Some time ago I was given the following system of equations:

$$a^2 + b^2 + ab = 25$$

$$b^2 + c^2 + bc = 49$$

$$c^2 + a^2 + ac = 64$$

with $a,b,c \gt 0$ and tasked with finding $(a+b+c)^2$.

Clearly, the expressions here show some symmetry, and they very well resemble the law of cosines. Hence I used geometry to construct a triangle that would, in some way, follow these equations. Consider a triangle $PQR$, where we take a point $Z$ in the interior of the triangle such that the angle subtended by each side at that point $Z$ is $120^\circ$.

This is better understood through the following diagram:


Now, it is clear that: $$[QZP] + [PZR] + [QZR] = [PQR]$$ $$\frac{\sqrt3}{4}(ab + bc + ca) = \sqrt{10\cdot5\cdot3\cdot2}$$ $$ab + bc + ca = 40$$

And then by summing all equations of the system I was given: $$2(a + b +c)^2 - 3(ab + bc + ca) = 138$$ Hence, $(a+b+c)^2 = 129$. But I was able to do it only due to how closely these equations correlated with the geometry I could construct when relating these equations with the law of cosines. I do not think that in all kinds of symmetric expressions I would find similar pattern where I could use the law of cosines to think of helpful geometry.

Hence my question arises: how should one deal with symmetric expressions? I am looking for some strategies that can be useful for the same, such as using geometry or completely algebraic approaches given they are not tedious. In this case, law of cosines faired very well.

How should one analyze these equations to come up with something useful? Like how a particular geometric law was useful here, are there other examples where geometrical interpretations can be made out of some system of equations, such as the law of sines?

Besides geometry, how could one solve such problems? I know manipulating equations is what should be done, but in most cases it is tough and difficult to figure out which operation to use with what equation to get anything of importance. What should one keep in mind when algebraically manipulating such equations, which often becomes cumbersome?

Consider the example I began the question with: is there a simple algebraic way to solve for $(a+b+c)^2$ or any way that avoids the use of law of cosines?

Clearly there is no definitive answer here, hence I am just looking for helpful advice. Such equations are bothersome to solve mechanically and most often it can just be solved with some simple approach.

  • $\begingroup$ There is indeed theory of symmetric polynomials where you can get some idea about how to approach these equations $\endgroup$
    – Tony Pizza
    Sep 23 at 17:43
  • 1
    $\begingroup$ You don't say that $a,b,c$ must be $\ge 0$ : it must be for this reason that you can assume they are sidelengths of triangles. If this is not the case : $a=0, b=5,c=-8$ is a solution. $\endgroup$
    – Jean Marie
    Sep 23 at 17:56
  • 1
    $\begingroup$ Sorry, I did miss that detail. $a,b,c$ were supposed to be positive here. What should have my approach been if say they were not positive? $\endgroup$ Sep 23 at 18:30
  • $\begingroup$ In M2 R=QQ[a,b,c,d] I=ideal(a^2+b^2+a*b-25,b^2+c^2+b*c-49,c^2+a^2+a*c-64,d-(a+b+c)^2) primaryDecomposition I -- {ideal(d-129,64*b-25*c,8*a-5*c,129*c^2-4096), ideal(d-9,c-8,b+5,a), ideal(d-9,c+8,b-5,a)}. $\endgroup$ Sep 24 at 6:28

3 Answers 3


Another way by using complex numbers:

Let $w=e^{\frac{2\pi i}3}$ and $z_1=a-wb$, $z_2=b-wc$, $z_3=c-wa$. Then we have $$|z_1|=5,|z_2|=7, |z_3|=8\,\,\text{and}\,\,$$$$z_1+wz_2+w^2z_3=0.\tag1$$ Notice that $a+b+c=\frac{z_1+z_2+z_3}{1-w}$ so that $a+b+c=\frac{|z_1+z_2+z_3|}{\sqrt3}$ for $a,b,c>0$.

I took $z_1=5$, $z_3=8$ in $(1)$ and obtained $z_2=\frac{13}{2}-\frac{3\sqrt3}{2}i.$ Then I found $|z_1+z_2+z_3|=|\frac{39}{2}-\frac{3\sqrt3}{2}i|=\sqrt{129}\sqrt3.$ My arbitrary choice worked.

  • 2
    $\begingroup$ [+1] Astute transformation. How have you come to this idea ? Maybe with (1) being a characterization of $w_1,w_2,w_3$ as an equilateral triangle ? $\endgroup$
    – Jean Marie
    Sep 24 at 4:05

When I have seen your question, my reaction has been "twofold" :

  • As there are three equations and three unknowns, a rule of thumb (not a theorem !) is that usualy, one gets a finite number of solutions $(a,b,c)$, that can be found by using a C.A.S. Having them, one can compute $(a+b+c)^2$ (I agree that it's brute force, but it works).

  • Do I have (same reaction as you !) a geometrical representation ? Answer : yes, but not the same one.

