# Who discovered integer triangles with one angle trisecting another?

When & who was the first mathematician to discover the following simple triangle with a unique property that it has one angle is equal to one third of another angle in the same triangle?

The triangle sides are the following: $$a^3, a(b^2 – a^2), b(b^2 – 2a^2)$$ where, $(b/a) > \sqrt{2}$, & $(a, b)$ are positive real constructible numbers

I know this seemingly a very simple question that may not be in Wikipedia!, also this is not a claim of solution of angle trisection

• @bassam karzeddin If we restrict to straightedge and compass constructions then angle trisection is certainly impossible, that was proved by Wentzel in 1837, although you may find it interesting that $\sqrt{2}$ can be constructed as long is one is allowed to use marked straightedge en.wikipedia.org/wiki/Doubling_the_cube. But it was known since antiquity that specific angles can be trisected even with unmarked straightedge and compass, e.g. trisecting $2\pi$ is equivalent to inscribing equilateral triangle into a circle, $\pi/2$ can also be trisected, and so on. Jan 7 '16 at 5:34

This is a neat observation. I was unable to find historical references for it in particular, but there is a rich history of solving this type of problems. The problem is clearly reminiscent of finding the Pythagorean triples, right triangles with integer sides, for which Elements X.29 gives a solution without a hint of how it was obtained (of course once the rule is known it is easy to check that it works). A method first appears in Diophantus's Arithmetica (c. 250 AD) when solving the famous problem II.8, "partition a given square into two squares", the one whose margins were too narrow in Fermat's copy of the book. Schappacher's article (pp.12-13) gives a nice commentary.

In modern retelling if $p,q,r$ are the sides then for $x:=p/r$, $y:=q/r$ we have $x^2+y^2=1$ by the Pythagorean theorem. Write $y^2=1-x^2$ and make substitution $y=t(1-x)$ then equation for $x$ is linear and gives $x=(t^2-1)/(t^2+1)$, $y=2t/(t^2+1)$. If $t$ is rational so are $x$ and $y$, and setting $t=n/m$ gives the Pythagorean triples $n^2-m^2$, $2nm$, $n^2-m^2$. The trick can be interpreted geometrically as intersecting the circle with lines of slope $-t$ passing through a point $(1,0)$, and it works because each line has exactly one other intersection point with the circle. Diophantus describes nothing of the sort of course. Schappacher's remarks (p.27) suggest that Poincare might have come up with this interpretation in 1901, it is now called rational parametrization in algebraic geometry.

To extend this to self-trisecting triangles we first need a "Pythagorean theorem". Just like having a right angle makes triangle's angles into $\alpha,\pi/2-\alpha,\pi/2$ the trisection condition makes them into $\alpha,3\alpha,\pi-4\alpha$. In the first case writing equations of the theorem of sines and eliminating $\alpha$ from them leads to the Pythagorean equation $p^2+q^2=r^2$, in the second case after some fun with more esoteric trig identities one gets $q^2r=(p+r)(p-r)^2$. This is the "Pythagorean theorem" for self-trisecting triangles. With the same notation as above the corresponding curve is a cubic $$y^2=(x+1)(x-1)^2.$$ Normally cubics do not have rational parametrizations, but this one is special because of the squared factor, it means that it has a node (self-intersection point) at $(1,0)$. Drawing lines through the node, i.e. making the substitution $y=t(x-1)$, immediately gives $x=t^2-1$ and $y=t(t^2-2)$. Setting $t=n/m$ gives us $$p=m(n^2-m^2),\ q=n(n^2-2m^2),\ r=m^3,$$ as in the OP up to notation. So not only does the formula work, it actually lists all self-trisecting triples. The nodal cubic looks something like the folium of Descartes (rotated by $135^\circ$ clockwise, scaled, and shifted by $1$ to the right so that the $x$-intercepts move to $(-1,0)$ and $(1,0)$), but it does not have an asymptote.

• Very rich answer and historical content indeed, even my simple understanding to the problem & deriving it was completely unaware of all those valuable knowledge that Greek Euclidean had, however there are much more to add to this basic simple triangle, with multi section of the angle in general Jan 11 '16 at 12:45
• @bassam karzeddin You mean finding integer sided triangles with one angle n-secting another? I suspect that there are at most finitely many for generic n. Jan 12 '16 at 2:18
• @bassam karzeddin I take it back, one can get self-n-secting families of triangles for any n by essentially the same method. Which means that the corresponding curves are highly degenerate and non-generic (that's why I thought otherwise). @ Franz Lemmermeyer Schappacher does suggest that Poincare gave geometric interpretation to what Diophantus did, albeit vaguely. Jan 15 '16 at 0:55
• Given any n, one can just start from any α with cosα∈Q. For such α, a triangle exists with angles α, nα, 2π−(n+1)α and with rational sides. Put β:=nα and γ:=2π−α−β. Let (a,b,c) be the sides of any triangle with angles α,β,γ. Then both a/b = sin α/sin β and b/c = sin β/sin γ are rational: indeed, they are rational functions in sin(α) and cos(α) by standard addition formulae for trig functions, and since they are also even functions of α, they can be written in terms of cos α alone. I think this is the easiest construction, although it's not clear to me that it gives all such triangles.
– RP_
Feb 6 '16 at 19:08
• @René Yep, that's how it goes, it does give all the triangles by the "converse" law of sines. I realized later that the detour through the implicit equation is unnecessary, and trig formulas give polynomial parametrization directly as Chebychev polynomials of the second kind with 2cos α as parameter. Feb 8 '16 at 23:25

Its the same triangle as here: https://math.stackexchange.com/a/2035612/4414 Where I used $y=(x-1)*\sqrt{x+1}$. Which came up by a similar question posted by the same OP.

With the additional benefit of a rational parametrization. For $x=t^2−1$ with $t$ rational we get indeed $y=t*(t^2-2)$. And from this parametrization it should give dense rational points.

But I don't see any classical constructive method behind it, which shouldn't exist for ruler and compass. Problem is how do we get from a triangle with cos(γ) to a triangle with x?

• In sci.math there is much more freedom to convey your ideas about an issue than this place, where one need not to waste his time in providing answers here, especially if it is new discoveries that clearly contradicts the common knowledge by the professionals, if you check my old posts you would certainly find your answer provided here was also there, as a direct subsequent result. the deep theme in angle trisection is that it is impossible to trisect some angles that were (proven in my posts) as non existent angles, exactly as I proved the nonexistence of any number with endless terms. Jan 19 '17 at 18:16