Distance from any point in the plane to vertices of a triangle related to its sides

Question

I was writing a program calculating all possible configurations of $n$ random circles with random radius and center coordinate in the plane and met this problem. Being puzzled by it for quite a long time, I decided to have a try at this site.

Problem statement. Given a triangle $ABC$, pick a random point $P$ in its plane (not restricting it inside the triangle). Known that three sides of $ABC$ are $a$, $b$ and $c$; the distance from $P$ to $A$, $B$ and $C$ are $a'$, $b'$ and $c'$ respectively.

Question: Find an equation $F(a, b, c, a', b', c') = 0$, with $F$ preferrably a polynomial in $a'$, $b'$ and $c'$.

Distance from any point to vertices of triangle related to its sides

After days of search, I actually came across an unnamed theorem quite close to what I was looking for:

$a'^2+b'^2+c'^2 = (a^2+b^2+c^2)/3 + 3*PG^2$,

where $PG$ represents the distance from $P$ to the centroid of triangle $ABC$.

I was a bit satisfied with this result but still wanting to get an explicit form without involving extra terms.

Considering that this problem is itself quite neat and probably of interest to some of you, I really want to get some hints from you guys.

Thanks for the help.

Answer 1

If the "extra term" in your expression for $F$ is the term involving the distance to the centroid, you might be able to write it using the data you have by following the wikipedia discussion of the centroid using trilinear coordinates.

PS There is no good way to

pick a random point $P$ in its plane

with no bound on where it might be.

Answer 2

$a',b',c'$ are the tripolar coordinates of $P$. MathWorld gives two equations $F$ by Euler, one of which is reproduced below (using $x,y,z$ for $a',b',c'$): $$(a^2+b^2-c^2)(x^2y^2+c^2z^2)+(a^2-b^2+c^2)(b^2y^2+x^2z^2)+(-a^2+b^2+c^2)(a^2x^2+y^2+z^2)-(a^2x^4+b^2y^4+c^2z^4)-a^2b^2c^2=0$$