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Ratio in nature and art

Ratio, rates and money · level 8, lesson 6 of 6

A question to start

Take a sheet of A4 paper, the standard size in most of the world. Fold it in half. The half-sheet is A5, and it has exactly the same shape as the A4 sheet, only smaller. Fold again: A6, same shape again. Put two A4 sheets side by side: A3, the same shape once more.

Most rectangles do not behave like this. Fold a square in half and you get a long thin strip; fold a North American letter-size sheet in half and you get a fatter shape. Which ratio of sides makes a rectangle keep its shape when halved? And what about the famous "golden ratio", said to be hidden in shells, faces and Greek temples? This last lesson of the topic looks at ratio as shape, with two special numbers, \(\sqrt2\) and the golden ratio, and an honest look at which of the stories about them are true.

The idea

Ratio as shape

Two rectangles have the same shape (they are similar, as in Similar triangles) exactly when the ratio of long side to short side is the same. So "keeping its shape" is a statement about one ratio.

A-series paper: the ratio \(1 : \sqrt2\)

Let a sheet have long side \(L\) and short side \(S\). Folding it in half across the long side gives a sheet with long side \(S\) and short side \(\tfrac L2\). For the same shape, the ratios must match:

\[\frac{L}{S} = \frac{S}{L/2} = \frac{2S}{L}.\]

Cross multiply: \(L^2 = 2S^2\), so \(\dfrac{L}{S} = \sqrt2 \approx 1.414\). Only the ratio \(\sqrt2 : 1\) keeps its shape when halved. Since \(\sqrt2\) is irrational (why), real sheets are rounded to whole millimetres.

The international standard ISO 216 starts from A0, a sheet of area exactly 1 square metre with sides in the ratio \(\sqrt2 : 1\). If \(S \times L = 1\) and \(L = \sqrt2\,S\), then \(\sqrt2\,S^2 = 1\), so

\[S = 2^{-1/4} \approx 0.8409 \text{ m}, \qquad L = 2^{1/4} \approx 1.1892 \text{ m}.\]

Rounded: A0 is 841 mm × 1189 mm. Each next size halves the longer side: A1 is 594 × 841, A2 420 × 594, A3 297 × 420, A4 210 × 297, A5 148 × 210 (each rounded down to the millimetre).

A1 A2 A3 A4 A5 A5

A photocopier's "141%" and "71%" buttons are \(\sqrt2\) and \(\tfrac{1}{\sqrt2}\): they turn A4 into A3 and A4 into A5 with nothing cut off. Lengths scale by \(\sqrt2\), so areas scale by 2, as they should for a sheet twice the size.

The golden ratio

Divide a length into two parts, a longer part \(a\) and a shorter part \(b\), so that the whole is to the longer part as the longer part is to the shorter:

\[\frac{a + b}{a} = \frac{a}{b}.\]

Call the ratio \(\varphi = \tfrac ab\) (the Greek letter phi). The left side is \(1 + \tfrac ba = 1 + \tfrac{1}{\varphi}\), so \(\varphi = 1 + \tfrac1\varphi\). Multiply by \(\varphi\):

\[\varphi^2 = \varphi + 1, \qquad \varphi^2 - \varphi - 1 = 0.\]

The quadratic formula gives \(\varphi = \tfrac{1 \pm \sqrt5}{2}\), and since a ratio of lengths is positive,

\[\varphi = \frac{1 + \sqrt5}{2} \approx 1.6180.\]

This is the golden ratio. Its equation gives it odd properties: \(\varphi^2 = \varphi + 1 \approx 2.618\) and \(\tfrac{1}{\varphi} = \varphi - 1 \approx 0.618\). Squaring adds 1; taking the reciprocal subtracts 1.

The golden rectangle

A golden rectangle has sides in the ratio \(\varphi : 1\). Cut a square off one end, and the piece left over is another golden rectangle. Proof: with sides \(\varphi\) and 1, removing a \(1 \times 1\) square leaves a rectangle with sides 1 and \(\varphi - 1\), whose ratio is \(\tfrac{1}{\varphi - 1} = \tfrac{1}{1/\varphi} = \varphi\). So you can keep cutting squares for ever.

1 × 1 φ ≈ 1.618 1

Fibonacci numbers lead to \(\varphi\)

The Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... add the two previous terms each time. The ratios of neighbours are

\[\tfrac21 = 2,\ \tfrac32 = 1.5,\ \tfrac53 \approx 1.667,\ \tfrac85 = 1.6,\ \tfrac{13}{8} = 1.625,\ \tfrac{21}{13} \approx 1.6154,\ \tfrac{34}{21} \approx 1.6190,\ \tfrac{55}{34} \approx 1.6176,\ \tfrac{89}{55} \approx 1.6182.\]

They swing above and below, closing in on 1.6180. Here is why. Write \(x_n = \tfrac{F_{n+1}}{F_n}\). Since \(F_{n+1} = F_n + F_{n-1}\),

\[x_n = 1 + \frac{F_{n-1}}{F_n} = 1 + \frac{1}{x_{n-1}}.\]

If the ratios settle to a limit \(L\), it must satisfy \(L = 1 + \tfrac1L\): the golden ratio's own equation. (Showing that they do settle takes a little more work; it is a classic exercise on sequences.)

