A bit of maths · the teacher's corner
Why are soap bubbles round?
Same string — which shape holds the most space?
Soap is lazy: its skin contracts to have the least surface possible. In 2D, the champion shape is the circle; in 3D, the sphere. The question that mathematician Yilin Wang asked as a child. ▶ watch the video (2 min)
The same string (12 cm) — add sides, the area climbs
Sides
4
Perimeter (fixed)
12 cm
Area enclosed
9.0 cm²
Score 4πA ⁄ P²
0.79
0 ——— the circle = 1.00 (the maximum)
Still corners. The string encloses 9.0 cm² — the circle would hold 11.5. Add more sides.
4πA ≤ P² — for a given perimeter P, the area A never beats the circle's. The score 4πA/P² is 1 for the circle alone, and less for everything else. The bubble "aims" at that 1 without any maths.
Yilin Wang's childhood question
Yilin Wang is a mathematician (Institut des Hautes Études Scientifiques, 2024 Salem Prize). In an interview, she tells how, as a child, she would ask her parents: "why are soap bubbles round?" A very simple question — and the doorway to a big mathematical idea.
She adds that her mother, top of her physics class and now an architect, showed her early on that gender makes no difference in science. A child's question, a calling: exactly the spirit of A bit of maths.
▶ Watch the interview (CIRM, in French)Soap is lazy — from the string to the real bubble
A bubble's skin is stretched: like an elastic band, it pulls from every side and tries to shrink. Less surface = less energy. In 2D (our string), the shape that encloses a given area with the least outline is the circle. In 3D, the real bubble encloses a fixed volume of air with the least skin: that's the sphere.
fixed volume → least surface → sphere
Why the round one always wins — the isoperimetric score
- 📐 A score for each shape. We compute q = 4πA/P²: it is 1 for the circle, 0.91 for the hexagon, 0.79 for the square, and collapses for flat shapes.
- 🔵 More sides, better score. As you round the string off (triangle → square → hexagon → …), the score climbs toward 1, but only reaches it in the limit: the circle.
- 🫧 The bubble "cheats"… with physics. It computes nothing: its surface tension pushes it straight to the shape with the highest score. The maths result, done by nature.
The same idea explains dewdrops, lead shot cooled as it falls, and why a drop of oil in water balls up.
🎯 The bubble challenges — from age 6 to 18
Hint: set the number of sides and read the area; blow the bubble and read the score — the machine checks for you.
Challenge 1
Age 6–7 — Your string is 12 cm long. You shape it into a perfect square. How long is each side?
See the working
12 ÷ 4 = 3 cm per side: the string splits into 4 equal parts.
Challenge 2
Age 8–9 — This square is 3 cm on each side. What is its area (the space inside)?
See the working
3 × 3 = 9 cm²: with 12 cm of string, the square encloses 9 cm².
Challenge 3
Age 9–10 — Same 12 cm string, but flattened into a 1 cm by 5 cm rectangle. What is its area?
See the working
5 × 1 = 5 cm². The flatter the shape, the less it holds: the square (9) already does better.
Challenge 4
Age 10–11 — Same string: the square encloses 9 cm², the flat rectangle 5 cm². How much more does the square hold?
See the working
9 − 5 = 4 cm². Same length of string, yet the square holds 4 cm² more than the flat rectangle.
Challenge 5
Age 11–12 — Set the machine to 6 sides (a hexagon). What area does the string enclose? (read the machine, to 0.1)
See the working
A regular hexagon with perimeter 12 encloses ≈ 10.4 cm²: more sides = rounder = more area.
Challenge 6
Age 12–13 — The circle with a 12 cm perimeter has radius r = 12 ÷ (2π). Give it to 0.1.
See the working
12 ÷ (2 × 3.14) = 12 ÷ 6.28 ≈ 1.9 cm: the radius of the circle with the same string as the square.
Challenge 7
Age 13–14 — Blow the bubble all the way (the most sides). What area does the near-circle enclose? (to 0.1)
See the working
A_max = 144 ÷ (4π) ≈ 11.5 cm²: no shape 12 cm around does better than the circle.
Challenge 8
Age 15 — Each shape gets a score q = 4πA ⁄ P². For the square (A = 9, P = 12), work out q to 0.01.
See the working
4π × 9 ÷ 144 = 36π ÷ 144 = π ÷ 4 ≈ 0.79: the square only reaches 0.79, not 1.
Challenge 9
Age 16–18 — The score q = 4πA ⁄ P² never goes above a certain value, reached only by the circle. What is that maximum score?
See the working
q_max = 1: that's the isoperimetric inequality 4πA ≤ P², with equality for the circle alone. The bubble 'aims' at that 1.
Teaching machine: we compare regular polygons at a fixed perimeter; their area A(n) = P²/(4n·tan(π/n)) tends to the circle's P²/(4π) as the number of sides grows. The isoperimetric inequality (the circle maximises area for a given perimeter; the sphere minimises surface for a given volume) is a theorem; the bubble "solves" it through physics — its surface tension minimises energy, hence surface. Yilin Wang's childhood anecdote is reported from her interview (linked): to be checked for exact wording.