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Difference between revisions of "2024 IMO Problems/Problem 4"
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Prove that <math>\angle KIL + \angle YPX = 180^{\circ}</math>
 
Prove that <math>\angle KIL + \angle YPX = 180^{\circ}</math>
 
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==Video Solution==
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https://youtu.be/WnZv3fdpFXo
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==Video Solution (AI solution)==
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https://youtu.be/cjnJ6EXKWW4
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==Video Solution(In Chinese)==
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https://youtu.be/QphkkutmY5M
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==Video Solution==
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Part 1: Derive tangent values <math>\angle AIL</math> and <math>\angle AIK</math> with trig values of angles <math>\frac{A}{2}</math>, <math>\frac{B}{2}</math>, <math>\frac{C}{2}</math>
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https://youtu.be/p_AmooMMln4
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Part 2: Derive tangent values <math>\angle XPM</math> and <math>\angle YPM</math> with side lengths <math>AB</math>, <math>BC</math>, <math>CA</math>, where <math>M</math> is the midpoint of <math>BC</math>
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https://youtu.be/MgrghZ2ESAg
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Part 3: Prove that <math>\angle AIL + \angle XPM = 90^\circ</math> and <math>\angle AIK + \angle YPM = 90^\circ</math>.
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https://youtu.be/iOp9mnmZyzU
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Comments: Although this is an IMO problem, the skills needed to solve this problem have all previously tested in AMC and its system math contests, such as HMMT.~ also proved by Kislay Kai
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Evidence 1: 2020 Spring HMMT Geometry Round Problem 8
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I used the property that because point <math>P</math> is on the angle bisector <math>AI</math>, <math>\triangle BPC</math> is isosceles. This is a crucial step to analyze <math>\angle XPY</math>. This technique was previously tested in this HMMT problem.
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Evidence 2: 2022 AMC 12A Problem 25
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The technique in this AMC problem can be easily and directly applied to this IMO problem to quickly determine the locations of points <math>X</math> and <math>Y</math>. If you read my solutions to both this AMC problem and this IMO problem, you will find that I simply took exactly the same approach to solve both.
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Video Solution with discussion of a generalized case==
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https://youtu.be/NJc79Ccg82E?si=J0YdHAz-46miJIO2
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==Solution simple==
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[[File:2024 AIME II 12 d.png|330px|right]]
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Let point <math>Z</math> be <math>YZ||AB, XZ||AC, G \in AC, IG||BC, G' \in AB, IG' || BC.</math>
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<math>F= AC \cap YZ, F' = AB \cap XZ.</math>
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<math>AFZF'</math> is the parallelogram with equal heights, so <math>AFZF'</math> is rhomb <math>\implies</math>
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<cmath>\angle CAZ = \angle AZY = \angle BCP, AI = IZ.</cmath>
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<cmath>AL = LC, AI = IZ \implies IL || ZC \implies \angle LIG = \angle BCZ.</cmath>
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<math>\angle AZY = \angle BCP = \angle YCP \implies</math> points <math>C,P, Y,</math> and <math>Z</math> are concyclic.
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Therefore <math>\angle ZPY = \angle LIG.</math>
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Similarly, <math>\angle ZPX = \angle KIG'. \blacksquare</math>
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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==Video Solution==
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https://youtu.be/6J7KH778ptg
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==See Also==
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{{IMO box|year=2024|num-b=3|num-a=5}}

Latest revision as of 20:48, 30 September 2024

Let $ABC$ be a triangle with $AB < AC < BC$. Let the incentre and incircle of triangle $ABC$ be $I$ and $\omega$, respectively. Let $X$ be the point on line $BC$ different from $C$ such that the line through $X$ parallel to $AC$ is tangent to $\omega$. Similarly, let $Y$ be the point on line $BC$ different from $B$ such that the line through $Y$ parallel to $AB$ is tangent to $\omega$. Let $AI$ intersect the circumcircle of triangle $ABC$ again at $P \neq A$. Let $K$ and $L$ be the midpoints of $AC$ and $AB$, respectively. Prove that $\angle KIL + \angle YPX = 180^{\circ}$ .

Video Solution

https://youtu.be/WnZv3fdpFXo

Video Solution (AI solution)

https://youtu.be/cjnJ6EXKWW4


Video Solution(In Chinese)

https://youtu.be/QphkkutmY5M

Video Solution

Part 1: Derive tangent values $\angle AIL$ and $\angle AIK$ with trig values of angles $\frac{A}{2}$, $\frac{B}{2}$, $\frac{C}{2}$

https://youtu.be/p_AmooMMln4

Part 2: Derive tangent values $\angle XPM$ and $\angle YPM$ with side lengths $AB$, $BC$, $CA$, where $M$ is the midpoint of $BC$

https://youtu.be/MgrghZ2ESAg

Part 3: Prove that $\angle AIL + \angle XPM = 90^\circ$ and $\angle AIK + \angle YPM = 90^\circ$.

https://youtu.be/iOp9mnmZyzU

Comments: Although this is an IMO problem, the skills needed to solve this problem have all previously tested in AMC and its system math contests, such as HMMT.~ also proved by Kislay Kai

Evidence 1: 2020 Spring HMMT Geometry Round Problem 8

I used the property that because point $P$ is on the angle bisector $AI$, $\triangle BPC$ is isosceles. This is a crucial step to analyze $\angle XPY$. This technique was previously tested in this HMMT problem.

Evidence 2: 2022 AMC 12A Problem 25

The technique in this AMC problem can be easily and directly applied to this IMO problem to quickly determine the locations of points $X$ and $Y$. If you read my solutions to both this AMC problem and this IMO problem, you will find that I simply took exactly the same approach to solve both.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution with discussion of a generalized case

https://youtu.be/NJc79Ccg82E?si=J0YdHAz-46miJIO2

Solution simple

2024 AIME II 12 d.png

Let point $Z$ be $YZ||AB, XZ||AC, G \in AC, IG||BC, G' \in AB, IG' || BC.$ $F= AC \cap YZ, F' = AB \cap XZ.$

$AFZF'$ is the parallelogram with equal heights, so $AFZF'$ is rhomb $\implies$ \[\angle CAZ = \angle AZY = \angle BCP, AI = IZ.\] \[AL = LC, AI = IZ \implies IL || ZC \implies \angle LIG = \angle BCZ.\] $\angle AZY = \angle BCP = \angle YCP \implies$ points $C,P, Y,$ and $Z$ are concyclic.

Therefore $\angle ZPY = \angle LIG.$

Similarly, $\angle ZPX = \angle KIG'. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Video Solution

https://youtu.be/6J7KH778ptg

See Also

2024 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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