Fermat Point - Another Proof

Another Proof

Another approach to find a point within the triangle, from where sum of the distances to the vertices of triangle is minimum, is to use one of the optimization (mathematics) methods. In particular, method of the lagrange multipliers and the law of cosines.

We draw lines from the point within the triangle to its vertices and call them X, Y and Z. Also, let the lengths of these lines be x, y, and z, respectively. Let the acute angle between X and Y be α, Y and Z be β. Then the angle between X and Z is (2π − α − β). Using the method of lagrange multipiers we have to find the minimum of the lagrangian, which is expressed as:

x + y + z +

λ₁ (x2 + y2 − 2xy cos(α) − a2) +

λ₂ (y2 + z2 − 2yz cos(β) − b2) +

λ₃ (z2 + x2 − 2zx cos(α + β) − c2) .

where a, b and c are the lengths of the sides of the triangle.

Calculating the partial derivatives δ/δx, δ/δy, δ/δz, δ/δα, δ/δβ gives a system of 5 equations:

δ/δx: 1 + λ₁(2x − 2y cos(α)) + λ₃(2x − 2z cos(α + β)) = 0

δ/δy: 1 + λ₁(2y − 2x cos(α)) + λ₂(2y − 2z cos(β)) = 0

δ/δz: 1 + λ₂(2z − 2y cos(β)) + λ₃(2z − 2x cos(α + β)) = 0

δ/δα: λ₁y sin(α) + λ₃z sin(α + β) = 0

δ/δβ: λ₂y sin(β) + λ₃x sin(α + β) = 0

After some algebraic manipulations equations for α and β separate from the rest of the parameters, giving:

sin(α) = sin(β)

sin(α + β) = −sin(β)

that gives: α = β = 120°

Q.E.D.

Note: if one of the vertices of triangle has angle not less than 120°, then the Fermat point is at that vertex.