Trapping π
The circle has diameter 1, so its circumference is exactly π. Slide n, or press double the sides, to fit a regular polygon just inside the circle (blue) and one just outside (orange). The inside perimeter is too small, the outside one too big, so π is trapped between them. Tick magnify to zoom in on the edge when the three curves become hard to tell apart. Notice how each doubling shrinks the gap by roughly a factor of 4.
| n | inside perimeter | outside perimeter | gap |
|---|
What's going on
A regular hexagon drawn inside a circle of diameter 1 is six equilateral triangles with sides ½, so its perimeter is exactly 3. The hexagon drawn outside has perimeter 2√3 ≈ 3.4641. The circle's circumference, which is π by definition, is longer than the inside path and shorter than the outside one. So 3 < π < 3.4641, and no measuring was needed.
Archimedes then doubled the number of sides four times, to 12, 24, 48 and 96, using only Pythagoras and square roots. If the inside polygon with n sides has side s (on a circle of diameter 2), the one with 2n sides has side √(2 − √(4 − s²)). No trigonometry, no decimals: he carried the square roots as fractions and rounded them carefully in the safe direction, so his bounds stayed true. From the 96-gons he announced 3 10/71 < π < 3 1/7, that is 3.1408 < π < 3.1429.
The gap between the two bounds is about π³/(2n²) for large n, which is why each doubling divides it by about 4. To gain one more decimal digit you need roughly √10 ≈ 3.2 times as many sides. Ludolph van Ceulen pushed this exact method to 2⁶² sides around 1600 and got 35 digits, a life's work. Today we have formulas that converge far faster, but the idea of squeezing an unknown between two things you can compute is still used all over mathematics.
Sheet: Trapping π between polygons. Related demo: π by throwing darts, a much slower way to find it.