Fermat point

It is so named because this problem was first raised by Fermat in a private letter to Evangelista Torricelli, who solved it.

The Fermat point of a triangle with largest angle at most 120° is simply its first isogonic center or X(13),[2] which is constructed as follows: An alternative method is the following: When a triangle has an angle greater than 120°, the Fermat point is sited at the obtuse-angled vertex.

In what follows "Case 1" means the triangle has an angle exceeding 120°.

"Case 2" means no angle of the triangle exceeds 120°.

Here is a proof using properties of concyclic points to show that the three lines RC, BQ, AP in Fig 2 all intersect at the point F and cut one another at angles of 60°.

By the converse of the inscribed angle theorem applied to the segment AF, the points ARBF are concyclic (they lie on a circle).

Using the inscribed angle theorem, this implies that the points BPCF are concyclic.

So, using the inscribed angle theorem applied to the segment BP, ∠BFP = ∠BCP = 60°.

So, the lines RC, BQ, AP are concurrent (they intersect at a single point).

This proof applies only in Case 2, since if ∠BAC > 120°, point A lies inside the circumcircle of △BPC which switches the relative positions of A and F. However it is easily modified to cover Case 1.

2 are perpendicular to the line segments AP, BQ, CR.

So, the lines joining the centers of the circles also intersect at 60° angles.

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.

As P is constrained to lie within △ABC, by the dogleg rule the length of this path exceeds

In other words, the Fermat point lies at the obtuse-angled vertex.

Then △CQD is a 60° rotation of △CPB about C so which shows that the sum of the distances sought is just the length of the path APQD from A to D along a piecewise linear line.

Now we show that if P is chosen to be the isogonic center of △ABC the path APQD lies on a straight line - and thus it is minimal.

Since P0 is the isogonic center, ∠AP0C = 120°; by construction ∠CP0Q0 = 60°, therefore A, P0 and Q0 are aligned on the line from A to D. (Also, △BCF is a 60° rotation of △BDA about B, so Q0 must lie somewhere on AD).

In other words, the Fermat point is coincident with the first isogonic center.

Note that |k| ≤ 1 because the angle between the unit vectors i, j is ∠C which exceeds 120°.

The proof now continues as above (adding the three inequalities and using i + j + k = 0) to reach the same conclusion that O (or in this case C) must be the Fermat point of △ABC.

Another approach to finding the point within a triangle, from which the sum of the distances to the vertices of the triangle is minimal, is to use one of the mathematical optimization methods; specifically, the method of Lagrange multipliers and the law of cosines.

We draw lines from the point within the triangle to its vertices and call them X, Y, Z.

Using the method of Lagrange multipliers we have to find the minimum of the Lagrangian L, which is expressed as: where a, b, c are the lengths of the sides of the triangle.

However the elimination is a long and tedious business, and the end result covers only Case 2.

This question was proposed by Fermat, as a challenge to Evangelista Torricelli.

He solved the problem in a similar way to Fermat's, albeit using the intersection of the circumcircles of the three regular triangles instead.