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

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==Solution==
 
==Solution==
 
[[File:2022 IMO 4.png|400px|right]]
 
[[File:2022 IMO 4.png|400px|right]]
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<cmath>TB = TD, TC = TE, BC = DE \implies \triangle TBC = \triangle TDE</cmath>
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<cmath>TB = TD, TC = TE, BC = DE \implies</cmath>
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<cmath> \implies \angle BTC = \angle DTE.</cmath>  
+
<cmath>\triangle TBC = \triangle TDE \implies \angle BTC = \angle DTE.</cmath>  
−
<cmath>\angle ABT = \angle TEA \implies  \triangle TQB \sim \triangle TSE \implies</cmath>  
+
<cmath>\angle BTQ = 180^\circ - \angle BTC = 180^\circ - \angle DTE = \angle STE</cmath>
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<cmath>\frac {QT}{ST}= \frac {TB}{TE} \implies QT \cdot TC = ST \cdot TD \implies</cmath>
+
<cmath>\angle ABT = \angle AET \implies  \triangle TQB \sim \triangle TSE \implies</cmath>
−
<math>CDQS</math> is cyclic <math>\implies \angle QCD = \angle QSD.</math>  
+
<cmath>\frac {QT}{ST}= \frac {TB}{TE} \implies QT \cdot TE =QT \cdot TC = ST \cdot TB= ST \cdot TD \implies</cmath>
 +
<math>\hspace{28mm}CDQS</math> is cyclic <math>\implies \angle QCD = \angle QSD.</math>  
 
<cmath>\angle QPR =\angle QPC = \angle QCD - \angle PQC =</cmath>
 
<cmath>\angle QPR =\angle QPC = \angle QCD - \angle PQC =</cmath>
 
<math>\angle QSD - \angle EST =  \angle QSR \implies</math>
 
<math>\angle QSD - \angle EST =  \angle QSR \implies</math>

Revision as of 17:46, 23 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.

Solution

2022 IMO 4.png

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

vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru

Solution

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

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