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2024 AMC 12B 第 19 题

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19.

边长为 1414 的等边三角形 △ABC\triangle ABC 绕其中心旋转角 θ\theta，其中 0<θ<60∘0 \lt \theta \lt 60^\circ 得到三角形 △DEF\triangle DEF。见图。六边形 ADBECFADBECF 的面积为 91391\sqrt3。求 tan⁡θ\tan\theta？

Equilateral △ABC\triangle ABC with side length 1414 is rotated about its center by angle θ,\theta, where 0<θ<60∘,0 \lt \theta \lt 60^\circ, to form △DEF.\triangle DEF. See the figure. The area of hexagon ADBECFADBECF is 913.91\sqrt3. What is tan⁡θ?\tan\theta?

34\dfrac{3}{4}

5311\dfrac{5\sqrt3}{11}

45\dfrac{4}{5}

1113\dfrac{11}{13}

7313\dfrac{7\sqrt3}{13}

答案：B
知识点：三角恒等式面积变换
难度评级：2040
解答：

六个顶点都在外接圆上，半径 R=143R = \dfrac{14}{\sqrt3}，所以 R2=1963R^2 = \dfrac{196}{3}。圆心角在 θ\theta 与 120∘−θ120^\circ - \theta 之间交替。六边形面积为 12R2(3sin⁡θ+3sin⁡(120∘−θ))=98(sin⁡θ+sin⁡(120∘−θ)). \begin{aligned} &\tfrac12 R^2\bigl(3\sin\theta + 3\sin(120^\circ - \theta)\bigr) \\ &= 98\bigl(\sin\theta + \sin(120^\circ - \theta)\bigr). \end{aligned}

由和差化积，sin⁡θ+sin⁡(120∘−θ)\sin\theta + \sin(120^\circ - \theta) =2sin⁡60∘cos⁡(θ−60∘)= 2\sin 60^\circ\cos(\theta - 60^\circ) =3cos⁡(60∘−θ)= \sqrt3\cos(60^\circ - \theta)。令面积为 91391\sqrt3，得 3cos⁡(60∘−θ)\sqrt3\cos(60^\circ - \theta) =91398=13314= \dfrac{91\sqrt3}{98} = \dfrac{13\sqrt3}{14}，所以 cos⁡(60∘−θ)=1314\cos(60^\circ - \theta) = \dfrac{13}{14}，且 sin⁡(60∘−θ)=3314\sin(60^\circ - \theta) = \dfrac{3\sqrt3}{14}。

于是 tan⁡(60∘−θ)=3313\tan(60^\circ - \theta) = \dfrac{3\sqrt3}{13}，并且 tan⁡θ=tan⁡(60∘−(60∘−θ))=3−33131+3⋅3313=103132213=5311. \begin{aligned} &\tan\theta = \tan\bigl(60^\circ - (60^\circ - \theta)\bigr) \\ &= \frac{\sqrt3 - \frac{3\sqrt3}{13}}{1 + \sqrt3\cdot\frac{3\sqrt3}{13}} \\ &= \frac{\frac{10\sqrt3}{13}}{\frac{22}{13}} \\ &= \frac{5\sqrt3}{11}. \end{aligned}

所以正确答案是 B。

The six vertices lie on the circumcircle of radius R=143,R = \dfrac{14}{\sqrt3}, so R2=1963.R^2 = \dfrac{196}{3}. Going around, the central angles alternate between θ\theta (three times) and 120∘−θ120^\circ - \theta (three times). The cyclic-hexagon area is 12R2(3sin⁡θ+3sin⁡(120∘−θ))=98(sin⁡θ+sin⁡(120∘−θ)). \begin{aligned} &\tfrac12 R^2\bigl(3\sin\theta + 3\sin(120^\circ - \theta)\bigr) \\ &= 98\bigl(\sin\theta + \sin(120^\circ - \theta)\bigr). \end{aligned}

By sum-to-product, sin⁡θ+sin⁡(120∘−θ)\sin\theta + \sin(120^\circ - \theta) =2sin⁡60∘cos⁡(θ−60∘)= 2\sin 60^\circ\cos(\theta - 60^\circ) =3cos⁡(60∘−θ).= \sqrt3\cos(60^\circ - \theta). Setting the area to 91391\sqrt3 gives 3cos⁡(60∘−θ)\sqrt3\cos(60^\circ - \theta) =91398=13314,= \dfrac{91\sqrt3}{98} = \dfrac{13\sqrt3}{14}, so cos⁡(60∘−θ)=1314\cos(60^\circ - \theta) = \dfrac{13}{14} and sin⁡(60∘−θ)=3314.\sin(60^\circ - \theta) = \dfrac{3\sqrt3}{14}.

Then tan⁡(60∘−θ)=3313,\tan(60^\circ - \theta) = \dfrac{3\sqrt3}{13}, and tan⁡θ=tan⁡(60∘−(60∘−θ))=3−33131+3⋅3313=103132213=5311. \begin{aligned} &\tan\theta = \tan\bigl(60^\circ - (60^\circ - \theta)\bigr) \\ &= \frac{\sqrt3 - \frac{3\sqrt3}{13}}{1 + \sqrt3\cdot\frac{3\sqrt3}{13}} \\ &= \frac{\frac{10\sqrt3}{13}}{\frac{22}{13}} \\ &= \frac{5\sqrt3}{11}. \end{aligned}

Thus, the correct answer is B.

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