Prove that circumference of circle with radius $r$ is $2r\pi$

Prove that circumference of circle with radius $r$ is $2r\pi$.

I tried this way: let $a$ be the edge of right $n$-th sided polygon inscribed in circle with radius $r$. Let $O$ be the center of this circle and $A,B$ be the two consecutive vertices of this polygon. We know that $\angle AOB$ is $\dfrac{360^\circ}n$. Then I aplied law of cosines and got $a^2=r^2+r^2-2rr\cos\dfrac{360^\circ}n$. From this, I got that $a=2r\sin\dfrac{\pi}n$. Circle have infinitely many sides, so $n\to\infty$. Now, we have: $$C=\lim_{n\to\infty}n2r\sin\dfrac{\pi}n=2r\lim_{n\to0}\dfrac{\sin n\pi}n=2r\lim_{n\to0}\dfrac{\pi\cos n\pi}1=2r\pi\cos0=2r\pi$$ Is my solution correct and acceptable? Is there any easier way to prove this?

• How did you define $\pi$?Why switch from degrees to radians? Radians essentially assume what you are trying to prove... Oct 14, 2014 at 13:23
• Finally, "circles have infinitely many sides" is just a nonsense phrase. Technically, you have to define what the length of a curve means. Oct 14, 2014 at 13:24
• How do you also show that the circumference is larger than the perimeter of an inscribed polygon? Doesn't that require arclengths? Oct 14, 2014 at 13:25
• @BruceZheng That just requires that the shortest path between two points is a line. But you still need some definition to the length of any curve. Oct 14, 2014 at 13:41
• @ThomasAndrews To put it more precisely: The perimeter of a circle is equal to the limit of the perimeters of $n$-gons as $n\to\infty$. Oct 14, 2014 at 13:43

2 Answers

Your approach does work, but there's an easier way.

The perimeter of an $n$-gon inscribed in a circle of radius $r$ is: $$P=2nr\sin\frac\pi n$$ Since "circles are regular polygons with an infinite number of sides" (more precisely: The perimeter of a circle is equal to the limit of the perimeter of $n$-gons as $n\to\infty$), you can just take the limit.

(Technically, that last sentence requires proof, but it shouldn't be too hard... in any case, it's intuitively obvious.)

However, when you evaluated the limit, you used the fact that $\sin'(x)=\cos(x)$. The proof of that, however, requires knowing that an angle in radians is equal to the arc length of a unit circle. (That is, in this photo, $s=\theta$ when $r=1$.)

If you admit that angles equal arc lengths of the unit circle, then it's easy to show that the circumference of a unit circle is just the angle of a circle, or $2\pi$. From there, it's only a small step to show that the circumference of any circle is $2\pi r$.

(P.S. Note that you used the "angle-of-a-circle-is-$2\pi$" fact implicitly in your proof when you changed $360^\circ$ to $2\pi$.)

• I was always under the impression that $\pi$ is defined by the ratio of the circumference to the diameter of a circle. Using that fact, all you need to write is $\pi d=2\pi r$. (Well, technically... you need to prove that the ratio is the same for all circles...) In any case, the harder problem is showing that—using the definition above—$A=\pi r^2$. The best proof I know involves your idea of calling a circle an $\infty$-gon. Oct 14, 2014 at 13:58

Another approach [based on intuition]:

Consider this drawing:

When $\theta$ goes to $0$, $d$ converges to the arc length contained by the angle $\theta$ and $sin\theta=\frac{d}{r}$

As known in geometry, for infinitely small angles $sin\theta = \theta$

So $\theta = \frac{d}{r}$ or $d=r\theta$

Again by intuition, the circumference of the circle is the sum of $d$'s for $2\pi$

$circ = r(\overbrace{\theta+\theta+\dots+\theta}^{ 2\pi})$

$circ = 2\pi r$