bar{b}, c, \bar{c} \right\} \text{ multiplied to } -\frac{1}{16} \text{ is} \left\{ b^2\bar{b}c, b\bar{b}^2\bar{c}, bc^2\bar{c}, \bar{b}c\bar{c}^2, b^4, c^4, \bar{b}^4, \bar{c}^4, b^2\bar{c}^2, \bar{b}^2c^2 \right\}\]

Notice that the denominator of the desired probability is $6^{12}$, therefore, the numerator of the desired probability will be the sum of the permutations of all combinations that multiply to $-1$.

Multiplying combinations of $\left\{ a, \bar{a} \right\}$ multiplied to $16$ with combinations of $\left\{ b, \bar{b}, c, \bar{c} \right\}$ multiplied to $-\frac{1}{16}$ gives the desired product $-1$. Similarly, multiplying combinations of $\left\{ a, \bar{a} \right\}$ multiplied to $-16$ with combinations of $\left\{ b, \bar{b}, c, \bar{c} \right\}$ multiplied to $\frac{1}{16}$ also gives the desired product $-1$.

Cases with combinations of $\left\{ a, \bar{a} \right\}$ multiplied to $16$ and combinations of $\left\{ b, \bar{b}, c, \bar{c} \right\}$ multiplied to $-\frac{1}{16}$: 

Case 1:
$\left\{ a^8, \bar{a}^8 \right\} \cdot \left\{ b^2\bar{b}c, b\bar{b}^2\bar{c}, bc^2\bar{c}, \bar{b}c\bar{c}^2 \right\}$
$\frac{12!}{8!2!} \cdot 2 \cdot 4 = \frac{12!}{8!} \cdot 4$

Case 2:
$\left\{ a^8, \bar{a}^8 \right\} \cdot \left\{ b^4, c^4, \bar{b}^4, \bar{c}^4 \right\}$
$\frac{12!}{8!4!} \cdot 2 \cdot 4 = \frac{12!}{8!} \cdot \frac13$ 

Case 3:
$\left\{ a^8, \bar{a}^8 \right\} \cdot \left\{ b^2\bar{c}^2, \bar{b}^2c^2 \right\}$
$\frac{12!}{8!2!2!} \cdot 2 \cdot 2 = \frac{12!}{8!}$ 

Case 4:
$\left\{ a^2\bar{a}^6, a^6\bar{a}^2 \right\} \cdot \left\{ b^2\bar{b}c, b\bar{b}^2\bar{c}, bc^2\bar{c}, \bar{b}c\bar{c}^2 \right\}$
$\frac{12!}{6!2!2!} \cdot 2 \cdot 4 = \frac{12!}{6!} \cdot 2$

Case 5:
$\left\{ a^2\bar{a}^6, a^6\bar{a}^2 \right\} \cdot \left\{ b^4, c^4, \bar{b}^4, \bar{c}^4 \right\}$
$\frac{12!}{6!2!4!} \cdot 2 \cdot 4 = \frac{12!}{6!} \cdot \frac16$ 

Case 6:
$\left\{ a^2\bar{a}^6, a^6\bar{a}^2 \right\} \cdot \left\{ b^2\bar{c}^2, \bar{b}^2c^2 \right\}$
$\frac{12!}{6!2!2!2!} \cdot 2 \cdot 2 = \frac{12!}{6!} \cdot \frac12$ 

Case 7:
$\left\{ a^4\bar{a}^4 \right\} \cdot \left\{ b^2\bar{b}c, b\bar{b}^2\bar{c}, bc^2\bar{c}, \bar{b}c\bar{c}^2 \right\}$
$\frac{12!}{4!4!2!} \cdot 4 = \frac{12!}{4!} \cdot \frac{1}{12}$

Case 8:
$\left\{ a^4\bar{a}^4 \right\} \cdot \left\{ b^4, c^4, \bar{b}^4, \bar{c}^4 \right\}$
$\frac{12!}{4!4!4!} \cdot 4 = \frac{12!}{4!} \cdot \frac{1}{144}$ 

