Consider an isosceles triangle $ABC$ with $AB = AC$. Let $\omega(XYZ)$ be the circumcircle of the triangle $XY Z$. The tangents to $\omega(ABC)$ through $B$ and $C$ meet at the point $D$. The point $F$ is marked on the arc $AB$ of $\omega(ABC)$ that does not contain $C$. Let $K$ be the point of intersection of lines $AF$ and $BD$ and $L$ the point of intersection of the lines $AB$ and $CF$. Let $T$ and $S$ be the centers of the circles $\omega(BLC)$ and $\omega(BLK)$, respectively. Suppose that the circles $\omega(BTS)$ and $\omega(CFK)$ are tangent to each other at the point $P$. Prove that $P$ belongs to the line $AB$.
Problem
Source: 2018 Brazil Ibero TST P4
Tags: geometry, isosceles, tangent, Isosceles Triangle, conditional geometry, Circumcenter
aqwxderf
02.07.2023 15:45
I think $L = AB \cap CK$ is the correct intersection.
aqwxderf
02.07.2023 16:13
Let $O $ be the center of $\odot CFK$. $\measuredangle KOC = 2 \cdot \measuredangle AFC = 2 \cdot \measuredangle ABC = \measuredangle KBC \Rightarrow KBOC$ - cyclic. Since $[OK] = [OC] \Rightarrow \overline{BO} $ bisects $ \measuredangle KBC$ externally $\Rightarrow \overline{BO} \perp \overline{BL}$. Check that $BSTO$ is also cyclic. $\odot CFKP$ and $\odot BSTOP$ - tangent $\Rightarrow \overline{OP}$ passes through the center of $\odot BSTOP \Rightarrow OP$ is a diameter in $\odot BSTOP$. $\measuredangle ABO = 90^{\circ} \Rightarrow P \in \overline{AB}$.
AndresSchwepps
02.07.2023 20:14
I've checked that the problem works with the original $L=AB\cap CF$. Let $X$ be the antipode of $A$ in $\omega(ABC)$, $G=ST\cap XF$, let $Q$ be the center of $\omega(CFK)$ and $R$ the center of $\omega(BTS)$. Claim 1: $F\in \omega(BLK)$. Proof:
$\angle KBL=\angle KBA=\angle BCA=\angle ABC=\angle AFC=\angle AFL$.
Claim 2: $K,S,T$ are collinear. Proof:
$K$ lies on the external bisector of $\angle LFB$ and on $\omega(LFB)$, therefore $KB=KL$, hence $K$ belongs to the perpendicular bisector of $BL$, which is the line $ST$.
Claim 3: $G$ belongs to the line $OB$ and to the circles $\omega(BLK)$ and $\omega(BXT)$. Proof:
Since $\angle GFK=90^o$, $G$ lies on $\omega(BLK)$, therefore $\angle KBG=90^o$, and $BG$ passes through $O$. Since triangle $OBX$ is isosceles with $OB=OX$ and $TG||BX$ (both are $\perp BL$), then $BXTG$ is an isosceles trapezoid, and thus is cyclic.
Claim 4: $O\in \omega(BTS)$. Proof:
$\angle TSB=\angle GSB=2\angle GFB=2\angle XFB=\angle XOB$.
Claim 5: $Q$ belongs to the line $OX$ and to the circle $\omega(BTS)$. Proof:
Let $\alpha=\angle CFB$, then $\angle CFK=90^o+\frac{\alpha}{2}$, therefore triangle $CQK$ is isosceles angles $\frac{\alpha}{2}$, and is directly similar to $CXB$. Consequently, $CXQ\sim CBK$ and $\angle CXQ=\angle CBK=180^o-\alpha=\angle CXB$, this proves that $Q$ lies on the line $BX$. The second assertion follows from
$$\angle BQS=\angle BXG=\angle BTG=\angle BTS.$$
Now, if $P'$ is the second intersection of line $AB$ and circle $\omega(BTS)$, since $\angle P'BQ=90^o$ we must have that $P'Q$ is a diameter of this circle, i.e. $R\in QP'$. If $P\neq P'$, we would have $\angle P'PQ=90^o$, so $PP'$ would be tangent to $\omega(CFK)$, and by hypothesis $PP'$ would be tangent to $\omega(BTS)$, which is impossible if $P\neq P'$, hence $P$ coincides with $P'$ and thus it belongs to the line $AB$.
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aqwxderf
02.07.2023 20:52
I was wrong. Pretty strange that the problem works for in the same way for a completely different point.