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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://www.youtube.com/watch?v=QphkkutmY5M&t=1s
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==Video Solution==
 
==Video Solution==
 
https://youtu.be/WnZv3fdpFXo
 
https://youtu.be/WnZv3fdpFXo

Revision as of 22:58, 26 July 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://www.youtube.com/watch?v=QphkkutmY5M&t=1s

Video Solution

https://youtu.be/WnZv3fdpFXo

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.

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)

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