Assignment #6

Fermat Point

by

Summer D. Brown


Problem: Consider any triangle ABC. Find a construction for a point P such that the sum of the distances from P to each of the three vertices is a minimum.

 

Construction:

Start with any triangle ABC.

 

Mark one of the vertices as a center of rotation. Rotate one of the adjacent sides of the triangle 60 degrees about the marked center.

Repeat a rotation by 60 degrees again using another vertex as the center of rotation and one of its adjacent sides.

Next, construct a line through points A' and C and another line through points C' and B.

The intersection point P of these two lines will be the point such that the sum of the distances from P to each of the vertices is a minimum. This point P is called the Fermat Point. Open a GSP script for the construction of the Fermat point for any triangle.

So, the sum of the distances AP, BP, and CP will be the minimum sum of distances from any point to each of the three vertices.

 

Proof:

Prove that P is the point such that the sum of the distances from P to each of the three vertices of triangle ABC is a minimum. Call such a point P the Fermat point of triangle ABC.

Take one of the triangles formed by the vertices of triangle ABC and point P. First, I will use triangle APB. Rotate triangle APB 60 degrees around vertex B.

Distances are preserved under a rotation. Thus, AP = A'P' and BP = BP'. The angle PBP' = 60 degrees since we rotated the triangle by 60 degrees. Since a(PBP') = 60 degrees and BP = BP', we can form the equilateral triangle PBP', which is outlined in green below.

We have formed a straight line segment A'C using the lengths of the segments AP, BP, and CP. We will have the minimum distance when the segments form a straight line. Thus, the point that will create this minimum distance must lie somwhere on this line.

Perform a similar rotation around vertex A with triangle APC to find another straight line containing all three segments. Equilateral triangle APP' is highlighted in blue.

Once again, we have formed a straight line segment BC' using the lengths of segments AP, BP, and CP. The Fermat point must lie somewhere on this line as well. Thus, the Fermat point must lie at the intersection of the lines A'C and BC'. Therefore, P is the point such that the sum of the distances from P to each of the three vertices of triangle ABC is a minimum.

Note that you can also use triangle BPC to rotate around vertex C in a similar fashion as above. The straight line segment AB' formed containing the lengths AP, BP, and CP will also pass through the Fermat point P.


Return to Summer's Main Page