Question

一个半径为 $r$ 的圆与边长为 $2$ 的正六边形同心，并且在正六边形外。若从圆上随机选一点，能看到正六边形的三个完整边的概率为 $1/2$ 。求 $r$ ？

A circle of radius $r$ is concentric with and outside a regular hexagon of side length $2.$ The probability that three entire sides of the hexagon are visible from a randomly chosen point on the circle is $1/2.$ What is $r?$

$2\sqrt{2} + 2\sqrt{3}$

$3\sqrt{3} + \sqrt{2}$

$2\sqrt{6} + \sqrt{3}$

$3\sqrt{2} + \sqrt{6}$

$6\sqrt{2} - \sqrt{3}$

答案：D

知识点：几何概率正多边形三角学

难度评级：2340

解答：

把正六边形放在圆的中心。每个顶点都对应一段圆弧，从该弧上只能完整看到在此顶点相交的两条边。这六段全等圆弧组成补事件，其概率为 $\tfrac{1}{2},$ 所以每段弧的度数为 $30^\circ.$

取以圆心 $O$ 到顶点 $A,$ 的射线为中心的那段弧，并设 $P$ 为它的上端点。则 $\angle POA = 15^\circ,$ 而在 $P$ 处第三条边刚好开始可见，所以 $P$ 位于该边所在的直线上。这条直线到 $O$ 的距离为边心距 $\sqrt{3}.$

因此 $\sqrt{3} = r\sin 15^\circ = r \cdot \dfrac{\sqrt{6} - \sqrt{2}}{4},$ 所以 $r = \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} = 3\sqrt{2} + \sqrt{6}.$

所以正确答案是 D。

Place the hexagon at the center of the circle. Corresponding to each vertex is an arc from which only the two sides meeting there are entirely visible. These six congruent arcs make up the complementary probability $\tfrac{1}{2},$ so each arc measures $30^\circ.$

Take the arc centered on the ray from the center $O$ through a vertex $A,$ and let $P$ be its upper endpoint. Then $\angle POA = 15^\circ,$ and at $P$ a third side is just becoming visible, so $P$ lies on that side's supporting line. Its distance from $O$ is the apothem $\sqrt{3}.$

Hence $\sqrt{3} = r\sin 15^\circ = r \cdot \dfrac{\sqrt{6} - \sqrt{2}}{4},$ giving $r = \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} = 3\sqrt{2} + \sqrt{6}.$

Thus, the correct answer is D.

2006 AMC 12A 第 22 题

答案：D

难度评级：2340

解答：

其他年份的第 22 题