Fermat point

by math_explorer, Oct 10, 2010, 1:27 AM

(the exceedingly obvious problem from which the current problem was generalized)
(given a triangle $\triangle ABC$, find the point $P$ that minimizes $PA + PB + PC$)

Here's a nice proof.

Lemma: "Equilateral Triangle Ptolemy"
If $\triangle XYZ$ is equilateral, then for any point $P$, $PX + PY \geq PZ$, with equality iff $P, X, Y, Z$ are concyclic and $P$ is on arc $XY$.
Proof: Guess which theorem you have to use. I think it begins with a P.

Anyways, given acute $\triangle ABC$, we construct $A'$ so that $\triangle BCA'$ is equilateral and outside of $\triangle ABC$. Then, for any point $P$, $PA + PB + PC \geq PA + PA' \geq AA'$, with equality iff $P$ is the intersection of $AA'$ and the circumcircle of $\triangle BCA'$. So that's the Fermat point.
$\angle BPC = \pi - \angle BA'C = 2\pi/3 = 120^\circ$
$\angle APC = \pi - \angle A'PC = \pi - \angle A'BC = 2\pi/3 = 120^\circ$
$\angle APB = \pi - \angle A'PB = \pi - \angle A'CB = 2\pi/3 = 120^\circ$

Extending this to 3D looks ridiculous. Oh well.

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