# Difference between revisions of "2015 IMO Problems/Problem 4"

## Problem

Triangle $ABC$ has circumcircle $\Omega$ and circumcenter $O$. A circle $\Gamma$ with center $A$ intersects the segment $BC$ at points $D$ and $E$, such that $B$, $D$, $E$, and $C$ are all different and lie on line $BC$ in this order. Let $F$ and $G$ be the points of intersection of $\Gamma$ and $\Omega$, such that $A$, $F$, $B$, $C$, and $G$ lie on $\Omega$ in this order. Let $K$ be the second point of intersection of the circumcircle of triangle $BDF$ and the segment $AB$. Let $L$ be the second point of intersection of the circumcircle of triangle $CGE$ and the segment $CA$.

Suppose that the lines $FK$ and $GL$ are different and intersect at the point $X$. Prove that $X$ lies on the line $AO$.

Proposed by Silouanos Brazitikos and Evangelos Psychas, Greece

## Solution

Lemma (On three chords). If two lines pass through different endpoints of two circles' common chord, then the other two chords cut by these lines on the circles are parallel.
Proof The second and the third chords are anti-parallel to the first (common) chord with respect to the given lines, so they are parallel to each other. $\Box$
To solve this problem, it is sufficient to apply the lemma 5 times. Indeed, let the lines $FD, GE, FK, GL$ meet $\Omega$ second time at $H, I, M, N$ respectively. One of the circles that figure in lemma is always $\Omega$, while the other is one of three other circles from the problem statement. Applying the lemma to the lines $FDH$ and $GEI$, $FKM$ and $BDC$, $FDH$ and $BKA$, $GLN$ and $CEB$, $GEI$ and $CLA$, we get $DE \parallel IH$, $KD \parallel MC$, $KD \parallel AH$, $LE \parallel NB$, $LE \parallel AI$, respectively. From this, $BC \parallel IH$, $MC \parallel AH$, $NB \parallel AI$. Therefore, $AN=IB=HC=AM$. This means that $N$ and $M$ are symmetric wrt $AO$, a diameter of $\Omega$ through $A$. So are $F$ and $G$, as $AF=AG$. Therefore, the lines $FM$ and $GN$ are symmetric wrt $AO$ and meet on it.