# A triangle's sides are integers, and its circumradius is a prime number. Prove that the triangle is right-angled.

The lengths of the sides of a triangle are integers, whereas the radius of its circumscribed circle is a prime number. Prove that the triangle is right-angled.

Solution: We'll use three well-known formulas for the area of $$\triangle ABC$$: $$[ABC]=\frac{abc}{4R}=\frac{ab\sin C}{2}=\sqrt{s(s-a)(s-b)(s-c)}.$$First of all, because $$a,b,c,R$$ are all integers, we know that $$[ABC]$$ is rational. But then $$4[ABC]=\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}\in\mathbb{Z}$$(since $$2s(2s-2a)(2s-2b)(2s-2c)$$ is an integer, its square root, if it is rational, must be an integer) so $$[ABC]=n/4$$ for some $$n\in\mathbb{Z}.$$ Then $$n=abc/R$$ is an integer, so $$R$$ divides one of $$a,b,c$$; WLOG $$a=kR$$ for some $$k\in\mathbb{Z}.$$ But now $$\frac{abc}{4R}=\frac{bc\sin A}{2}\implies\sin A=\frac{a}{2R}=\frac{k}{2}$$and since $$\sin A\in[0,1]$$ we have $$k\in\{1,2\}.$$ If $$k=1$$ then we have from Law of Cosines $$a^2=R^2=b^2+c^2-2bc\cos A=b^2+c^2-bc\sqrt{3},$$which is absurd because $$R,b,c$$ are all integers. So $$k=2\implies\sin A=1\implies A=90^{\circ}$$ and we are done.

Is there a cooler solution than this? I found it slightly ridiculous