# Difference between revisions of "2009 IMO Problems/Problem 2"

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''Author: Sergei Berlov, Russia'' | ''Author: Sergei Berlov, Russia'' | ||

+ | |||

+ | == Solution == | ||

+ | ===Diagram=== | ||

+ | <asy> | ||

+ | dot("O", (50, 38), NW); | ||

+ | dot("A", (40, 100), N); | ||

+ | dot("B", (0, 0), S); | ||

+ | dot("C", (100, 0), S); | ||

+ | dot("Q", (24, 60), W); | ||

+ | dot("P", (52, 80), E); | ||

+ | dot("L", (62, 30), SE); | ||

+ | dot("M", (38, 70), N); | ||

+ | dot("K", (27, 42), W); | ||

+ | draw((100, 0)--(24, 60), dotted); | ||

+ | draw((0, 0)--(52, 80), dashed); | ||

+ | draw((0, 0)--(100, 0)--(40, 100)--cycle); | ||

+ | draw((24, 60)--(52, 80)); | ||

+ | draw((27, 42)--(38, 70)--(62, 30)--cycle); | ||

+ | draw(circle((49, 49), 23)); | ||

+ | label("$\Gamma$", (72, 49), E); | ||

+ | draw(circle((50, 38), 63)); | ||

+ | label("$\omega$", (-13, 38), NW); | ||

+ | </asy> | ||

+ | Diagram by qwertysri987 | ||

+ | ---------------------------- | ||

+ | By parallel lines and the tangency condition, <cmath>\angle APM\cong \angle LMP \cong \angle LKM.</cmath> Similarly, <cmath>\angle AQP\cong \angle KLM,</cmath> so AA similarity implies <cmath>\triangle APQ\sim \triangle MKL.</cmath> Let <math>\omega</math> denote the circumcircle of <math>\triangle ABC,</math> and <math>R</math> its circumradius. As both <math>P</math> and <math>Q</math> are inside <math>\omega,</math> | ||

+ | |||

+ | <cmath>\begin{align*} | ||

+ | R^2-QO^2&=\text{Pow}_{\omega}(Q)\\ | ||

+ | &=QB\cdot AQ \\ | ||

+ | &=2AQ\cdot MK\\ | ||

+ | &=2AP\cdot ML\\ | ||

+ | &=AP\cdot PC\\ | ||

+ | &=\text{Pow}_{\omega}(P)\\ | ||

+ | &=R^2-PO^2. | ||

+ | \end{align*}</cmath> It follows that <math>OP=OQ.</math> <math>\blacksquare</math> |

## Latest revision as of 10:41, 22 July 2020

## Problem

Let be a triangle with circumcentre . The points and are interior points of the sides and respectively. Let and be the midpoints of the segments and , respectively, and let be the circle passing through and . Suppose that the line is tangent to the circle . Prove that .

*Author: Sergei Berlov, Russia*

## Solution

### Diagram

Diagram by qwertysri987

By parallel lines and the tangency condition, Similarly, so AA similarity implies Let denote the circumcircle of and its circumradius. As both and are inside

It follows that