Iran TST 2015 2
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
$I_b$ 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.
Because destroying everything mindlessly isn't cool enough.
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: Iran TST 2015 2
Thanic Nur Samin wrote:
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: Iran TST 2015 2
There is a simple solution by it.Zawadx wrote:
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.
Re: Iran TST 2015 2
Ah, that's pretty cool. Maybe I'd catch that with a perfect figure; then it seems to be more natural.
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- Posts:12
- Joined:Sat Jun 13, 2015 1:46 pm
- Location:Halishahar, Chittagong
Re: Iran TST 2015 2
Just angle chasing is enough :p