Two circles $\omega_1,\omega_2$ intersect each other at points $A,B$. Let $PQ$ be a common tangent line of these two circles with $P \in \omega_1$ and $Q \in \omega_2$. An arbitrary point $X$ lies on $\omega_1$. Line $AX$ intersects $ \omega_2$ for the second time at $Y$ . Point $Y'\ne Y$ lies on $\omega_2$ such that $QY = QY'$. Line $Y'B$ intersects $ \omega_1$ for the second time at $X'$. Prove that $PX = PX'$. Proposed by Morteza Saghafian
Problem
Source: Iranian Geometry Olympiad 2018 IGO Advanced p1
Tags: geometry, circles, equal segments
Anaskudsi
19.09.2018 23:18
I solved it in the test with only one minute. This is my solution : We have :$\angle{QY'Y}=\angle{QYY'}=\angle{Y'QP}$ $\Longrightarrow YY' \mid\mid PQ$. We have: $\angle{Y'YX} =\angle{Y'BA} =\angle{X'XY}$ $\Longrightarrow YY' \mid\mid XX'$ $\Longrightarrow XX' \mid\mid PQ \Longrightarrow \angle{XX'P} =\angle{X'PQ} =\angle{X'XP} \Longrightarrow PX=PX'$ So we are done. $\blacksquare$
MarkBcc168
20.09.2018 12:41
By Reim's theorem, $XX'\parallel YY'$. But $QY=QY'\implies YY'\parallel PQ$ hence $XX'\parallel PQ\implies PX=PX'$.
Vrangr
20.09.2018 20:49
$XX' \parallel YY'$ and $YY' \parallel PQ$, so, $XX' \parallel PQ$. Therefore, $PX = PX'$.
Eka01
16.01.2025 12:23
It is easy to see that the tangent to the midpoint of an arc is parallel to the chord making that arc, so $YY' || PQ$ and it needs to be shown that $XX' || PQ || YY'$ but $XX' || YY'$ is a simple consequence of reim's theorem.