Three positive real numbers $a,b,c$ satisfy the equations $a^2+\sqrt3 ab+b^2=25$, $b^2+c^2=9$ and $a^2+ac+c^2=16$ .Then find the value of $ab+ 2cb+\sqrt3 ac$?
Is there some way to find the desired value without actually finding values of $a,b,c$ or any other smart method to find $a,b,c$
The given system is $$a^2+\sqrt 3ab+b^2=25\\b^2+c^2=9\\a^2+ac+c^2=16$$ Then $$9+16=25\iff a^2+ac+c^2+ b^2+c^2= a^2+\sqrt 3ab+b^2\iff c(a+2c)=\sqrt 3 ab$$ It follows $$ab+2cb+\sqrt 3 ac=b(a+2c)+\sqrt 3ac=\frac{\sqrt 3\space a(b^2+c^2)}{c}=\frac {9\sqrt 3\space a}{c}=X$$ Hence one needs to calculate $\frac ac$. The successive following steps are clear to understand: $$a^2+\sqrt 3\space ab+(9-c^2)=25\iff b=\frac {16+c^2-a^2}{\sqrt 3\space a}=\frac{ac+2c^2}{\sqrt 3\space a}$$ It follows $$ \left(\frac{ac+2c^2}{\sqrt 3\space a}\right)^2+c^2=9$$ So the system $$4c^4+4ac^3+4a^2c^2=27a^2\\a^2+ac+c^2=16$$ Hence $$4c^2(16)=27a^2\Rightarrow \frac ac=\frac{8}{3\sqrt 3}$$ Consequently $$X=\frac{9\sqrt 3\cdot 8}{3\sqrt 3}=\color{red}{24}$$
Here is a geometrical solution:
by rewriting the equations as
$$\left\{ \begin{aligned} 5^2&=a^2+b^2-2ab\cos\frac{5\pi}{6}\\ 3^2&=b^2+c^2-2bc\cos\frac{\pi}{2}\\ 4^2&=c^2+a^2-2ca\cos\frac{2\pi}{3} \end{aligned} \right. $$
and evaluating
$$4\left(\frac{1}{2}ab\sin\frac{5\pi}{6}+\frac{1}{2}bc\sin\frac{\pi}{2}+\frac{1}{2}ca\sin\frac{2\pi}{3}\right)=4S_{\Delta ABC}=24.$$
Sketch of a existence proof: Since $3^2+4^2=5^2$, $\Delta ABC$ exists. Consider $\overset{\mmlToken{mo}{⏜}}{HC}$, i.e., the segment that lies between $AB$ and $CA$ of a circle, for which $BC$ is a diameter. For any $P$ on the arc, $\angle BPC=\pi/2$ while $\angle CPA$ increases continuously from $\pi/2$ to $\pi$ (from $H$ to $C$), so there exists $P$ so that $\angle CPA=2\pi/3$.
In general, if such configuration does exist, the point where three segments meet is the intersection of three arcs. Each arc results the given opening angle with respect to each side of the triangle and they sum up to $2\pi$.
What a prowess, @Piquito ! Here is another way of treating this issue.
Its "pros": it is systematic. Its "cons": it necessitates a computer algebra system (I used Mathematica).
Let us give the following numbers to the 3 equations:
$$\begin{cases}a^2+\sqrt 3ab+b^2&=&25 \ \ &(1)\\b^2+c^2&=&9 \ \ \ \ \ &(2) \\a^2+ac+c^2&=&16 \ \ \ &(3)\end{cases}$$
Let us consider equations (1) and (3) as quadratics in $a$.
They must have a common root. This can be expressed by setting their resultant to $0$ (https://en.wikipedia.org/wiki/Resultant):
This gives:
$$81 - 66b^2 + b^4 + 41\sqrt{3}bc - \sqrt{3}b^3c - 7c^2 + 2b^2c^2 - \sqrt{3}bc^3 + c^4=0 \ \ \ (4)$$
Now, we use constraint (2) meaning that point $(b/3,c/3)$ is on the unit circle, a constraint that we can translate into the following one (classical parameterization of the unit circle https://en.wikipedia.org/wiki/Tangent_half-angle_formula):
$$b = 3\dfrac{1 - t^2}{1 + t^2}, \ \ c = 3\dfrac{2 t}{1 + t^2} \ \ \ (5)$$ for a certain $t \in (-\infty,+\infty)$. Plugging (5) into (4) gives:
$$12 - 16\sqrt{3}t - 35t^2 + 16\sqrt{3}t^3 + 12t^4=0 \ \ \ (6)$$
This antipalindromic (https://en.wikipedia.org/wiki/Reciprocal_polynomial) 4th degree polynomial has four explicit (real) solutions
$$t=\dfrac{1}{12}(\pm9-4\sqrt{3}\pm\sqrt{273-72 \sqrt{3}}) \ \ \ (7)$$
(the two $\pm$ signs are independant: their four combinations are valid).
The objective of finding the value of
$$X = ab+ 2cb+\sqrt3 ac$$
is now within reach, because we are able to express it as an expression of $t$ alone, because:
this is the case for $b$ and $c$ (formulas (5)).
Concerning $a$, being the solution of quadratic equation (3), can be expressed as a function of $c$, itself function of $t$.
Then, it remains to check that, whatever the root chosen in (7), one gets the same result $X=24$. This is the case.
