AMC12 2001 A

AMC12 2001 A · Q17

AMC12 2001 A · Q17. It mainly tests Circle theorems, Geometric probability (basic).

A point $P$ is selected at random from the interior of the pentagon with vertices $A = (0,2)$, $B = (4,0)$, $C = (2 \pi + 1, 0)$, $D = (2 \pi + 1,4)$, and $E=(0,4)$. What is the probability that $\angle APB$ is obtuse?

从五边形内部随机选取一点 $P$，五边形的顶点为 $A = (0,2)$，$B = (4,0)$，$C = (2 \pi + 1, 0)$，$D = (2 \pi + 1,4)$，以及 $E=(0,4)$。$\angle APB$ 为钝角的概率是多少？

(A) $\frac{1}{5}$ $\frac{1}{5}$

(B) $\frac{1}{4}$ $\frac{1}{4}$

(D) $\frac{3}{8}$ $\frac{3}{8}$

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

Answer

Correct choice: (C)

正确答案：（C）

Solution

The angle $APB$ is obtuse if and only if $P$ lies inside the circle with diameter $AB$. (This follows for example from the fact that the inscribed angle is half of the central angle for the same arc.) The area of $AFB$ is $[AFB] = \frac {AF\cdot FB}2 = 4$, and the area of $ABCDE$ is $CD\cdot DE - [AFB] = 4\cdot (2\pi+1) - 4 = 8\pi$. From the Pythagorean theorem the length of $AB$ is $\sqrt{2^2 + 4^2} = 2\sqrt{5}$, thus the radius of the circle is $\sqrt{5}$, and the area of the half-circle that is inside $ABCDE$ is $\frac{ 5\pi }2$. Therefore the probability that $APB$ is obtuse is $\frac{ \frac{ 5\pi }2 }{ 8\pi } = \boxed{\text{(C) } \frac 5{16}}$.

角 $APB$ 为钝角当且仅当点 $P$ 位于以 $AB$ 为直径的圆内。（例如可由同弧所对的圆周角等于圆心角的一半推出。）三角形 $AFB$ 的面积为 $[AFB] = \frac {AF\cdot FB}2 = 4$，而五边形 $ABCDE$ 的面积为 $CD\cdot DE - [AFB] = 4\cdot (2\pi+1) - 4 = 8\pi$。由勾股定理，$AB$ 的长度为 $\sqrt{2^2 + 4^2} = 2\sqrt{5}$，因此该圆的半径为 $\sqrt{5}$，且落在 $ABCDE$ 内的半圆面积为 $\frac{ 5\pi }2$。因此 $APB$ 为钝角的概率为 $\frac{ \frac{ 5\pi }2 }{ 8\pi } = \boxed{\text{(C) } \frac 5{16}}$。

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