Let $ABC$ be a scalene triangle and $l_0$ be a line that is tangent to the circumcircle of $ABC$ at point $A$. Let $l$ be a variable line which is parallel to line $l_0$. Let $l$ intersect segment $AB$ and $AC$ at the point $X$, $Y$ respectively. $BY$ and $CX$ intersects at point $T$ and the line $AT$ intersects the circumcirle of $ABC$ at $Z$. Prove that as $l$ varies, circumcircle of $XYZ$ passes through a fixed point.
Problem
Source: 2023 Turkey Egmo Tst P6
Tags: geometry, circumcircle
qscweadzx
23.03.2023 20:12
Let $XY$, $AZ$ intersects $BC$ at $S$, $D$, respectively. Let line $SZ$ intersects the circumcircle of $ABC$ at K. We claim that $K$ is the fix point. First, since $SK\cdot SZ=SB\cdot SC=SX \cdot SY$, we have $K$ lies on circle $(XYZ)$. Second, since $(B,C;D,S)=-1$, we know that $MD\cdot SD=BD\cdot CD=AD\cdot DZ$ , hence $AMSZ$ are on a circle and $\angle MAD=\angle MSZ$. On the other hand, $SK\cdot SZ=SB\cdot SC=SM\cdot SD$ shows that $MDKZ$ are on a circle and $\angle MDA =\angle MKS$. Therefore, $\angle AMD=\angle DMK$, $AM$ and $MK$ are symmetric about line $BC$, $K$ is a fix point.
giannis2006
01.07.2024 17:31
Firstly, $l \parallel l_0 => BXYC$ is cyclic. Let $E = AT \cap BC$, $K = XY \cap BC$ and $P = KZ \cap (ABC)$. By power of point we get that $ KX \times KY = KB \times KC = KZ \times KP$ and hence $XYPZ$ is cyclic. It is well known that $(B,C;K,E)=-1$ and hence $B=ZB \cap (ABC), C = ZC \cap (ABC), P = ZK \cap (ABC)$ and $A = ZE \cap (ABC)$ form an harmonic quadrilateral, from where we conclude that $AP$ is symmedian in triangle $ABC$ and hence point $P$ is fixed, which completes the proof.
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