Fermat-Torricelli theorem

Theorem (Fermat-Torricelli).  Let all angles of a triangle ABC be at most 120.  Then the inner point F of the triangle which makes the sum AF+BF+CF as little as possible, is the point from which the angle of view of every side is 120.

Proof.  Let’s perform the rotation of 60 about the point A.  When P is the image of the point C, the triangle ACP is equilateral and its angles are 60.  Let F be any inner point of the triangle ABC and Q its image in the rotation.  We infer that if the sides of the triangle ABC are all seen from F in the angle 120, then the points B, F, Q, P lie on the same line.


Generally, the triangles APQ and ACF are congruent, whence  CF=QP.  From the equilateral trianglesMathworldPlanetmath we obtain:


Here, the right hand side is minimal when the points B, F, Q, P are collinearMathworldPlanetmath, in which case

CFA=PQA= 180-AQF= 120,
AFB= 180-QFA= 120,
BFC= 360-240= 120.

Remark.  The point F is called the Fermat pointMathworldPlanetmath of the triangle ABC.


  • 1 Tero Harju: Geometria. Lyhyt kurssi.  Matematiikan laitos. Turun yliopisto, Turku (2007).
Title Fermat-Torricelli theorem
Canonical name FermatTorricelliTheorem
Date of creation 2013-03-22 19:36:39
Last modified on 2013-03-22 19:36:39
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 16
Author pahio (2872)
Entry type Theorem
Classification msc 51M04
Classification msc 51F20
Related topic CenterOfATriangle
Defines Fermat point