Is there a pythagorean triple such that all angles of the corresponding triangle are simple fractions of $\pi$?

Solution 1:

This is a consequence of the fact that the unit group of $\Bbb Z[i]$ is $\{\pm1,\pm i\}$.

For, take a Pythagorean triple: $a^2+b^2=c^2$, with the angle $\beta$ opposite the side of length $b$. You are asking whether it is possible for $n\beta\equiv0\pmod{2\pi}$.

Now $z=\frac a c+\frac b ci$ is an element of $\Bbb Q(i)$ on the unit circle, and its argument is $\arctan(b/a)=\beta$. The argument of $z^n$ is $n\beta\pmod{2\pi}$.

But if $n\beta\equiv0\pmod{2\pi}$, then $z^n=1$, and so $z$ is a root of unity, of which there are only the powers of $i$ in $\Bbb Q(i)$.

Solution 2:

Your three angles are rational multiples of $\pi$. If your three sides are also rational $($since this is what the expression Pythagorean triple implies$)$, then this would mean that the sine and cosine of rational multiples of $\pi$ are also rational. But, according to Niven's theorem, this can only happen for the $30^\circ-60^\circ-90^\circ$ triangle.

Solution 3:

Let $a,b,c$ the sides of a pythagorean triangle with correlative angles $\alpha$ and $\beta$ ($\gamma$ is always $\frac{\pi}{2}$). It is known, by trascendental number theory, that $\sin\space x$ and $\cos\space x$ are transcendental when $x\ne 0$ is algebraic. Consequently $\alpha$ and $\beta$ are necessarily transcendental because $\sin\space \alpha=\frac{a}{c}$ and $\sin\space \beta=\frac{b}{c}$.

Besides $\alpha + \beta= \frac{\pi}{2}$ but this do not mean necessarily that $\alpha$ and $\beta$ are rational fractions of $\pi$.

Your question is interesting. I think the most probably, with exception of $(\alpha, \beta)$=$(\frac {\pi}{3},\frac{\pi}{6})$ is that$(\alpha, \beta)$= $(\frac{\pi}{2}-h,\frac {\pi}{2}+h)$ for some "disturbed" $h$.

I'll try to get a better answer.

NOTE.-The triangle with angles 90, 60 and 30 degrees is not Pythagorean. There are consequently no exception for pythagorean triangles. And there is just one exception for rectangle triangles according to the Niven's Theorem cited in other answer here by @Lucian.

Solution 4:

Let $A$ be the angle which is $\frac{\pi}{2}$ and assume that another angle $B=\frac{m}{n}\pi$.

Then $2 \cos(B) =e^{\frac{m}{n}\pi i}+e^{-\frac{m}{n}\pi i}$ is an algebraic integer, as the sum of two algebraic integer. This implies that $2 \cos(B)$ is an integer, and since we are in the first quadrant we have $$2 \cos(B) \in \{ 0 ,1, 2\}$$

It is easy to check all three cases.