AoPS - Problem

Source: Tuymaada 2001, day 2, problem 3.

Tags: geometry, circumcircle, power of a point, radical axis, geometry proposed

mathmanman

30.04.2007 23:07

$ABCD$ is a convex quadrilateral; half-lines $DA$ and $CB$ meet at point $Q$; half-lines $BA$ and $CD$ meet at point $P$. It is known that $\angle AQB=\angle APD$. The bisector of angle $\angle AQB$ meets the sides $AB$ and $CD$ of the quadrilateral at points $X$ and $Y$, respectively; the bisector of angle $\angle APD$ meets the sides $AD$ and $BC$ at points $Z$ and $T$, respectively. The circumcircles of triangles $ZQT$ and $XPY$ meet at point $K$ inside the quadrilateral. Prove that $K$ lies on the diagonal $AC$. Proposed by S. Berlov

First we can see that in fact $ AC$ is the radical axis of the circles $ PXY$ and $ QZT$, therefore their intersection in the interior of the quadrilateral will trivially lie on $ AC$. Now, let's prove that $ AC$ is their axis. We will prove only that $ A \in d$, where $ d$ is the radical axis, for $ C$ beeing the same thing. But $ A \in d \iff AZ \cdot AQ = AX \cdot AP$, (the power of $ A$ w.r.t to the circles), but we have that $ AD \cdot AQ = AB \cdot AP$, because the quadrilateral $ BPQD$ is cyclic, therefore we only need to prove that : $ \frac {AD}{AZ} = \frac {AB}{AX}\iff \frac {PD}{AP} = \frac {AQ}{QB}$ $ \iff \frac {PD}{AQ} = \frac {AP}{QB}$, which is obvious because the triangles $ \triangle{PAD}$ and $ \triangle{QAB}$ are similar.