AMC10 2015 A

AMC10 2015 A · Q25

AMC10 2015 A · Q25. It mainly tests Probability (basic), Geometric probability (basic).

Let $S$ be a square of side length $1$. Two points are chosen independently at random on the sides of $S$. The probability that the straight-line distance between the points is at least $\frac12$ is $\frac{a-b\pi}{c}$, where $a$, $b$, and $c$ are positive integers and $\gcd(a,b,c)=1$. What is $a+b+c$?

设$S$为边长为$1$的正方形。在$S$的边上独立随机选取两个点。两点间的直线距离至少为$\frac12$的概率为$\frac{a-b\pi}{c}$，其中$a,b,c$为正整数，且$\gcd(a,b,c)=1$。求$a+b+c$。

(A) 59 59

(B) 60 60

(D) 62 62

(E) 63 63

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Let the square have vertices (0,0), (1,0), (1,1), and (0,1), and consider three cases. Case 1: The chosen points are on opposite sides of the square. In this case the distance between the points is at least $\frac{1}{2}$ with probability 1. Case 2: The chosen points are on the same side of the square. It may be assumed that the points are (a,0) and (b,0). The pairs of points in the ab-plane that meet the requirement are those within the square $0 \le a \le 1,\ 0 \le b \le 1$ that satisfy either $b \ge a+\frac{1}{2}$ or $b \le a-\frac{1}{2}$. These inequalities describe the union of two isosceles right triangles with leg length $\frac{1}{2}$, together with their interiors. The area of the region is $\frac{1}{4}$, and the area of the square is 1, so the probability that the pair of points meets the requirement in this case is $\frac{1}{4}$. Case 3: The chosen points are on adjacent sides of the square. It may be assumed that the points are (a,0) and (0,b). The pairs of points in the ab-plane that meet the requirement are those within the square $0 \le a \le 1,\ 0 \le b \le 1$ that satisfy $\sqrt{a^2+b^2} \ge \frac{1}{2}$. These inequalities describe the region inside the square and outside a quarter-circle of radius $\frac{1}{2}$. The area of this region is $1-\frac{1}{4}\pi\left(\frac{1}{2}\right)^2 = 1-\frac{\pi}{16}$, which is also the probability that the pair of points meets the requirement in this case. Cases 1 and 2 each occur with probability $\frac{1}{4}$, and Case 3 occurs with probability $\frac{1}{2}$. The requested probability is \[ \frac{1}{4}\cdot 1+\frac{1}{4}\cdot\frac{1}{4}+\frac{1}{2}\left(1-\frac{\pi}{16}\right)=\frac{26-\pi}{32}, \] and $a+b+c=59$.

答案（A）：设正方形的顶点为 (0,0)、(1,0)、(1,1)、(0,1)，并考虑三种情况。情况 1：所选点在正方形的相对两边上。在这种情况下，两点间距离至少为 $\frac{1}{2}$ 的概率为 1。情况 2：所选点在正方形的同一边上。不妨设两点为 (a,0) 与 (b,0)。在 $ab$ 平面内满足条件的点对，是位于正方形 $0 \le a \le 1,\ 0 \le b \le 1$ 中且满足 $b \ge a+\frac{1}{2}$ 或 $b \le a-\frac{1}{2}$ 的那些点对。这些不等式描述了两块直角等腰三角形（直角边长为 $\frac{1}{2}$）及其内部的并集。该区域面积为 $\frac{1}{4}$，正方形面积为 1，因此此情况下点对满足要求的概率为 $\frac{1}{4}$。情况 3：所选点在正方形的相邻两边上。不妨设两点为 (a,0) 与 (0,b)。在 $ab$ 平面内满足条件的点对，是位于正方形 $0 \le a \le 1,\ 0 \le b \le 1$ 中且满足 $\sqrt{a^2+b^2} \ge \frac{1}{2}$ 的那些点对。这些不等式描述了正方形内、半径为 $\frac{1}{2}$ 的四分之一圆外的区域。该区域面积为 $1-\frac{1}{4}\pi\left(\frac{1}{2}\right)^2 = 1-\frac{\pi}{16}$，这也就是此情况下点对满足要求的概率。情况 1 和情况 2 各以概率 $\frac{1}{4}$ 发生，情况 3 以概率 $\frac{1}{2}$ 发生。所求概率为 \[ \frac{1}{4}\cdot 1+\frac{1}{4}\cdot\frac{1}{4}+\frac{1}{2}\left(1-\frac{\pi}{16}\right)=\frac{26-\pi}{32}, \] 并且 $a+b+c=59$。

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