What type of triangle satisfies: $8R^2 = a^2 + b^2 + c^2 $? In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which type ?
 A: This essentially means $\sin^2(A)+\sin^2(B)+\sin^2(C)=2$. (This follows from $\sin$ rule.)
Replace $C = \pi - (A+B)$ to get $\sin^2(A+B) = \cos^2(A) + \cos^2(B)$. 
Expand $\sin(A+B)$ and do the manipulations to get 
$2\cos^2(A)\cos^2(B) = 2\sin(A)\sin(B)\cos(A)\cos(B)$ 
which means $\cos(A) = 0$ or $\cos(B) = 0$ or $\cos(A)\cos(B) = \sin(A)\sin(B) \Rightarrow \cos(A+B) = 0 \Rightarrow \cos(C) = 0$. 
Hence either $A = \pi/2$ or $B = \pi/2$ or $C = \pi/2$. 
So the triangle is a right-angled triangle.
A: In general, in $\triangle{ABC}$
If $\sin^2 A + \sin^2 B + \sin^2 C \gt 2$ then $\triangle{ABC}$ is acute angled.
If $\sin^2 A + \sin^2 B + \sin^2 C  = 2$ then $\triangle{ABC}$ is right angled.
If $\sin^2 A + \sin^2 B + \sin^2 C \lt 2$ then $\triangle{ABC}$ is obtuse angled.
Assume $A \le B \lt \pi/2$ and $ A \le B \le C$.
Basically, if $k = \sin^2 A + \sin^2 B + \sin^2 C$
then we have that
$3-2k = \cos 2A + \cos 2B + \cos (2A+2B)$
i.e
$3-2k = 2\cos(A+B)\cos(A-B) + 2\cos^2(A+B) -1 $
i.e
$4-2k = 4\cos(A+B)\cos A\cos B$
So if $k > 2$, then $\cos(A+B) \lt 0$ hence acute.
$k = 2$, then $\cos(A+B) = 0$ hence right triangled.
$k < 2$, then $\cos(A+B) \gt 0$, hence obtuse.
In fact, we can go further and show that the maximum possible value of $k$ is $k = \frac{9}{4}$ which corresponds to $\triangle{ABC}$ being equilateral, as follows:
$4\sin^2 A + 4\sin^2 B + 4\sin^2 C = 9  + \delta$
i.e.
$(2 - 2\cos2A) + (2-2\cos 2B) + 4(1- \cos^2 (A+B)) = 9 + \delta$
i.e.
$1 + 2\cos2A + 2\cos 2B + 4\cos^2(A+B) = -\delta$
i.e.
$1 + 4\cos(A+B)\cos(A-B) + 4\cos^2(A+B) = -\delta$
i.e.
$\sin^2(A-B) + \cos^2(A-B) + 4\cos(A+B)\cos(A-B) + 4\cos^2(A+B) = -\delta$
i.e.
$\sin^2(A-B) + (\cos (A-B) + 2\cos(A+B))^2 = -\delta$.
Hence $\delta \le 0$ and so $\sin^2 A + \sin^2 B + \sin^2 C \le \frac{9}{4}$
The case $\delta = 0$ gives us $\sin(A-B) = 0$ and $\cos(A+B) = \frac{-1}{2}$.
Hence $A=B=C$.
Thus the max value of $\sin^2 A + \sin^2 B + \sin^2 C$ is $\frac{9}{4}$ and is achieved when $A=B=C$.
A: $$\sin^2A+\sin^2B+\sin^2C$$
$$=1-(\cos^2A-\sin^2B)+1-\cos^2C$$
$$=2-\cos(A+B)\cos(A-B)-\cos C\cdot\cos C$$
$$=2-\cos(\pi-C)\cos(A-B)-\cos\{\pi-(A+B)\}\cdot\cos C$$
$$=2+\cos C\cos(A-B)+\cos(A+B)\cdot\cos C\text{ as }\cos(\pi-x)=-\cos x$$
$$=2+\cos C\{\cos(A-B)+\cos(A+B)\}$$
$$=2+2\cos A\cos B\cos C$$
$(1)$ If $2+2\cos A\cos B\cos C=2, \cos A\cos B\cos C=0$
$\implies $ at least one of $\cos A,\cos B,\cos C$ is $0$ which needs the respective angles $=\frac\pi2$
But we can have at most one angle $\ge \frac\pi2$ 
So, here we shall have exactly one angle $=\frac\pi2$
$(2)$ If $2+2\cos A\cos B\cos C>2, \cos A\cos B\cos C>0$
Either all of  $\cos A,\cos B,\cos C$ must be $>0\implies$ all the angles are acute
or exactly two cosine ratios $<0$ which needs the respective angles $> \frac\pi2,$ which is impossible for a triangle
$(3)$ If $2+2\cos A\cos B\cos C<2, \cos A\cos B\cos C<0$
Either all the ratios  $<0$ which needs the respective angles $> \frac\pi2,$ which is impossible fro a triangle
or exactly one of the cosine ratios is $<0\implies $  the respective angle $> \frac\pi2,$ 
