Four Centroids and Parallels: What Is This About?
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Copyright © 1996-2012 Alexander BogomolnyExplanation
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
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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
P and G are images of each other under that homothety implying that the center, say J of homothety lies on GP such that
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Copyright © 1996-2012 Alexander Bogomolny| 41143735 |
