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My teacher challenged my class to prove that the area is

$$A=\pi r^2.$$

We recently learned about Riemann sums, so I thought it would be possible to apply them to them to deriving the formula for the area of the circle. I know similar proofs exist, but this is one that I genuinely came up with on my own and I am wondering if it is valid. Please tell me if there is anything invalid about this proof or how it can be improved.

Imagine splitting a circle into an infinite number of isosceles triangles, where two legs extend from a vertex at the center of the circle to the edge of the circle. The central angle that each triangle makes can be represented as $\frac{2\pi}{n}$, where $n$ is the number of triangles in the circle.

The area of a triangle is $A=\frac{1}{2}ab\sin{C}$. Since the legs of each of the triangles extends from the center of the circle to the edge, that means that $a=b=r$, the radius of the circle. Therefore, the summation of every triangle in the circle as the number of triangles approaches infinity can be represented as:

$\lim_{ n\to\infty} \sum_{i=1}^{n} \frac{1}{2}r^2\sin(\frac{2\pi}{n})$

which can be rewritten as:

$\lim_{ n\to\infty} \frac{1}{2}r^2n\sin(\frac{2\pi}{n}) = 2\pi(\frac{1}{2})r^2 = \pi r^2$

Snaw
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TreeGuy
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    I believe this is similar to what Archimedes did. However there are a few technical issues with such proofs. First, what you have shown is that Riemann sums using a specific kind of partition for the circle converge to a certain value, but this does not prove that all Riemann sums using partitions with maximal diameter going to $0$ will converge to the same value (i.e. that the area is well-defined). – Tob Ernack Mar 16 '22 at 23:30
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    The second point is what definition of $\pi$ and $\sin$ you used. This may or may not be a problem, but you should make sure to avoid a circular argument (pun intended!). – Tob Ernack Mar 16 '22 at 23:30
  • Radians are defined using sectional areas so this proof seems circular. – CyclotomicField Mar 16 '22 at 23:30
  • @CyclotomicField Radians are, in my experience, much more commonly defined using sectional arc lengths. But this is indeed something that needs clarifying, like Tob Ernack points out above. – Arthur Mar 16 '22 at 23:35
  • @Arthur So when you're abstracting to hyperbolic trig you want to use sectional areas instead of arc length. I didn't know there was another definition until just now. – CyclotomicField Mar 16 '22 at 23:38
  • @CyclotomicField Does this mean that the using radians is still a circular argument? – TreeGuy Mar 16 '22 at 23:39
  • @PythonTron I don't think so because arclength is different than area and you're using the definition of $\pi$ directly this way. – CyclotomicField Mar 16 '22 at 23:46
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    Addition to my first comment: regarding the partition, there is also the problem that when you inscribe a polygon in the circle, you still leave some uncovered area around the edges of the polygon, and you would have to show that this remainder area goes to zero as the number of sides of the polygon increases. This can be done with a bit more work. I think Archimedes actually bounded the circle with inscribed and circumscribed polygons and used "squeezing". But ultimately, all these technical issues are resolved with the full theory of integration (or measures). – Tob Ernack Mar 17 '22 at 00:07
  • @TobErnack I guess it was just intuition that allowed me to know that as the number of sides of the polygon increased, the leftover space decreased. Is that not common sense? – TreeGuy Mar 17 '22 at 00:19
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    It's certainly obvious intuitively, I agree. But a complete proof would have to address that issue. But don't view my comments as being too harsh, your proof is still a very nice idea that shows why the area of the circle is what it is. I'm just pointing these things out because despite "area" being an intuitive concept, it requires a surprising amount of machinery to make it rigorous. – Tob Ernack Mar 17 '22 at 00:26
  • What about Riemann-summing up circumferences using the radius $r$ as variable? It works: $$\sum_{k\leq n}2\pi x_k\Delta x=\sum_{k\leq n}2\pi\frac{rk}{n}\frac{r}{n}\to 2\pi \int_0^r xdx=\pi r^2$$ Look here for details. – Snoop Mar 17 '22 at 00:33
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    It's a very nice way to discover the formula for the area of a circle. – littleO Mar 17 '22 at 00:38
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    @PythonTron To show you an example of how a similar argument can fail if you are not careful: here is a fairly famous "meme" proof that $\pi = 4$. – Tob Ernack Mar 17 '22 at 00:38
  • @TobErnack I greatly appreciate the feedback. I love learning about all the fine details of proofs. – TreeGuy Mar 17 '22 at 00:39
  • @Snoop That's really interesting. I love seeing how the same concept can be applied multiple ways. – TreeGuy Mar 17 '22 at 00:40
  • @TobErnack I've watched a video on this exact thing. It explained how simply approaching something does not necessarily make it equal something. https://www.youtube.com/watch?v=lCOlS_qn8RQ – TreeGuy Mar 17 '22 at 00:42

