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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.

Jean Marie
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Ilovemath
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    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? – Calvin Lin Sep 04 '21 at 06:33
  • Another hint: The result to be reached is $24\sqrt{3}$ (found using a Computer Algebra System) – Jean Marie Sep 04 '21 at 09:31
  • @Calvin Lin I did try what you suggested but I couldn't manage. – Ilovemath Sep 04 '21 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? – Math Lover Sep 04 '21 at 11:21
  • @Math Lover I am not sure this can be done in the plane. I think that it has to be developed in 3D space with OABC a tetrahedron. – Jean Marie Sep 04 '21 at 11:44
  • @JeanMarie Angles add to $360^0$. I just added an answer. – Math Lover Sep 04 '21 at 12:16
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    @Math Lover: You are right ! [+1]! – Jean Marie Sep 04 '21 at 12:22

2 Answers2

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Please note that

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

Math Lover
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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

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

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

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

giuseppe mancò
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