mscroggs.co.uk
mscroggs.co.uk

subscribe

Blog

 2019-04-09 
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 rectanglesPolynomialSolution
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.

Similar posts

Dragon curves II
TMiP 2019 treasure punt
Christmas card 2018
World Cup stickers 2018, pt. 3

Comments

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.
Reply
(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.
Reply
g0mrb
 Add a Comment 


I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li>
To prove you are not a spam bot, please type "nogaced" backwards in the box below (case sensitive):

Archive

Show me a random blog post
 2019 

Sep 2019

A non-converging LaTeX document
TMiP 2019 treasure punt

Jul 2019

Big Internet Math-Off stickers 2019

Jun 2019

Proving a conjecture

Apr 2019

Harriss and other spirals

Mar 2019

realhats

Jan 2019

Christmas (2018) is over
 2018 
▼ show ▼
 2017 
▼ show ▼
 2016 
▼ show ▼
 2015 
▼ show ▼
 2014 
▼ show ▼
 2013 
▼ show ▼
 2012 
▼ show ▼

Tags

wool mathsjam chess pythagoras platonic solids reuleaux polygons game of life people maths logic accuracy the aperiodical error bars manchester dates european cup realhats interpolation stickers london plastic ratio a gamut of games nine men's morris chebyshev braiding world cup trigonometry reddit cambridge games pac-man craft talking maths in public captain scarlet matt parker speed hats coins countdown palindromes big internet math-off puzzles books propositional calculus ternary tennis rugby tmip folding paper arithmetic hexapawn frobel twitter folding tube maps triangles weather station royal baby go curvature machine learning sport light golden spiral harriss spiral mathsteroids bodmas programming binary inline code final fantasy electromagnetic field graph theory christmas approximation draughts oeis chalkdust magazine game show probability noughts and crosses national lottery news flexagons polynomials javascript asteroids raspberry pi gerry anderson video games dataset latex fractals christmas card football golden ratio map projections dragon curves python probability london underground cross stitch estimation pizza cutting manchester science festival misleading statistics php geometry radio 4 sorting rhombicuboctahedron menace sound bubble bobble martin gardner data statistics mathslogicbot

Archive

Show me a random blog post
▼ show ▼
© Matthew Scroggs 2019