We want to find angles $x^\circ$, $y^\circ$ and $z^\circ$ so that $$x^2+y^2+z^2=180\;.$$

We know that the largest angle must be smaller than $180^\circ$ and bigger than $180^\circ/3 = 60^\circ$. So the largest angle must be $169^\circ, 144^\circ, 121^\circ, 100^\circ, 81^\circ$ or $64^\circ$.

It can't be $169^\circ$ or $144^\circ$, because then the other two angles would have to add up to $11^\circ$ or $36^\circ$, respectively, which no two non-zero squares do. (Note that we want our angles to be non-zero in order to get a legitimate triangle.)

Moreover, the largest angle also can't be $121^\circ$, because then the other two angles would have to add up to $59^\circ$ and you can check that no two squares do.

If the largest angle is $100^\circ$ then the other two angles add up to $80^\circ$, so the second largest angle must be either $64^\circ$ or $49^\circ$. The latter gives $31^\circ$ as the smallest angle, which is not a square, whereas the foremost yields $16^\circ$, which is a square. So we found $$ 10^2+8^2+4^2 = 180\;.$$

If the largest angle is $81^\circ$ then the other two squares must add up to $99 ^\circ$, so the next largest angle must be either $81^\circ$ or $64^\circ$, giving the smallest angle as $18^\circ$ or $35^\circ$, neither of which are square numbers.

If the largest angle is $64^\circ$, then the other two square must add up to $116 ^\circ$, so the second largest angle must also be $64^\circ$, giving the smallest angle as $52^\circ$, which is not a square number.

So there is exactly one triangle with all three angles perfect squares, viz a triangle with angles $$\left(10^2,8^2,6^2\right)\;.$$

*This problem is taken from the UKMT Mathematical Challenges.*

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