# Fermat's Problem

 Problem Statement: In a given acute -angled triangle ABC, locate a point P whose distances from A, B, and C have the smallest possible sum.

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.

In order to begin this proof, we must first consider an arbitrary point say P inside triangle ABC.  Then, join this point to points A, B, and C respectively.

Choosing point B as our point of rotation, rotate triangle APB 60o about point B.  See next diagram.

We have now obtained the new triangle C'P'B.

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 = 180o - angle BPP' = 120o , and angle APB = angle C'P'B = 180o - angle PP'B = 120oTherefore, 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 120o
QED.

Use GSP to explore this problem.