Ask Uncle Colin: A Rational Triangle (of sorts)

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 $A=\frac{\pi }{2}(1-\cos (A))$.

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.