Asymmetric Propeller: a Generalization


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Explanation

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Copyright © 1996-2012 Alexander Bogomolny


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This is a variant of the Asymmetric Propeller theorem, a simple generalization and the last one to deal with three equilateral triangles. The applet attempts to illustrate the proof by L. Bankoff, P. Erdös and M. S. Klamkin.

The three equilateral triangles with a shared vertex O may be conveniently denoted as OAA', OBB' and OCC'. Two triangles ABC and A'B'C' are equal, similarly oriented and are obtained from each other by a rotation through 60o. If O is the center of a coordinate system and d a rotation through 60o in the positive direction, then A' = dA, B' = dB, and C' = dC. For j = d2, a rotation through 120o, a well-known criterion is used to verify whether a triangle is equilateral or not. In our case, we should check whether

j2(A + B') + j(B + C') + (C + A') = 0.

Let's see. Since j2 = d4 = -d,

j2(A + B') + j(B + C') + (C + A') = -d(A + dB) + d2(B + dC) + (C + dA)
  = -dA - d2B + d2B - C + C + dA
  = 0.

So indeed, the triangle in question is equilateral.

Hubert Shutrick found an elegant geometric argument that could be described as a "nondiscrete induction": if the statement of the theorem holds for one configuration of three equilateral triangles then it also holds for a modification of the latter. Since one can always start with a symmetric configuration of three congruent equilateral triangles for which the statement obviously holds, it holds for any configuration.

Asymmetric Propeller

  1. Asymmetric Propeller (An Interactive Gizmo)
  2. Asymmetric Propeller: a Generalization
  3. A Case of Similarity
  4. Napoleon's Propeller
  5. Asymmetric Propeller and Napoleon's Theorem
  6. Asymmetric Propeller by Plane Tiling

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2012 Alexander Bogomolny

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