AoPS - Problem

Source: Peru IMO TST 2017 p9

Tags: geometry, circumcircle, cyclic quadrilateral, equal angles, tangent circles

parmenides51

14.05.2019 13:04

Let $ABCD$ be a cyclie quadrilateral, $\omega$ be it's circumcircle and $M$ be the midpoint of the arc $AB$ of $\omega$ which does not contain the vertices $C$ and $D$. The line that passes through $M$ and the intersection point of segments $AC$ and $BD$, intersects again $\omega$ in $N$. Let $P$ and $Q$ be points in the $CD$ segment such that $\angle AQD = \angle DAP$ and $\angle BPC = \angle CBQ$. Prove that the circumcircle of $NPQ$ and $\omega$ are tangent to each other.

Firstly, note that by Sine Ceva in $\triangle ABN$ with cevians $NM, AC, BD$, we have that $\frac{DN}{NC} = \frac{AD}{BC}.$ As $\triangle ADQ \sim \triangle PDA$ by the condition, we've $DA^2 = DQ \cdot DP.$ Analogously, $CB^2 = CP \cdot CQ.$ Therefore, we know that $$\frac{DQ}{QC} \cdot \frac{DP}{PC} = \frac{AD^2}{BC^2} = \frac{ND^2}{NC^2}.$$This is known to imply that $\angle CNP = \angle QND.$ Therefore, if we let $P', Q'$ be the second intersections of $NP, NQ$ with $\omega$, we have that $P'Q' || PQ$. Hence, as $\triangle NPQ$ is homothetic to $\triangle NP'Q'$, their circumcircles are tangent to each other.