AMC12 2017 A

AMC12 2017 A · Q25

AMC12 2017 A · Q25. It mainly tests Complex numbers (rare), Combinations.

The vertices $V$ of a centrally symmetric hexagon in the complex plane are given by $V = \{ \sqrt{2}i, -\sqrt{2}i, (1/\sqrt{8})(1 + i), (1/\sqrt{8})(-1 + i), (1/\sqrt{8})(1 - i), (1/\sqrt{8})(-1 - i) \}$. 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 $P = -1$?

中心对称六边形的顶点 $V$ 在复平面中由 $V = \{ \sqrt{2}i, -\sqrt{2}i, (1/\sqrt{8})(1 + i), (1/\sqrt{8})(-1 + i), (1/\sqrt{8})(1 - i), (1/\sqrt{8})(-1 - i) \}$ 给出。对于每个 $j$，$1 \leq j \leq 12$，独立地从 $V$ 中随机选择元素 $z_j$。令 $P = \prod_{j=1}^{12} z_j$ 为 $12$ 个数的乘积。$P = -1$ 的概率是多少？

(A) $5 \cdot 11 / 3^{10}$ $5 \cdot 11 / 3^{10}$

(B) $5^2 \cdot 11^2 / 3^{10}$ $5^2 \cdot 11^2 / 3^{10}$

(D) $5 \cdot 7 \cdot 11^2 / 3^{10}$ $5 \cdot 7 \cdot 11^2 / 3^{10}$

(E) $2^2 \cdot 5 \cdot 11 / 3^{10}$ $2^2 \cdot 5 \cdot 11 / 3^{10}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): If $z_j$ is an element of the set $A=\{\sqrt{2}i,-\sqrt{2}i\}$, then $|z_j|=\sqrt{2}$. Otherwise $z_j$ is an element of $$ B=V\setminus A=\left\{\frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i)\right\} $$ and $|z_j|=\frac{1}{2}$. It follows that $|P|=\prod_{j=1}^{12}|z_j|=1$ exactly when 8 of the 12 factors $z_j$ are in $A$ and 4 of the factors are in $B$. The product of 8 complex numbers each of which is in $A$ is a real number, either $16$ or $-16$. The product of 4 numbers each of which is in $B$ is one of $\frac{1}{16}$, $\frac{1}{16}i$, $-\frac{1}{16}$, or $-\frac{1}{16}i$. Thus a product $P=\prod_{j=1}^{12} z_j$ is $-1$ exactly when 8 of the $z_j$ are from $A$, 4 of the $z_j$ are from $B$, and the last of the 4 elements from $B$ is chosen so that the product is $-1$ rather than $i$, $-i$, or $1$. Because the probability is $\frac{1}{3}$ that a particular factor $z_j$ is from $A$, the probability is $\frac{2}{3}$ that a particular factor $z_j$ is from $B$, and the probability is $\frac{1}{6}$ that a particular factor $z_j$ is a specific element of $V$, the probability that the product $P$ will be $-1$ is given by $$ \binom{12}{4}\left(\frac{1}{3}\right)^8\left(\frac{2}{3}\right)^3\left(\frac{1}{6}\right) =\frac{12\cdot 11\cdot 10\cdot 9}{4\cdot 3\cdot 2\cdot 1}\cdot\frac{1}{3^8}\cdot\frac{2^3}{3^3}\cdot\frac{1}{6} =\frac{2^2\cdot 5\cdot 11}{3^{10}}. $$

答案（E）：如果 $z_j$ 属于集合 $A=\{\sqrt{2}i,-\sqrt{2}i\}$，则 $|z_j|=\sqrt{2}$。否则 $z_j$ 属于 $$ B=V\setminus A=\left\{\frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i)\right\} $$ 并且 $|z_j|=\frac{1}{2}$。因此，$|P|=\prod_{j=1}^{12}|z_j|=1$ 当且仅当 12 个因子 $z_j$ 中有 8 个在 $A$ 中、4 个在 $B$ 中。8 个都在 $A$ 中的复数之积是实数，可能为 $16$ 或 $-16$。4 个都在 $B$ 中的数之积是 $\frac{1}{16}$、$\frac{1}{16}i$、$-\frac{1}{16}$ 或 $-\frac{1}{16}i$ 之一。因此，乘积 $P=\prod_{j=1}^{12} z_j$ 等于 $-1$ 当且仅当：8 个 $z_j$ 来自 $A$，4 个 $z_j$ 来自 $B$，并且这 4 个来自 $B$ 的元素中最后一个选取得使得整体乘积为 $-1$，而不是 $i$、$-i$ 或 $1$。由于某个特定因子 $z_j$ 来自 $A$ 的概率为 $\frac{1}{3}$，来自 $B$ 的概率为 $\frac{2}{3}$，且某个特定因子 $z_j$ 取为 $V$ 中某个指定元素的概率为 $\frac{1}{6}$，所以乘积 $P$ 为 $-1$ 的概率为 $$ \binom{12}{4}\left(\frac{1}{3}\right)^8\left(\frac{2}{3}\right)^3\left(\frac{1}{6}\right) =\frac{12\cdot 11\cdot 10\cdot 9}{4\cdot 3\cdot 2\cdot 1}\cdot\frac{1}{3^8}\cdot\frac{2^3}{3^3}\cdot\frac{1}{6} =\frac{2^2\cdot 5\cdot 11}{3^{10}}. $$

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