Sharygin's group of triangles

I recently noticed an elementary result in Euclidean geometry that I found so pretty, I just had to blog about it. It also provided the perfect excuse to write a little React for the first time since I left Intercom.

The intriangle

Given a triangle \(ABC\), let \(A', B', C'\), be the points where the angle bisectors meet opposite sides, as shown below:

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

Let's call \(A'B'C'\) the intriangle of \(ABC\).

Isosceles intriangles

Now ask yourself: when is the intriangle isosceles? Clearly it is isosceles if the same is true of the original triangle, so the potentially interesting question is whether any scalene triangles have isosceles intriangle. I came across this question in [1]. The following proposition, which is a slight extension of remarks in these notes, is the first hint that this might be a fun question to ask:

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:

  1. \(a' = b'\), i.e., the intriangle is isosceles about C',
  2. \(p(a, b, c) = 0\) where: $$ p(a, b, c) = (a^3 + a^2b + ab^2 + b^3) + (a^2 + ab + b^2)c - (a + b)c^2 - c^3, $$
  3. the points \(A', B', C', C\) are cyclic,
  4. \(\cos(C) = -\frac{c}{2(a+b+c)}\).
Furthermore, when any (and hence all) of the above conditions hold: $$ \cos(C') = -\cos(C) \in \left(\frac{2}{5 + \sqrt{17}}, \frac{1}{4}\right), $$ and so: $$ C \in \left(102.5^{\circ}, 104.5^{\circ}\right). $$ Finally, such triangles do exist, even rational ones, for example: $$ a = 1481089;\quad b = 18800081;\quad c = 19214131. $$

Proof (sketch) I do not have a beautiful proof but for what's it's worth, here's one way to get there. [UPDATE 2020-12-10: Alexander Belopolsky has written an excellent account of Sharygin's beautiful proof. He has also provided a slick derivation of the integer solution in the proposition statement.]

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!

The elliptic curve

The most significant part of proposition 1 is that the expression in property (ii) is a non-singular cubic. Let's call this the Sharygin cubic. Furthermore the point \(O = [1, -1, 0]\) is naturally distinguished so we have an elliptic curve. In the affine \(ab\)-plane (\(c=1\)) the curve looks like this:

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.

Can we do better than an open set?

This is all very well but isn't it a bit of a shame that we can only interpret points in an open set of this elliptic curve as geometric objects (triangles)? If we could somehow give meaning to the points when \(a\) or \(b\) is negative, we might expect some nice interplay between the arithmetic of the curve and the geometry of triangles.

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.

The triangle inequality

Here is a picture of our elliptic curve in the affine \(ab\)-plane (\(c=1\)) in which we have shaded the regions satisfying the triangle inequalities: $$ \begin{align} |a| + |b| &> 1,\\ |a| + 1 &> |b|,\\ |b| + 1 &> |a|. \end{align} $$

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 elliptic curve. Let's remember this for later.


The intriangle is constructed by taking angle bisectors, more precisely internal angle bisectors. What about the external angle bisectors? Giving ourselves the freedom to choose internal or external bisectors at each of the three angles means we have eight cases to consider a piori. However recall the following:

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:

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.

Isosceles extriangles

So points on our elliptic curve in which exactly one of \(a, b\) are negative correspond to triangles with a distinguished side, and triangles with a distinguished side have a natural extriangle. It turns out this fits together well:

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:

  1. \(a' = b'\), i.e., the \(A\)-extriangle is isosceles about C',
  2. \(p(-a, b, c) = 0\), where \(p\) is Sharygin the cubic,
  3. 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.

Interactive triangle arithmetic

At Intercom I was lucky to have an opportunity to learn the basics of front-end web development from legends such as Peter McKenna and Eoin Hennessy. Now that I've moved on, I find myself looking for excuses to dive back into React and so I ended up creating the little app below.

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

You can find the source here on GitHub. I am indebted and extremely grateful to Alexander Belopolsky who submitted a pull request in which he introduced logic to handle the case when the blue and green points coincide. Thank you again Alexander!

A geometric interpretation of the group law?

Given two triangles with isosceles intriangles / extriangles we know there is a naturally associated third, their sum. It seems very likely there is a natural geometric construction for this. I gave this a few hours but ended up empty handed.

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:

The group over \(\mathbb{Q}\)

I was introduced to the ideas here when I came across [1]. The central claim of these notes is that the group of the curve over the rationals is \(\mathbb{Z}\oplus \mathbb{Z}_2\). I did not have the appetite to go through the arguments in these notes but thanks to some helpful words of Álvaro Lozano-Robledo here I learned that algorithms for determining the group of an elliptic curve are implemented in Magma, and that they succeed for this curve. Furthermore, there is even an online Magma calculator here and the following script:
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);
Abelian Group isomorphic to Z/2
Defined on 1 generator
    2*$.1 = 0
1 true
consistent with the claim in [1].

The j-invariant

For completeness: it is the curve with j-invariant \(11^6/2^2 3^2 17\).

Poncelet's porism

You know you've made it when you've got a porism named after you. Of course it probably does help to be called Poncelet.

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?

Other comments

Aside from finding a geometric realisation of the group law, I have a few other questions I might think about next time I'm stuck in an airport.

A bit about Sharygin

Considering isosceles intriangles, extriangles must surely have been done long ago. I have followed [1] and named things for Igor Fedorovich Sharygin. In my ignorance, I had not previously heard of Sharygin. Based on these words I now understand him to have been an influential and prolific contributor to education as well as a person who suffered for admirable principles, difficult to uphold in 20th century Russia.

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].


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