IMO-2011-G1

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Tahmid Hasan
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IMO-2011-G1

Unread post by Tahmid Hasan » Thu Jul 26, 2012 1:26 pm

Let $ABC$ be an acute triangle. Let $\omega$ be a circle whose centre $L$ lies on the side $BC$. Suppose that $\omega$ is tangent to $AB$ at $B'$ and $AC$ at $C'$. Suppose also that the circumcentre $O$ of triangle $ABC$ lies on the shorter arc $B'C'$ of $\omega$. Prove that the circumcircle of $ABC$ and $\omega$ meet at two points.
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Tahmid Hasan
Posts:665
Joined:Thu Dec 09, 2010 5:34 pm
Location:Khulna,Bangladesh.

Re: IMO-2011-G1

Unread post by Tahmid Hasan » Thu Jul 26, 2012 1:29 pm

My solution
Let $H$ be the orthocenter of $\triangle ABC$.It is well-known that $\angle BHC=180^{\circ}-\angle A$.So the reflection$(H')$ of $H$ in $BC$ lies on $\odot ABC$.
Let $O'$ be the reflection of $O$ in $BC$.If $O'$ is outside $\odot ABC$,then $\odot ABC,\omega$ intersect at two points.
But then $\angle BO'C<\angle BH'C$
$\Rightarrow \angle BOC<\angle BHC$
$\Rightarrow 2\angle A<180^{\circ}-\angle A$
$\Rightarrow \angle A<60^{\circ}$.
So it suffices to prove that $\angle A<60^{\circ}$.
Since $\angle B,\angle C$ are acute and $LB' \bot AB,LC' \bot AC$;$B',C'$ lie on the segments $AB,AC$ respectively.
So $\angle BOC<\angle B'OC'$
$\Rightarrow 2\angle A<\frac {1}{2}(360^{\circ}-\angle B'LC')$
$=180^{\circ}-\frac {1}{2}(180^{\circ}-\angle A)$[The last arguement follows from the fact that $AB'LC'$ is cyclic.]
$=90^{\circ}+\frac {1}{2}\angle A$
$\Rightarrow A<60^{\circ}$
So the two circles intersect at two points.
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