Fermat Point - Geometry

Geometry

Given any Euclidean triangle ABC and an arbitrary point P let d(P) = PA+PB+PC. The aim of this section is to identify a point P₀ such that d(P₀) < d(P) for all P ≠ P₀. If such a point exists then it will be the Fermat point. In what follows Δ will denote the points inside the triangle and will be taken to include its boundary Ω.

A key result that will be used is the dogleg rule which asserts that if a triangle and a polygon have one side in common and the rest of the triangle lies inside the polygon then the triangle has a shorter perimeter than the polygon.

Let P be any point outside Δ. Associate each vertex with its remote zone; that is, the half-plane beyond the (extended) opposite side. These 3 zones cover the entire plane except for Δ itself and P clearly lies in either one or two of them. If P is in two (say the B and C zones intersection) then setting P' = A implies d(P') = d(A) < d(P) by the dogleg rule. Alternatively if P is in only one zone, say the A-zone, then d(P') < d(P) where P' is the intersection of AP and BC. So for every point P outside Δ there exists a point P' in Ω such that d(P') < d(P).

Case 1. The 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 Δ (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 d(P) = CP+PQ+QF which is simply the length of the path CPQF. As P is constrained to lie within ABC, by the dogleg rule the length of this path exceeds AC+AF = d(A). Therefore d(A) < d(P) for all P є Δ, P ≠ A. Now allow P to range outside Δ. From above a point P' є Ω exists such that d(P') < d(P) and as d(A) ≤ d (P') it follows that d(A) < d(P) for all P outside Δ. Thus d(A) < d(P) for all P ≠ A which means that A is the Fermat point of Δ. In other words the Fermat point lies at the obtuse angled vertex.

Case 2. The triangle has no angle ≥ 120°.

Let P be any point inside Δ and construct the equilateral triangle CPQ. Then CQD is a 60° rotation of CPB about C so d(P) = PA+PB+PC = AP+PQ+QD 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. By the angular restriction P₀ lies inside Δ moreover BCF is a 60° rotation of BDA about B so Q₀ must lie somewhere on AD. Since CDB = 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 P' є Ω exists such that d(P') < d(P) and as d(P₀) ≤ d(P') it follows that d(P₀) < d(P) for all P outside Δ. That means P₀ is the Fermat point of Δ. In other words the Fermat point is coincident with the first isogonic center.