\(AA'\) bisects the angle at \(A\); similarly for \(B, C\).

Let's call \(A'B'C'\) the intriangle of \(ABC\).
**Proposition 1**
*
For a scalene (Euclidean) triangle ABC, with side lengths \(a, b, c\),
and intriangle side lengths \(a', b', c'\), the following are equivalent:
*

- \(a' = b'\), i.e., the intriangle is isosceles about C',
- \(p(a, b, c) = 0\) where: $$ p(a, b, c) = (a^3 + a^2b + ab^2 + b^3) + (a^2 + ab + b^2)c - (a + b)c^2 - c^3, $$
- the points \(A', B', C', C\) are cyclic,
- \(\cos(C) = -\frac{c}{2(a+b+c)}\).

**Proof** (sketch)
I do not have a beautiful proof but for what's it's worth, here's one
way to get there.

Using the cosine rule based at \(A\) for the two triangles \(AB'C'\) and \(ABC\), eliminating the \(\cos(A)\) between the two equations, and noting that \(AB' = bc/(a+c)\), \(AC' = bc/(a+b)\) we get: $$ a'^2 = b^2c^2\left( \frac{1}{(a+c)^2} +\frac{1}{(a+b)^2}\right) - \frac{bc}{(a+b)(a+c)}\left( b^2 + c^2 - a^2 \right). $$ Taking the similar expression for \(b'^2\) and factorising we obtain: $$ a'^2 - b'^2 = \frac{abc(a-b)p(a,b,c)} {(a+b)(b+c)^2(a+c)^2} $$ This shows the equivalence of properties (i) and (ii).

The equivalence of (ii) and (iv) follows since p(a,b,c) is the numerator of: $$ \frac{a^2 + b^2 - c^2}{ab} + \frac{c}{a + b + c}. $$

This just leaves property (iii). To see that (iii) implies (i), note that the angles \(C'B'A'\) and \(C'CA'\) are equal (same circle arc), as are \(C'A'B'\) and \(C'CB'\) and then use the fact that \(CC'\) is a bisector. Finally (iii) follows from (ii) by further applications of the cosine rule, along similar lines to those above, this time taking the angle \(C'\) in the triangle \(A'B'C'\).

This shows that the four properties are equivalent. The stated bound on the value of \(\cos(C')\) follows by noting that the minimum is approached as we approach \(a = b\) and the maximum when either \(a\) or \(b\) approaches zero. ∎

What a weirdly narrow range of possible angles for \(C\): less than 2 degrees!

Red points define triangles with isosceles intriangle.

Points with both \(a, b > 0\), coloured red above, correspond to triangles
with isosceles intriangle. We admit exactly one non-scalene triangle,
\(a = b = (1+\sqrt{17})/8, c=1\), into this family of triangles and so obtain a family of
triangles with isosceles intriangles parameterised by an open set of an
elliptic curve.

Here's the good news: using *external* angle bisectors, we can extend
the interpretation of points on the elliptic curve as triangles to include
points with negative values of \(a, b\). This was the simple observation
that pleased me so much when it occurred to me.

Red points define triangles if we discard signs.

Bearing in mind that any line can meet a cubic at most three times
(including at infinity) it's easy to see that the identity component of the
elliptic curve, shown in red, lies entirely inside the shaded region.
This means that if we discard signs, all points in this component define
triangles, provided we admit the two degenerate triangles when
\(a\) or \(b\) vanishes.
But wait, there's more! The triangles corresponding to points where either
\(a\) or \(b\) are negative are decorated with one extra piece of
information: they have a *distinguished side*, namely the side which
corresponds to a negative coordinate on the ellptic curve. Let's remember
this for later.

**Classical result**
*
For a scalene triangle ABC, bisect each angle either internally or
externally and
let \(A', B', C'\) be the three points where these bisectors meet opposite
sides. Then \(A', B', C'\) are collinear iff exactly one or three angles
were bisected externally.
*

(This follows easily from Menelaus's theorem; see exercies 1, 2 in [6]
section 3.4.)

You can't build an honest triangle out of collinear points, so the cases we're interested in are:

- the case when all bisectors are internal, i.e., the intriangle,
- the three cases when exactly two bisectors are external and one is internal, i.e., the three extriangles.

Here is a possible picture:

\(AA'\) bisects the angle at \(A\) internally;

\(BB'\) and \(CC'\) bisect the angles at \(B\) and \(C\) externally

Note that a triangle with a distinguished side has a natural extriangle:
use external bisectors for the angles at either end of the distinguished
side and internal bisector for the other angle. The picture
above illustrates the extriangle of triangle \(ABC\) with
distinguished side opposite angle \(A\). Let's call this the
\(A\)-extriangle.
\(BB'\) and \(CC'\) bisect the angles at \(B\) and \(C\) externally

**Proposition 2**
*
For a scalene (Euclidean) triangle ABC with side lengths \(a, b, c\),
and A-extriangle side lengths \(a', b', c'\) the following are
equivalent:
*

- \(a' = b'\), i.e., the \(A\)-extriangle is isosceles about C',
- \(p(-a, b, c) = 0\), where \(p\) is Sharygin the cubic,
- the points \(A', B', C', C\) are cyclic,

**Proof** (sketch)
The proof is completely analogous to that of proposition 1. Indeed
essentially all correponding formulae hold except with \(a\) replaced by
\(-a\).
∎

Part (ii) means that we have indeed identified exactly those triangles which correspond to the points with negative \(a, b\) on the identity component on our elliptic curve.

