# Finding three unknowns from three equations

Let $$a$$,$$b$$ and $$c$$ be three positive real numbers such that

$$\begin{cases}3a^2+3ab+b^2&=&75\\ b^2+3c^2&=&27\\c^2+ca+a^2&=&16\end{cases}$$

Find the value of $$ab+2bc+3ca$$.

My attempt: I observed that $$3 . 16+27=75$$. Then on replacing $$16$$ by $$c^2+ca+a^2$$, $$27$$ by $$b^2+3c^2$$ and $$75$$ by $$3a^2+3ab+b^2$$, I got $$2c^2+ca=ab$$.

But after this I am unable to proceed. Is there a way to proceed from here?

Any constructive hint is appreciated.

• Hint/suggestion: Let $O$ be the origin, and consider 3 points that are length $\sqrt{3}a, b, \sqrt{3}c$ away from it. Apply Cosine rule. What happens? Sep 4, 2021 at 6:33
• Another hint: The result to be reached is $24\sqrt{3}$ (found using a Computer Algebra System) Sep 4, 2021 at 9:31
• @Calvin Lin I did try what you suggested but I couldn't manage. Sep 4, 2021 at 10:15
• @Ilovemath the point is that if you take a $\triangle PQR$ with $PQ = \sqrt{75}, QR = \sqrt {27}, RP = \sqrt{48}$ and there is a point $O$ inside the triangle such that, $OP = \sqrt3 a, OQ = b, OR = \sqrt3 c$. What angles do they make? Sep 4, 2021 at 11:21
• @Math Lover: You are right ! [+1]! Sep 4, 2021 at 12:22

$$(\sqrt3 a)^2 + b^2 - 2 (\sqrt3 a) b \cos 150^0 = 3a^2 + b^2 + 3ab = 75$$

$$b^2 + (\sqrt3 c)^2 - 2 b (\sqrt3 c) \cos 90^0 = b^2 + 3c^2 = 27$$

$$(\sqrt3 a)^2 + (\sqrt3 c)^2 - 2 (\sqrt3 a) (\sqrt3 c) \cos 120^0 = 3a^2 + 3 c^2 + 3 a c = 48$$

Angles add to $$360^0$$ so there is a point $$O$$ inside $$\triangle PQR$$ with $$OP = \sqrt3 a, OQ = b, OR = \sqrt3 c$$ and $$PQ = \sqrt{75}, QR = \sqrt{27}$$ and $$PR = \sqrt{48}$$

Next observe that $$PQ^2 = QR^2 + PR^2$$ which means $$\triangle PQR$$ is a right triangle.

$$\displaystyle S_{\triangle PQR} = \frac{1}{2} \cdot \sqrt{27} \cdot \sqrt{48} = 18$$

But $$\displaystyle S_{\triangle PQR} = S_{\triangle POR} + S_{\triangle QOR} + S_{\triangle POQ}$$

As we know area of a triangle is $$\frac{1}{2} a b \sin \theta$$ where $$a, b$$ are two sides with angle between them being $$\theta$$.

Adding individual areas we get to,

$$\frac{\sqrt3}{4} (ab + 2bc + 3 ac) = 18$$

So, $$ab + 2bc+3ac = 24 \sqrt3$$

There are two values of the expression.

The value of

$$ab+2bc+3ca=-24\sqrt{3}$$

and

$$ab+2bc+3ca=+24\sqrt{3}$$.

The respective values of $$a$$, $$b$$ and $$c$$ are:

$$a=+\sqrt{\frac{24576\sqrt{3}}{6553}+\frac{93184}{6553}}$$,

$$b=+\sqrt{\frac{58995}{6553}-\frac{31104\sqrt{3}}{6553}}$$,

$$c=-\sqrt{\frac{10368\sqrt{3}}{6553}+\frac{39312}{6553}}$$;

and

$$a=-\sqrt{\frac{93184}{6553}-\frac{24576\sqrt{3}}{6553}}$$,

$$b=-\sqrt{\frac{58995}{6553}+\frac{31104\sqrt{3}}{6553}}$$,

$$c=-\sqrt{\frac{-10368\sqrt{3}}{6553}+\frac{39312}{6553}}$$.