14th Turkish Mathematical Olympiad

By the law of sines and the law of cosines, the sines and cosines of the angles of such a triangle are also rational. Therefore it suffices to prove that if $\theta$ is an angle whose degree measure is a rational number that is not an integer multiple of $90$, then both $\sin \theta$ and $\cos \theta$ cannot be rational.

Assume otherwise. Let $u,v$ and $w>1$ be pairwise relatively prime integers such that $\sin \theta =u/w$, $\cos \theta =v/w$. $w$ cannot be even as $u^2+v^2\equiv w^2\equiv 0(\mathrm{mod}4)$ (mod $4$) is not possible.

We will prove by induction that there exist integers $k_n$ such that $w^n\cos n\theta =2^{n-1}{v}^n+k_n{w}^2$ for $n\ge 1$. $k_1=0$ works for $n=1$. If $n=2$, then $w^2\cos 2\theta =w^2(2{\cos }^2\theta -1)=2v^2-w^2$, and $k_2=-1$ will do. For $n>1$, $$\begin{array}{rl}{w}^n\cos n\theta & =w^n(2\cos \theta \cos (n-1)\theta -\cos (n-2)\theta )\\ & =2v(2^{n-2}{v}^{n-1}+k_{n-1}{w}^2)-w^2(2^{n-3}{v}^{n-2}+k_{n-2}{w}^2)\\ & =2^{n-1}{v}^n+(2v{k}_{n-1}-(2^{n-3}{v}^{n-2}+k_{n-2}{w}^2))w^2,\end{array}$$ and we can take $k_n=2v{k}_{n-1}-(2^{n-3}{v}^{n-2}+k_{n-2}{w}^2)$.

Now choose a positive integer $n$ such that $\cos n\theta =1$. Then $w^n=2^{n-1}{v}^n+k_n{w}^2$. But, as $(w,2v)=1$ and $w>1$, this is a contradiction.