Let $I$ be the incenter of a triangle $ABC$. A point $X$ satisfies the conditions $XA+XB=IA+IB$, $XA+XC=IA+IC$. The points $Y,Z$ are defined similarly. Show that the lines $AX,BY,CZ$ are concurrent or parallel to each other.

My friend discovered this problem when he was drawing random ellipses for fun. But we have no idea how to solve such a problem because we literally know nothing about ellipses (except its definition). So I can't post where I'm stuck here. We're just curious to see the solution, whether or not it's elementary.

We do not know what kind of tags we should add because we do not know what methods are to be used. Please edit the tagging.


  • $\begingroup$ nice question (+1) $\endgroup$ – tired Aug 19 '16 at 13:45
  • $\begingroup$ same opinion (+1) $\endgroup$ – Jean Marie Aug 20 '16 at 0:55

For the moment, I will just sketch an approach, and develop the calculations later. It will be clear soon that this approach has to work, so we may also leave the non-interesting computation part to a CAS. The key idea is highlighted.

  1. Concurrency and collinearity are straightforward to check through Ceva's theorem, hence we just have to compute the trilinear coordinates of $X,Y,Z$, then check that an associated determinant vanishes;
  1. The ellipses $\Gamma_A$ with foci at $B,C$ and $\Gamma_B$ with foci at $A,C$ are two conics meeting at two points. We may write down their trilinear equations and exploit Vieta's theorem to get the trilinear coordinates of $Z$, since the trilinear coordinates of $I$ are just $[1;1;1]$;
  1. In order to write down the trilinear equation of our ellipses it is enough to recall that $$ AI^2 = bc-4rR $$ and so on. By Vieta's theorem we do not even need all the coefficients of the trilinear equation, so computations are not that painful.

  • $\begingroup$ Thanks for your solution. But I wonder if there is any solution without tedious calculations? $\endgroup$ – Colescu Aug 19 '16 at 14:27
  • $\begingroup$ @YuxiaoXie: that is a good question. Maybe there is a tricky way for exploiting Marden's theorem, but a pure synthetic solution still eludes me. It would be interesting also to understand which points of $ABC$ lead to a concurrency as above, since the incenter is not the only one. $\endgroup$ – Jack D'Aurizio Aug 19 '16 at 14:31
  • $\begingroup$ that would be interesting to know. But we haven't found another one yet. $\endgroup$ – Colescu Aug 19 '16 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.