﻿ Limit of perimeter of n-sided inscribed regular polygons

# Limit of perimeter of n-sided inscribed regular polygons

A regular polygon is a polygon that has all their interior angles and sides in the same measure. So the perimeter of a regular polygon is the length of the side multiplying the number of sides. With the increase of the number of sides, the perimeter of a regular polygon will be approximation of the perimeter of the circumcircle of the polygon.

In this article, we will discuss the limit of perimeter of n-sided inscribed regular polygon. At the end of article, we developed a regular polygon generator that can visualize the conclusion.

## Perimeter of regular polygons

Given n-sided inscribed regular polygon and its circumcircle, join the center of the circle with the vertices, we know the regular polygon is made of n isosceles triangles. These isosceles triangles have two equal sides of length R . The included angle of the two sides or the central angle is,

\dfrac{2\pi }{n}
(1)

Construct altitude from point O to the side, which bisects the central angle.

In one of the right triangles, we get

\sin \dfrac{\pi}{n} = \dfrac{\dfrac{s}{2} }{R}=\dfrac{s}{2R}

So we get the formula of single side of the regular polygon

s = 2R \sin \dfrac{\pi}{n}
(2)

Multiply n, we get the perimeter of the regular polygon,

P = 2Rn \sin \dfrac{\pi}{n}
(3)

So the limit of the perimeter is to get the limit of

\lim\limits_{n\to \infty} n \sin \dfrac{\pi}{n}
(4)

## The limit of nsinPi/n

In the unit circle, which means the radius is equal to 1, the central angle \theta meet

-\dfrac{\pi }{2} ≤ \theta ≤ \dfrac{\pi }{2}

One arm of the central angle OC intersects the circle at point B and the tangent AC at point C .

Let the area of sector OAB , \triangle OAB and \triangle OAC be A_s , A_0 and A_1

They have the following relationship,

A_0 ≤ A_s ≤ A_1
(5)

in which,

A_0 = \dfrac{1}{2}AO\cdotp BD =\dfrac{1}{2} | \sin \theta|

A_s = \pi\cdotp 1^2\cdotp \dfrac{\theta}{2\pi} = \dfrac{|\theta|}{2}

A_1 = \dfrac{1}{2} AO\cdotp AC = \dfrac{1}{2}|\tan \theta|

Therefore,

\dfrac{1}{2} |\sin \theta| ≤ \dfrac{|\theta|}{2} ≤ \dfrac{1}{2}|\tan \theta|

Multiply 2 and divide by |\sin \theta| , we get,

1 ≤ |\dfrac{\theta}{\sin \theta}| ≤ \dfrac{1}{|\cos \theta|}

Take the reciprocals and reverses the inequality

1 ≥ |\dfrac{\sin \theta}{\theta}| ≥ |\cos \theta|

Since f (\theta) =\dfrac{ \sin \theta}{\theta } is monotone and \cos \theta ≥ 0 in the domain of [ -\dfrac{\pi }{2}, \dfrac{\pi }{2}] ,

we can get rid of the absolute sign. So,

1 ≥ \dfrac{\sin \theta}{\theta} ≥ \cos \theta

When θ \to 0 , \cos θ = 1 ,

1 ≥ \dfrac{\sin \theta}{\theta} ≥1

\therefore \dfrac{\sin \theta}{\theta} = 1
(6)

Come back to expression (4)

\lim\limits_{n\to \infty} n\sin \dfrac{\pi}{n} =\lim\limits_{n\to \infty} \pi\cdotp \dfrac{n}{\pi} \sin \dfrac{\pi}{n} =\lim\limits_{n\to \infty} \pi\cdotp \dfrac{\sin \dfrac{\pi}{n}}{ \dfrac{\pi}{n}}

When n\to \infty , \dfrac{\pi}{n} \to 0 , \lim\limits_{n\to \infty} n\sin \dfrac{\pi}{n} = \pi

and the perimeter of the regular polygon is 2\pi r according to equation (3).

## Regular Polygon Generator

To verify the conclusion, just input the number of sides in the input box of the tool and click the generate button. The larger the number, the more the polygon will be the approximation of a circle.

To avoid crash of browser, the maximum of n is set as 100.

Steven Zheng posted 7 months ago

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