Let $\gamma$ be the incircle in the triangle $A_0A_1A_2$. For all $i\in\{0,1,2\}$ we make the following constructions (all indices are considered modulo 3): $\gamma_i$ is the circle tangent to $\gamma$ which passes through the points $A_{i+1}$ and $A_{i+2}$; $T_i$ is the point of tangency between $\gamma_i$ and $\gamma$; finally, the common tangent in $T_i$ of $\gamma_i$ and $\gamma$ intersects the line $A_{i+1}A_{i+2}$ in the point $P_i$. Prove that a) the points $P_0$, $P_1$ and $P_2$ are collinear; b) the lines $A_0T_0$, $A_1T_1$ and $A_2T_2$ are concurrent.
Problem
Source: Romanian IMO TST 2006, day 3, problem 3
Tags: geometry, incenter, circumcircle, power of a point, radical axis, geometry proposed
ashwinrk_jain
16.05.2006 20:46
thnks for telling me. now i can post the solution more comfertabally
maky
16.05.2006 20:57
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
nsato
16.05.2006 22:53
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
Valentin Vornicu
16.05.2006 23:08
Of course it is.
maky
16.05.2006 23:14
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
maky
17.05.2006 20:02
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
xirti
01.06.2006 08:35
Can anyone give a simple proof for the steinbart theorem, using only projective geometry?
ssilwa
24.07.2013 08:45
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.
stephcurry
25.07.2015 01:12
Sorry for the revive, but the link posted by nsato is broken can anyone post a new link?