AMC12 2024 B

AMC12 2024 B · Q19

AMC12 2024 B · Q19. It mainly tests Triangles (properties), Trigonometry (basic).

Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta$, where $0 < \theta < 60^{\circ}$, to form $\triangle DEF$. See the figure. The area of hexagon $ADBECF$ is $91\sqrt{3}$. What is $\tan\theta$?

边长为 $14$ 的正三角形 $\triangle ABC$ 绕其中心旋转角度 $\theta$，其中 $0 < \theta < 60^{\circ}$，形成 $\triangle DEF$。见图。六边形 $ADBECF$ 的面积为 $91\sqrt{3}$。求 $\tan\theta$？

(A) \frac{3}{4} \frac{3}{4}

(B) \frac{5\sqrt{3}}{11} \frac{5\sqrt{3}}{11}

(D) \frac{11}{13} \frac{11}{13}

(E) \frac{7\sqrt{3}}{13} \frac{7\sqrt{3}}{13}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let O be circumcenter of the equilateral triangle Easily get $OF = \frac{14\sqrt{3}}{3}$ $2 \cdot (\triangle(OFC) + \triangle(OCE)) = OF^2 \cdot \sin(\theta) + OF^2 \cdot \sin(120 - \theta)$ \[= \frac{14^2 \cdot 3}{9} ( \sin(\theta) + \sin(120 - \theta) )\] \[= \frac{196}{3} ( \sin(\theta) + \sin(120 - \theta) )\] \[= 2 \cdot {\frac{1}{3} } \cdot(ADBECF) = 2\cdot \frac{91\sqrt{3}}{3}\] \[\sin(\theta) + \sin(120 - \theta) = \frac{13\sqrt{3}}{14}\] \[\sin(\theta) + \frac{ \sqrt{3}}{2}\cos( \theta) +\frac{ \sqrt{1}}{2}\sin( \theta) = \frac{13\sqrt{3}}{14}\] \[\sqrt{3} \sin( \theta) + \cos( \theta) = \frac{13 }{7}\] \[\cos( \theta) = \frac{13 }{7} - \sqrt{3} \sin( \theta)\] \[\frac{169 }{49} - \frac{26\sqrt{3} }{7} \sin( \theta) + 4 \sin( \theta)^2 = 1\] \[\sin( \theta) = \frac{5\sqrt{3} }{14} or \frac{4\sqrt{3} }{7}\] $\frac{4\sqrt{3} }{7}$ is invalid given $\theta \leq 60^\circ$ , $\sin(\theta ) < \sin( 60^\circ ) = \frac{\sqrt{3} }{2} = \frac{\sqrt{3} \cdot 3.5}{7}$ \[\cos( \theta) = \frac{11 }{14}\] \[\tan( \theta) = \frac{5\sqrt{3} }{11} \boxed{B }\]

设 O 为正三角形的 circumcenter 容易得到 $OF = \frac{14\sqrt{3}}{3}$ $2 \cdot (\triangle(OFC) + \triangle(OCE)) = OF^2 \cdot \sin(\theta) + OF^2 \cdot \sin(120 - \theta)$ \[= \frac{14^2 \cdot 3}{9} ( \sin(\theta) + \sin(120 - \theta) )\] \[= \frac{196}{3} ( \sin(\theta) + \sin(120 - \theta) )\] \[= 2 \cdot {\frac{1}{3} } \cdot(ADBECF) = 2\cdot \frac{91\sqrt{3}}{3}\] \[\sin(\theta) + \sin(120 - \theta) = \frac{13\sqrt{3}}{14}\] \[\sin(\theta) + \frac{ \sqrt{3}}{2}\cos( \theta) +\frac{ \sqrt{1}}{2}\sin( \theta) = \frac{13\sqrt{3}}{14}\] \[\sqrt{3} \sin( \theta) + \cos( \theta) = \frac{13 }{7}\] \[\cos( \theta) = \frac{13 }{7} - \sqrt{3} \sin( \theta)\] \[\frac{169 }{49} - \frac{26\sqrt{3} }{7} \sin( \theta) + 4 \sin( \theta)^2 = 1\] \[\sin( \theta) = \frac{5\sqrt{3} }{14} or \frac{4\sqrt{3} }{7}\] $\frac{4\sqrt{3} }{7}$ 在 $\theta \leq 60^\circ$ 时无效，$\sin(\theta ) < \sin( 60^\circ ) = \frac{\sqrt{3} }{2} = \frac{\sqrt{3} \cdot 3.5}{7}$ \[\cos( \theta) = \frac{11 }{14}\] \[\tan( \theta) = \frac{5\sqrt{3} }{11} \boxed{B }\]

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