Ukrajina 2008

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 $△ALK$. Since trapeziums $ABKD$ and $BLDA$ share one of the diagonals, their diagonals have equal length. Therefore, $AL=AK$ and $△ALK$ is isosceles.

As $BQ$ is a bisector of the isosceles 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 $△ALK$.

Fig.7