Asymmetric Propeller: What Is It?
A Mathematical Droodle

Explanation

Copyright © 1996-2010 Alexander Bogomolny

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 60^o. If O is the center of a coordinate system and d a rotation through 60^o in the positive direction, then A' = dA, B' = dB, and C' = dC. For j = d², a rotation through 120^o, a well-known criterion is used to verify whether a triangle is equilateral or not. In our case, we should check whether

Let's see. Since j² = d⁴ = -d,

So indeed, the triangle in question is equilateral.

Hubert Shutrick found a geometric proof based on two lemmas, where all triangles are assumed equilateral:

Lemma 1

If ΔA₁B₁O, ΔA₂B₂O and ΔA₃B₃O satisfy the theorem, then so do ΔA'₁B'₁O, ΔA₂B₂O and ΔA₃B₃O where A₁, A'₁, O collinear and so are B₁, B'₁, O.

Lemma 2

If ΔA₁B₁O, ΔA₂B₂O and ΔA₃B₃O satisfy the theorem, then so do ΔA'₁B'₁O, ΔA₂B₂O and ΔA₃B₃O where ΔA'₁B'₁O is obtained from ΔA₁B₁O by a rotation about O.

Since the theorem is obviously true when the three triangles are equal and equally spaced and the two lemmas show that any configuration can be transformed into the latter in a finite number of steps, one only needs to prove the two lemmas.

Proof of Lemma 1

In the diagram, the M's are the midpoints and we note that 2M₂M'₂ = A₁A'₁= B₁B'₁ = M₃M'₃ and that M₂M''₂ is parallel with A₁A'₁ and M₃M'₃ is parallel with B₁B'₁. It's enough to prove that the triangles M₁M₂M'₂ and M₁M₃M'₃ are congruent and we already have two of their sides equal. The parallelism gives that the quadrilateral M₁NM₂M₃ is cyclic so the angles NM₂M₁ and NNM₃M₁ are equal and the lemma is proved.

Proof of Lemma 2

Once again, M₂M'₂ = M₃M'₃ but now M₂M'₂ is parallel with A₁A'₁ and M₃M'₃ is parallel with B₁B'₁. Since the line A₁A'₁ goes to B₁B'₁ by rotating 60° about O, the line M₂M'₂ will go to M₃M'₃ by rotating 60° about M₁ and since M₂M'₂ = M₃M'₃ the point M'₂ will go to M'₃. Therefore, M₁M₂M'₂ and M₁M₃M'₃ are congruent as required.

Copyright © 1996-2010 Alexander Bogomolny