Two circles $\omega_1$ and $\omega_2$ meeting at point $A$ and a line $a$ are given. Let $BC$ be an arbitrary chord of $\omega_2$ parallel to $a$, and $E$, $F$ be the second common points of $AB$ and $AC$ respectively with $\omega_1$. Find the locus of common points of lines $BC$ and $EF$.
Problem
Source: Sharygin Finals 2023 8.8
Tags: geometry, Sharygin Geometry Olympiad, Sharygin 2023
Rg230403
02.08.2023 12:16
Let $D$ be the other intersection of the two circles. Now, $D$ is clearly the miquel point of $BCFE$. Thus, if $X=BC\cap EF$ then $X\in (DBE)$. Thus, $\angle (DX, a)=\angle DXB=\angle DEB=\angle DEA$. Thus, $X$ lies on the line $\ell$ through $D$ such that $\angle (\ell, a)$ is the same as arc $DA$ subtends in $\omega_1$. Clearly, we can also do it in reverse from any point on $\ell$.
InterLoop
20.07.2024 08:18
Note by spiral similarity that if $M = \omega_1 \cap \omega_2$, then $MCGF$ and $MBEG$ are concyclic. $G$ always lies on $BC$, $\angle MGC = \angle MFA$, which implies the locus is a line through $M$.
Mahdi_Mashayekhi
16.08.2024 11:04
Let $K$ be the second intersection of $\omega_1$ and $\omega_2$. Let $KC$ meet $\omega_2$ at $T$. Let the line parallel to $BC$ through $A$ meet $\omega_2$ at $L$. Let $KL$ meet $EF$ at $D$. Applying pascal on $EFALKT$ implies $D$ and $C$ and intersection of $AL,TE$ are collinear. Note that $\angle TEA = \angle TKA = \angle CKA = \angle CBA$ so $TE \parallel BC \parallel AL$ so $DC \parallel BC$ which implies $D$ lies on $BC$. so the locus of common points of $BC$ and $EF$ is $KL$.