<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] We use the well-known formula $R = \frac{abc}{4A}$, where $A$ is the area of the triangle.
Thus $8R^2 = \frac{8a^2b^2c^2}{16A^2}$, and by Heron's formula
$$16A^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c).$$
Now assume that $a^2 + b^2 + c^2 = 8R^2$. Then we have
$$8a^2b^2c^2 - (a^2 + b^2 + c^2)(a+b+c)(-a+b+c)(a-b+c)(a+b-c) = 0.$$
Expanding this gives
$$a^6 + b^6 + c^6 - a^2b^4 - b^2c^4 - c^2a^4 - a^4b^2 - b^4c^2 - c^4a^2 + 2 a^2 b^2 c^2 = 0,$$
which can be factored as
$$(a^2 + b^2 - c^2)(b^2 + c^2 - a^2)(c^2 + a^2 - b^2) = 0.$$
Thus the claim follows from Pythagorean theorem.
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