Different Cases of Fermat Point

For different triangles, Fermat Point can be inside or on the boundary of the triangle under consideration. Let’s discuss the following two cases of Fermat Point.

• Case 1: Triangle has an Angle ≥ 120°
• Case 2: Triangle has no Angle ≥ 120°

Case 1: Triangle has an Angle ≥ 120°

• Without loss of generality, suppose that the angle at A is ≥ 120°.
• Construct the equilateral triangle AFB and for any point P in a triangle (except A itself) construct Q so that the triangle AQP is equilateral and has the orientation shown.
• Then the triangle ABP is a 60° rotation of the triangle AFQ about A so these two triangles are congruent and it follows that [Tex]\displaystyle d(P)=|CP|+|PQ|+|QF|     [/Tex]which is simply the length of the path CPQF.
• As P is constrained to lie within triangle ABC, by the dogleg rule the length of this path exceeds [Tex]\displaystyle |AC|+|AF|=d(A). [/Tex]

Therefore, d(A) < d(P) for all [Tex]P\in \Delta ,P\neq A. [/Tex]

Now allow P to range outside the triangle. From above a point [Tex] P’ \in \Omega   [/Tex] exists such that d(P’)<d(P) and as d(A) ≤ d(P’) it follows that d(A) < d(P) for all P outside triangle. Thus d(A) < d(P) for all [Tex]P\neq A   [/Tex] which means that A is the Fermat point of the triangle. In other words, the Fermat point lies at the obtuse-angled vertex.

Case 2: Triangle has no Angle ≥ 120°

• Construct the equilateral triangle BCD, let P be any point inside the triangle, and construct the equilateral triangle CPQ. Then triangle CQD is a 60° rotation of triangle CPB about C so

[Tex]\displaystyle d(P)=|PA|+|PB|+|PC|=|AP|+|PQ|+|QD| [/Tex]

which is simply the length of the path APQD.

• Let P₀ be the point where AD and CF intersect. This point is commonly called the first isogonic center. Carry out the same exercise with P₀ as you did with P, and find the point Q₀.
• By the angular restriction, P₀ lies inside triangle ABC. Moreover, triangle BCF is a 60° rotation of triangle BDA about B, so Q₀ must lie somewhere on AD. Since angle CBD = 60°, it follows that Q₀ lies between P₀ and D which means AP₀Q₀D is a straight line so d(P₀) = |AD|.
• Moreover, if P ≠ P₀ then either P or Q won’t lie on AD which means d(P₀) = |AD| < d(P)
• Now allow P to range outside Δ. From above a point [Tex]\displaystyle P’\in \Omega   [/Tex] exists such that d(P’) < d(P) and as d(P₀) ≤ d(P) it follows that d(P₀) < d(P) for all P outside triangle BDA.
• That means P₀ is the Fermat point of the triangle. In other words, the Fermat point is coincident with the first isogonic center.

Fermat Point is the point that Pierre de Fermat, the 17th-century French mathematician, posed as a challenge to his compatriot Evangelista Torricelli to geometrically determine, is named in Fermat’s honour as the solution that would minimize the total combined distance from each vertex of a triangular figure to any single internal locus. Torricelli solved the problem, therefore other than Fermat Point, it is also known as the Fermat Point or the Torricelli Point or Fermat Torricelli Point.

Although various methods exist to locate the Fermat point, connecting the vertices of the original triangle to those of the equilateral triangles constructed on each of its sides furnishes a straightforward technique. The intersection of these segments is the Fermat point.

The Fermat point gives a solution to the geometric median and Steiner tree problems for three points.