Line passes through the fix point

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Line passes through the fix point

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moony_

#1 h Apr 11, 2025, 1:44 PM

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Let $ABC$ be a triangle. $P$ and $Q$ are points, such that $PA = PB$ , $QA$ = $QC$ and $\angle{PBC} =\angle{QCB}$ ( $P$ - inside $\triangle{ABC}$ and $Q$ - oitside). Proove that line $PQ$ passes through the fix point.

This post has been edited 5 times. Last edited by moony_, Apr 11, 2025, 1:46 PM

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#2 h Apr 11, 2025, 4:51 PM

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JBMO shortlist 2022 G6 In this problem triangle was given to be right but it is not a necessary condition.

GioOrnikapa wrote:

Let $\omega_1 = (D, DB), \omega_2 = (E, EC).$ $B \neq F = \omega_1 \cap \overline{BC}, G \neq C = \omega_2 \cap \overline{BC}$ . $\theta := \angle{DBC} = \angle{BCE}$ .
Claim 1. $TA$ is tangent to $(AFG)$ .
Proof. $\angle{AFG} = \angle{AFB} = \frac{\angle{ADB}}{2} = 90^{\circ} - \angle{ABD} = 90^{\circ} - \angle{B} + \theta$ and $\angle{TAG} = \angle{TAC} - \angle{GAC} = \angle{C} + \angle{A} - (90^{\circ} - \angle{ECG}) = 180^{\circ} - \angle{B} - 90^{\circ} + \theta = 90^{\circ} - \angle{B} + \theta.$ Thus, $\angle{AFG} = \angle{TAG}$ and we are done with this claim. $\ \blacksquare$
Claim 2. $T$ is similitude center of $\omega_1$ and $\omega_2$ .
Proof. Consider the inversion from $T$ with radius $TA$ . Due to the first claim, we have $TA^2 = TG \cdot TF$ . Therefore, $G$ and $F$ will swap the places. Clearly, $B$ and $C$ will also swap the places and $A$ will stay the same. Thus, $\omega_1 = (ABF)$ and $\omega_2 = (ACG)$ will swap the places. It is well-known that if there exists an inversion that swaps two circles, then center of the inversion is also one of the centers of similitude of those two circles and thus we are done. $\ \blacksquare$
Claim 2 immediately implies that $T, D, E$ are collinear. $\square$

Remark. Triangle being right was not necessary.

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