AMC12 2006 A

AMC12 2006 A · Q22

AMC12 2006 A · Q22. It mainly tests Triangles (properties), Geometric probability (basic).

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

半径为 $r$ 的圆与边长为 $2$ 的正六边形同心且位于其外部。从圆上随机选取一点，能看到正六边形的三条完整边的概率为 $1/2$。求 $r$。

(A) $2 + 2\sqrt{3}$ $2 + 2\sqrt{3}$

(B) $3\sqrt{3} + \sqrt{2}$ $3\sqrt{3} + \sqrt{2}$

(D) $3\sqrt{2} + \sqrt{6}$ $3\sqrt{2} + \sqrt{6}$

(E) $6\sqrt{2} - \sqrt{3}$ $6\sqrt{2} - \sqrt{3}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Project any two non-adjacent and non-opposite sides of the hexagon to the circle; the arc between the two points formed is the location where all three sides of the hexagon can be fully viewed. Since there are six such pairs of sides, there are six arcs. The probability of choosing a point is $1 / 2$, or if the total arc degree measures add up to $\frac{1}{2} \cdot 360^{\circ} = 180^{\circ}$. Each arc must equal $\frac{180^{\circ}}{6} = 30^\mathrm{\circ}$. Error creating thumbnail: Unable to save thumbnail to destination Call the center $O$, and the two endpoints of the arc $A$ and $B$, so $\angle AOB = 30^{\circ}$. Let $P$ be the intersections of the projections of the sides of the hexagon corresponding to $\overline{AB}$. Notice that $\triangle APO$ is an isosceles triangle: $\angle AOP = 15^{\circ}$ and $\angle OAP = OAB - 60^{\circ} = \frac{180^{\circ}-30^{\circ}}{2} - 60^{\circ} = 15^{\circ}$. Since $OA$ is a radius and $OP$ can be found in terms of a side of the hexagon, we are almost done. If we draw the altitude of $APO$ from $P$, then we get a right triangle. Using simple trigonometry, $\cos 15^{\circ} = \frac{\frac{r}{2}}{2\sqrt{3}} = \frac{r}{4\sqrt{3}}$. Since $\cos 15^{\circ} = \cos (45^{\circ} - 30^{\circ}) = \frac{\sqrt{6} + \sqrt{2}}{4}$, we get $r = \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) \cdot 4\sqrt{3} = 3\sqrt{2} + \sqrt{6} \Rightarrow \boxed{D}$.

将正六边形中任意两条既不相邻也不相对的边投影到圆上；这两点之间的圆弧就是能够完整看到六边形三条边的位置。由于这样的边对共有六组，因此共有六段圆弧。选到这些弧上的点的概率为 $1/2$，也就是说这些弧的总圆心角为 $\frac{1}{2} \cdot 360^{\circ} = 180^{\circ}$。因此每段弧对应的圆心角为 $\frac{180^{\circ}}{6} = 30^\mathrm{\circ}$。设圆心为 $O$，弧的两个端点为 $A$ 和 $B$，则 $\angle AOB = 30^{\circ}$。令 $P$ 为与弦 $\overline{AB}$ 对应的六边形两边投影线的交点。注意到 $\triangle APO$ 是等腰三角形：$\angle AOP = 15^{\circ}$，且 $\angle OAP = OAB - 60^{\circ} = \frac{180^{\circ}-30^{\circ}}{2} - 60^{\circ} = 15^{\circ}$。由于 $OA$ 是半径，且 $OP$ 可用六边形的边长表示，我们几乎完成了。若从 $P$ 向 $AO$ 作 $\triangle APO$ 的高，则得到一个直角三角形。用简单三角函数，$\cos 15^{\circ} = \frac{\frac{r}{2}}{2\sqrt{3}} = \frac{r}{4\sqrt{3}}$。又因为 $\cos 15^{\circ} = \cos (45^{\circ} - 30^{\circ}) = \frac{\sqrt{6} + \sqrt{2}}{4}$，所以 $r = \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) \cdot 4\sqrt{3} = 3\sqrt{2} + \sqrt{6} \Rightarrow \boxed{D}$。

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