Fermat point
In geometry, the Fermat point of a triangle, also called the Torricelli point or Fermat–Torricelli point, is a point such that the total distance from the three vertices of the triangle to the point is the minimum possible. It is so named because this problem is first raised by Fermat in a private letter to Evangelista Torricelli, who solved it.
The Fermat point gives a solution to the geometric median and Steiner tree problems for three points.
The Fermat point of a triangle with largest angle at most 120° is simply its first isogonic center or X(13), which is constructed as follows:
An alternate 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 red lines in Fig 1 are concurrent and cut one another at angles of 60°.
The triangles RAC and BAQ are congruent because the second is a 60° rotation of the first about A. Hence ∠ARF = ∠ABF and ∠AQF = ∠ACF. By converse of angle in the same segment, ARBF and AFCQ are both concyclic. Thus ∠AFB = ∠AFC = ∠BFC = 120°. Because ∠BFC and ∠BPC add up to 180°, BPCF is also concyclic. Hence ∠BFP = ∠BCP = 60°. Because ∠BFP + ∠BFA = 180°, AFP is a straight line.
This proof only applies in Case 2 since if ∠BAC > 120° 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. Then ∠AFB = ∠AFC = 60° hence ∠BFC = ∠AFB = ∠AFC = 120° which means BPCF is concyclic so ∠BFP = ∠BCP = 60° = ∠BFA. Therefore A lies on FP.
Clearly any 3 lines perpendicular to the red ones in Fig 1, in particular those joining the centres of the circles, must also cut at angles of 60° and thereby form an equilateral triangle. This curiosity is known as Napoleon's Theorem.
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