Difference between revisions of "2022 IMO Problems/Problem 4"

Revision as of 23:17, 3 September 2023

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.

vladimir.shelomovskii@gmail.com, vvsss

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Difference between revisions of "2022 IMO Problems/Problem 4"

Revision as of 23:17, 3 September 2023

Problem

Video Solution

Solution

@@ Line 9: / Line 9: @@
 ==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==