Another Irregular tiling using Phi: the Ammann Chair

The Ammann Chair Tiling

 

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A single tile which produces an "irregular" tiling was found by Robert Ammann in 1977. Its dimensions involve powers of the square root of Phi (Φ):

which means it can be split into two smaller tiles of exactly the same shape. They all depend on the property of Φ that

By splitting the larger tile, and any others of identical size, in the same way, we can produce more and more tiles all of exactly the same shape, but the whole tiling will never have any periodic pattern.
Each time we split a tile, we make one that is

times the size and one that is

times the size. There are only two sizes of tile in each dissection.

The buttons under the tile show successive stages as the dissections get more numerous.

This tiling is irregular or aperiodic which means that no part of it will appear as an indefinitely recurring pattern as in the regular tilings.

Enrique Zeleny's Ammann Chair Mathematica demonstration shows these dissections either animated on the page or using the free Mathematica Player, and in colour too.

Tilings and Patterns B Gr�nbaum, G C Shephard (a Dover paperback of the original 1987 edition is due out in 2009) ISBN 0486469816 is encyclopaedic in its depth and range of tilings and patterns. It mentions this Ammann tiling on page 553.
Penrose Tiles to Trapdoor Ciphers: And the Return of Dr Matrix, M Gardner, (The Mathematical Association of America; Revised edition 1997), the second chapter gives more on Ammann's work but omits this simplest of aperiodic tilings given above. As with all mathematical books by Martin Gardner, they are excellent and I cannot recommend them highly enough!
Aperiodic Tiles, R.Ammann, B. Gr�nbaum, G.C. Shephard, Discrete and Computational Geometry (1992), pages 1-25.

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More Geometrical Gems

Here are some ore little gems of geometry showing Phi in unexpected places.
The first is about cutting three triangles from a rectangle. The next two are about balancing a flat shape on a pin head or pencil point. The final one is about areas of flat circular shapes.

First the two balancing puzzles.
The centre of gravity of a flat shape is the point at which the shape balances horizontally on a point such as the tip of a pencil. If you are careful, you can get the shape to spin on that point, also called the pivot(al) point.

A Rectangle-Triangle dissection Problem

The problem is, given a rectangle, to cut off three triangles from the corners of the rectangle so that all three triangles have the same area. Or, expressed another way, to find a triangle inside a given rectangle (any rectangle) which when it is removed from the rectangle leaves three triangles of the same area.
As shown here, the area of the leftmost triangle is x(w+z)/2.
The area of the top-right triangle is yw/2.
The area of the bottom triangle is (x+y)z/2.
Making these equal means:

The first equation tells us that x = yw/(w+z).
The second equation, when we multiply out the brackets and cancel the zx terms on each side, tells us that xw = zy. This means that y/x = w/z.

Or in other words, we have our first deduction that

Returning to xw = zy, we put x = yw/(w+z) into it giving yw²/(w+z)=zy.
We can cancel y from each side and rearrange it to give w² = z² + zw.
If we divide by z² we have a quadratic equation in w/z: (w/z)² = 1 + w/z
Let X=w/z then X² = 1 + X
The positive solution of this is X = Phi, that is, w/z = Phi or w = z Phi.
Since we have already seen that y/x = w/z then:

The Golden Section strikes again!

This puzzle appeared in J A H Hunter's Triangle Inscribed In a Rectangle in The Fibonacci Quarterly, Vol 1, 1963, Issue 3, page 66.
A follow-up article by H E Huntley entitled Fibonacci Geometry in volume 2 (1964) of the Fibonacci Quarterly on page 104 shows that, if the rectangle is itself a golden rectangle (the ratio of the longer side to the shorter one is Phi) then the triangle is both isosceles and right-angled!

Balancing an "L" shape

	In this first problem, suppose we take a square piece of card. Where will the pivot point be? That should be easy to guess - at the centre of the square.
	Now suppose we remove a small square from one corner to make an "L" shape. Where will the pencil point be now if the "L" is to balance and turn freely? The centre of gravity this time will be close to the centre but down a little on one of the square's diagonals.
	If we took off a very large square to make the "L" shape quite thin, then the centre of gravity would lie outside the L shape and we could not balance it at all!
	So we are left with the question What is the largest size of square that we can remove so that we can still just balance the L shape? Nick Lord found it was when the ratio of the big square's side and the removed square's side are in the golden ratio phi!

Note 79.59: Balancing and Golden Rectangles N Lord, Mathematical Gazette vol 79 (1995) pages 573-4.

Balancing a Crescent

Here is a similar problem but this time using circles: to remove the largest circle and still have a pivotal point:

The full circle balances at its centre point

Removing a small circle at the top moves the balance point down a little

Removing a large circle moves the balance point outside the crescent shape

The limit point is when the radii of the two circles are in the golden ratio and the centre of gravity is just on the edge of the inner circle