If we also admit the two equilateral triangles where \(a = -b = \pm 1\) (whose extriangles have vertices at infinity) as well as the degenerate triangle corresponding to the point \(O\) at infinity then, topologically we have the full circle.

Use the sliders below to drag points along the elliptic curve and see the resulting "equation of triangles":

There is some room for improvement, most notably with regard to floating point precision when numbers involved in the calculations get very large or small.

You can find the source here on GitHub. In a fit of wild optimism I have even opened an issue concerning the cases when the blue and green points coincide in the hope that somebody might submit a PR for it.

Aside from artistic interest, there is another reason to seek a geometric expression of the group law: group addition is associative. This means that whatever the geometric expression of the group law, the fact of its associativity should itself be a potentially-interesting statement in Euclidean geometry.

I think I even have a pretty good guess that associativity should essentially correspond either to Pappus's hexagaon theorem or perhaps to Desargues's theorem (which are in fact closely related in Euclidean geometry) since:

- Pappus's theorem is a special case of the Cayley-Bacharach theorem which essentially corresponds to the fact that elliptic curve addition is associative.
- Under the projective plane construction, alternative division rings correspond to Moufang planes (see, e.g., page 154 of [4]) and the plane satisfies Desargues's theorem iff the ring is associative. (See also "Moufang Consequences Theorem" on page 34 of [5].)

P2<A, B, C> := ProjectiveSpace(Rationals(), 2); X := Curve(P2, A^3 + A^2*B + A*B^2 + B^3 + A^2*C + A*B*C + B^2*C - A*C^2 - B*C^2 - C^3); O := X![1, -1, 0]; E := EllipticCurve(X, O); TorsionSubgroup(E), Rank(E);yields:

Abelian Group isomorphic to Z/2 Defined on 1 generator Relations: 2*$.1 = 0 1 trueconsistent with the claim in [1].

The question of identifying a geometric construction for the group law of our triangles reminds me of a question I once wondered about in relation to Poncelet's porism. There too an elementary question about Euclidean geometry has an underlying connection to elliptic curves. A key difference is that the curve one obtains for a particular instance of the problem does not have a distinguished identity point, and so a choice must be made to obtain an elliptic curve but with some care to account for this, the question of identifying a geometric construction associated to the group operation remains. Again, I do not know the answer but I recall it was discussed some years ago here by David Speyer and Gavin Wraith. Perhaps they know the answer?

- The group-theoretic inverse of a triangle is the reflection that interchanges \(a, b\). Can we find geometric interpretations for other arithmetic operations in the elliptic curve? E.g., what is twice or \(n\) times a triangle?
- Horizontal or vertical lines in the \(ab\)-plane that meet the identity component three times pick out triples of triangles satisfying \(P + Q = R\) in which all three triangles have two side lengths in common. Is there anything special about these?
- Points on the non-identity component of the curve do not define any obvious geometric object. Triangles are out since an open set of them do not satisfy the triangle inequality. Is it possible to find a geometric object to which they correspond?
- A regular 7-gon has diagonals of two different lengths. Taking these together with the side length of the 7-gon we can form a triangle. In [1] it is noted that this triangle has isosceles intriangle. Does this have any special arithmetic properties?
- The question of isosceles intriangles makes sense in non-Euclidean geometry. What happens in hyperbolic space?

Unfortunately for me, I cannot enjoy Sharygin's discussions of the ideas here because I do not speak Russian. Nevertheless, thanks to Ivan Smirnov (who supplied corrected versions of the citations in [1]) I can supply references which may be enjoyed by those who are not so linguistically limited.

The geometry is apparently discussed in problem 58 on page 154 of Sharygin's book [2] (with solution page 157) as well as in a 1983 issue of Kvant [3].

- I. V. Netay and A.V. Savvateev,
*"Sharygin triangles and elliptic curves"*, (2016). - И. Ф. Шарыгин [I. F. Sharygin], "Задачи по геометрии" ["Problems in geometry"], М., Наука [Nauka, Moscow], (1982).
- И. Ф. Шарыгин [I. F. Sharygin], "Вокруг биссектрисы" ["Around the bisector"], Kvant, No.8, pp.32--36 (1983).
- D. R. Hughes and F. Piper,
*"Projective planes"*, Springer-Verlag (1973). Graduate Texts in Mathematics, No. 6. - K. McCrimmon,
*"A taste of Jordan algebras"*, Springer (2004). - H. S. M. Coxeter and S. L. Geitzer,
*"Geometry revisited"*, MAA (1967).