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 $\angle OMT = \frac{1}{2}
\angle BAC$.
7ᵗʰ Iranian Geometry Olympiad (Intermediate) P2
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- Anindya Biswas
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Re: 7ᵗʰ Iranian Geometry Olympiad (Intermediate) P2
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
— John von Neumann