Find the value of $ab+ 2cb+\sqrt3 ac$?

Solution 1:

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 $$X=\frac{9\sqrt 3\cdot 8}{3\sqrt 3}=\color{red}{24}$$

Solution 2:

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 $2\pi$.