mscroggs.co.uk
mscroggs.co.uk
Click here to win prizes by solving the mscroggs.co.uk puzzle Advent calendar.
Click here to win prizes by solving the mscroggs.co.uk puzzle Advent calendar.

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
Christmas card 2019
TMiP 2019 treasure punt
Christmas card 2018

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 "ratio" in the box below (case sensitive):

Archive

Show me a random blog post
 2019 

Dec 2019

Christmas card 2019

Nov 2019

Christmas (2019) is coming!

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

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

Archive

Show me a random blog post
▼ show ▼
© Matthew Scroggs 2012–2019