[OGC1] Online Geometry Camp: Day 3

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nayel
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[OGC1] Online Geometry Camp: Day 3

Unread post by nayel » Tue Aug 27, 2013 11:41 am

Post all the day 3 discussion threads links here.
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Tusher Chakraborty
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Re: [OGC1] Online Geometry Camp: Day 3

Unread post by Tusher Chakraborty » Tue Aug 27, 2013 12:46 pm

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asif e elahi
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Re: [OGC1] Online Geometry Camp: Day 3

Unread post by asif e elahi » Tue Aug 27, 2013 4:52 pm

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

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

Unread post by SANZEED » Tue Aug 27, 2013 5:31 pm

In $\triangle OMN$, $OM=ON$ and $\angle MOR=\angle NOR$, so $OR$ is the perpendicular bisector of $MN$.
Now for $\triangle AMN, R$ is on the perpendicular bisector of $MN$, and at the same time, it is on the angle bisector of $\angle MAN$. So it is a well-known fact that $R$ is on $\bigodot AMN$. Let $AR$ intesect $BC$ at $E$.
Note that $180^{\circ}-\angle MRE=\angle MRA=\angle MNA=180^{\circ}-\angle MNC= \angle MBC=\angle MBE$. This implies that $M,B,E,R$ are concyclic. Similarly, we can show that $N,C,E,R$ are concyclic. Thus $E$ is the point where $\bigodot BMR$ and $\bigodot CNR$ intersect and it is on $BC$.

Comment: This is IMO-2004 G1. :|
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asif e elahi
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Re: [OGC1] Online Geometry Camp: Day 3

Unread post by asif e elahi » Tue Aug 27, 2013 6:34 pm

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.

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asif e elahi
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Re: [OGC1] Online Geometry Camp: Day 3

Unread post by asif e elahi » Tue Aug 27, 2013 10:50 pm

When we will discuss with problems? :evil:

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

Unread post by SANZEED » Wed Aug 28, 2013 4:34 pm

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.
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