Evaluate ab+bc+ca Given Real Numbers

[anemone]

Given real numbers $a,\,b,\,c$ such that

$a^2+ab+b^2=2$, $b^2+bc+c^2=3$, $c^2+ca+a^2=5$

Evaluate $ab+bc+ca$.

 

[Saitama]

Given real numbers $a,\,b,\,c$ such that

$a^2+ab+b^2=2$, $b^2+bc+c^2=3$, $c^2+ca+a^2=5$

Evaluate $ab+bc+ca$.

Spoiler

Rewrite the given equations as follows:
$$\frac{a^2+b^2-(\sqrt{2})^2}{2ab}=\frac{-1}{2}$$
$$\frac{b^2+c^2-(\sqrt{3})^2}{2bc}=\frac{-1}{2}$$
$$\frac{a^2+c^2-(\sqrt{5})^2}{2ca}=\frac{-1}{2}$$

The above three equations geometrically represent the following triangle:

(As it can be clearly seen, the bigger triangle is a right-angled triangle.)

The areas of three smaller triangles add to give the area of bigger triangle. Hence,
$$\frac{1}{2}ab\sin(120^{\circ})+\frac{1}{2}bc\sin(120^{\circ})+\frac{1}{2}ca\sin(120^{\circ})=\frac{1}{2}\sqrt{2}\sqrt{3}$$
$$\Rightarrow ab+bc+ca=\frac{\sqrt{6}}{\sin(120^{\circ})}=\boxed{2\sqrt{2}}$$
$\blacksquare$

 

[anemone]

Thanks for participating, Pranav! Your answer is correct, well done!:)