In a triangle ABC, $\begin {vmatrix} a&b&c \\b&c&a \\c&a&b \end {vmatrix}=0$, then find $\sin A\sin B+\sin B\sin C+\sin C\sin A$

Question

From the given determinant, $$a^3+b^3+c^3-3abc=0$$

Which implies $a+b+c=0$ or $a^2+b^2+c^2 -ab-bc-ac=0$

Since the former isn’t possible, $$a^2+b^2+c^2=ab+bc+ac$$ $$\sin A\sin B+\sin B\sin C +\sin C\sin A=\sin^2A +\sin^2B+\sin ^2C$$

The answer for this is either $\cos^2A+\cos^2 B+\cos^2C$ or one of numbers $0, 1, \frac 94$

How do I proceed from here?

Answer 1

The sides $a,b,c$ of the triangle satisfy $$ a^2+b^2+c^2=ab+bc+ac \\ \begin{matrix} \implies \dfrac{1}{2}(2a^2+2b^2+2c^2-2ab-2bc-2ac) & =0 \\ (a-b)^2+(b-c)^2+(c-a)^2 &=0 \end{matrix}$$ This straightaway means that $a=b=c$ (i.e. and equilateral triangle) otherwise the inequality cannot hold.

Therefore, $\sin A\sin B+\sin B\sin C+\sin C\sin A=\dfrac{3}{4}*3=\dfrac{9}{4}$

Answer 2

The cubic need not be factorised, as side lengths are positive. By AM-GM the triangle is equilateral, leading to @AnindyaPrithvi's calculation of the desired trigonometry.