Ask Uncle Colin: A Rational Triangle (Of Sorts)

Dear Uncle Colin,

Is it possible to have a triangle where the ratio of side lengths $a:b:c$ is equal to the ratio of angles $A:B:C$? (Ignore the equilateral triangle, it’s trivial).

- Right-Angled Triangles Included, Obviously

Hi, RATIO, and thanks for your message!

First things first: it certainly can’t be done with a right-angled triangle, but explaining why is useful for setting up the method.

In any right-angled triangle including an angle of $A$, the side are in the ratio $\sin(A):\cos(A):1$ and the angles are in ratio $A: \frac{\pi}{2} - A : \frac{\pi}{2}$. That makes the ratio constant explicit and we can write down the equations:

  • $A = \frac{\pi}{2}\sin(A)$
  • $\frac{\pi}{2}-A = \frac{\pi}{2}\cos(A)$

… and the second becomes $A = \frac{\pi}{2}\left( 1 - \cos(A) \right)$.

Eliminating the $A$ and dividing by $\frac{\pi}{2}$ gives $\sin(A) = (1-\cos(A))$, and the only plausible angles for which that’s true are $A=0$ and $A=\frac{\pi}{2}$, both of which give us only a degenerate triangle.

But what if we remove the right-angle restriction?

Well, we get into the long grass fairly quickly. Let’s assume a triangle exists, with angles $A$, $B$ and $\pi - A -B$ and sides $a$, $b$ and $c$ opposite those in order, such that $A:B:\pi-A-B = a:b:c$.

We can break out the sine rule and state that $a:b:c = \sin(A):\sin(B):\sin(\pi-A-B)$, which means we can now completely ignore the side lengths. If we also notice that $\sin(\pi - A - B) \equiv \sin(A+B)$, we get $A:B:\pi - A-B = \sin(A):\sin(B):\sin(A+B)$. Can we make that work?

We could (by which I mean, I did) try to solve that. It’s a mess. Instead, let’s try special cases – the first that springs to mind is, what if $A=B$?

If $A=B$, then we have $A: \pi-2A = \sin(A) : \sin(2A)$, which is already a much simpler thing to solve. We can write it as $\frac{\pi-2A}{A} = \frac{\sin(2A)}{A}$, or even $\frac{\pi}{A} - 2 = 2\cos(A)$.

That rearranges to $A = \frac{\pi}{2+2\cos(A)}$, which has solutions… but unfortunately, only when $A = \frac{\pi}{3}$ – the equilateral triangle – or $A = \frac{\pi}{2}$ – which is degenerate. Curses!

What if $A = 2B$? Then we’ve got $A: 2A: \pi - 3A = \sin(A): \sin(2A):\sin(3A)$, which looks quite sleek. Sadly, that’s not going to work, either. If $\frac{A}{\sin(A)} = \frac{2A}{\sin(2A)}$, then $\frac{A}{\sin{A}} = \frac{A}{\sin{A}\cos{A}}$ and we have $\cos(A)=1$ again – and a degenerate triangle.

A smart thing to do would be to investigate the expression $\frac{x}{\sin(x)}$. That has a derivative of $\frac{\sin(x) - x\cos(x)}{\sin^2(x)}$, which is positive for $0 < x < \pi$ – it’s only zero when $x = \tan(x)$.

In particular, it’s an increasing function, and therefore one-to-one – which means that $\frac{A}{\sin(A)} = \frac{B}{\sin(B)}$ implies $A=B$. As a result, the only triangles that can possibly work have all three angles the same.

Neat question! I hope that helps.

- Uncle Colin

  • Thanks to Andrew Buhr for letting me know about a linking problem in this post.