Problem Statement:


Throughout his lifetime, Pierre
Fermat worked on some of the best known mathematical problems in history.
We will begin with a proof of Fermat's Problem, and then explore some extensions,
and related questions.
Choosing point B as our point of rotation, rotate triangle APB 60^{o} about point B. See next diagram.
Now, triangles ABC' and PBP' are equilateral. as shown in the diagram below.
Therefore, AP + BP + CP = C'P' + P'P + PC, which creates a path from C' to C which is usually a broken line with angles at P' and P. Also, this path is a minimum when it is straight.
Now angle BPC = 180^{o}
 angle BPP' = 120^{o} , and angle APB = angle C'P'B = 180^{o}
 angle PP'B = 120^{o}. Therefore, we can conclude that
the desired point point P, for which AP + BP + CP is minimal,
is the point from which each of the sides BC, CA, and AB subtends
an angle of
120^{o}
QED.