A Point in the Triangle

Date: 02/12/99 at 04:37:50
From: john allen
Subject: A Point in the Triangle

Hello Dr. Math,

Can you help me find the point in the plane of the triangle ABC, for 
which the sum of the distance from the vertices is a minimum? Can you 
please include the complete solution to this problem?

Thank you,
John

Date: 02/12/99 at 10:19:31
From: Doctor Floor
Subject: Re: A Point in the Triangle

Fermat first suggested the problem you send in, and Torricelli solved 
it in 1659. Therefore, the point we are looking for is known as the 
Fermat point, Torricelli point or Fermat-Torricelli point.

This point minimizes the distances only if all angles are smaller than 
120 degrees. The point is also known as the 1st isogonic (iso=equal, 
gon=angle) center, because it turns out that this point F has the 
property that angle(AFB) = angle(BFC) = angle(CFA) = 120 degrees.

The point F can be constructed in the following way: construct points 
A', B' and C' in such a way that A'BC, AB'C and ABC' are equilateral 
triangles that point outward triangle ABC. F is found as the 
intersection of AA', BB' and CC'.

You can find some information on the Fermat point, and a lot of other 
triangle centers, at the site of Professor Clark Kimberling (Univ. of 
Evansville):

   http://www.evansville.edu/~ck6/tcenters/

A result using F can be found at:
 
   A regular hexagon in a triangle - Floor van Lamoen
   http://home.wxs.nl/~lamoen/wiskunde/hexagon.htm

Now I will give you a proof that the constructed point is indeed the 
point that minimizes the distances to the vertices of ABC, published 
by Hofmann (1929):

For your reference, here is a picture of the situation described: 



Let D be a point inside triangle ABC. We give the distances names: 
d1 = AD, d2 = BD and d3 = CD. Construct an equilateral triangle DCG, 
such that G is pointing towards BC. So DG = CG = CD = d3.

Construct a point H such that triangle CGH is congruent to triangle 
CDB. In fact, we see that triangle CGH is a rotation of triangle CDB 
over 60 degrees (C is the rotation center). We see that GH = DB = d2. 
Also, we see that CH = CB and angle BCH = 60 degrees, so CBH is an 
equilateral triangle.

This means that, whatever point D was, we will always end up in the 
same point H: the third vertex of an equilateral triangle on side BC. 
Next we see that AD+DG+GH = d1+d2+d3. We can minimize AD+DG+GH by 
making it into a straight line, and the point that minimizes this must 
lie on AH. So the point minimizing d1+d2+d3 is at the intersection of 
lines AA', BB', and CC' as described above.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/