Let $[AB]$ be a chord of the circle $\Gamma$ not passing through its center and let $M$ be the midpoint of $[AB].$ Let $C$ be a variable point on $\Gamma$ different from $A$ and $B$ and $P$ be the point of intersection of the tangent lines at $A$ of circumcircle of $CAM$ and at $B$ of circumcircle of $CBM.$ Show that all $CP$ lines pass through a fixed point.
Problem
Source: Turkey JBMO Team Selection Test Problem 3
Tags: geometry, circumcircle, geometry proposed
RSM
29.05.2012 13:12
Note that, $ \angle PAB=\angle ACM $ and $ \angle PBA=\angle BCM $. So $ P $ lies on $ \odot ABC $ and $ CP $ is the A-symmedian of $ ABC $. So if the tangents at $ A,B $ on $ \odot ABC $ meet at $ X $, then $ CP $ always passes thru $ X $. So done.
ArT_Ty
01.05.2013 16:48
It's easy to show that $P$ lie on circumcircle of triangle $ABC$. From $\bigtriangleup AMC \sim %Error. "bigtriangle" is a bad command. PBC$ and $PB \cdot AC+AP \cdot BC =AB \cdot PC =2AM \cdot PC$ We'll get that $\frac{AM}{AC}=\frac{PB}{PB}$ which implies that $AM \cdot PC =PB \cdot AC$ So $\frac{1}{2} AB \cdot PC = BC \cdot AP$ $\therefore$ $APBC$ is a Harmonic Quadrilateral such that $CP$ always pass through point of intersection between tangent line of circle at $A$ and $B$
ThisIsART
10.06.2014 04:34
Why $ \bigtriangleup AMC \sim %Error. "bigtriangle" is a bad command. PBC $????
itslumi
27.11.2019 14:46
CP is A-symmedian.Interesting problem !
hakN
03.05.2021 12:05
Let the tangent to $(CAM)$ at $A$ intersect $\Gamma$ at $P'$. $\angle P'CB = \angle P'AB = \angle ACM \implies \angle MCB = \angle ACP' = \angle DBA$. So $BP'$ is also tangent to $(CBM)$ and so $P' = P$. Let the tangents to $\Gamma$ at $A$ and $B$ intersect at $F$. Because $CAPB$ is a harmonic quadrilateral since $CP$ is a symmedian in $\triangle CAB$, we must have $C, P , F$ are collinear and since $F$ is fixed we are done.$\square$