Four Centroids and Parallels

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

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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 G_c, G_a, G_b, respectively. The lines through these centroids parallel to CP, AP, BP, respectively, meet in G, the centroid of ΔABC. Additionally, triangles ABC and G_aG_bG_c 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,
LG_a : LP = 1 : 3,

so that the homothety with center L and coefficient 3 maps G on A and G_a on P making GG_a||AP. Similarly, GG_b||BP and GG_c||CP.

Let U be the midpoint of AG. In ΔBCU, CG_b : G_bU = 2 : 1 and BG_c : G_cU = 2 : 1 making G_bG_c||BC and G_bG_c : BC = 1 : 3. Similar claims regarding G_aG_b and G_aG_c 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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