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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.

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Figure

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

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  • $\begingroup$ 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

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