thnks for telling me. now i can post the solution more comfertabally

for a) because of the power of $P_{0}$ related to $\gamma_{i}$ we get that $P_{0}T_{0}^{2}=PA_{1}\cdot PA_{2}$, but left side is the power of $P_{0}$ related to $\gamma$ (the incircle) and right side is the power of $P_{0}$ related to the circumcircle. Thus $P_{0}$ is on the radical axis of the two circles, and this is true for $P_{1}$ and $P_{2}$ also in the exam i realised this AFTER i have shown that the projection of $P_{i}$ on $OI$ is the same independant of $i$. i am very curios to see the solution at b) because no one i talked to solved it

Isn't part (b) just problem 2 here? http://www.mathlinks.ro/Forum/contest/4th%20Edition/4th%20edition%20-%20round%207.pdf Link to MathLinks Contest discussion: https://artofproblemsolving.com/community/c60h24007p152422

Of course it is.

i can`t find the solution (too many buttons ) is it here too ? edit : i found it very nice, but very hard to find, especially in the exam

well..but using that steinbart theorem (i hope i write the name ok) it is easy let $S_{i}$ be the intersection between the incircle $\gamma$ and the side $A_{i+1}A_{i+2}$. Then $S_{i}T_{i}$ is the polar of $P_{i}$ related to $\gamma$. But $P_{i}$ are collinear, and by duality we have that the polars have all three one common point. then, steinbart and we`re done. i hope i didn`t say a big stupid thing

Can anyone give a simple proof for the steinbart theorem, using only projective geometry?

A simple proof for part a follows from the fact that each $P_i$ lies on the radical axis of $\gamma$ and $\gamma_i$ (from $P_iT_i$ is tangent to $\gamma$). From the previous fact, it follows that since $\gamma$ is the incenter, the radical axis of $\gamma$ and the circumcenter contain each of $P_i$ so they are collinear as desired.

Sorry for the revive, but the link posted by nsato is broken can anyone post a new link?