APMO 2018 Problem 1

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APMO 2018 Problem 1

Unread post by samiul_samin » Thu Jan 10, 2019 11:03 pm

Let $H$ be the ortho center of the triangle $ABC$.Let $M$ and $N$ be the midpoint of the sides $AB$ and $AC$ ,respectively.Assume that $H$ lies inside the quadrilateral $BMNC$ and that the circumcircles of triangles $BMH$ and $CNH$ are tangent to each other. The line through $H$ parallel to $BC$ intersects the circumcircles of the triangles $BMH$ and $CNH$ in the points $K$ and $L$, respectively. Let $F$ be the intersection point of $MK$ and $NL$ and let $J$ be the incenter of triangle $MHN$. Prove that $F J = F A$.

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Re: APMO 2018 Problem 1

Unread post by thczarif » Fri Jul 19, 2019 12:11 pm

Let $P$ be a point on the common tangent to $\left(BMH\right)$ and $\left(CMH\right)$ such that $P$ is further from $BC$ than $H$. Then \[\angle{MHP}=\angle{MKH}=\angle{MBH}=\frac{\pi}{2}-A\]and similarly $\angle{NHP}=\frac{\pi}{2}-A$, so $\angle{MHN}=\pi-2A$. Then $\angle{MJN}=\frac{\pi}{2}+\frac{\pi-2A}{2}=\pi-A$, so $AMJN$ is cyclic. Now observe that since $MN$ is parallel to $KL$, \[\angle{FMN}=\angle{FKL}=\angle{MKH}=\frac{\pi}{2}-A\]and similarly $\angle{FNM}=\frac{\pi}{2}-A$, but these angle relations are satisfied for $F$ inside $\triangle{AMN}$ only when $F$ is the circumcenter of $\triangle{AMN}$, so $F$ is the circumcenter of cyclic quadrilateral $AMJN$. It follows that $FJ=FA$.

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