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1996 AMC 12 第 30 题

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30.

一个内接于圆的六边形有连续三条边的长度均为 33，另有连续三条边的长度均为 55。一条弦把六边形分成两个梯形，其中一个梯形有三条长度均为 33 的边，另一个有三条长度均为 55 的边。该弦长等于 mn\frac{m}{n}，其中 mm 和 nn 是互质正整数。求 m+nm+n。

A hexagon inscribed in a circle has three consecutive sides each of length 33 and three consecutive sides each of length 5.5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 33 and the other with three sides each of length 5,5, has length equal to mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m+n.

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答案：E
知识点：cyclic polygons三角学triple-angle identities
难度评级：2330
小提示：

设 α\alpha 和 β\beta 分别为长度 33 和 55 的边所对圆心角的一半

Let α\alpha and β\beta be the half-central angles subtended by sides 33 and 55

大提示：

利用 α+β=60∘\alpha+\beta=60^\circ 和弦长之比求 sin⁡2α\sin^2\alpha，再应用 sin⁡3αsin⁡α\frac{\sin3\alpha}{\sin\alpha}

Use α+β=60∘\alpha+\beta=60^\circ and the chord ratio to find sin⁡2α\sin^2\alpha, then apply sin⁡3αsin⁡α\frac{\sin3\alpha}{\sin\alpha}

解答：

设圆的半径为 RR，并设 α,β\alpha,\beta 分别为长度 3,53,5 的边所对半圆心角。则 3=2Rsin⁡α3=2R\sin\alpha、5=2Rsin⁡β5=2R\sin\beta，且 α+β=60∘\alpha+\beta=60^\circ。因此 53=sin⁡(60∘−α)sin⁡α\frac53=\frac{\sin(60^\circ-\alpha)}{\sin\alpha} =32cot⁡α−12=\frac{\sqrt3}{2}\cot\alpha-\frac12，所以 tan⁡α=3313\tan\alpha=\frac{3\sqrt3}{13}，且 sin⁡2α=27196\sin^2\alpha=\frac{27}{196}。分割弦跨过连续三条长度为 33 的边，所以 L3=sin⁡3αsin⁡α\frac L3=\frac{\sin3\alpha}{\sin\alpha} =3−4sin⁡2α=3−2749=12049=3-4\sin^2\alpha=3-\frac{27}{49}=\frac{120}{49}。因此 L=36049L=\frac{360}{49}，所以 m+n=409m+n=409。正确答案是 E。

Let the circle have radius R,R, and let α,β\alpha,\beta be the half-central angles for the sides 3,5.3,5. Then 3=2Rsin⁡α,3=2R\sin\alpha, 5=2Rsin⁡β,5=2R\sin\beta, and α+β=60∘.\alpha+\beta=60^\circ. Thus 53=sin⁡(60∘−α)sin⁡α\frac53=\frac{\sin(60^\circ-\alpha)}{\sin\alpha} =32cot⁡α−12,=\frac{\sqrt3}{2}\cot\alpha-\frac12, so tan⁡α=3313\tan\alpha=\frac{3\sqrt3}{13} and sin⁡2α=27196.\sin^2\alpha=\frac{27}{196}. The dividing chord spans the three consecutive sides of length 3,3, so L3=sin⁡3αsin⁡α\frac L3=\frac{\sin3\alpha}{\sin\alpha} =3−4sin⁡2α=3−2749=12049.=3-4\sin^2\alpha=3-\frac{27}{49}=\frac{120}{49}. Therefore L=36049,L=\frac{360}{49}, so m+n=409.m+n=409. The correct answer is E.

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