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



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