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

(Diagram)
 
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== Solution ==
 
== Solution ==
 
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===Diagram===
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<asy>
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dot("O", (50, 38), NW);
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dot("A", (40, 100), N);
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dot("B", (0, 0), S);
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dot("C", (100, 0), S);
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dot("Q", (24, 60), W);
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dot("P", (52, 80), E);
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dot("L", (62, 30), SE);
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dot("M", (38, 70), N);
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dot("K", (27, 42), W);
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draw((100, 0)--(24, 60), dotted);
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draw((0, 0)--(52, 80), dashed);
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draw((0, 0)--(100, 0)--(40, 100)--cycle);
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draw((24, 60)--(52, 80));
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draw((27, 42)--(38, 70)--(62, 30)--cycle);
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draw(circle((49, 49), 23));
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label("$\Gamma$", (72, 49), E);
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draw(circle((50, 38), 63));
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label("$\omega$", (-13, 38), NW);
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</asy>
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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>  
 
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>  
  

Latest revision as of 10:41, 22 July 2020

Problem

Let $ABC$ be a triangle with circumcentre $O$. The points $P$ and $Q$ are interior points of the sides $CA$ and $AB$ respectively. Let $K,L$ and $M$ be the midpoints of the segments $BP,CQ$ and $PQ$, respectively, and let $\Gamma$ be the circle passing through $K,L$ and $M$. Suppose that the line $PQ$ is tangent to the circle $\Gamma$. Prove that $OP=OQ$.

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, \[\angle APM\cong \angle LMP \cong \angle LKM.\] Similarly, \[\angle AQP\cong \angle KLM,\] so AA similarity implies \[\triangle APQ\sim \triangle MKL.\] Let $\omega$ denote the circumcircle of $\triangle ABC,$ and $R$ its circumradius. As both $P$ and $Q$ are inside $\omega,$

\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*} It follows that $OP=OQ.$ $\blacksquare$

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