In childhood, when we were taught circles for the first time, our teacher always told us that a circle is like a polygon which has infinite sides. But how to prove it?
A regular polygon's interior angle is given by $\frac\pi n(n-2)$ and when we use limits, $$\lim\limits_{n\to\infty}\frac\pi n(n-2)=\pi$$ But now how do we use it to prove that this polygon is a circle in fact.
Edit: We have to use this formula to prove that a polygon with infinite size is a circle.


1 Answer 1


A proof depends by the definition of a circle that we use.

If the definition is:

the locus of points equidistant from a given point called center $C$

consider a regular polygon with $C$ as center of symmetry. For a point on the polygon the distance $d$ from $C$ is such that: $$ r\le d\le r\cos \dfrac{\theta}{2} $$ where: $r$ is the distance of a vertex form $C$ and $\theta$ is the angle of vertex $C$ subtended by a side.

If the number of sides $n \rightarrow \infty$ than $\theta \rightarrow 0$ and : $$ \lim_{\theta \to 0}r\cos \dfrac{\theta}{2}=r $$ so, at the limit, all points of the polygon have the same distance $r$ from $C$.

if you want use the internal angle $\alpha=\dfrac{\pi}{n}(n-2)$, note that $\theta = \pi -\alpha$ and: $$ n \rightarrow 0 \iff \alpha \rightarrow \pi \iff \theta \rightarrow 0 $$

or use:

$$ r\le d\le r \sin \dfrac {\alpha}{2} $$

  • $\begingroup$ Thank you for an approach. But my question is different. This doesn't answer mine. I will edit to explain how $\endgroup$ Oct 6, 2015 at 13:01
  • $\begingroup$ The interior angle of a polygon is $alpha=\pi(n-2)/n$. So you formula is not correct, and see the add to my answer. $\endgroup$ Oct 6, 2015 at 13:11
  • $\begingroup$ What is $\theta$? $\endgroup$ Oct 6, 2015 at 13:18
  • $\begingroup$ Defined in my answer: the angle at center $C$ that subtends a side of the polygon. $\endgroup$ Oct 6, 2015 at 13:19
  • $\begingroup$ If $\theta \to 0$ how it is a circle? $\endgroup$ Oct 6, 2015 at 13:20

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