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

Revision as of 00:34, 19 November 2023

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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

@@ Line 2: / Line 2: @@
 Prove that the circumcircles of triangles <math>KQH</math> and <math>FKM</math> are tangent to each other.
+==Solution==
+{{solution}}
 ==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

Page

Toolbox

Search

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

Revision as of 00:34, 19 November 2023

Solution

See Also