# 2022 IMO Problems/Problem 4

## Problem

Let $ABCDE$ be a convex pentagon such that $BC = DE$. Assume that there is a point $T$ inside $ABCDE$ with $TB = TD$, $TC = TE$ and $\angle ABT = \angle TEA$. Let line $AB$ intersect lines $CD$ and $CT$ at points $P$ and $Q$, respectively. Assume that the points $P, B, A, Q$ occur on their line in that order. Let line $AE$ intersect lines $CD$ and $DT$ at points $R$ and $S$, respectively. Assume that the points $R, E, A, S$ occur on their line in that order. Prove that the points $P, S, Q, R$ lie on a circle.

## Video Solution

https://www.youtube.com/watch?v=-AII0ldyDww [Video contains solutions to all day 2 problems]

https://youtu.be/WpM0mLyPyLg?si=yi9AZPVdYSPMCcHa [Video Solution by little fermat]

## Solution

$$TB = TD, TC = TE, BC = DE \implies$$ $$\triangle TBC = \triangle TDE \implies \angle BTC = \angle DTE.$$ $$\angle BTQ = 180^\circ - \angle BTC = 180^\circ - \angle DTE = \angle STE$$ $$\angle ABT = \angle AET \implies \triangle TQB \sim \triangle TSE \implies$$ $$\angle PQC = \angle EST, \hspace{18mm}\frac {QT}{ST}= \frac {TB}{TE} \implies$$ $$QT \cdot TE =QT \cdot TC = ST \cdot TB= ST \cdot TD \implies$$ $\hspace{28mm}CDQS$ is cyclic $\implies \angle QCD = \angle QSD.$ $$\angle QPR =\angle QPC = \angle QCD - \angle PQC =$$ $$\angle QSD - \angle EST = \angle QSR \implies$$ $\hspace{43mm}PRQS$ is cyclic.