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Nice proof! The only property of $\pi$ and the circle you seem to be relying on is that a central angle of $\alpha$ radians corresponds to an arclength of $\alpha 2\pi r$, that is, you rely on the formula for the circumference of a circle, but not on the formula for its area, so I don't see a circularity here. Note though that you've given an argument which does not use the standard mathematical definition of area given by definite integrals, since you've used specific triangles instead of arbitrary rectangles, but I wouldn't count that against your proof. I like your trick as it makes things simple computationally.

You can also use the standard approach via Riemann sums, where we fit rectangles inside. Let $f(x)=\sqrt{r^2-x^2}$. The graph of this function is the upper part of a circle of radius $r$ centered at the origin. When we fit an infinite number of rectangles we get $\int_{-r}^r \sqrt{r^2-x^2}\,dx$ which can be calculated using the fundamental theorem of calculus and the substitution $x=r\sin t$:

\begin{align*} \int_{-r}^r \sqrt{r^2-x^2}\,dx &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{r^2-r^2\sin^2t}\cdot r\cos t \, dt = r^2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2 t\,dt\\ &= \frac{r^2}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1+\cos(2t)\,dt=\frac{r^2}{2}\left[t+\frac{\sin(2t)}{2}\right]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\\ &= \frac{\pi r^2}{2} \end{align*} so that the area of the whole circle is $2\cdot\frac{\pi r^2}{2}=\pi r^2$.

Your idea of taking isosceles triangles, instead of the more standard rectangles used in integration, utilizes the symmetry of the circle and definitely makes the calculations easier in this case!

Snaw
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    It's also worth mentioning that the integration can be done in polar coordinates by a change of variables, which becomes analogous to the method using triangles. – Tob Ernack Mar 16 '22 at 23:52
  • @TobErnack Good observation! – Snaw Mar 16 '22 at 23:53
  • Is there any particular reason as to why the substitution x=rsin(t) was used other than it was simply found to work? Are other trig substitutions possible? Keep in mind that I'm in AP Calc BC, and we don't learn trig substitution anymore, so my knowledge on it is lacking. – TreeGuy Mar 17 '22 at 00:14
  • @PythonTron Geometrically what we're doing is equivalent to parametrizing the upper half circle ${(x,y) ~|~ y=\sqrt{r^2−x^2}}$ using trigonometric functions: ${(r\cos\alpha,r\sin\alpha) ~|~ \alpha\in[0,\pi]}$. There are other (less obvious) trigonometric substitutions to solve integrals like $\int\sqrt{a^2+x^2},dx$ and similar ones, see in wiki. Basically they work because of the trig identities that simplify the expressions, but probably they can all also be looked at from a geometric perspective. – Snaw Mar 17 '22 at 00:30
  • @Snaw I just watched a video on trig substitution and I get it now. Thank you. – TreeGuy Mar 17 '22 at 00:35
  • I'm personally more of a fan of the Onion method, because it requires less knowledge on integration than the naïve "integrate $\sqrt{r^2-x^2}$" method above, and all you need to understand is the power rule for integration. Also, it makes it a bit more intuitive as to why the factor of $2$ disappears from the area of a circle formula. – Kyan Cheung Mar 17 '22 at 00:48
  • @Kyky Agreed, both the "onion method" and the "triangles method" of OP are better for getting an intuition for $\pi r^2$ compared to $\int \sqrt{r^2-x^2}, dx$. They make the formula appear much clearer. Specifically $\int_0^r 2\pi t , dt = \pi r^2$ is definitely very sleek. (However, it does require slightly more justification to make the proof completely rigorous: the onion method needs double integrals as explained in the link, and the triangles method requires some justification for why the areas of the leftover parts necessarily tend to zero, as explained in the comments above by Tob). – Snaw Mar 17 '22 at 01:14