What is the non-trivial, general solution of these equal ratios? Provide non-trivial solution of the following:

$$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$$

$a=?, b=?, c=?$
The solution should be general.
 A: HINT:
$\displaystyle\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+(c+a)+(a+b)}$
Similarly, 
$\displaystyle F=\frac{a}{b+c}=\frac{b}{c+a}=\frac{a-b}{b+c-(c+a)}$
Either $a=b$ or $\displaystyle F=-1$
Hope these should help
A: Here is how you proceed from Nilan's system of equations. Write down the system in matrix form, $\mathbf{A}\mathbf{x}=0$ where $\mathbf{x}=(a,b,c)$ and $\mathbf{A}$ is the matrix of coefficients in terms of $t$.
This is a homogeneous system: 


*

*if $\mathbf{A}$ is non-singular ($\det \mathbf{A}\neq0$), it has only one solution: the
trivial one ($a=b=c=0$);

*if $\mathbf{A}$ is singular ($\det \mathbf{A}=0$), it has an
infinite number of solutions.


We are therefore interested in the case where $\mathbf{A}$ is singular. Work out the determinant of $\mathbf{A}$, which will turn out to be $2t^3+3t^2-1$. Find the roots of this polynomial, which will be $t=-1$ and $t=\frac{1}{2}$.
Solve the system for each of these two cases. You will get:


*

*$t=1$ $\rightarrow$ $a+b+c=0$.

*$t=\frac{1}{2}$ $\rightarrow$ $a=b=c$.
These are the general solutions; any setting of $a,b,c$ that does not satisfy one of the above two conditions is not a solution. 

More Generality
It turns out that this generalizes to any number of variables, i.e. write the problem as:
$$
\frac{x_i}{\sum x - x_i} = \frac{x_j}{\sum x - x_j} \quad \forall i,j=1,2,\dots,n
$$
Then the only solutions are:


*

*$\sum x = 0$;

*all $x$'s are equal.


To see this, you can either generalize the linear-algebraic approach above (see here), or more simply, look at the problem formulation and realize that either all the denominators must be identical (which corresponds to Condition 1) or $\sum x = 0$ so that the (magnitudes of the) denominators cancel out with the numerators for all $i$'s.
A: $$\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=t$$
Then $$-a+bt+ct=0$$ $$at-b+ct=0$$ $$at+bt-c=0$$ Try to solve them. This would be a help for you.
A: Besides to the good answers posted for this question (this answer and this one), I think finding eigenvalues and eigenvectors of such system will help you to find the solution.
$$\begin{array}{l}\frac{a}{{b + c}} = \frac{b}{{c + a}} = \frac{c}{{a + b}} = t\\\left[ {\begin{array}{*{20}{c}}1&{ - t}&{ - t}\\{ - t}&1&{ - t}\\{ - t}&{ - t}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\\\left[ C \right]\left\{ X \right\} = \left\{ 0 \right\}\\\det \left[ {C - \lambda I} \right] = 0\\\det \left[ {\begin{array}{*{20}{c}}{1 - \lambda }&{ - t}&{ - t}\\{ - t}&{1 - \lambda }&{ - t}\\{ - t}&{ - t}&{1 - \lambda }\end{array}} \right] = 0\\\left( {1 - \lambda } \right)\left\{ {{{\left( {1 - \lambda } \right)}^2} - {{\left( { - t} \right)}^2}} \right\} = 0\\\lambda  = 1,1 - t,1 + t\\\begin{array}{*{20}{c}}{\lambda  = 1}\\{\lambda  = 1 - t}\\{\lambda  = 1 + t}\end{array}\left| {\begin{array}{*{20}{c}}{ + \left( {1 - \lambda } \right)a - \left( t \right)b - \left( t \right)c = 0}\\{ - \left( t \right)a + \left( {1 - \lambda } \right)b - \left( t \right)c = 0}\\{ - \left( t \right)a - \left( t \right)b + \left( {1 - \lambda } \right)c = 0}\end{array}} \right\rangle \left\{ {\left. {\begin{array}{*{20}{c}}{a = b = c = 0}\\{a = b = c = 0}\\{a + b + c = 0}\end{array}} \right\rangle } \right.\left. {\underline {\, 
 {a + b + c = 0} \,}}\! \right| \end{array}$$
As it seems the answer is compatible with the other answers to this question. 
