# Archimedes' Approximation of Pi

Archimedes' approximation of Pi is by calculating the perimeter of regular polygons to determine the range of Pi. The more the number of sides, the more the perimeter of the polygon close to the circumference of its inscribed or circumscribed circle.

Through doubling the sides of a hexagon to a 96-gon regular polygon, Archimedes found the ratio of the circumference of any circle to its diameter is less than 22/7 but greater than 223/71, i,.e.,

\dfrac{223}{71} < \pi < \dfrac{22}{7}

The lower limit of the approximation \dfrac{223}{71} is determined by the ratio of the perimeter of an inscribed 96-gon regular polygon to the diameter of its circumscribed circle. The upper limit \dfrac{22}{7} is determined by the ratio of a circumscribed 96-gon regular polygon to its inscribed circle. The method lies in 3 theorems. First, the ratio of the circumference of any circle to its diameter is constant. Second, the perimeter of an inscribed polygon is less than the circumference of its circumscribed circle. Third, the perimeter of a circumscribed polygon is greater than the circumference of its inscribed circle. So to prove the above approximation, implies proof of the following inequality,

\dfrac{223}{71} < \dfrac{P_{96}}{R} <\pi < \dfrac{p_{96}}{r} < \dfrac{22}{7}

Here we are going to skip the three preconditions, prove the above approximation directly.

## Proof of Upper limit

AB is the half of the side of regular hexagon (6-gon). Point O is the center of its circumcircle. OC bisects \angle AOB . So,

\angle AOC = \angle BOC = 30\degree

In the right triangle \triangle OAB , we get

\dfrac{OB}{AB}=2 and \dfrac{OA}{AB}=\sqrt{3}>\dfrac{265}{153}

Add the two equations, we get

On the other hand, according to Angle Bisector Theorem,

\dfrac{OB}{OA}=\dfrac{BC}{AC}

\dfrac{OA+OB}{OA}=\dfrac{AC+BC}{AC}=\dfrac{AB}{AC}

\therefore \dfrac{OA+OB}{AB} = \dfrac{OA}{AC}

Plug in the inequity (1), we get

\because OC^2 = AC^2+OA^2

\dfrac{OC^2}{AC^2} =1+\Big( \dfrac{OA}{AC} \Big) ^2>1+\Big( \dfrac{571}{153}\Big) ^2=\dfrac{349450}{23409}

Let OD bisect \angle AOC , now AD is the half side of a 24-gon regular polygon. According to Angle Bisector Theorem,

\dfrac{OC}{OA}=\dfrac{CD}{AD}

\dfrac{OC+OA}{OA}=\dfrac{AD+CD}{AD} =\dfrac{AC}{AD}

\dfrac{OC+OA}{AC} = \dfrac{OA}{AD}

Therefore,

\because OD^2 = AD^2+OA^2

\dfrac{OD^2 }{AD^2} = 1+\Big( \dfrac{OA}{AD} \Big) ^2 >1+\Big(\dfrac{1162\dfrac{1}{8} }{153} \Big)^2=\dfrac{1373943\dfrac{33}{64} }{23409}

Let OE bisect \angle AOD , now AE is the half side of a 48-gon regular polygon. According to Angle Bisector Theorem,

\dfrac{OD}{OA}=\dfrac{DE}{AE}

\dfrac{OD+OA}{OA}=\dfrac{DE+AE}{AE}=\dfrac{AD}{AE}

\dfrac{OD+OA}{AD} = \dfrac{OA}{AE}

Therefore,

\dfrac{OA}{AE} = \dfrac{OD}{AD}+\dfrac{OA}{AD}

>\dfrac{1172\dfrac{1}{8} }{153} +\dfrac{1162\dfrac{1}{8} }{153}

Therefore,

\because OE^2 = AE^2+OA^2

\dfrac{OE^2}{AE^2} = 1+\Big(\dfrac{OA}{AE} \Big) ^2>1+\Big( \dfrac{2334\dfrac{1}{4} }{153} \Big)^2=\dfrac{5472132\dfrac{1}{4} }{23409}

Let OF bisect \angle AOE , now AF is the half side of a 96-gon regular polygon. According to Angle Bisector Theorem,

\dfrac{OE}{OA}=\dfrac{EF}{AF}

\dfrac{OE+OA}{OA}=\dfrac{EF+AF}{AF}=\dfrac{AE}{AF}

\dfrac{OE+OA}{AE} =\dfrac{OA}{AF}

Therefore,

\dfrac{OA}{AF} = \dfrac{OE}{AE}+\dfrac{OA}{AE}>\dfrac{2339\dfrac{1}{4} }{153} +\dfrac{2334\dfrac{1}{4} }{153}

Take the reciprocal of the inequality

\dfrac{AF}{OA} <\dfrac{153}{4673\dfrac{1}{2} }

Let the perimeter of 96-gon p_{96} , diameter of the inscribed circle d

\dfrac{ p_{96}}{d} <\dfrac{153\times 96 }{4673\dfrac{1}{2}} =\dfrac{14688}{4673\dfrac{1}{2}} = 3+\dfrac{667\dfrac{1}{2} }{4673\dfrac{1}{2}} <\dfrac{22}{7}

## Proof of Lower limit \pi > \dfrac{223}{71}

We are going to prove the perimeter of 96-gon regular polygon is greater than \dfrac{223}{71}

In the figure, a n-sided regular polygon and 2n-sided regular polygon share the same circumscribed circle.

R = 1

AB = s_n

AC = s_{2n}

OD = \sqrt{1-(\dfrac{s_n}{2})^2 }

DC = 1-OD= 1- \sqrt{1-(\dfrac{s_n}{2})^2 }

AC^2= AD^2+DC^2

s_{2n}^2 = \Big(\dfrac{s_n}{2}\Big) ^2 +\Bigg( 1- \sqrt{1-(\dfrac{s_n}{2})^2 }\Bigg) ^2

Therefore

Iterating the function, we get,

\begin{array}{|c|c|c|c|c|} \hline n&6&12&24&48\\ \hline 2n&12&24&48&96\\ \hline s_n&1&0.517638&0.261052&0.130806\\ \hline s_2n&0.517638&0.261052&0.130806&0.065438\\ \hline P_{2n}&6.211656&6.265248&6.278688&6.282048\\ \hline P_{2n}/D&3.105828&3.132624&3.139344&3.141024\\ \hline \end{array}

\because \dfrac{223}{71} ≈ 3.14084507 <3.1410

\therefore P_{96} > \dfrac{223}{71}

In the end, we have proved Archimedes' Approximation of Pi as

\dfrac{223}{71} < \pi < \dfrac{22}{7}