## [OGC1] Online Geometry Camp: Day 3

Discussion on Bangladesh National Math Camp
nayel
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Joined: Tue Dec 07, 2010 7:38 pm
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### [OGC1] Online Geometry Camp: Day 3

Post all the day 3 discussion threads links here.
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

Tusher Chakraborty
Posts: 27
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### Re: [OGC1] Online Geometry Camp: Day 3

আজকের টপিক হলঃ
বৃত্ত সংক্রান্ত আলোচনা।
উপপাদ্য-৩৩ হতে ৫০(মাধ্যমিক জ্যামিতি বই)
বিকালে আমরা প্রবলেম নিয়ে আলোচনা করবো।
"Your present circumstances don't determine where you can go; they merely determine where you start." -Nido Qubein

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm
Location: Sylhet,Bangladesh

### Re: [OGC1] Online Geometry Camp: Day 3

Let $ABC$ be an acute angled triangle with $AB$ and $AC$ are not equal.The circle with diameter $BC$ intersects $AB$ and $AC$ at $M$ and $N$.$O$ is the midpoint of $BC$. The bisectors of $\angle BAC$ and $\angle MON$ intersects at $R$.Prove that circumcircles of $\triangle BMR$ and $\triangle CNR$ have a common point on $BC$.
Last edited by Phlembac Adib Hasan on Wed Aug 28, 2013 3:59 pm, edited 1 time in total.
Reason: Latexed

SANZEED
Posts: 550
Joined: Wed Dec 28, 2011 6:45 pm
Location: Mymensingh, Bangladesh

### Re: [OGC1] Online Geometry Camp: Day 3

$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm
Location: Sylhet,Bangladesh

### Re: [OGC1] Online Geometry Camp: Day 3

Another problem.
Let $ABC$ be an acute angle triangle.And let $P$ and $Q$ be points on segment $BC$.Construct point $C'$ such a way that the convex quadrilateral $APBC'$ is cyclic,$QC'$ and $CA$ parallel,and $C'$ and $Q$ lie on opposite sides of $AB$.Construct $B'$ similarly.Prove $B'C'PQ$ cyclic.

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm
Location: Sylhet,Bangladesh

### Re: [OGC1] Online Geometry Camp: Day 3

When we will discuss with problems?

SANZEED
Posts: 550
Joined: Wed Dec 28, 2011 6:45 pm
Location: Mymensingh, Bangladesh

### Re: [OGC1] Online Geometry Camp: Day 3

Let $C'A\cap QB'=X$. Now $180^{\circ}-\angle C'PQ=\angle C'PB=\angle C'AB=\angle C'XQ$ implies $C',X,Q,P$ are concyclic. Let $C'Q\cap PX=E, AC\cap PX=S$. Then $\angle PEQ=\angle C'ES=\angle CSE=\angle ASX$. Also,
$\angle PQE=\angle PQC'=\angle PXC'=\angle SXA$. Thus $\triangle PEQ\sim \triangle ASX$.
So, $\angle XAS=\angle QPE\Rightarrow \angle XAC=\angle XPC\Rightarrow X\in \bigodot APC$. But $B'\in \bigodot APC$ implies $B'\equiv X$,and $B',C',P,Q$ are concyclic.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$