Fermat$-$Torricelli theorem

Theorem (Fermat$-$Torricelli).  Let all angles of a triangle $ABC$ be at most $120^{\circ}$.  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 $120^{\circ}$.

Proof.  Let’s perform the rotation of $60^{\circ}$ about the point $A$.  When $P$ is the image of the point $C$, the triangle $ACP$ is equilateral and its angles are $60^{\circ}$.  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 $120^{\circ}$, 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

 $\displaystyle\angle CFA\;=\;\angle PQA\;=\;180^{\circ}\!-\!\angle AQF\;=\;120^% {\circ},$ $\displaystyle\angle AFB\;=\;180^{\circ}\!-\!\angle QFA\;=\;120^{\circ},$ $\displaystyle\angle BFC\;=\;360^{\circ}\!-\!240^{\circ}\;=\;120^{\circ}.$

Remark.  The point $F$ is called the of the triangle $ABC$.

References

• 1 Tero Harju: Geometria. Lyhyt kurssi.  Matematiikan laitos. Turun yliopisto, Turku (2007).
Title Fermat$-$Torricelli theorem FermatTorricelliTheorem 2013-03-22 19:36:39 2013-03-22 19:36:39 pahio (2872) pahio (2872) 16 pahio (2872) Theorem msc 51M04 msc 51F20 CenterOfATriangle Fermat point