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.


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


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.

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