The Fermat Point (AKA the Torricelli Point; AKA the isogenic center; AKA the Steiner Point) of a triangle ABC is a point P such that the sum of the segments AP, BP and CP is minimized. A good overview of Fermat Points is available at various sites around the web including [url=https://en.wikipedia.org/wiki/Fermat_point]Wikipedia[/url] and [url=http://mathworld.wolfram.com/FermatPoints.html]Wolfram MathWorld[/url]. This page introduces you to the core concepts of Fermat Points and their role in network optimization in a hands-on way.[br][br]To start, try moving the point P in the applet below to shorten the network AP+BP+CP which connects A, B and C.

You probably can't get the exact minimum by just playing around, but you should have noticed that the optimal location for P appears to be fairly close to point B. You may also have noticed that it appears that the three angles formed at P by the network are approximately 120 degrees. I've zoomed way in on P and added the angles at P so you can really get a sense for this second observation.

It is true that when P forms three 120 degree angles and is inside triangle ABC, the network AP+BP+CP is minimized. The point P that achieves this is called the [b]Fermat Point[/b] of triangle ABC.[br][br]In the next two applets we will show:[br] [br][list][*]how to construct the Fermat Point of a Triangle when all angles of triangle ABC are less than 120 degrees.[/*][*]that the Fermat Point of a triangle does indeed minimize the network required to connect the vertices of a triangle ABC when all angles are less than 120 degrees. [/*][/list]Later we'll consider the (easier) case of triangle ABC has one angle greater than 120 degrees.

The above construction merely shows how to create a point P such that all three angles formed by P and the vertices A, B and C are 120 degrees. We still need to show that P as constructed does in fact minimize the network formed by AP, BP and CP. That proof is found in the interactive applet below. [br][br]Read the proof, and then note how moving P to the intersection of BB' and CC' minimizes the route from B to B', and therefore also the network AP, BP and CP. This critical observation is the key step of the proof. The point of the proof is merely to make it more transparent that locating P amounts to choosing the straightest route from B to B', and this happens to occur at the Fermat Point of triangle ABC.

The last thing to consider is the case when triangle ABC has a 120 degree or larger angle. In this case, the construction above places P outside triangle ABC. When this occurs, the optimal network is simply the two segments forming the 120 degree angle. The below applet illustrates this. Adjust B so it is "passes" P, the Fermat Point of triangle ABC. Notice that the optimal network (highlighted in red) switches from being AP+BP+CP to simply AB+BC. Also notice, this switch occurs precisely when the angle at B exceeds 120 degrees, which is also precisely when B "passes" P.

[b]Summary:[/b] [br][br][list][*]The Fermat Point of a triangle ABC minimizes the network required to connect points A, B and C when all angles of the triangle are less than 120 degrees. [/*][*]The Fermat Point of a triangle with all angles less than 120 degrees can be constructed by producing equilateral triangles on two legs, and then intersecting the segments formed by the equilateral triangles new vertex and the third vertex of the original triangle.[/*][*]When a triangle ABC has an angle that exceeds 120 degrees, the legs of this angle form the optimal network connecting points A, B and C.[/*][/list]