A Point in the TriangleDate: 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/ |
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