<s>[INST] For a triangle $ABC$, $a^2+b^2+c^2=8R^2$ then it is a right triangle? $ABC$ is a triangle, $a^2+b^2+c^2=8R^2$ then how do we prove it is a right triangle?
 [/INST] Assuming that $R$ means circumradius we can use the formula
$$R=\sqrt{\frac{a^2+b^2+c^2}{8(1+\cos A\cos B\cos C)}},$$
where $a$, $b$, and $c$ are the sidelengths and $A$, $B$, and $C$ are the angles. Plugging in $a^2+b^2+c^2=8R^2$, we reduce the above to 
$$1+\cos A\cos B\cos C=1,$$
whence one of $\cos A$, $\cos B$, or $\cos C$ must be $0$ - meaning that we have a right triangle. 
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