Show that point lies on bisector

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Show that point lies on bisector

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Source: 63 Polish MO 2012 Finals - Problem 3

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#1 h Apr 23, 2018, 4:48 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Triangle $ABC$ with $AB = AC$ is inscribed in circle $o$ . Circles $o_1$ and $o_2$ are internally tangent to circle $o$ in points $P$ and $Q$ , respectively, and they are tangent to segments $AB$ and $AC$ , respectively, and they are disjoint with the interior of triangle $ABC$ . Let $m$ be a line tangent to circles $o_1$ and $o_2$ , such that points $P$ and $Q$ lie on the opposite side than point $A$ . Line $m$ cuts segments $AB$ and $AC$ in points $K$ and $L$ , respectively. Prove, that intersection point of lines $PK$ and $QL$ lies on bisector of angle $BAC$ .

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#2 h Apr 29, 2018, 11:34 AM • 4 Y

Y by Amir Hossein, ayamgabut, Adventure10, Mango247

mruczek wrote:

Triangle $ABC$ with $AB = AC$ is inscribed in circle $o$ . Circles $o_1$ and $o_2$ are internally tangent to circle $o$ in points $P$ and $Q$ , respectively, and they are tangent to segments $AB$ and $AC$ , respectively, and they are disjoint with the interior of triangle $ABC$ . Let $m$ be a line tangent to circles $o_1$ and $o_2$ , such that points $P$ and $Q$ lie on the opposite side than point $A$ . Line $m$ cuts segments $AB$ and $AC$ in points $K$ and $L$ , respectively. Prove, that intersection point of lines $PK$ and $QL$ lies on bisector of angle $BAC$ .

Let $o_3$ be the incircle of triangle $AKL$ . Note that $K$ is the insimilicenter of circles $o_1$ and $o_3$ (since $K$ is the intersection of their common internal tangents). Note also that $P$ is the exsimilicenter of circles $o_1$ and $(ABC)$ . So, by Monge d'Alembert we get that $PK$ passes through insimilicenter of $o_3$ and $(ABC)$ . Similarly, $QL$ also passes through insimilicenter of $o_3$ and $(ABC)$ . Since $AB=AC$ , it can be seen that this point is located on bisector of angle $BAC$ , so the conclusion follows.

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#3 h Jun 28, 2018, 4:08 AM • 2 Y

Y by Adventure10, Mango247

Another way to do this is to note that $PQ,KL,o_1o_2$ are concurrent by Monge,so by Deasargue's theorem, $X=PK \cap QL ,O=Po_1 \cap Qo_2,Y=o_1K \cap o_2L$ are collinear,but $O,Y$ already lie on the angle bisector of $A$ (they are respectively circumcenter of $(ABC)$ and incenter of $AKL$ ),so $X$ is also on the angle bisector.

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#4 h Mar 7, 2019, 4:50 AM • 1 Y

Y by Adventure10

Angle chasing and Trig Ceva in $\triangle AKL$ also works...

This post has been edited 1 time. Last edited by SinaQane, Mar 7, 2019, 4:51 AM

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#5 h Jul 31, 2023, 3:39 PM

Y by

This has a very nice solution by homothety. Relabel the circles as $\omega_b$ , $\omega_c$ , and $\Omega$ , and consider the incircle $\omega$ of triangle $AKL$ . Note that $\omega$ and $\omega_b$ are homothetic at $K$ , and considering the similar case yields that $O = \overline{B'K} \cap \overline{C'L}$ is the homothety center between $\omega$ and $\Omega$ . Then $O \in \overline{O_\omega O_\Omega}$ , but this is precisely the $A$ -bisector.

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