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Junior Balkan Mathematical Olympiad

Geometry Difficulty 6.1 National Olympiad Prove it JBMO

Problem:

Let $ABC$ be a triangle with $\angle C = 90^{\circ}$ and $D \in CA$ , $E \in CB$ , and $k_1, k_2, k_3, k_4$ semicircles with diameters $CA$ , $CB$ , $CD$ , $CE$ respectively, which have common part with the triangle $ABC$ . Let also,
$k_1 \cap k_2 = \{C, K\},\quad k_3 \cap k_4 = \{C, M\},\quad k_2 \cap k_3 = \{C, L\},\quad k_1 \cap k_4 = \{C, N\}$
Prove that $K, L, M$ and $N$ are cocyclic points.

Solution

Solution:

The points $K, L, M, N$ belong to the segments $AB$ , $BD$ , $DE$ , $EA$ respectively, where $CK \perp AB$ , $CL \perp BD$ , $CM \perp DE$ , $CN \perp AE$ . Then quadrilaterals $CDLM$ and $CENM$ are inscribed. Let $\angle CAE = \varphi$ , $\angle DCL = \theta$ . Then $\angle EMN = \angle ECN = \varphi$ and $\angle DML = \angle DCL = \theta$ . So $\angle DML + \angle EMN = \varphi + \theta$ and therefore $\angle LMN = 180^{\circ} - \varphi - \theta$ . The quadrilaterals $CBKL$ and $CAKN$ are also inscribed and hence $\angle LKC = \angle LBC = \theta$ , $\angle CKN = \angle CAN = \varphi$ , and so $\angle LKN = \varphi + \theta$ , while $\angle LMN = 180^{\circ} - \varphi - \theta$ , which means that $KLMN$ is inscribed.

Note that $KLMN$ is convex because $L, N$ lie in the interior of the convex quadrilateral $ADEB$ .

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