# An Epilogue to “Truth vs. Intellect”

This post illustrates an alternative of compute the approximate value of $\pi$.

We begin with a circle whose radius is $r$, and let $L_{n}, L_{n+1}$ denotes the side’s length of regular polygon inscribed in the circle with $2^n$ and $2^{n+1}$ sides respectively, $n=2, 4, ....$

Fig. 1

On one hand, we see the area of $\Delta ABC$ as $\frac{1}{2}\cdot AB\cdot BC = \frac{1}{2}\cdot AB\cdot L_{n+1}$.

On the other hand, it is also $\frac{1}{2}\cdot AC\cdot BE = \frac{1}{2}\cdot 2r\cdot \frac{L_n}{2}=\frac{1}{2}\cdot r\cdot L_n.$

Therefore, $\frac{1}{2}AB\cdot L_{n+1}= \frac{1}{2}r\cdot L_n.$

Or, $AB^2\cdot L_{n+1}^2 = r^2\cdot L_n^2\quad\quad\quad(1)$

where by Pythagorean theorem, $AB^2= (2r)^2 - L_{n+1}^2.\quad\quad\quad(2)$

Substituting (2) into (1) gives $(4r^2-L_{n+1}^2)L_{n+1}^2 = L_n^2\implies 4r^2L_{n+1}^2 - L_{n+1}^4 = r^2 L_n^2.$

That is, $L_{n+1}^4-4r^2L_{n+1}^2+r^2 L_n^2 = 0.$

Let $p = L_{n+1}^2$, we have $p^2-4r^2 p + r^2 L_n^2=0.\quad\quad\quad(3)$

Solving (3) for $p$ yields $p = 2r^2 \pm r \sqrt{4 r^2-L_n^2}.$

Since $L_n^2$ must be greater than $L_{n+1}^2$ (see Exercise 1), it must be true (see Exercise 2) that $L_{n+1}^2=2r^2 - r \sqrt{4r^2-L_n^2}.\quad\quad\quad(4)$

Notice when $r=\frac{1}{2}$, we obtain (5) in “Truth vs. Intellect“.

With increasing $n$, $L_n\cdot 2^n \approx \pi\cdot 2r \implies \pi \approx \frac{L_n 2^n}{2r}.\quad\quad\quad$

We can now compute the approximate value of $\pi$ from any circle with radius $r$:

Fig. 2 $r=2$

Fig. 3 $r=\frac{1}{8}$

Exercise 1 Explain $L_{n}^2 > L_{n+1}^2$ geometrically.

Exercise 2 Show it is $2r^2-r\sqrt{4r^2-L_n^2}$ that represents $L_{n+1}^2.$