Consider the elliptic curve defined by the cubic: $$ a^3 + a^2b + ab^2 + b^3 + (a^2 + ab + b^2)c - (a+b)c^2 - c^3 = 0 $$ in $\mathbb{P^2}$ with distinguished point $[1, -1, 0]$ as identity.

Recently I learned that the real points of the identity component of this curve naturally parameterise Euclidean triangles with a neat property. Specifically they are scalene triangles such that the triangle formed by intersecting angle bisectors with opposite sides is isosceles. Here's an example:

An isosceles extriangle

(The triangle in red, constructed from one internal and two external angle bisectors, is isosceles.)

Let's call this property $P$. See here for a fuller discussion, with more pictures and a little history.

This means that given two triangles satisfying property $P$, there is a naturally associated third: their sum under elliptic curve addition. My questions are:

  1. Can we find a geometric construction for the sum of two of these triangles ?

  2. Is there is a natural family of geometric objects parameterised by the non-identity component of the elliptic curve ?

Regarding the second question, as discussed here it seems like triangles are out but it seems plausible we could find something. E.g., an idea with the sort of flavour I have in mind is as follows: a triangle satisfying property $P$, has a distinguished side. We can regard the other two sides as a singular quadric. Perhaps admitting non-singular quadrics gives us room to find an interpretation for points on this other component.

  • $\begingroup$ as for the second question, to see the "complex triangles" you can "rotate" $y$ and plot $(x,iy)$ instead of $(x,y)$ to get a picture. Circles become hyperbolas, but the angle "bissection" becomes a bit strange to describe. The real triangles can be plotted normally with exterior bissectors. $\endgroup$
    – mercio
    Jan 12, 2017 at 16:34
  • $\begingroup$ you can also plot $(x,y^2)$ instead, which will have the added bonus that whatever construction works for the addition procedure will work the same on both components. $\endgroup$
    – mercio
    Jan 12, 2017 at 18:18
  • $\begingroup$ Thanks for these very interesting remarks @mercio. I don't think I quite follow you though, as I can't quite see what your $x, y$ are. My first guess was $x = b/c, y = a/c$ but this doesn't seem to fit with your remarks. Perhaps you could give an example of a candidate object corresponding to the point $[a,b,c] = [1-\sqrt{17}, 1-\sqrt{17},8]$? $\endgroup$ Jan 12, 2017 at 20:14
  • $\begingroup$ I mean, if you try to plot the triangles as in your demo, with a fixed horizontal segment $[AB]$ on the $x$ axis, and with $C$ varying so that $A'C' = B'C'$, what happens when the triangle inequality is not satisfied is that the $y$ coordinate of everyone is purely imaginary. So if instead of plotting $(x,y)$ you plot $(x,iy)$ or $(x,y^2)$ you are suddenly looking at real things. $\endgroup$
    – mercio
    Jan 12, 2017 at 21:51
  • $\begingroup$ @KeD The expression of the group law in coordinates is complicated but if you see anything please do share it. $\endgroup$ Jan 13, 2017 at 12:34

1 Answer 1


This is very nice! For now, I only have a little note to add that is too long for a comment. You can bring the curve $$C: a^3 + a^2b + ab^2 + b^3 + (a^2 + ab + b^2)c - (a+b)c^2 - c^3=0$$ to Weierstrass form $$E: y^2 + 1/3xy = x^3 + 7/9x^2 + 5/27x + 1/81$$ via $C\to E$ with coordinates $$(a/3 + b/3, -a/9,-2a - 2b - c).$$ And then you can easily bring $E$ to a minimal model $y^2 + xy = x^3 + x^2 - 2x$, and compute its Mordell-Weil group: it has rank $1$ and torsion $\mathbb{Z}/2\mathbb{Z}$. The group of rational points in the minimal model is generated by the point $(0,0)$ of order $2$ and the point $(2,2)$ of infinite order. (All this I have done with Magma.)

  • $\begingroup$ Thanks very much for this, delighted to find someone else who thinks this is a nice thing to think about :) $\endgroup$ Jan 12, 2017 at 8:48
  • $\begingroup$ Incidentally, I came across this problem in this arXiv posting which is mostly a calculation of the group over $\mathbb{Q}$. I didn't quite have the inclination to go through their argument so it's nice to have the confidence boost that the claimed result is correct, provided by your Magma calculations. I'm not sure if you already looked but I give a little more context about this problem here. $\endgroup$ Jan 12, 2017 at 8:55

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