Weber Points - Geometric Mean of an Hexagon I want to compute the Weber Point (Geometric mean) of an hexagon ($abcdef$) which is symmetric with respect to some axis $L$.
The axis of symmetry $L$ is such that ($a$) is symmetric to ($f$), ($b$) to ($e$) and ($c$) to ($d$).
We know that the Weber point lies in $L$ but how to compute it ?
Thank you.
 A: Rotate the position so that $L$ is the $x$-axis, for ease of exposition. We can forget points $d$, $e$, and $f$, and look for $x$ which minimises the sum of the distances from $(x,0)$ to $a$, $b$, and $c$. If $x$ varies by $\delta x$, then the distance from $(x,0)$ to $a$ (say) varies by $-\delta x \cos A$, where $A$ is the angle that the line from $a$ to $(x,0)$ makes with the positive $x$-axis. So we want the point $(x,0)$ such that $\cos A + \cos B + \cos C = 0$.
I can't see an explicit formula for $x$ (though I won't be surprised if somebody else can). But you can find this point to any required accuracy by a simple binary search. If you need a faster method, then Newton will help.
A: There is probably no explicit formula to compute the Fermat Weber point.
But you already reduced the problem to a 1-dimensional one: you could
write down the line L in parameter form $(l_1,l_2)*t+(c_1,c_2)$. For a sufficiently
fine grid $t_1,t_2,..,t_n$ you could then compute the sum of the distances of
the corresponding point on L to the points a,b,c finally chosing those
$t^*$ minimizing the sum of the distances. This should work sufficiently fast
and give you a good approximation of the FW-point.
H
