The now famous Dutch graphic artist M. C. Escher was fond of and made a frequent use of plane tessellations, sometimes by the copies of the same figure. One particular tessellation with a special kind of hexagon has been mentioned in his notebooks and now is part of Escher's theorem (after J. F. Rigby.) Here that theorem with a slight change of notations.
- Let A'B'C' be an equilateral triangle and B any point. Let C be the point such that
A'B = A'Cand ∠CA'B = 120°.Let A be such that B'A = B'Cand ∠CB'A = 120°.Then C'A = C'Band ∠AC'B = 120°.
- Congruent copies of hexagon AC'BA'CB' can be used to tessellate the plane.
- The line AA', BB', CC' are concurrent.
As Rigby observed, the theorem was not stated accurately because no assumption had been made as to the orientation of the three 120° angles used in the construction nor the orientation of ΔA'B'C'. For the theorem to work those orientations must be the same. (They are the same in the applet below.) In all likelihood, Escher had this in mind but did not mention in his Notes.
|What if applet does not run?|
To prove the first part, consider Napoleon triangles ABC'', AB''C abd A''BC. By Napoleon's theorem, their centers form an equilateral triangle. A' and B' are two of the vertices of that equilateral triangle and since ΔA'B'C' is equilateral, C' is the third vertex.
The applet serves to illustrate the second part of the claim. The tessellation can be obtained from the Napoleon's tessellation we discussed elsewhere bu joining vertices of the equlateral triangles to their centers. (To see that in the applet, check both Show tessellation and Hint boxes.)
The third part is the content of Kiepert's theorem.
- J. F. Rigby, Napoleon, Escher, and Tessellations, Mathematics Magazine, Vol. 64, No. 4, (Oct., 1991), pp. 242-246
Copyright © 1996-2018 Alexander Bogomolny