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Ukrainian National Mathematical Olympiad, 4th Round

Geometry Difficulty 5.7 AIME, harder Prove it Ukraine

In a parallelogram $ABCD$ $\angle ABC = 105^\circ$ . It is known that inside this parallelogram there is a point $M$ , such that the triangle $BMC$ is equilateral and $\angle CMD = 135^\circ$ . Let $K$ be a midpoint of the side $AB$ . Find $\angle BKC$ .

Solution

Drop the perpendicular $CL$ to the line $DM$ . Then in the triangle $MCL$ we know: $\angle MLC = 90^\circ$ , $\angle LMC = 45^\circ$ . Therefore, $\angle LCM = 45^\circ$ and $CL = \frac{1}{\sqrt{2}} CM$ (fig. 36).

Since $\angle LCD = 60^\circ$ , from the right triangle $LCD$ we find that $CD = 2CL = \sqrt{2} CM$ . So, $AB = CD = \sqrt{2} CM = \sqrt{2} BM$ , and since $\angle ABM = 45^\circ$ , we obtain that $\angle BMA = 90^\circ$ .

Then $MK$ is a median from the vertex of the right angle in the right triangle $AMB$ , and so $KM = BK = AK$ . Therefore, the quadrilateral $KBCM$ is a deltoid, its diagonals are perpendicular and intersect at the point $O$ .

Thus in the triangle $BOK$ we have that $\angle BOK = 90^\circ$ and $\angle OBK = \angle ABC - \angle MBC = 105^\circ - 60^\circ = 45^\circ$ . This implies that $\angle BKO = 45^\circ$ .

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