Why bubbles are round
Tonight's idea
A soap film shrinks as far as it can, so a bubble wraps its air in the least possible surface: a sphere. The flat version: a loop of string 12 units long should enclose as much area as possible. Try regular polygons. More sides, more area, and the circle is the ceiling.
| sides | 3 | 4 | 6 | 12 | 100 | circle |
|---|---|---|---|---|---|---|
| area | 6.928 | 9 | 10.392 | 11.196 | 11.455 | 11.459 |
Isoperimetric theorem. Of all shapes with the same perimeter, the circle encloses the most area.
Jakob Steiner found moves that improve any shape that is not a circle. Two of them:
- A dent. Join its ends with a straight line and flip the dent across it. Same perimeter, more area.
- Lopsided. Cut the shape with a line that halves its perimeter. Keep the half with more area and replace the other by its mirror image. Same perimeter, more area.
A third move (the stretch) handles the rest. So no shape except the circle can be best. But be careful. This does not by itself prove that a best shape exists. Perhaps shapes can be improved for ever and none is best. That gap was closed later, by other methods.
Watch one: where the table comes from
Cut a regular polygon with \(n\) sides and perimeter 12 into \(n\) triangles from its centre. Each has base \(\tfrac{12}{n}\) and height \(h\), the distance from the centre to the middle of a side. The total area is
\[n \times \tfrac12 \times \tfrac{12}{n} \times h = 6h.\]
That is half the perimeter times \(h\). Halve one triangle: you get a right triangle with angle \(\tfrac{180°}{n}\) at the centre and opposite side \(\tfrac{6}{n}\). So \(h = \dfrac{6/n}{\tan(180°/n)}\).
Hexagon: \(h = \dfrac{1}{\tan 30°} = \sqrt3\), area \(6\sqrt3 \approx 10.392\). Circle: \(h\) is the radius \(\tfrac{12}{2\pi}\), area \(\tfrac{36}{\pi} \approx 11.459\).
Your turn
- A rectangle with perimeter 12 has sides \(x\) and \(6 - x\). Write its area as 9 minus a square. Which rectangle is best, and what is its area?
- Find \(h\) and the area for a regular octagon with perimeter 12, to three decimal places. What percentage of the circle's area is that?
- Turn the question round. A square and a circle each have area 100. Find both perimeters. Which shape needs less fence?
- "Every whole number bigger than 1 is smaller than its own square, so it can be improved. So no whole number except 1 can be the largest. Therefore 1 is the largest whole number." Which step fails? What does this say about Steiner's argument?
Stretch: Steiner's third move
Half a shape sits on a line, with ends \(A\) and \(B\). Pick \(C\) on its curve and put a hinge there. The two caps outside triangle \(ACB\) stay rigid. The triangle has area \(\tfrac12 \cdot CA \cdot CB \cdot \sin C\). Which angle \(C\) is best? Which curve has that angle at every point?
Answers
- \(x(6 - x) = 6x - x^2 = 9 - (x - 3)^2\). The square is never negative, so the area is at most 9, when \(x = 3\). The best rectangle is the 3 by 3 square.
- \(h = \dfrac{6/8}{\tan 22.5°} = \dfrac{0.75}{0.41421} \approx 1.811\), and the area is \(6h \approx 10.864\). The circle has 11.459, so the octagon reaches about 94.8% of it.
- The square has side 10 and perimeter 40. The circle has \(\pi r^2 = 100\), so \(r = \sqrt{100/\pi} \approx 5.642\) and perimeter \(2\pi r \approx 35.45\). The circle needs less fence. Most area for a given perimeter and least perimeter for a given area are the same fact.
- The first two sentences are true. The last step fails: it quietly assumes that a largest whole number exists. All the argument shows is that if there is a largest, it is 1, and there is none. Steiner's moves show in the same way that if a best shape exists, it must be the circle. That a best shape exists needs a separate proof.
Stretch. Opening or closing the hinge keeps the lengths \(CA\) and \(CB\), the two caps, and the length of the curve. Only \(\sin C\) changes, and it is largest when \(C = 90°\). So a best half-shape has a right angle at every point \(C\) of its curve. By the converse of Thales' theorem, each such \(C\) lies on the circle with diameter \(AB\). The half is a semicircle, and with its mirror image the whole shape is a circle.