In Euclidean geometry, the Fermat point of a triangle, also called the Torricelli point or Fermat–Torricelli point, is a point such that the sum of the three distances from each of the three vertices of the triangle to the point is the smallest possible or, equivalently, the geometric median of the three vertices. 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 gives a solution to the geometric median and Steiner tree problems for three points. Construction 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: Construct an equilateral triangle on each of two arbitrarily chosen sides of the given triangle. Draw a line from each new vertex to the opposite vertex of the original triangle. The two lines intersect at the Fermat point. An alternative method is the following: On each of two arbitrarily chosen sides, construct an isosceles triangle, with base the side in question, 30-degree angles at the base, and the third vertex of each isosceles triangle lying outside the original triangle. For each isosceles triangle draw a circle, in each case with center on the new vertex of the isosceles triangle and with radius equal to each of the two new sides of that isosceles triangle. The intersection inside the original triangle between the two circles is the Fermat point. 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°. Location of X(13) Fig. 2 shows the equilateral triangles attached to the sides of the arbitrary triangle . Here is a proof using properties of concyclic points to show that the three lines in Fig 2 all intersect at the point and cut one another at angles of 60°. The triangles are congruent because the second is a 60° rotation of the first about . Hence and . By the converse of the inscribed angle theorem applied to the segment , the points are concyclic (they lie on a circle). Similarly, the points are concyclic. , so , using the inscribed angle theorem. Similarly, . So . Therefore, . Using the inscribed angle