BMO 2019 Shortlist

Denote $\varphi =\widehat{XAB}=\widehat{YAC}$, $\alpha =\widehat{CAX}=\widehat{BAY}$. Then, because the quadrilaterals $ABSK$ and $ACTL$ are cyclic, we have $$\widehat{BSK}+\widehat{BAK}=180^{\circ }=\widehat{BSK}+\varphi =\widehat{LAC}+\widehat{LTC}=\widehat{LTC}+\varphi ,$$ so, due to the 90-degree angles formed, we have $\widehat{KSL}=\widehat{KTL}$. Thus, $KLST$ is cyclic. Figure 6: G6 Consider $M$ to be the midpoint of $BC$ and $K^{\prime }$ to be the symmetric point of $K$ with respect to $M$. Then, $BKC{K}^{\prime }$ is a parallelogram, and so $BK\parallel C{K}^{\prime }$. But $BK\parallel CT$, because they are both perpendicular to $AX$. So, $K^{\prime }$ lies on $CT$ and, as $\widehat{KT{K}^{\prime }}=90^{\circ }$ and $M$ is the midpoint of $K{K}^{\prime }$, $MK=MT$. In a similar way, we have that $MS=ML$. Thus, the center of $(KLST)$ is $M$. Consider $D$ to be the foot of altitude from $A$ to $BC$. Then, $D$ belongs in both $(ABKS)$ and $(ACLT)$. So, $$\widehat{ADT}+\widehat{ACT}=180^{\circ }=\widehat{ABS}+\widehat{ADS}=\widehat{ADT}+90^{\circ }-\alpha =\widehat{ADS}+90^{\circ }-\alpha ,$$

and $\overline{AD}$ is the bisector of $\widehat{SDT}$. Because $DM$ is perpendicular to $AD$, $DM$ is the external bisector of this angle, and, as $MS=MT$, it follows that $DMST$ is cyclic. In a similar way, we have that $DMLK$ is also cyclic. So, we have that $ST$, $KL$ and $DM$ are the radical axes of these three circles, $(KLST)$, $(DMST)$, $(DMKL)$. These lines are, therefore, concurrent, and we have proved the desired result. □

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Alternative solution.

We continue after proving that $M$ is the center of $(KLST)$. If $D$ is the foot of perpendicular from $A$ to $BC$, then $ASDKB$ is cyclic, as well as $ATDLC$. The radical axes of those two circles and $(KLST)$ are concurrent, thus $KS$ and $LT$ intersect on point $Q\in AD$. So, if $P$ is the intersection point of $KL$ and $TS$, due to Brokard's theorem, $AQ$ is perpendicular to $MP$. This is, of course, equivalent to proving that $P$ belongs on $BC$. □