Given an isosceles triangle $ABC$ with $AB=AC$. let $\omega(XYZ)$ be the circumcircle of a triangle $XYZ$. Tangents to $\omega(ABC)$ at $B$ and $C$ meet at $D$. Point $F$ is marked on the arc $AB$ (opposite to $C$). Let $K$, $L$ be the intersection points of $AF$ and $BD$, $AB$ and $CF$, respectively. Prove that if circles $\omega(BTS)$ and $\omega(CFK)$ are tangent to each other, the their tangency point belongs to $AB$. (Here $T$ and $S$ are the centers of the circles $\omega(BLC)$ and $\omega(BLK)$, respectively.)
Problem
Source: 2017 Belarus Team Selection Test 6.3
Tags: geometry, circumcircle
Sugiyem
01.04.2019 10:46
Here is my sketch of solution: (1) First of all, define $O$ to be the circumcenter of $ABC$, $N$ be circumcenter of $BTS$, $X$ be a point other than $B$ on circumcircle of $BTS$ such that $BX$ is perpendicular to $TS$ and $M$ be a point other than $B$ on circumcircle of $BTS$ such that $BM$ is parallel to $TS$. (2) Prove that $BLFK$ is cyclic, this is just angle chasing (3) Prove that $BSOT$ is cyclic, we can solve it by length bashing. We only need to prove that $BS\sin{\angle OBT}$+$BT\sin{\angle OBS}$= $BO\sin{\angle SBT}$ (4) Prove that $MS$ is perpendicular to $KF$, again this is just angle chasing (5) Prove that $MO$ is perpendicular to $CF$, by angle chasing (6) By (4) and (5) we can get that $M$ is the circumcenter of triangle $CFK$, therefore $\angle MBX=\angle (BX,TS)=90^{\circ}$. So, $M,N,X$ are collinear. This gives us that if circumcircle of $BTS$ and $CFK$ are tangent, then the tangency point lies on $AB$.
cjfev4
29.05.2019 16:14
Sugiyem wrote: Here is my sketch of solution: (1) First of all, define $O$ to be the circumcenter of $ABC$, $N$ be circumcenter of $BTS$, $X$ be a point other than $B$ on circumcircle of $BTS$ such that $BX$ is perpendicular to $TS$ and $M$ be a point other than $B$ on circumcircle of $BTS$ such that $BM$ is parallel to $TS$. (2) Prove that $BLFK$ is cyclic, this is just angle chasing (3) Prove that $BSOT$ is cyclic, we can solve it by length bashing. We only need to prove that $BS\sin{\angle OBT}$+$BT\sin{\angle OBS}$= $BO\sin{\angle SBT}$ (4) Prove that $MS$ is perpendicular to $KF$, again this is just angle chasing (5) Prove that $MO$ is perpendicular to $CF$, by angle chasing (6) By (4) and (5) we can get that $M$ is the circumcenter of triangle $CFK$, therefore $\angle MBX=\angle (BX,TS)=90^{\circ}$. So, $M,N,X$ are collinear. This gives us that if circumcircle of $BTS$ and $CFK$ are tangent, then the tangency point lies on $AB$. How to showing the proof of length bashing (number 3) on this problem?