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Napoleon's TheoremOn each side of a triangle, erect an equilateral triangle, lying exterior to the original triangle. Then the segments connecting the centroids of the three equilateral triangles themselves form an equilateral triangle. By Dr. Scott Brodie, M.D., Ph.D.
Since the centroid of a triangle lies along each median, 2/3 of the distance from the vertex to the midpoint of the opposite side, we have u = (2/3)·sqrt(3)/2 · b = b/sqrt(3), and (1) becomes
Expanding(*) the cosine of the sum, and recalling that cos(60o) = 1/2; sin(60o) = sqrt(3)/2, we have
Substituting (3) into (2) yields
Now apply the Law of Cosines to
and recall (as in the derivation of the Law of Sines):
Substituting (5) and (6) into (4) gives
Since (7) is symmetrical in a, b, and c, it follows that the triangle connecting the three centroids is equilateral, QED.  (*) Michael Lambrou has suggested a different way to proceed after obtaining (2). Apply the Law of Cosines to triangles ABE and BCE to express the side BE in two different ways: By (2), the left hand side equals 3 times GI. Similarly, the right hand side equals 3 times IH, wherefrom As a bonus, we get AD = BE = CF.
Now consider the six triangles that converge on point A. Three
of them (ABD, ACF, and A.EE.BB) are equilateral. Recollect that the angles of a triangle
sum to 180o, while the angles around a point sum to 360o.
Since
Copyright © 1996-2008 Alexander Bogomolny |
ABC, A denotes both the vertex A and the correposnding 
A, a is both BC and its length.
In addition, let G denote the centroid
of the equilateral triangle on side AB, and I denote the centroid
of the equilateral triangle on side AC, etc. Let s denote the length of
segment GI, t the length of segment AG, and u the length of segment AI.
(
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