Let the circles $\omega$ and $\omega^ \prime$ intersect in $A$ and $B$. Tangent to circle$\omega$ at $A$ intersects$\omega^ \prime$ in $C$ and tangent to circle $\omega^ \prime$ at $A$ intersects $\omega$ in $D$. Suppose that $CD$ intersects$\omega$ and $\omega^ \prime$ in $E$ and $F$, respectively (assume that $E$ is between $F$ and $C$). The perpendicular to $AC$ from $E$ intersects $\omega^ \prime$ in point $P$ and perpendicular to $AD$ from $F$ intersects$\omega$ in point $Q$ (The points $A, P$ and $Q$ lie on the same side of the line $CD$). Prove that the points $A, P$ and $Q$ are collinear. Proposed by Mahdi Etesami Fard
Problem
Source: IGO 2016,Advanced level,P1
Tags: geometry, geometry proposed
Math_CYCR
13.09.2016 07:39
Let $X= EP \cap (AEC)$ and let $Y= FQ \cap (AFD)$. Easily we get $\triangle YDA \sim \triangle XAC$ and $\angle DQA= \angle CPA$ Consider an homothety that sends $\triangle XAC$ to $\triangle X'A'C'$ such that $\triangle X'A'C' \cong \triangle YDA$. Since $P'$ and $Q$ are points on the heights from $X'$ and $Y$ respectively, and $\angle DQA= \angle A'P'C'$ we get $P' =Q$. Thus $\triangle DQA \cong \triangle A'P'C'$. Furthermore $\triangle DQA \sim \triangle APC$. $\Longrightarrow \angle QAD + \angle PAC= \angle QAD+ \angle QDA= \angle ADC+ \angle ACD= 180 - \angle DAC$
Afo
11.12.2020 14:36
Solution. Claim 1. $AF=AE$. Proof. $\angle AFE=\angle FAD+\angle FDA = \angle CDA +\angle DCA$. It follows that by symmetry that $\angle AFE =\angle AEF$. Claim 2. $PQFE$ is a cyclic quadrilateral with center $A$. Proof. Since $AF=AE$, by symmetry it's enough to show that $AE=AP$. This is true since $\angle APC = \pi - \angle AFE = \pi-\angle AEF=\angle AEC$. Now $\angle QAP= \angle QAF + \angle FAE+\angle EAP = 2\angle ACD + (\pi - 2\angle ADC-2\angle ACD) + 2\angle ADC = \pi$.
FarrukhKhayitboyev
26.08.2024 20:03
When points $E$ and $F$ are below the point $B$, $A,P,Q$ will be collinear only if $F$ is between $E$ and $C$. I wondered why it was impossible, looking at the diagram of Afo, I understood. Also, here is the sketch where $E$ and $F$ are below the point B.
Attachments:
