# Proof: Pi is Constant

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$\pi$, defined as the ratio of a circle's circumference to its diameter, is constant. That is, it does not vary given circles of differing size. This claim is equivalent to saying "all circles are proportional" or "all circles are similar."

## Prerequisites

• Laws of similar triangles
• Laws of limits

## Proof

Let Circle2 represent a circle of radius r2, and construct it. Let Circle1 represent a circle of radius r1, such that r2 > r1 and Circle2 is concentric with Circle1, and construct it. Choose an integer $n \ge 3$ and let $\theta = 360^\circ/n$. Construct an isosceles triangle, $\triangle T_2$, such that each equal leg has length r2, the vertex adjacent to both equal legs (v2) lies at the center of Circle2, and the measure of the angle at v2 is equal to $\theta$. Let s2 represent the length of the side opposite $\theta$. By this, it can be seen that filling Circle2's interior with n triangles congruent to $\triangle T_2$ such that all share a vertex at the center of Circle2, the equal legs extend outward from that vertex, and none overlap, a regular polygon is created whose perimeter approximates the circumference of Circle2.

By the above constructions, a similar isosceles triangle, $\triangle T_1$, is made such that each equal leg has length r1, the vertex adjacent to both equal legs (v1) lies at the center of Circle1, and the measure of the angle at v1 is equal to $\theta$. Let s1 represent the length of the side opposite $\theta$ in $\triangle T_1$. Let c2 represent the circumference of Circle2, and c1 represent the circumference of Circle1. Then:

$c_2\approx ns_2$
$c_1\approx ns_1$

Further, let n increase without bound. Then:

$c_2 = \lim_{n \to \infty}ns_2$
$c_1 = \lim_{n \to \infty}ns_1$

and

 ${c_2\over 2r_2}={\lim_{n \to \infty}ns_2\over 2r_2}=\lim_{n \to \infty}{ns_2\over 2r_2}$ (the ratio of Circle2's circumference to its diameter) ${c_1\over 2r_1}={\lim_{n \to \infty}ns_1\over 2r_1}=\lim_{n \to \infty}{ns_1\over 2r_1}$ (the ratio of Circle1's circumference to its diameter)

To prove the proposition, it suffices to show that:

${\lim_{n \to \infty}}{ns_2\over 2r_2}=\lim_{n \to \infty}{ns_1\over 2r_1}$

By the laws of similar triangles:

${s_1\over r_1}={s_2\over r_2}$

Substituting:

$\lim_{n \to \infty} \frac{ns_2}{2r_2}=\lim_{n \to \infty}\frac{ns_2}{2r_2}$

QED