### Archimedes simplified

In the third century BC, Archimedes calculated an amazing approximation to π, bettering the old value of 256/81 worked out by 17th century Egyptians.

How did he do it? He circumscribed and inscribed a hexagon around a circle, then bisected each hexagon’s central angles to make a dodecagon, then a 24-gon, 48-gon, and finally a 96-gon. The perimeters of the polygons bound π above and below, and each pair of perimeters is related to the ones before.

Consider a pair of polygons being bisected:

If the radius AO = 1, then AB = tan α and AC = tan ½α.

If it has *n* sides, the outer perimeter *P* = 2*n* tan α and the new polygon has perimeter *P’* = 4*n* tan ½α. The inner perimeter is *p* = 2*n* sin α, with the new polygon’s perimeter being *p’* = 4*n* sin ½α.

Now, 1/*P* + 1/*p* = 1/(2*n* tan α) + 1/(2*n* sin α) = (sin α + tan α)/(2*n* sin α tan α) = (1 + sec α)/(2*n* tan α) = (cos α + 1)/(2*n* sin α) = (cos² ½α – sin² ½α + 1)/(4*n* sin ½α cos ½α) = (2 cos² ½α – 1 + 1)/(4*n* sin ½α cos ½α) = (1/2*n*) cot ½α = 2/*P’*

And, *P’ p* = (4*n* tan ½α)(2*n* sin α) = 16*n*² tan ½α sin ½α cos ½α = 16*n*² sin² ½α = *p’*²

So, we have easy ways to calculate successive perimeters from our basic hexagons, with *p* = 6, *P* = 4 √3, using only simple arithmetic and square roots.

n |
P |
p |
---|---|---|

6 | 6.928203 | 6 |

12 | 6.430781 | 6.211657 |

24 | 6.319320 | 6.265257 |

48 | 6.292172 | 6.278700 |

96 | 6.285429 | 6.282064 |

which puts 3.1427 > π > 3.1410.

Now, Archimedes didn’t have trigonometric functions to utilize in his calculations, so he calculated the perimeters using similar triangles, a much more complicated process. But it gives the same values. He also lacked decimals, so he worked with rationals instead, producing the bounds 3 1/7 > π > 3 10/71.