What is true in nature

  • Sunflowers, pinecones, pineapples. The spirals you can see in a sunflower head usually come in two families, and the numbers of spirals are very often neighbouring Fibonacci numbers, such as 34 and 55. A good explanation is that each new seed grows at about \(137.5°\) from the last, the golden angle \(\tfrac{360°}{\varphi^2}\), which packs seeds evenly with no gaps lining up. This is well supported by measurement and by models, though not every plant follows it.
  • The nautilus shell. It is a real logarithmic spiral, but measurements show its growth ratio is not the golden ratio, and it varies from shell to shell. The "golden nautilus" is a myth.

What is doubtful in art and architecture

  • The Parthenon in Athens is often said to be built on golden rectangles. The claim is doubtful: the result depends on which edges and points you choose to measure, and there is no evidence that its builders used the golden ratio.
  • Similar claims about the Great Pyramid, the Mona Lisa and the "perfect face" rest on the same loose measuring.
  • Some modern designers really did use it on purpose. The architect Le Corbusier based his 20th-century "Modulor" system of proportions on the golden ratio. That is a true story about a choice, not a law of beauty.

Why are so many false claims made? Because with enough lengths to choose from, some ratio near 1.6 is almost always there to be found. Anything between about 1.57 and 1.67 is within 3% of \(\varphi\), and many everyday shapes fall in that range. A claim is convincing only if the points are chosen before measuring and the match is close.

Worked examples

1. A4 from A0. Each halving divides the long side by 2 and swaps the sides. Two halvings divide both sides by 2. So A4, four halvings from A0, has sides A0's divided by 4: \(841 \div 4 \approx 210.25\) and \(1189 \div 4 \approx 297.25\), which round to the official 210 mm × 297 mm. Check the ratio: \(297 \div 210 \approx 1.414\). ✓

2. A sheet's mass. Office paper is often 80 g/m². A4 is \(\tfrac{1}{16}\) of A0, so it has area \(\tfrac{1}{16}\) m² and mass \(80 \div 16 = 5\) grams. (A0 has area 1 m² for exactly this convenience.)

3. The golden rectangle check. A rectangle is 1.618 by 1. Remove a 1 × 1 square: 1 by 0.618 is left, and \(1 \div 0.618 \approx 1.618\). ✓

4. A doubtful match. Someone measures a building 31 m wide and 19 m tall and announces the golden ratio. \(31 \div 19 \approx 1.632\), within 1% of \(\varphi\). But had they measured to the roof edge instead, 31 by 20, they would get 1.55. A ratio near 1.6 between two numbers someone has chosen after the fact is weak evidence of anything.

Try it

  1. Find the ratio of the sides of A3 (297 × 420 mm) and compare with \(\sqrt2\).
  2. What are the dimensions of A6, rounded down to the millimetre, starting from A5 (148 × 210)?
  3. What enlargement, as a percent, turns A5 into A3?
  4. Check numerically that \(\varphi^2 = \varphi + 1\) and \(\tfrac1\varphi = \varphi - 1\).
  5. Find \(\tfrac{144}{89}\) and \(\tfrac{233}{144}\). Which side of \(\varphi\) is each on?
  6. A rectangle is 8 cm by 5 cm. Is it a golden rectangle? Can any rectangle with whole-number sides be exactly golden?
  7. Letter-size paper is 8.5 by 11 inches. Fold it in half. Does it keep its shape? Find both ratios.
  8. (Harder.) Show that \(\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cdots}}\) follows from \(\varphi = 1 + \tfrac1\varphi\), and that stopping this continued fraction after a few steps gives ratios of Fibonacci numbers.
Answers
  1. \(420 \div 297 \approx 1.4141\), against \(\sqrt2 \approx 1.4142\). Equal to 3 decimal places.
  2. Halve the long side: \(210 \div 2 = 105\). A6 is 105 × 148 mm.
  3. Two steps up, each \(\times\sqrt2\): lengths \(\times 2\), so 200%.
  4. \(\varphi \approx 1.618034\); \(\varphi^2 \approx 2.618034 = \varphi + 1\) ✓; \(\tfrac1\varphi \approx 0.618034 = \varphi - 1\) ✓.
  5. \(144 \div 89 \approx 1.61798\), just below \(\varphi\); \(233 \div 144 \approx 1.61806\), just above.
  6. \(8 \div 5 = 1.6\): close (they are Fibonacci numbers) but not golden. No whole-number rectangle can be exactly golden, because \(\varphi = \tfrac{1 + \sqrt5}{2}\) is irrational, so it is not a ratio of two whole numbers.
  7. \(11 \div 8.5 \approx 1.294\). The half-sheet is 8.5 by 5.5: \(8.5 \div 5.5 \approx 1.545\). Different ratios, so it does not keep its shape.
  8. Replace the \(\varphi\) in \(1 + \tfrac1\varphi\) by \(1 + \tfrac1\varphi\) again and again to get the continued fraction. Stopping early: \(1\), \(1 + \tfrac11 = 2\), \(1 + \tfrac12 = \tfrac32\), \(1 + \tfrac{1}{3/2} = \tfrac53\), \(1 + \tfrac{3}{5} = \tfrac85\), ... each step is \(x \mapsto 1 + \tfrac1x\), which is exactly the Fibonacci ratio rule, so the answers are \(\tfrac{F_{n+1}}{F_n}\).

Common mistake

Seeing the golden ratio everywhere. Measure enough things and some ratio will land near 1.618. Ask which points were chosen, whether they were chosen before measuring, and how close the match really is.

Thinking every paper size halves neatly. Only the ratio \(\sqrt2 : 1\) keeps its shape when halved. Letter-size paper does not, which is why reducing two letter pages onto one sheet leaves margins.

Where it leads