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?
 A: 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.
A: 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. 
A: Let $S$ be the area of the triangle.
Since we have $4RS=abc$ and
$$\begin{align}16S^2&=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\\&=-(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2)\end{align}$$
if $a^2+b^2+c^2=8R^2$, then we have
$$\begin{align} 0 &=\frac{a^2+b^2+c^2}{R^2}-8\\&=\frac{a^2+b^2+c^2}{a^2b^2c^2}(-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2a^2)-8\\&=\frac{(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)}{a^2b^2c^2}.\end{align}$$
$$\Rightarrow (-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)=0$$
So, the triangle has to be a right triangle.
P.S. Actually, $a^2+b^2+c^2=8R^2\iff\text{The triangle $ABC$ is a right triangle}$.
Also, we can see the followings :
$$a^2+b^2+c^2\gt 8R^2\iff \text{The triangle $ABC$ is an acute‐angled triangle}.$$
$$a^2+b^2+c^2\lt 8R^2\iff \text{The triangle $ABC$ is an obtuse triangle}.$$
