Difference between revisions of "2015 IMO Problems/Problem 3"

Revision as of 12:53, 1 June 2024

Let $ABC$ be an acute triangle with $AB>AC$ . Let $\Gamma$ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$ . Let $M$ be the midpoint of $BC$ . Let $Q$ be the point on $\Gamma$ such that $\angle HKQ=90^\circ$ . Assume that the points $A$ , $B$ , $C$ , $K$ , and $Q$ are all different, and lie on $\Gamma$ in this order.

Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

Solution

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 ==Solution==
 {{solution}}
-== The Actual Problem ==
-Let <math>ABC</math> be an acute triangle with <math>AB > AC</math>. Let <math>Γ</math> be its circumcircle, <math>H</math> its orthocenter, and <math>F</math> the foot of the altitude from <math>A</math>. Let <math>M</math> be the midpoint of <math>BC</math>. Let <math>Q</math> be the point on <math>Γ</math> such that <math>\angle HQA = 90◦</math> and let <math>K</math> be the point on <math>Γ</math> such that <math>\angle HKQ = 90◦</math> . Assume that the points <math>A</math>, <math>B</math>, <math>C</math>, <math>K</math> and <math>Q</math> are all different and lie on <math>Γ</math> in this order. Prove that the circumcircles of triangles <math>KQH</math> and <math>FKM</math> are tangent to each other.
 ==See Also==

2015 IMO (Problems) • Resources
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
All IMO Problems and Solutions

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Difference between revisions of "2015 IMO Problems/Problem 3"

Revision as of 12:53, 1 June 2024

Solution

See Also