# 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 ?

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Yesterday, itself i told you that whenever you see this type of things please apply the sine rule. – anonymous Nov 17 '10 at 7:53

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.

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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 $A=B=C$.

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$$\sin^2A+\sin^2B+\sin^2C$$ $$=1-(\cos^2A-\sin^2B)+1-\cos^2C$$ $$=2-\cos(A+B)\cos(A-B)-\cos C\cdot\cos C$$ $$=2-\cos(\pi-C)\cos(A-B)-\cos\{\pi-(A+B)\}\cdot\cos C$$ $$=2+\cos C\cos(A-B)+\cos(A+B)\cdot\cos C\text{ as }\cos(\pi-x)=-\cos x$$ $$=2+\cos C\{\cos(A-B)+\cos(A+B)\}$$ $$=2+2\cos A\cos B\cos C$$

$(1)$ 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 $> \frac\pi2,$

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