2
$\begingroup$

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.

$\endgroup$

closed as off-topic by fedja, David Handelman, Jan-Christoph Schlage-Puchta, Neil Strickland, Dima Pasechnik Dec 9 '17 at 13:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – fedja, Neil Strickland
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ There's nothing Euclidean in this question, it's purely affine $\endgroup$ – YCor Dec 7 '17 at 18:00
  • $\begingroup$ @YCor ok, I removed the tag "Euclidean geometry" $\endgroup$ – user101163 Dec 7 '17 at 18:05
  • 1
    $\begingroup$ What's the point of posting a question, deleting it 20 minutes later and then reposting it in exactly the same form up to minor notation changes and a stupid misprint in the last equation? Any normal person would conclude that you are just trying to cheat on a geometry test. Anyway, the answer to the question as posed is "Yes, I do". $\endgroup$ – fedja Dec 7 '17 at 22:09
  • $\begingroup$ @fedja I wrongfully originally asked the question with an account I don't use on mathoverflow (indeed it has reputation score of 1), so I reposted it with this account. If this is not allowed please suggest me what I shall do. Could you please briefly explain in what consists your solution to the problem? $\endgroup$ – user101163 Dec 8 '17 at 8:07
  • $\begingroup$ You are right, i fixed the error $\endgroup$ – user101163 Dec 8 '17 at 21:36
2
$\begingroup$

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.

$\endgroup$