Note: I used degrees in the title for the sake of brevity, but am using radians in the body of the question for clarity. Sorry for any confusion this causes.

Here's the question: are there any right triangles $\triangle ABC$ such that $C=\frac{\pi}{2}$, all side lengths $(a, b, c)$ are rational, and $A/\pi, B/\pi \in \mathbb{Q}$?

The first thing I noticed is that this is an equivalent problem to finding some $\theta$ such that $$\sin(\theta), cos(\theta), \frac{\theta}{\pi}\in \mathbb{Q}\setminus\{0\}$$

Intuitively I'd expect this to have no solutions, I have no specific reason for this, just that my gut tells me this is impossible.

My only idea for how I could look into this problem would be to show that $\Re (e^{i\theta})$ and $\Im(e^{i\theta})$would give an irrational value for $\frac{\theta}{\pi}\in \mathbb{Q}\setminus\{0\}$ and somehow algabraicly manipulating this into a form that can be proven transcendental (which would imply that it's also irrational) by the Gelfond-Schneider theorem, however I haven't come up with anything yet.

Keep in mind that this is just a little puzzle that I've been throwing around in my head, so for all I know it might not even have an answer yet. The most advanced math that I can say I understand confidentally is multivariable calculus and I kind of know the basics undergraduate level courses such as abstract algebra, real analysis, and complex analysis. From what I understand problems to do with irrationality tend to be unexpectedly difficult, so if it does require more advanced math than that I'd be fine with a more high level description of the proof.

  • $\begingroup$ Are $A,B$ angles in radians? Should they be in $\Bbb Z$? $\endgroup$ Sep 23, 2020 at 14:26
  • 4
    $\begingroup$ See Niven's Theorem $\endgroup$
    – lulu
    Sep 23, 2020 at 14:28
  • $\begingroup$ @DietrichBurde sorry for the confusion, the title was in degrees, the body of the question is using radians. I edited my question to make this clear. $\endgroup$
    – Snacc
    Sep 23, 2020 at 14:36
  • 1
    $\begingroup$ I've done some further tweaking since '$A$ rational' and '$A$ a rational multiple of $\pi$' really aren't quite the same thing. $\endgroup$ Sep 23, 2020 at 16:14

1 Answer 1


Suppose you had such a triangle with legs $a$ and $b$ and hypotenuse $c$, which (by scaling) we may assume are relatively prime integers. Then the argument of the complex number $a+bi$ would be one of the angles of the triangle, and thus would be a rational multiple of $\pi$. This means that there is some nonzero integer $n$ such that $(a+bi)^n$ is real, and thus equal to $(a-bi)^n$.

Now observe that $$c^2=(a+bi)(a-bi)$$ so $$c^{2n}=(a+bi)^n(a-bi)^n=(a+bi)^{2n}.$$ Considering the unique factorization of both sides over the Gaussian integers, this implies that $c$ and $a+bi$ are associate in the Gaussian integers, and in particular $c$ divides $a+bi$. But since $a,b,$ and $c$ are relatively prime, this is only possible if $c=1$. In that case $a$ or $b$ must be $0$, and so our "triangle" is degenerate.

(Or, in more abstract language: $\frac{a+bi}{c}$ is a root of unity and in particular an algebraic integer, so since $\mathbb{Z}[i]$ is integrally closed, it must be in $\mathbb{Z}[i]$.)

Here's an alternate argument which uses the irreducibility of cyclotomic polynomials instead of factorization of Gaussian integers. As above, given such a triangle, $\zeta=\frac{a+bi}{c}$ will be a root of unity; say it is a primitive $m$th root of unity. Then $\mathbb{Q}(\zeta)\subseteq\mathbb{Q}(i)$ and in particular $\mathbb{Q}(\zeta)$ has degree at most $2$ over $\mathbb{Q}$. This means the minimal polynomial of $\zeta$, i.e. the cyclotomic polynomial $\Phi_m$, has degree at most $2$. But $\deg\Phi_m=\varphi(m)$, and $\varphi(m)\leq 2$ is true only for $m=1,2,3,4,6$. You can then check that none of these values of $m$ actually give a nondegenerate triangle with rational sides.


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