Fermat Problem

Theorem
Let $\triangle ABC$ be a triangle

Let the vertices of $\triangle ABC$ all have angles less than $120 \degrees$.

Let $\triangle ABG$, $\triangle BCE$ and $\triangle ACF$ be equilateral triangles constructed on the sides of $ABC$.

Let $AE$, $BF$ and $CG$ be constructed.

Let $P$ be the point at which $AE$, $BF$ and $CG$ meet.


 * FermatPointConstruction.png

Then $P$ is the Fermat-Torricelli point of $\triangle ABC$.

If one of vertices of $\triangle ABC$ be of $120 \degrees$ or more, then that vertex is itself the Fermat-Torricelli point of $\triangle ABC$.

Proof
The sum of the distances will be a minimum when the lines $PA$, $PB$ and $PC$ all meet at an angle of $120 \degrees$.

This is a consequence of the Fermat problem being a special case of the Steiner Tree Problem.

Consider the circles which circumscribe the $3$ equilateral triangles $\triangle ABG$, $\triangle BCE$ and $\triangle ACF$.

Consider quadrilaterals formed by $\triangle ABG$, $\triangle BCE$ and $\triangle ACF$ along with another point on each of those circumscribing circles.

Because these are cyclic quadrilaterals, the angle formed with these new points is $120 \degrees$.

It follows that $\Box APBG$, $\Box BPCE$ and $\Box APCF$ are those cyclic quadrilaterals.

Hence $\angle APC = \angle APB = \angle BPC = 120 \degrees$ and the result follows.

Also known as
The Fermat problem is also recognised as a special case of the Steiner Tree Problem for $3$ nodes.

Some refer to it as the Steiner problem, but there are a number of such problems, and it is a good idea to be able to distinguish between them.

Also see

 * Definition:Extremum Problem