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Difference between revisions of "2024 IMO Problems/Problem 4"

(→‎Video Solution)
(→‎Video Solution)
 
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https://youtu.be/iOp9mnmZyzU
 
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.
 +
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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.
 +
 +
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.
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Latest revision as of 01:45, 24 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://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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