Sharygin(Russia) Geometry Olympiad 2016 grade 9/1

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Thamim Zahin
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Sharygin(Russia) Geometry Olympiad 2016 grade 9/1

Unread post by Thamim Zahin » Sun Jan 15, 2017 10:58 pm

The diagonals of a parallelogram $ABCD$ meet at point $O$. The tangent to the circumcircle of triangle $BOC$ at $O$ meets ray $CB$ at point $F$. The circumcircle of triangle $FOD$ meets $BC$ for the second time at point $G$. Prove that $AG=AB$.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

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Thamim Zahin
Posts:98
Joined:Wed Aug 03, 2016 5:42 pm

Re: Sharygin(Russia) Geometry Olympiad 2016 grade 9/1

Unread post by Thamim Zahin » Sun Jan 15, 2017 11:23 pm

We know that, $\angle DBC =\angle ADB =\angle FOC =\angle a $[last one is for tangency]

$\angle CDB = \angle ABD =\angle b$

$\angle AGB = \angle DAG =\angle i$

$\angle GAC = \angle u$
$\angle AGD =\angle o$

and,$\angle DAC+\angle GAC =\angle DAC = \angle ACB =\angle i+\angle u$

Now from triangle $\triangle BCO$, we get $\angle ACB+\angle BCA=\angle COD$ [$O,A$ are on same line]

So, $\angle COD=\angle a+\angle i+\angle u$ or,$\angle FOD=\angle i+\angle u$

We know that, $ODGF$ is cyclic. So, $(\angle i+\angle o)+(\angle i+\angle u) =180^o$

That give us that $ADGC$ is cyclic. So, $\angle ADC=\angle CGA=\angle a+\angle b$. and $\angle ABG =\angle a+\angle b$ [$c,g$ are on same line]

Or, $\angle AGB=\angle ABG$

So, $AB=AG$

$[Done]$
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

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