The Fermat Point

or

where to build the Airport/Mall

Suppose three cities (Albany, Schenectady, & Troy?) form a triangle, and they want to build an airport (or a mall) near their common center.  In order to save money they decide to make the access roads as short as possible.  A committee (with a mathematician on it?) decides that the best way to achieve this is to find a point where the total of the distances to each city’s center is a minimum.
     So the question then is ‘Where to put the airport?’.
    This point is called the Fermat Point, because Fermat posed this problem to Torricelli, who solved it in 1659 (does this make sense?).
    Thus we have a triangle ABC (let’s assume it is acute to start with) and a point P in the interior with connecting lines to the vertices, and we wish to minimize the total length PA+PB+PC.
    If we could ‘unfold’ the lengths PA, PB, & PC and make a straight line we could then maybe use the same method as that used in finding the minimum inscribed triangle.
    Luckily this can be done.
    Consider the triangle APC interior to ABC, and rotate it 60° counterclockwise about A.  Then connect P to P’ and notice that APP’ forms an equilateral triangle?
    Now consider the polygonal line BP-PP’-P’C’, and the straight line segment BC.  Move the point P to this line BC’ so that P’ is also on BC’.
    Then BP-PP’-P’C’ coincides with BC’, so it is the shortest distance from B to C’, and thus PA+PB+PC is at a minimum, and we have found P?
    Notice that the angle AP’C’ is 120°?, so this means that P is that point where the interior angles APC, APB, & APC are all 120°?
    Also note that if the triangle ABC has an angle of 120° or larger, then the point P will coincide with that vertex where this angle occurs?
    This problem is easily illustrated using Geometer’s SketchPad (see the sketchpad sketch ‘Fermat Pt.’).