7ᵗʰ Iranian Geometry Olympiad (Intermediate) P2

For discussing Olympiad level Geometry Problems
IftakharTausifFarhan
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7ᵗʰ Iranian Geometry Olympiad (Intermediate) P2

Unread post by IftakharTausifFarhan » Sat Dec 12, 2020 4:09 pm

Let $ABC$ be an isosceles triangle $(AB = AC)$ with its circumcenter $O$. Point $N$ is
the midpoint of the segment $BC$ and point $M$ is the reflection of the point $N$ with respect to the
side $AC$. Suppose that $T$ is a point so that $ANBT$ is a rectangle. Prove that $\angle OMT = \frac{1}{2}
\angle BAC$.

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Anindya Biswas
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Re: 7ᵗʰ Iranian Geometry Olympiad (Intermediate) P2

Unread post by Anindya Biswas » Tue Jan 05, 2021 2:07 pm

IGO 2020 Intermediate P2.png
IGO 2020 Intermediate P2.png (39.36KiB)Viewed 5306 times
Let's connect $O,C$ and $O,T$ and $C,M$ and $A,N$.
In $\triangle OCM$ and $\triangle OAT$,
$OC=OM$
$CM=CN=NB=AT$
and $\angle OCM=\angle OCA+\angle ACM=\angle OAC+\angle ACN=90^\circ=\angle OAT$
$\therefore \triangle OCM \cong \triangle OAT$
$\therefore OM=OT, \angle MOC=\angle TOA$

Now, in $\triangle TOM$ and $\triangle AOC$,
$\frac{TO}{AO}=\frac{OM}{OC}$
$\angle TOM=\angle TOA+\angle AOM=\angle AOM+\angle MOC=\angle AOC$
So, $\triangle TOM\sim\triangle AOC$

$\therefore \angle OMT=\angle OCA=\angle OAC=\frac12\angle BAC$.
Q.E.D.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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