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Difference between revisions of "2024 IMO Problems/Problem 4"
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==Video Solution==
 
==Video Solution==
 
https://youtu.be/WnZv3fdpFXo
 
https://youtu.be/WnZv3fdpFXo
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 +
==Video Solution==
 +
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>
 +
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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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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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 01:18, 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 $BC$

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


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

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