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

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that the points <math>R, E, A, S</math> occur on their line in that order. Prove that the points <math>P, S, Q, R</math> lie on
 
that the points <math>R, E, A, S</math> occur on their line in that order. Prove that the points <math>P, S, Q, R</math> lie on
 
a circle.
 
a circle.
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==Video Solution==
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https://www.youtube.com/watch?v=-AII0ldyDww [Video contains solutions to all day 2 problems]
  
 
==Solution==
 
==Solution==
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'''vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru'''
 
'''vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru'''
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==Solution==
 
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https://www.youtube.com/watch?v=-AII0ldyDww [Video contains solutions to all day 2 problems]
 

Revision as of 08:36, 25 July 2022

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]

Solution

2022 IMO 4.png

\[TB = TD, TC = TE, BC = DE \implies\] \[\triangle TBC = \triangle TDE \implies \angle BTC = \angle DTE.\] \[\angle BTQ = 180^\circ - \angle BTC = 180^\circ - \angle DTE = \angle STE\] \[\angle ABT = \angle AET \implies \triangle TQB \sim \triangle TSE \implies\] \[\angle PQC = \angle EST, \hspace{18mm}\frac {QT}{ST}= \frac {TB}{TE} \implies\] \[QT \cdot TE =QT \cdot TC = ST \cdot TB= ST \cdot TD \implies\] $\hspace{28mm}CDQS$ 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, www.deoma–cmd.ru

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