CMO 2013#5

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Phlembac Adib Hasan
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CMO 2013#5

Unread post by Phlembac Adib Hasan » Wed Apr 03, 2013 9:03 pm

Let $O$ denote the circumcentre of an acute-angled triangle $ABC$. Let point $P$ on side $AB$ be such that $\angle BOP = \angle ABC$, and let point $Q$ on side $AC$ be such that $\angle COQ = \angle ACB$. Prove that the reflection of $BC$ in the line $PQ$ is tangent to the circumcircle of triangle $APQ$.

Comment: মোটামুটি মাপের প্রবলেম। বেশি সুন্দরও না, বেশি বিশ্রীও না।

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FahimFerdous
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Re: CMO 2013#5

Unread post by FahimFerdous » Wed Apr 03, 2013 10:23 pm

প্রবলেমটা আমার বেশ ভাল লাগসে। একগাদা প্রোপার্টি আছে। কিন্তু শেষে আবারও আমাকে একবার সাইন ল মারতে হইল। :(
Last edited by Phlembac Adib Hasan on Thu Apr 04, 2013 8:23 am, edited 1 time in total.
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Tahmid Hasan
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Re: CMO 2013#5

Unread post by Tahmid Hasan » Wed Apr 03, 2013 11:50 pm

Phlembac Adib Hasan wrote:Let $O$ denote the circumcentre of an acute-angled triangle $ABC$. Let point $P$ on side $AB$ be such that $\angle BOP = \angle ABC$, and let point $Q$ on side $AC$ be such that $\angle COQ = \angle ACB$. Prove that the reflection of $BC$ in the line $PQ$ is tangent to the circumcircle of triangle $APQ$.

Comment: মোটামুটি মাপের প্রবলেম। বেশি সুন্দরও না, বেশি বিশ্রীও না।
Let $\odot ABC \cap \odot APQ=D \neq A,AO \cap BC=R$.
$\angle POQ=360^{\circ}-\angle BOC-\angle BOP-\angle COQ=360^{\circ}-2\angle A-\angle B-\angle C$
$=180^{\circ}-\angle A \Rightarrow O \in \odot APQ$.
$\angle BOP=\angle ABR,\angle OBP=\angle BAR \Rightarrow \triangle BOP \sim \triangle ABR \Rightarrow \frac{BP}{AR}=\frac{BO}{AB}$.
Similarly $\triangle COQ \sim \triangle ACR \Rightarrow \frac{CQ}{AR}=\frac{CO}{AC}$.
Hence $\frac{BP}{CQ}=\frac{AC}{AB}$.
From lemma-3 Yufei Zhao, $\frac{BP}{CQ}=\frac{BD}{CD}$. So $\frac{BD}{CD}=\frac{AC}{AB}$.
Let $D' \in \odot ABC$ such that $AD' \parallel BC$. Then $\frac{BD'}{CD'}=\frac{AC}{AB}$.
Now using the fact that for any positive real $r \exists$ a unique point $P$ on arc $BAC$ of $\odot ABC$ such that $\frac{BP}{CP}=r$, we get $D=D'$. $ABCD$ is an isosceles trapezoid.
Now draw a tangent $\ell$ to $\odot APQ$ at $D$. Let $\ell \cap BC=S$.
Now $\angle QDS=\angle DPQ=\angle DBC=\angle ACB \Rightarrow DQCS$ is cyclic.
$\angle DQS+\angle DQP=\angle DCS+\angle DBC=180^{\circ} \Rightarrow P,Q,S$ are collinear.
$\angle AOD=\angle AQD \Rightarrow 2\angle ACD=\angle QDC+\angle QCD \Rightarrow QSC=\angle QSD \Rightarrow \ell$ is the reflection of $BC$ in $PQ$, done!
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Phlembac Adib Hasan
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Re: CMO 2013#5

Unread post by Phlembac Adib Hasan » Thu Apr 04, 2013 8:34 am

FahimFerdous wrote:প্রবলেমটা আমার বেশ ভাল লাগসে। একগাদা প্রোপার্টি আছে। কিন্তু শেষে আবারও আমাকে একবার সাইন ল মারতে হইল। :(
আমার সাইন ল লাগে নাই! :D
Solution:
$\angle OPA=\angle OQC=\angle B+\pi /2 -\angle C$. So $A,P,Q,O$ concyclic. Suppose the line passing through $O$ and perpendicular to $PQ$ intersect $\bigcirc APQ$ again at $L'$ and $BC$ at $L$. We claim the reflection of $BC$ across $PQ$ touches $\bigcirc APQ$ at $L'$.

WLOG we may assume $\angle B\ge \angle C$.
$\angle QOL'=\pi/2-\angle PQO=\angle C\Longrightarrow O,Q,C,L$ concyclic. So $\angle OL'Q=\angle OAQ=\angle OCQ=\angle OLQ$. Since $LL'\perp PQ$ we find that $L'$ is the reflection of $L$ across $PQ$. Suppose after that reflection $C$ maps to $C'$.
\[\angle C'L'Q=\angle CLQ=\angle COQ=\angle ACB=\angle QOL'\] So $C'L'$ is the tangent of $\bigcirc APQ$ as desired.
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*Mahi*
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Re: CMO 2013#5

Unread post by *Mahi* » Thu Apr 04, 2013 12:17 pm

My solution is a linear combination of Adib and Tahmid's one, though I used a little different method for proving $ADQOP$ concyclic.
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