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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.

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.

Figure

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  • $\begingroup$ nice question (+1) $\endgroup$
    – tired
    Aug 19, 2016 at 13:45
  • $\begingroup$ same opinion (+1) $\endgroup$
    – Jean Marie
    Aug 20, 2016 at 0:55
  • $\begingroup$ it's not listed in the ETC. $\endgroup$
    – brainjam
    Jun 26, 2021 at 22:09
  • 1
    $\begingroup$ Now it is - X(46705) $\endgroup$
    – ivan
    Feb 18, 2022 at 11:03

1 Answer 1

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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? $\endgroup$
    – Yuxiao Xie
    Aug 19, 2016 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. $\endgroup$ Aug 19, 2016 at 14:31
  • $\begingroup$ that would be interesting to know. But we haven't found another one yet. $\endgroup$
    – Yuxiao Xie
    Aug 19, 2016 at 14:54

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