Method of Exhaustion

The method of exhaustion is a method of finding the area of a shape by inscribing within it a sequence of polygons whose areas converge to the area of the shape. This method was most famously used by Archimedes around 250 BC to estimate the area of a circle and to then derive $\pi$. He did this by using an inscribed and circumscribed polygon and letting the number of sides go to infinity to approach the circumference of the circle.

Archimedes already knew how to calculate the area of a polygon. Because the area of the inscribed and circumscribed polygons ($P_{n}^{in}$​ and $P_{n}^{out}$, with $n$ sides) approaches that of the circle as the number of sides grows, the following holds:

$$\frac{a(P_{n}^{in})}{r^2} < \pi < \frac{a(P_{n}^{out})}{r^2}$$ $$\lim_{n \to \infty} \frac{a(P_{n}^{in})}{r^2} = \pi = \lim_{n \to \infty} \frac{a(P_{n}^{out})}{r^2}$$

Since $\pi$ is trapped between two bounds that converge to it, $\pi$ is squeezed to a single value as the number of sides grows. This method was so effective that it remained the most accurate way to compute $\pi$ for almost 1900 years, until it was finally surpassed by infinite series in the 17th century.

Archimedes used a 96-sided polygon to estimate $\pi$. Working entirely by hand and using only geometry, he arrived at the bounds:

$$3.1408 < \pi < 3.1429$$

The method presented here reaches the same goal by a different route. Where Archimedes relied on geometry, this approach uses modern algebra, calculus and trigonometry, which were unknown in Archimedes’ day.

Computing $\pi$ this way first requires a method for finding the area of a regular polygon with $n$ sides, both inscribed in and circumscribed about the circle. Each is derived below.

Calculating the area of an inscribed polygon

To construct an inscribed polygon, draw $n$ radii with length $r$, equally spaced around the centre so that neighbouring radii are an angle $\gamma = \frac{360}{n}$ apart. Connect each radius’s end to its neighbours.

These radii divide the polygon into $n$ isosceles triangles; call each one $T_i$. The polygon’s area is the sum of their areas:

$$a(P_n^{in}) = \sum_{i=1}^{n}{a(T_i)}$$

Each triangle has two legs of length $r$ meeting at the same apex angle $\gamma$, so by Side-Angle-Side all $n$ triangles are congruent. The sum therefore reduces to:

$$a(P_n^{in}) = n \cdot a(T)$$

What remains is to express $a(T)$ using the known quantities: the leg length $r$ and the apex angle $\gamma$. The area of a triangle is $\frac{1}{2} \cdot b \cdot h$. Taking one leg as the base $b = r$ leaves only the height $h$: the perpendicular distance from the opposite vertex to that leg.

$$\frac{h}{r} = \sin\gamma \;\overset{\times\, r}{\Longrightarrow}\ h = r\sin\gamma$$

Substituting this height and base gives:

$$a(T) = \frac{1}{2} \cdot r^2 \cdot \sin{\gamma}$$

Finally, writing $\gamma = \frac{360}{n}$ and inserting this into $a(P_n^{in}) = n \cdot a(T)$ yields:

$$a(P_n^{in}) = \frac{n \cdot r^2 \cdot \sin{\frac{360}{n}}}{2}$$

Calculating the area of a circumscribed polygon

To construct a circumscribed polygon, again draw $n$ radii of length $r$, equally spaced around the centre so that neighbouring radii are an angle $\phi = \frac{360}{n}$ apart. At each point where a radius meets the circle, draw a tangent line; together these tangents form the polygon.

Now connect each corner of the polygon to the centre $M$; call each such segment $s$. At any corner, the two triangles on either side of $s$ share the hypotenuse $s$, each have a leg $r$, and each have a right angle at their point of tangency. By RHS (SSR) they are congruent, so $s$ bisects the angle $\phi$ between the two radii, giving each triangle an angle $\frac{1}{2}\phi$ at $M$. By symmetry every corner is the same, so all $2n$ right triangles are congruent. Therefore:

$$a(P_n^{out}) = 2n \cdot a(T)$$

It remains to express $a(T)$ in known quantities. In each right triangle the side adjacent to $\frac{1}{2} \phi$ is the radius $r$, so the opposite side is $r \cdot \tan{\frac{1}{2}\phi}$. The area is therefore:

$$a(T) = \frac{1}{2} \cdot r^2 \cdot \tan{\frac{1}{2}\phi}$$

Finally, writing $\phi = \frac{360}{n}$ and substituting back into $a(P_n^{out}) = 2n \cdot a(T)$ gives:

$$a(P_n^{out}) = n \cdot r^2 \cdot \tan{\frac{180}{n}}$$

To conclude

To approximate the value of $\pi$, an inscribed and circumscribed polygon can be used. As the number of sides approaches infinity, each polygon’s area approaches that of the circle, $\pi r^2$. Expressing each area in terms of the number of sides $n$ and the radius $r$ and dividing by $r^2$ yields the following where the inscribed polygon bounds $\pi$ from below and the circumscribed polygon from above:

$$\frac{n \cdot \sin{\frac{360}{n}}}{2} < \pi < n \cdot \tan{\frac{180}{n}}$$

As $n$ grows, both bounds close in on $\pi$ from either side, until in the limit they meet exactly:

$$\lim_{n\rightarrow\infty} \frac{n \cdot \sin{\frac{360}{n}}}{2} = \pi = \lim_{n\rightarrow\infty} n \cdot \tan{\frac{180}{n}}$$

Squeezed between two bounds that both converge to it, $\pi$ is pinned to a single value and can be computed to any accuracy desired by taking $n$ large enough.