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Slovenia — National Math Olympiad

Geometry Difficulty 5.4 AIME, harder Prove it Slovenia

Let $ABCD$ be a square and let $E$ and $F$ be two points outside the square, such that $BEC$ and $CFD$ are equilateral triangles. Let $G$ be the intersection of the lines $BE$ and $FD$ , and let $H$ be a point, such that the quadrilateral $CEHF$ is a rhombus. Prove that the points $G$ , $E$ , $H$ and $F$ lie on the same circle.

Solution

Since $ABCD$ is a square and the triangles $BEC$ and $CFD$ are equilateral we have $|CE| = |CB| = |CD| = |CF|$ . The triangle $FCE$ is isosceles and
$\begin{aligned} \angle ECF &= 2\pi - \angle FCD - \angle DCB - \angle BCE \\ &= 2\pi - \frac{\pi}{3} - \frac{\pi}{2} - \frac{\pi}{3} = \frac{5\pi}{6}. \end{aligned}$

So, $\angle CEF = \frac{\pi}{12}$ and $\angle EFC = \frac{\pi}{12}$ . Since the quadrilateral $CEHF$ is a rhombus, we have $\angle FEH = \angle HFE = \frac{\pi}{12}$ . So, $\angle BEH = \frac{\pi}{3} + 2 \cdot \frac{\pi}{12} = \frac{\pi}{2}$ and $\angle DFH = \frac{\pi}{2}$ . In the quadrilateral $GEHF$ we have $\angle GEH + \angle HFG = \frac{\pi}{2} + \frac{\pi}{2} = \pi$ , so the points $G$ , $E$ , $H$ and $F$ lie on the same circle.

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