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Belarusian Mathematical Olympiad

Geometry Difficulty 6.7 National Olympiad Prove it Belarus

Two circles $\Omega$ and $\Gamma$ are internally tangent at the point $B$ . The chord $AC$ of $\Gamma$ is tangent to $\Omega$ at the point $L$ , and the segments $AB$ and $BC$ intersect $\Omega$ at the points $M$ and $N$ . Let $M_1$ and $N_1$ be the reflections of $M$ and $N$ about the line $BL$ ; and let $M_2$ and $N_2$ be the reflections of $M$ and $N$ about the line $AC$ . The lines $M_1M_2$ and $N_1N_2$ intersect at the point $K$ .
Prove that the lines $BK$ and $AC$ are perpendicular.

Solution

By the Archimedes's lemma, $BL$ is the bisector of $\angle ABC$ , so $M_1 \in BC$ and $N_1 \in AB$ . Denote $MM_1 \cap LB = T$ and $MM_2 \cap LA = P$ . Since $\angle MTL = \angle MPL = 90^\circ$ , the quadrilateral $MTLP$ is cyclic with the diameter $ML$ . Hence
$\angle MM_1M_2 = \angle MTP = \angle MLP = \angle MBL = \angle TBM_1 = 90^\circ - \angle TM_1B.$
Thus $\angle M_2M_1B = 90^\circ$ and, similarly $\angle N_2N_1B = 90^\circ$ , which implies that the quadrilateral $BN_1KM_1$ is cyclic. Homotety with center $B$ , mapping $\Omega$ to $\Gamma$ , maps $\triangle MBN$ to $\triangle ABC$ , hence these triangles are similar. Since the lines $MN$ and $M_1N_1$ are antiparallel with respect to $\angle ABC$ , we can write (in oriented angles):

$\begin{aligned} \angle(BK, AC) &= \angle(BK, N_1K) + \angle(N_1K, AB) + \angle(AB, AC) = \\ &= \angle(BM_1, M_1N_1) + 90^\circ + \angle(MB, MN) = 90^\circ. \end{aligned}$

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