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

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Problem 4. Let ABCDE be a convex pentagon such that BC = DE. Assume that there is a
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==Problem==
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point T inside ABCDE with TB = TD, TC = TE and ∠ABT = ∠TEA. Let line AB intersect
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Let <math>ABCDE</math> be a convex pentagon such that <math>BC = DE</math>. Assume that there is a
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lines CD and CT at points P and Q, respectively. Assume that the points P, B, A, Q occur on their
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point <math>T</math> inside <math>ABCDE</math> with <math>TB = TD</math>, <math>TC = TE</math> and <math>\angle ABT = \angle TEA</math>. Let line <math>AB</math> intersect
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line in that order. Let line AE intersect lines CD and DT at points R and S, respectively. Assume
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lines <math>CD</math> and <math>CT</math> at points <math>P</math> and <math>Q</math>, respectively. Assume that the points <math>P, B, A, Q</math> occur on their
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that the points R, E, A, S occur on their line in that order. Prove that the points P, S, Q, R lie on
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line in that order. Let line <math>AE</math> intersect lines <math>CD</math> and <math>DT</math> at points <math>R</math> and <math>S</math>, respectively. Assume
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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
 
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]
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https://youtu.be/WpM0mLyPyLg?si=yi9AZPVdYSPMCcHa
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[Video Solution by little fermat]
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==Solution==
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[[File:2022 IMO 4.png|400px|right]]
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<cmath>TB = TD, TC = TE, BC = DE \implies</cmath>
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<cmath>\triangle TBC = \triangle TDE \implies \angle BTC = \angle DTE.</cmath>
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<cmath>\angle BTQ = 180^\circ - \angle BTC = 180^\circ - \angle DTE = \angle STE</cmath>
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<cmath>\angle ABT = \angle AET \implies  \triangle TQB \sim \triangle TSE \implies</cmath>
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<cmath>\angle PQC = \angle EST, \hspace{18mm}\frac {QT}{ST}= \frac {TB}{TE} \implies</cmath>
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<cmath>QT \cdot TE =QT \cdot TC = ST \cdot TB= ST \cdot TD \implies</cmath>
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<math>\hspace{28mm}CDQS</math> is cyclic <math>\implies \angle QCD = \angle QSD.</math>
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<cmath>\angle QPR =\angle QPC = \angle QCD - \angle PQC =</cmath>
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<cmath>\angle QSD - \angle EST =  \angle QSR \implies</cmath>
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<math>\hspace{43mm}PRQS</math> is cyclic.
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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==See Also==
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{{IMO box|year=2022|num-b=3|num-a=5}}

Latest revision as of 00:55, 19 November 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

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

See Also

2022 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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