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Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Let $ABC$ be an equilateral triangle. A regular hexagon $P X Q Y R Z$ of side length $2$ is placed so that $P$ , $Q$ , and $R$ lie on segments $\overline{BC}$ , $\overline{CA}$ , and $\overline{AB}$ , respectively. If points $A$ , $X$ , and $Y$ are collinear, compute $BC$ .

Solution

Notice that $\angle QAR = 60^{\circ}$ , and $\triangle YAR$ is isosceles with base angle $120^{\circ}$ . This implies that $Y$ is the circumcenter of $\triangle AQR$ . Thus, $YA = YR = YQ = 2$ . We have $\angle AYR = 90^{\circ}$ , so $AR = 2\sqrt{2}$ . Moreover, $\angle AYQ = 150^{\circ}$ , so $\angle YAQ = 15^{\circ}$ , which implies that $AQ = 4 \sin 75^{\circ} = \sqrt{6} + \sqrt{2}$ . By symmetry, we also get that $BR = AQ = \sqrt{6} + \sqrt{2}$ . Hence, the answer is $AR + BR = \sqrt{6} + 3\sqrt{2}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.