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Geometry Difficulty 7.2 National olympiad, round 2 Prove it Ukraine

Side $BC$ of the parallelogram $ABCD$ is extended beyond the point $C$ and point $K$ is placed on the extension so that $\triangle CDK$ is isosceles triangle with base $CK$ . Side $DC$ is extended beyond the point $C$ and point $L$ is placed on the extension so that $CBL$ is isosceles triangle with base $CL$ . Bisectors of angles $\angle LBC$ and $\angle CDK$ intersect at point $Q$ . Find radius of a circle circumscribed about triangle $ALK$ , if $\angle BQD = \alpha$ and $KL = a$ .

Answer: $\frac{a}{2\sin 2\alpha}$ .

Solution

Since trapezium $ABKD$ is equilateral, points $A, B, K, D$ are on one circle. Thereafter, points $B, L, D, A$ reside on one circle too. Thus all the five points $B, L, D, A, K$ are on the same circle $w$ circumscribed about the triangle $\triangle ALK$ . Since trapeziums $ABKD$ and $BLDA$ share one of the diagonals, their diagonals have equal length. Therefore, $AL = AK$ and $\triangle ALK$ is isosceles.

As $BQ$ is a bisector of the isosceles triangle $\triangle LBC$ (fig.7), $BQ \perp LC$ and $LC \parallel AB$ . Therefore, $BQ \perp AB$ and $\angle QBA = 90^\circ$ , similarly $\angle QDA = 90^\circ$ . This implies that points $Q, B, D, A$ are on one circle, and this circle is $w$ . As angles $\angle BQD$ and $\angle ALK$ rest upon chords of equal length: $BD = AK$ , $\angle ALK = \alpha$ and $\angle LAK = 180^\circ - 2\alpha$ . According to the law of sines, $R = \frac{LK}{2\sin \angle LAK} = \frac{a}{2\sin 2\alpha}$ for $\triangle ALK$ .

Fig.7

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