Four Centroids and Parallels: What Is This About?
A Mathematical Droodle

 

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Explanation

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Copyright © 1996-2018 Alexander Bogomolny

Explanation

The applet attempts to suggest a simple result from C. Bradley's article, The Fermat Point Configuration, in the Math Gazette (2008, pp. 214-222.)

A point P in the plane of ΔABC, forms three triangles ABP, BCP, CAP with centroids Gc, Ga, Gb, respectively. The lines through these centroids parallel to CP, AP, BP, respectively, meet in G, the centroid of ΔABC. Additionally, triangles ABC and GaGbGc are homothetic, with center of the homothety incident to PG and divides the latter in proportion 1:3.

 

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Since the configuration is that of centroids, hence of medians, let's consider the midpoint L of BC. BC is a side in both ΔABC and ΔBCP. Thus we have

  LG : LA = 1 : 3 and,
LGa : LP = 1 : 3,

so that the homothety with center L and coefficient 3 maps G on A and Ga on P making GGa||AP. Similarly, GGb||BP and GGc||CP.

Let U be the midpoint of AG. In ΔBCU, CGb : GbU = 2 : 1 and BGc : GcU = 2 : 1 making GbGc||BC and GbGc : BC = 1 : 3. Similar claims regarding GaGb and GaGc also hold, making the two triangles homothetic.

P and G are images of each other under that homothety implying that the center, say J of homothety lies on GP such that GJ : JP = 1 : 3.

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Copyright © 1996-2018 Alexander Bogomolny
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