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2019-04-09

## Harriss and other spirals

In the latest issue of Chalkdust, I wrote an article with Edmund Harriss about the Harriss spiral that appears on the cover of the magazine. To draw a Harriss spiral, start with a rectangle whose side lengths are in the plastic ratio; that is the ratio $$1:\rho$$ where $$\rho$$ is the real solution of the equation $$x^3=x+1$$, approximately 1.3247179.
A plastic rectangle
This rectangle can be split into a square and two rectangles similar to the original rectangle. These smaller rectangles can then be split up in the same manner.
Splitting a plastic rectangle into a square and two plastic rectangles.
Drawing two curves in each square gives the Harriss spiral.
A Harriss spiral
This spiral was inspired by the golden spiral, which is drawn in a rectangle whose side lengths are in the golden ratio of $$1:\phi$$, where $$\phi$$ is the positive solution of the equation $$x^2=x+1$$ (approximately 1.6180339). This rectangle can be split into a square and one similar rectangle. Drawing one arc in each square gives a golden spiral.
A golden spiral

### Continuing the pattern

The golden and Harriss spirals are both drawn in rectangles that can be split into a square and one or two similar rectangles.
The rectangles in which golden and Harriss spirals can be drawn.
Continuing the pattern of these arrangements suggests the following rectangle, split into a square and three similar rectangles:
Let the side of the square be 1 unit, and let each rectangle have sides in the ratio $$1:x$$. We can then calculate that the lengths of the sides of each rectangle are as shown in the following diagram.
The side lengths of the large rectangle are $$\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1$$ and $$\frac1{x^2}+\frac1x+1$$. We want these to also be in the ratio $$1:x$$. Therefore the following equation must hold:
$$\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1=x\left(\frac1{x^2}+\frac1x+1\right)$$
Rearranging this gives:
$$x^4-x^2-x-1=0$$ $$(x+1)(x^3-x^2-1)=0$$
This has one positive real solution:
$$x=\frac13\left( 1 +\sqrt[3]{\tfrac12(29-3\sqrt{93})} +\sqrt[3]{\tfrac12(29+3\sqrt{93})} \right).$$
This is equal to 1.4655712... Drawing three arcs in each square allows us to make a spiral from a rectangle with sides in this ratio:
A spiral which may or may not have a name yet.

### Continuing the pattern

Adding a fourth rectangle leads to the following rectangle.
The side lengths of the largest rectangle are $$1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4}$$ and $$1+\frac2x+\frac1{x^2}+\frac1{x^3}$$. Looking for the largest rectangle to also be in the ratio $$1:x$$ leads to the equation:
$$1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4} = x\left(1+\frac2x+\frac1{x^2}+\frac1{x^3}\right)$$ $$x^5+x^4-x^3-2x^2-x-1 = 0$$
This has one real solution, 1.3910491... Although for this rectangle, it's not obvious which arcs to draw to make a spiral (or maybe not possible to do it at all). But at least you get a pretty fractal:

### Continuing the pattern

We could, of course, continue the pattern by repeatedly adding more rectangles. If we do this, we get the following polynomials and solutions:
 Number of rectangles Polynomial Solution 1 $$x^2 - x - 1=0$$ 1.618033988749895 2 $$x^3 - x - 1=0$$ 1.324717957244746 3 $$x^4 - x^2 - x - 1=0$$ 1.465571231876768 4 $$x^5 + x^4 - x^3 - 2x^2 - x - 1=0$$ 1.391049107172349 5 $$x^6 + x^5 - 2x^3 - 3x^2 - x - 1=0$$ 1.426608021669601 6 $$x^7 + 2x^6 - 2x^4 - 3x^3 - 4x^2 - x - 1=0$$ 1.4082770325090774 7 $$x^8 + 2x^7 + 2x^6 - 2x^5 - 5x^4 - 4x^3 - 5x^2 - x - 1=0$$ 1.4172584399350432 8 $$x^9 + 3x^8 + 2x^7 - 5x^5 - 9x^4 - 5x^3 - 6x^2 - x - 1=0$$ 1.412713760332943 9 $$x^{10} + 3x^9 + 5x^8 - 5x^6 - 9x^5 - 14x^4 - 6x^3 - 7x^2 - x - 1=0$$ 1.414969877544769
The numbers in this table appear to be heading towards around 1.414, or $$\sqrt2$$. This shouldn't come as too much of a surprise because $$1:\sqrt2$$ is the ratio of the sides of A$$n$$ paper (for $$n=0,1,2,...$$). A0 paper can be split up like this:
Splitting up a piece of A0 paper
This is a way of splitting up a $$1:\sqrt{2}$$ rectangle into an infinite number of similar rectangles, arranged following the pattern, so it makes sense that the ratios converge to this.

### Other patterns

In this post, we've only looked at splitting up rectangles into squares and similar rectangles following a particular pattern. Thinking about other arrangements leads to the following question:
Given two real numbers $$a$$ and $$b$$, when is it possible to split an $$a:b$$ rectangle into squares and $$a:b$$ rectangles?
If I get anywhere with this question, I'll post it here. Feel free to post your ideas in the comments below.

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Comments in green were written by me. Comments in blue were not written by me.
2019-05-02
@g0mrb: CORRECTION: There seems to be no way to correct the glaring error in that comment. A senior moment enabled me to reverse the nomenclature for paper sizes. Please read the suffixes as (n+1), (n+2), etc.

(anonymous)
2019-05-02
I shall remain happy in the knowledge that you have shown graphically how an A(n) sheet, which is 2 x A(n-1) rectangles, is also equal to the infinite series : A(n-1) + A(n-2) + A(n-3) + A(n-4) + ... Thank-you, and best wishes for your search for the answer to your question.

g0mrb