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

(Created page with "Let ABC be a triangle with AB < AC < BC. Let the incentre and incircle of triangle ABC be I and ω, respectively. Let X be the point on line BC different from C such that the...")
 
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Let ABC be a triangle with AB < AC < BC. Let the incentre and incircle of triangle
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Let <math>ABC</math> be a triangle with <math>AB < AC < BC</math>. Let the incentre and incircle of triangle
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ABC be I and ω, respectively. Let X be the point on line BC different from C such that the line
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<math>ABC</math> be <math>I</math> and <math>\omega</math>, respectively. Let <math>X</math> be the point on line <math>BC</math> different from <math>C</math> such that the line
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through X parallel to AC is tangent to ω. Similarly, let Y be the point on line BC different from
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through <math>X</math> parallel to <math>AC</math> is tangent to <math>\omega</math>. Similarly, let <math>Y</math> be the point on line <math>BC</math> different from
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B such that the line through Y parallel to AB is tangent to ω. Let AI intersect the circumcircle of
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<math>B</math> such that the line through <math>Y</math> parallel to <math>AB</math> is tangent to <math>\omega</math>. Let <math>AI</math> intersect the circumcircle of
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triangle ABC again at P ̸= A. Let K and L be the midpoints of AC and AB, respectively.
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triangle <math>ABC</math> again at <math>P \neq A</math>. Let <math>K</math> and <math>L</math> be the midpoints of <math>AC</math> and <math>AB</math>, respectively.
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Prove that ∠KIL + ∠Y P X = 180◦
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Prove that <math>\angle KIL + \angle YPX = 180^{\circ}</math>
 
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Revision as of 00:16, 19 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 $\angle KIL + \angle YPX = 180^{\circ}$ .

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