Four Centroids and Parallels: What Is This About?
A Mathematical Droodle
What if applet does not run? |
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 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 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_{a}G_{b}G_{c} are homothetic, with center of the homothety incident to PG and divides the latter in proportion 1:3.
What if applet does not run? |
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_{b}U = 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
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny68278856