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 $a,b,c$

  • $\begingroup$ If you don't know the values of $a, b, c$, how do you expect to know the value of that expression? $\endgroup$ – ÍgjøgnumMeg Apr 22 '16 at 22:07
  • 2
    $\begingroup$ @Ed_4434 By some form of algebraic manipulation of the three equations. Whether that is possible here is the question.. $\endgroup$ – MathematicianByMistake Apr 22 '16 at 22:15
  • $\begingroup$ "Without actually finding values of $a, b, c$" $\endgroup$ – ÍgjøgnumMeg Apr 22 '16 at 22:21
  • $\begingroup$ I don't know what surprises you, @Ed_4434, theory of elimination has its roots in such calculations. The issue presented here could be expressed in term of ideals, Gröbner basis, and so on... but that's not this kind of abstract mathematics that H.P. Das desires... $\endgroup$ – Jean Marie Apr 22 '16 at 22:28
  • $\begingroup$ Can you say us what is the origin of your (interesting) problem ? $\endgroup$ – Jean Marie Apr 24 '16 at 14:58

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 $2\pi$.

  • $\begingroup$ Well, this configuration does exist and it is easy to prove the existence. Maybe you should mention that. $\endgroup$ – Batominovski Jun 27 '16 at 23:03
  • $\begingroup$ Splendid solution ! $\endgroup$ – Jean Marie Jun 28 '16 at 4:28

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.

  • $\begingroup$ Thanks for the compliment, Jean Marie. Before my answer I have also used roots of quadratic equation but could not go further. I think, as on many occasions for math problems, what was needed was rather "open well our eyes". Best regards. $\endgroup$ – Piquito Apr 23 '16 at 11:31

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