2006 AMC 12A 第 22 题

先试着解答 2006 AMC 12A 第 22 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2006 AMC 12A 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

22.

一个半径为 rr 的圆与边长为 22 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 12\frac{1}{2}。rr 是多少?
A circle of radius rr is concentric with and outside a regular hexagon of side length 2.2. The probability that three entire sides of the hexagon are visible from a randomly chosen point on the circle is 12.\frac{1}{2}. What is r?r?
22+232\sqrt{2} + 2\sqrt{3}
33+23\sqrt{3} + \sqrt{2}
26+32\sqrt{6} + \sqrt{3}
32+63\sqrt{2} + \sqrt{6}
62−36\sqrt{2} - \sqrt{3}
答案:D
知识点:几何概率正多边形三角学
小提示:
只能看到两条完整边的那些互补圆弧合起来占半个圆;由对称性,每段这样的圆弧都是 30∘30^\circ。
The complementary arcs, from which only two whole sides are visible, total half the circle; by symmetry each measures 30∘30^\circ
大提示:
中心到一条边的距离(边心距)是 3\sqrt{3},且 3=rsin⁡15∘\sqrt{3} = r\sin 15^\circ。
The distance from the center to a side (the apothem) is 3,\sqrt{3}, and 3=rsin⁡15∘\sqrt{3} = r\sin 15^\circ
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 12\tfrac{1}{2},所以每段弧的度数为 30∘30^\circ。
取以圆心 OO 到顶点 AA 的射线为中心的那段弧,并设 PP 为它的上端点。则 ∠POA=15∘\angle POA = 15^\circ,而在 PP 处第三条边刚好开始可见,所以 PP 位于该边所在的直线上。这条直线到 OO 的距离为边心距 3\sqrt{3}。
因此 3=rsin⁡15∘=r⋅6−24\sqrt{3} = r\sin 15^\circ = r \cdot \dfrac{\sqrt{6} - \sqrt{2}}{4},所以 r=436−2=32+6。 r = \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} = 3\sqrt{2} + \sqrt{6}\text{。}
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 12,\tfrac{1}{2}, so each arc measures 30∘.30^\circ.
Take the arc centered on the ray from the center OO through a vertex A,A, and let PP be its upper endpoint. Then ∠POA=15∘,\angle POA = 15^\circ, and at PP a third side is just becoming visible, so PP lies on that side’s supporting line. Its distance from OO is the apothem 3.\sqrt{3}.
Hence 3=rsin⁡15∘=r⋅6−24,\sqrt{3} = r\sin 15^\circ = r \cdot \dfrac{\sqrt{6} - \sqrt{2}}{4}, giving r=436−2=32+6. r = \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} = 3\sqrt{2} + \sqrt{6}.
Thus, the correct answer is D.
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