Iran TST 2015 2

For discussing Olympiad level Geometry Problems
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Thanic Nur Samin
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Iran TST 2015 2

Unread post by Thanic Nur Samin » Fri Aug 05, 2016 9:24 pm

$I_b$ is the $B$-excentre of $\triangle ABC$ and $\omega$ is its circumcircle. $M$ is the midpoint of the arc $BC$ not containing $A$. $MI_b$ meets $\omega$ at point $T$ other than $M$. Prove that $TB.TC=TI_b^2$
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

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Thanic Nur Samin
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Re: Iran TST 2015 2

Unread post by Thanic Nur Samin » Fri Aug 05, 2016 9:24 pm

Can you see a symmedian?
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

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Zawadx
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Re: Iran TST 2015 2

Unread post by Zawadx » Sat Aug 06, 2016 11:54 am

Thanic Nur Samin wrote:
Can you see a symmedian?
I don't need your silly symmedians 8-)
You should know the lemma surrounding the points mentioned in the problem. Then this solution follows naturally:
Let $I_bB \cap \omega = X$ and $I_bC \cap \omega = Y$.

Then by the Incentre/Excentre Lemma, $X$ is the midpoint of arc $AC$ not containing $B$ and $y$ is the midpoint of arc $AB$ containing $C$.

Now, $TM$ bisects $\angle BTC = \angle A$, so it is enough to prove that $\angle TBI_b + \angle TCI_b = \frac {\angle A}2$. This is equivalent to proving that arc $XY$ subtends $ \frac {\angle A}2$ at the circumference, which we can pove easily by the Incentre/Excentre lemma (QED).
I liked the prob, tho that may be because I solved it easily :P

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Thanic Nur Samin
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Re: Iran TST 2015 2

Unread post by Thanic Nur Samin » Sat Aug 06, 2016 12:42 pm

Zawadx wrote:
I don't need your silly symmedians 8-)
There is a simple solution by it.
Let $N$ be the antipode of $M$. Now, it is the circumcentre of $\triangle BCI_b$. Since $NB\perp MB$ and $NC\perp MC$, $I_bM$ is a symmedian. Again, $\angle NTM=\dfrac{\pi}{2}$, So the reflection of $I_b$ across $T$ will lie on the circumcircle of $\triangle BCI_b$. Now, use the symmedian intersects circumcircle midpoint lemma(is there any actual name?), and we find that $\triangle TCI_b$ is similar to $\triangle TI_bB$. The proof follows immediately.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

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Zawadx
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Re: Iran TST 2015 2

Unread post by Zawadx » Sat Aug 06, 2016 3:33 pm

Ah, that's pretty cool. Maybe I'd catch that with a perfect figure; then it seems to be more natural.

Mehedi Hasan Nowshad
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Re: Iran TST 2015 2

Unread post by Mehedi Hasan Nowshad » Sun Aug 07, 2016 9:13 pm

Just angle chasing is enough :p

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