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Ukrainian National Mathematical Olympiad

Geometry Difficulty 5.5 AIME, harder Prove it Ukraine

Inside the triangle $ABC$ there exists a point $O$ such that $\angle BOC = 90^\circ - \angle BAC$ . The rays $BO$ and $CO$ intersect the sides $AC$ and $AB$ at the points $K$ and $L$ respectively. The points $K_1$ and $L_1$ are selected on the segments $LC$ and $BK$ respectively, so that $BK_1 = K_1K$ and $CL_1 = L_1L$ . Let $M$ be the middle of the side $BC$ . Prove that $\angle K_1ML_1$ is a right angle.

(Anton Trygub)

Fig. 18

Solution

Let $K_2$ and $L_2$ be the midpoints of the segments $BK$ and $CL$ respectively (fig. 18). Then $MK_2 \parallel AC$ and $ML_2 \parallel AB$ . Hence, $\angle K_2ML_2 = \angle BAC$ . From the isosceles triangles $BK_1K$ and $CL_1L$ we have that $\angle L_1L_2K_1 = \angle K_1K_2L_1 = 90^\circ$ and then $L_1, L_2, K_1, K_2$ lie on the same circle with diameter $K_1L_1$ . Also note that
$\angle K_2K_1L_2 = \angle K_2L_1L_2 = 90^\circ - \angle BOC = \angle BAC = \angle K_2ML_2.$
So, $M$ also lies on a circle with diameter $K_1L_1$ . Hence, $\angle K_1ML_1 = 90^\circ$ .

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