As there are 3 variables, I consider solving the system of 3 equations with 3 unknowns as determining the intersection of 3 surfaces in an $(a,b,c)$ coordinates system ; more precisely, the intersection of 3 elliptical cylinders (blue, green, red on my figure), because in this case, there are solutions ; but such a system of equations could have no real-valued solutions ; this is the case for example with RHS values $1,9,25$ instead of $25, 49, 64$...

enter image description here

Remarks :

  1. With values such as $1,9,25$, the fact that you are asked the sum $(a+b+c)^2$ could hide the fact that all $a,b,c$ are complex-valued...

  2. Solutions come by pairs, due to the fact that the LHS expression are homogeneous with degree $2$ : if $(x,y,z)$ is a solution, then $(-x,-y,-z)$ is a solution as well.

I have obtained the solutions of the system and the figure by writing a "SAGE" program (public domain software). See it below ; I strongly advise you to execute it by just calling https://sagecell.sagemath.org/ : an editing window will open ; copy-paste the program in it, then activate button "Evaluate". Wait for some seconds only. Remark : the figure can be animated by mouse-dragging.

Edit : A different approach is possible, more in the spirit of symmetrical expressions. It is obtained by recognizing in the different LHSides a factor of a known identity. More precisely if both sides of the first equality are multiplied by $(a-b)$, its LHS becomes :


Doing the same on the 2 other equations, then adding, we get the linear constraint:


from which we can extract :


(which, by the way, has a barycentrical form : whatever the solution $(a,b,c)$, $a$ can be written as the weighted mean of $b$ and $c$ with, always, the same weights $24/39$ and $15/39$).

It remains to constitute a system of 2 equations with (only) the 2 unknowns $b$ and $c$ for example by

  • plugging (1) into the first equation and

  • "twinning" it with the second equation.

var('a b c')
[show(sol[k]) for k in range(len(sol))]
from sage.plot.plot3d.implicit_surface import ImplicitSurface
G1=ImplicitSurface(a^2+b^2+a*b-25,(a,-k, k), (b,-k, k),(c,-k,k),plot_points=60,color='blue',opacity=op);
G2=ImplicitSurface(c^2+b^2+c*b-49,(a,-k, k), (b,-k, k),(c,-k, k),plot_points=60,color='red',opacity=op);
G3=ImplicitSurface(c^2+a^2+c*a-64,(a,-k, k), (b,-k, k),(c,-k, k),plot_points=60,color='green',opacity=op);
G4=ImplicitSurface(a+b+c-sqrt(129),(a,-k, k), (b,-k, k),(c,-k,k),plot_points=60,color='green',opacity=op);
G5=ImplicitSurface(a+b+c+sqrt(129),(a,-k, k), (b,-k, k),(c,-k, k),plot_points=60,color='yellow',opacity=op);
  • $\begingroup$ Please see my edit. $\endgroup$
    – Jean Marie
    Sep 24 at 4:01
  • $\begingroup$ +1, very detailed and well-written answer. Thank you $\endgroup$ Sep 24 at 6:48

Some thoughts.

Algebraically manipulating equations by hand is cumbersome. For the OP, if we first eliminate $c$, second eliminate $b$, we will get $81a^2(129a^2 - 1600) = 0$. This process can be described as follows. Let \begin{align*} f_1 &= a^2 + b^2 + ab - 25, \\ f_2 &= b^2 + c^2 + bc - 49, \\ f_3 &= c^2 + a^2 + ca - 64,\\ g_1 &= 117\,{a}^{4}-162\,{a}^{3}b-27\,{a}^{2}{b}^{2}+72\,a{b}^{3}+4131\,{a}^{ 2}-648\,ab,\\ g_2 &= 117\,{a}^{3}b-117\,{a}^{3}c-45\,{a}^{2}{b}^{2}+45\,{a}^{2}bc-72\,a{b}^ {3}+72\,a{b}^{2}c-1755\,{a}^{2}-1080\,ab, \\ g_3 &= -117\,{a}^{4}+45\,{a}^{3}b+117\,{a}^{3}c+72\,{a}^{2}{b}^{2}-45\,{a}^{2 }bc-72\,a{b}^{2}c+1755\,{a}^{2}+1080\,ab. \end{align*} We have $$g_1f_1 + g_2f_2 + g_3f_3 = 81a^2(129a^2 - 1600).$$

If we use CAS (Computer Algebra System), we can use Resultant:
$f_4 := \mathrm{Res}(f_2, f_3, c)$,
$f_5 := \mathrm{Res}(f_1, f_4, b)$.
Then $f_5 = 81a^2(129a^2 - 1600)$.

  • $\begingroup$ [+1] Very thorough "algebraic geometry" approach. $\endgroup$
    – Jean Marie
    Sep 24 at 4:08
  • $\begingroup$ @JeanMarie Thanks. $\endgroup$
    – River Li
    Sep 24 at 5:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .