Problem 35: Balkan MO 2016, Problem 2

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Joined: Mar 10, 2015

by henderson, Jul 9, 2016, 12:37 PM

$\color{red}\bf{Problem \ 35}$ Let $ABCD$ be a cyclic quadrilateral with $AB<CD$ . The diagonals intersect at the point $F$ and lines $AD$ and $BC$ intersect at the point $E$ . Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $AD$ and $BC$ respectively, and let $M$ , $S$ and $T$ be the midpoints of $EF$ , $CF$ and $DF$ respectively. Prove that the second intersection point of the circumcircles of triangles $MKT$ and $MLS$ lies on the segment $CD$ .
$\color{red}\bf{My \ solution}$ $\color{green}\textbf{Claim.}$ The circumcircles of the triangles $MKT$ and $MLS$ intersect at the midpoint of $CD.$

$\color{green}\textbf{Proof.}$ Let $P$ be the midpoint of $CD$ and $\angle ADB=\alpha,$ $\angle BDC=\beta$ and $\angle ACD=\theta.$ Then angle chasing gives $\angle KTP=\angle PSL=2\alpha+\beta+\theta.$ Note that, we also have $TP=FS=LS$ and $SP=FT=KT.$ Combining these facts, we obtain that $\triangle KTP \sim \triangle PSL.$ So, $KP=LP.$ Since we also have $MK=ML,$ we conclude that $MKPL$ is a kite $\color{red}\Longrightarrow$ $MP$ is the angle-bisector of $\angle KML.$ So, $\angle KMP=\angle LMP=\frac{\angle KML}{2}=\angle KEL=180^\circ-(2\alpha+\beta+\theta)=180^\circ-\angle KTP=180^\circ-\angle PSL.$ Now, it's clear that $MKTP$ and $MLSP$ are both cyclic quadrilaterals, as desired. $\quad {\square}$