Given any Euclidean triangle of vertexes $v\_1,v\_2,v\_3$ and any point $P$ in the interior of the triangle, consider the following system of 6 real equations in $c\_{12},c\_{13},c\_{23},d\_1,d\_2,d\_3$:


$$c\_{12}(v\_1-v\_2)+c\_{13}(v\_1-v\_3)=d\_1(v\_1-P)$$
$$c\_{12}(v\_2-v\_1)+c\_{23}(v\_2-v\_3)=d\_2(v\_2-P)$$
$$c\_{23}(v\_3-v\_2)+c\_{13}(v\_3-v\_1)=d\_3(v\_3-P)$$


I am trying to prove the existence of a positive solution (i.e. such that $c\_{12}>0,c\_{13}>0,c\_{23}>0,d\_1>0,d\_2>0,d\_3>0$) to such system for every triangle, but until now I have not succeeded.


Notice that it is not sufficient to infer that couples of these vectors are linearly independent. Indeed, solving the first equation, for each value $c\_{12}>0$ one finds $c\_{13}(c\_{12})>0$ (i.e. the value of $c\_{13}$ depending on $c\_{12}$). Solving the third equation given $c\_{13}(c\_{12})$ one finds $c\_{23}(c\_{12})$ and lastly one has to prove that such $c\_{23}(c\_{12})$ solves the second equation.


Do you know how to find such $c\_{12}>0,c\_{13}>0,c\_{23}>0,d\_1>0,d\_2>0,d\_3>0$ (or at least prove their existence)? I would be grateful if you could explain it to me.



Question on strictly positive solutions of a system involving an Euclidean triangle
Ok, here goes.


WLOG, $P=0$. Then there exist positive $a\_k$ such that $\sum\_k a\_kv\_k=0$ (this is just a restatement of the condition that the origin is inside). We want to find $c\_{km}=c\_{mk}>0$ and $d\_k>0$ such that $\sum\_{m:m\ne k}c\_{km}(v\_k-v\_m)=d\_kv\_k$. Projecting to the hyperplane (line in $\mathbb R^2$) orthogonal to $v\_k$, we see that we really have no freedom in the choice for $c\_{km}$: they should be proportional to $a\_m$ because it is the only linear dependence available. Fortunately, this can be arranged: just put $c\_{km}=a\_ka\_m$. Then 
$$
\sum\_{m:m\ne k}c\_{km}(v\_k-v\_m)=a\_k\left[\left(\sum\_{m:m\ne k}a\_m\right)v\_k-\sum\_{m:m\ne k}a\_mv\_m\right]=a\_k\left(\sum\_{m}a\_m\right)v\_k
$$
so $d\_k=a\_k\sum\_{m}a\_m$.


That's it. 


