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

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''Author: Sergei Berlov, Russia''
 
''Author: Sergei Berlov, Russia''
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== Solution ==
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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>
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<cmath>\begin{align*}
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R^2-QO^2&=\text{Pow}_{\omega}(Q)\\
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&=QB\cdot AQ \\
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&=2AQ\cdot MK\\
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&=2AP\cdot ML\\
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&=AP\cdot PC\\
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&=\text{Pow}_{\omega}(P)\\
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&=R^2-PO^2.
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\end{align*}</cmath> It follows that <math>QO=PO.</math> <math>\blacksquare</math>

Revision as of 12:03, 1 April 2016

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

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 $QO=PO.$ $\blacksquare$

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