AMC12 2010 B

AMC12 2010 B · Q18

AMC12 2010 B · Q18. It mainly tests Probability (basic), Coordinate geometry.

A frog makes $3$ jumps, each exactly $1$ meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than $1$ meter from its starting position?

一只青蛙做 $3$ 次跳跃，每次恰好 $1$ 米长。每次跳跃的方向独立且随机选择。青蛙最终位置与起始位置的距离不超过 $1$ 米的概率是多少？

(A) \frac{1}{6} \frac{1}{6}

(B) \frac{1}{5} \frac{1}{5}

(D) \frac{1}{3} \frac{1}{3}

(E) \frac{1}{2} \frac{1}{2}

Answer

Correct choice: (C)

正确答案：（C）

Solution

We will let the moves be complex numbers $a$, $b$, and $c$, each of magnitude one. The starts on the origin. It is relatively easy to show that exactly one element in the set \[\{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}\] has magnitude less than or equal to $1$. (Can you show how?) Hence, the probability is $\boxed{\text{(C)} \frac {1}{4}}$.

令三次位移分别为复数 $a$、$b$、$c$，且每个的模都为 $1$。起点在原点。可以较容易证明集合 \[\{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}\] 中恰有一个元素的大小不超过 $1$。（你能证明吗？）因此所求概率为 $\boxed{\text{(C)} \frac {1}{4}}$。

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