Let γ be the incircle in the triangle A0A1A2. For all i∈{0,1,2} we make the following constructions (all indices are considered modulo 3): γi is the circle tangent to γ which passes through the points Ai+1 and Ai+2; Ti is the point of tangency between γi and γ; finally, the common tangent in Ti of γi and γ intersects the line Ai+1Ai+2 in the point Pi. Prove that a) the points P0, P1 and P2 are collinear; b) the lines A0T0, A1T1 and A2T2 are concurrent.
Problem
Source: Romanian IMO TST 2006, day 3, problem 3
Tags: geometry, incenter, circumcircle, power of a point, radical axis, geometry proposed
16.05.2006 20:46
thnks for telling me. now i can post the solution more comfertabally
16.05.2006 20:57
for a) because of the power of P0 related to γi we get that P0T20=PA1⋅PA2, but left side is the power of P0 related to γ (the incircle) and right side is the power of P0 related to the circumcircle. Thus P0 is on the radical axis of the two circles, and this is true for P1 and P2 also in the exam i realised this AFTER i have shown that the projection of Pi 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
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
16.05.2006 23:08
Of course it is.
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
17.05.2006 20:02
well..but using that steinbart theorem (i hope i write the name ok) it is easy let Si be the intersection between the incircle γ and the side Ai+1Ai+2. Then SiTi is the polar of Pi related to γ. But Pi are collinear, and by duality we have that the polars have all three one common point. then, steinbart and weredo≠.ihopeididnt say a big stupid thing
01.06.2006 08:35
Can anyone give a simple proof for the steinbart theorem, using only projective geometry?
24.07.2013 08:45
A simple proof for part a follows from the fact that each Pi lies on the radical axis of γ and γi (from PiTi is tangent to γ). From the previous fact, it follows that since γ is the incenter, the radical axis of γ and the circumcenter contain each of Pi so they are collinear as desired.
25.07.2015 01:12
Sorry for the revive, but the link posted by nsato is broken can anyone post a new link?