Tangent circles

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Source: 2017 Belarus Team Selection Test 6.3

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#1 h Mar 31, 2019, 6:20 PM • 1 Y

Y by Adventure10

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.)

This post has been edited 2 times. Last edited by Vlados021, Mar 31, 2019, 6:21 PM

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#2 h Apr 1, 2019, 7:46 AM • 3 Y

Y by Smkh, Adventure10, Mango247

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 $AB$ .

This post has been edited 2 times. Last edited by Sugiyem, Apr 1, 2019, 7:48 AM

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#3 h May 29, 2019, 1:14 PM • 2 Y

Y by Adventure10, Mango247

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?

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