AoPS - Problem

Source: 2020 Balkan MO shortlist G5

Tags: geometry, circumcircle

parmenides51

14.09.2021 20:04

Let $ABC$ be an isosceles triangle with $AB = AC$ and $\angle A = 45^o$. Its circumcircle $(c)$ has center $O, M$ is the midpoint of $BC$ and $D$ is the foot of the perpendicular from $C$ to $AB$. With center $C$ and radius $CD$ we draw a circle which internally intersects $AC$ at the point $F$ and the circle $(c)$ at the points $Z$ and $E$, such that $Z$ lies on the small arc $BC$ and $E$ on the small arc $AC$. Prove that the lines $ZE$, $CO$, $FM$ are concurrent. Brazitikos Silouanos, Greece

I give a sketch of the official soln. Let $OC$ meet $ZE$ at $X$. We want $X$ to lie on $FM$. Obviously $X$ is midpoint of $ZE$ and $ZDFE$ is isosceles trapezoid. It is sufficient to prove that $ZMEF$ is parallelogram. Let $ZE$ and $AC$ meet at $K$. Note that $ADMC$ is cyclic, and by power bash of $K$ wrt $(ABC)$ and $(C)$ + Pythagorean theorem for $ADC$, $K$ is center of $(ADMC)$, so $ZK$ is perpendicular bisector of $DM$, done (use $MZ=ZD=FE$ and that $DM$ and $OC$ are parallel because $<BMD=<BCO=45$; $ZM$ and $FE$ are parallel because $DZM$ is isosceles) Comment: The tricky thing was proving the center of $(ADMC)$ is on $ZE$ and noting the parallelogram.