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 triangles we obtain:
AF+BF+CF=FQ+BF+QP=BFQP |
Here, the right hand side is minimal when the points B, F, Q, P are collinear, in which case
∠CFA=∠PQA= 180∘-∠AQF= 120∘, | ||
∠AFB= 180∘-∠QFA= 120∘, | ||
∠BFC= 360∘-240∘= 120∘. |
Remark. The point F is called the Fermat point of the triangle ABC.
References
- 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 |