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Mongolian Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it Mongolia

Draw median $AM$ and bisector $AL$ of a scalene triangle $ABC$ . Tangent lines of circumcircle of the triangle $ABC$ at points $B$ , $C$ intersect at point $T$ and line $AT$ intersects the circumcircle at point $K$ which is different from $A$ . The line $AM$ intersects circumcircle of the triangle $AKL$ at point $P$ . Prove that $\angle APH = 90^\circ$ , where $H$ is orthocenter of the triangle $ABC$ .

Solution

Note that $AF$ , $FK$ are sim medians of triangles $\Delta ABF$ , $\Delta BKF$ respectively.
Therefore we get $\angle CMK = \angle CMP$ . Since $LK$ angle bisector and quadrilateral
$APLK$ is inscribed in a circle, $\angle LPM = \angle LKM$ and from this follows
$PM = MK$ . Since $\angle MAC \sim \angle MCK$ , we get $AM/CM = MC/MK \Rightarrow AM \cdot PM = AM \cdot MK = MC^2 = a^2/4$ . On the other hand $AM \cdot AP = AM^2 - AM \cdot PM = (1/4)(2c^2+2b^2-a^2) - (1/4)a^2 = bc \cos \alpha$ . The fact that quadrilateral $IHNB$ implies
$AH \cdot AN = AI \cdot AB = cb \cos \alpha = AM \cdot PM$ . Hence quadrilateral $HNMP$ can
be inscribed in a circle and it implies $\angle APM = 90^\circ \Rightarrow \angle HPM = \angle APH = 90^\circ$ . The proof completed.

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