Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.7 AIME, harder Prove it Estonia

Let $ABCDE$ be a regular pentagon and let $c$ be the circle with diameter $AB$ . Diagonals $AC$ and $AD$ intersect the circle $c$ at points $F$ and $G$ , respectively. Line $FG$ intersects the side $AE$ at point $H$ . Let $K$ be the midpoint of the side $DE$ . Prove that points $F, H, E$ , and $K$ are concyclic.

Solution

As $AB$ is a diameter of $c$ , $\angle AFB = 90^\circ$ (Fig. 27) and $BF$ is an altitude of triangle $ABC$ . From $AB = BC$ , $BF$ is a median and $F$ is the midpoint of $AC$ . By symmetry, point $K$ lies on $BF$ and $\angle FKE = 90^\circ$ . Notice that $\angle BAC = \angle CAD = \angle DAE$ , as $BAC, CAD$ , and $DAE$ are inscribed angles that subtend to equal arcs of the circumcircle of the regular pentagon $ABCDE$ . Points $A, B, F$ and $G$ lie on circle $c$ in this order, thus $\angle ABF = 180^\circ - \angle AGF = \angle AGH$ . Hence, $ABF$ and $AGH$ are similar from two angles, implying $\angle AHG = \angle AFB = 90^\circ$ . Therefore, the opposite angles at $K$ and $H$ of quadrilateral $FHEK$ are right angles and it is cyclic.

Fig. 27

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