Case 9:
$\left\{ a^4\bar{a}^4 \right\} \cdot \left\{ b^2\bar{c}^2, \bar{b}^2c^2 \right\}$
$\frac{12!}{4!4!2!2!} \cdot 2 = \frac{12!}{4!} \cdot \frac{1}{48}$

Cases with combinations of $\left\{ a, \bar{a} \right\}$ multiplied to $-16$ and combinations of $\left\{ b, \bar{b}, c, \bar{c} \right\}$ multiplied to $\frac{1}{16}$: 

Case 10:
$\left\{ a\bar{a}^7, a^7\bar{a} \right\} \cdot \left\{ b^2\bar{b}^2, c^2\bar{c}^2, b^2c^2, \bar{b}^2\bar{c}^2 \right\}$
$\frac{12!}{7!2!2!} \cdot 2 \cdot 4 = \frac{12!}{7!} \cdot 2$

Case 11:
$\left\{ a\bar{a}^7, a^7\bar{a} \right\} \cdot \left\{ b\bar{b}c\bar{c} \right\}$
$\frac{12!}{7!} \cdot 2$ 

Case 12:
$\left\{ a\bar{a}^7, a^7\bar{a} \right\} \cdot \left\{ b^3\bar{c}, b\bar{c}^3, \bar{b}^3c, \bar{b}c^3 \right\}$
$\frac{12!}{7!3!} \cdot 2 \cdot 4 = \frac{12!}{7!} \cdot \frac{4}{3}$ 

Case 13:
$\left\{ a^5\bar{a}^3, a^3\bar{a}^5 \right\} \cdot \left\{ b^2\bar{b}^2, c^2\bar{c}^2, b^2c^2, \bar{b}^2\bar{c}^2 \right\}$
$\frac{12!}{5!3!2!2!} \cdot 2 \cdot 4 = \frac{12!}{6!} \cdot 2$

Case 14:
$\left\{ a^5\bar{a}^3, a^3\bar{a}^5 \right\} \cdot \left\{ b\bar{b}c\bar{c} \right\}$
$\frac{12!}{5!3!} \cdot 2 = \frac{12!}{6!} \cdot 2$ 

Case 15:
$\left\{ a^5\bar{a}^3, a^3\bar{a}^5 \right\} \cdot \left\{ b^3\bar{c}, b\bar{c}^3, \bar{b}^3c, \bar{b}c^3 \right\}$
$\frac{12!}{5!3!3!} \cdot 2 \cdot 4 = \frac{12!}{6!} \cdot \frac{4}{3}$

Sum of the permutations of all combinations that multiply to $-1$: \begin{align*} &\frac{12!}{8!} \cdot 4 + \frac{12!}{8!} \cdot \frac13 + \frac{12!}{8!} + \frac{12!}{6!} \cdot 2 + \frac{12!}{6!} \cdot \frac16 + \frac{12!}{6!} \cdot \frac12 + \frac{12!}{4!} \cdot \frac{1}{12} + \frac{12!}{4!} \cdot \frac{1}{144} + \frac{12!}{4!} \cdot \frac{1}{48} + \frac{12!}{7!} \cdot 2 + \frac{12!}{7!} \cdot 2 + \frac{12!}{7!} \cdot \frac{4}{3}\\ &+ \frac{12!}{6!} \cdot 2 + \frac{12!}{6!} \cdot 2 + \frac{12!}{6!} \cdot \frac{4}{3} \\ &= \frac{12!}{7!} \left( \frac{1}{2} + \frac{1}{24} + \frac{1}{8} + 14 + \frac{7}{6} + \frac{7}{2} + \frac{35}{2} + \frac{35}{24} + \frac{35}{8} + 2 + 2 + \frac{4}{3} + 14 + 14 + \frac{28}{3} \right)\\ &= \frac{12!}{7!} \cdot \frac{256}{3} \end{align*}

\[\frac{ \frac{12!}{7!} \cdot \frac{256}{3} }{ 6^{12} } = \boxed{\textbf{(E)} ~\frac{2^2\cdot 5\cdot 11}{3^{10}}}\]

~isabelchen

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=CfO4zIeYrSY

See Also

2017 AMC 12A (Problems • Answer Key • Resources)
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