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I have looked all over the internet and have not seen someone really in detail describe what about circles give them the properties of pi. It is obvious to see why a square's area is the side length squared (a square with a side length of two has 2 rows of two square units giving it an area of 4) but not with circles. Why is pi the "magic number" of circumference/diameter? Why does it narrow down to that specific number for all circles? I know there are other articles on this page about this but none of them seemed to answer the question. They just said that if you rotated a wheel all the way around it travels pi times its radius. It didn't explain why this was so.

user372217
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  • You probably want to ask why the ratio of diameter to circumference is the same for all circles (otherwise, your question is rather pointless, and could just as well be asked for any other "magic value"). – barak manos Sep 25 '16 at 16:28
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    Every circle is the image of any other circle by a homothetic transformation. A given homothetic transformation scales every distance by the same multiple hence the ratio of the new circumference over the new diameter stays the same. – Did Sep 25 '16 at 17:02
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    Related: http://math.stackexchange.com/questions/3198/proof-that-pi-is-constant-the-same-for-all-circles-without-using-limits – Hans Lundmark Sep 25 '16 at 19:19

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Great question! This is indeed generally taken for granted (and is a bit of a pet peeve of mine - when I was learning geometry, I got in trouble for asking this!).

Here's an intuitive explanation. This isn't a rigorous proof, obviously, but hopefully it explains what's going on; and with a bit of work it can be turned into a rigorous proof, so it isn't misleading.

Let's say I have a circle $O$ of diameter $d$ and circumference $c$, and I "zoom in" on it by a factor of $k$. Can I tell the result from a (un-zoomed) circle $O'$ of diameter $kd$? Well, if you think about it for a bit you should see that the answer is "no" (note that the circle itself has no thickness, so I don't see it get "thicker" when I zoom in).

So what? Well, when I "zoom in" on $O$, the diameter becomes $kd$ - and the circumference becomes $kc$. That means that the circumference of $O'$ is also $kc$ (because I can't tell $O'$ from the "zoomed in" version of $O$). But $${kc\over kd}={c\over d},$$ so the ratio of circumference to radius stays the same.

This is in the same spirit as similar triangles - basically, I'm saying that any two circles are similar. Indeed, to turn the informal argument above into a genuine proof, we expand on this idea (this is what Did's comment above, mentioning homothety, is all about). But hopefully this helps make the picture clear.

Noah Schweber
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Not a proof per se, but another way to think at it is that sizes are relative. Suppose you are in the 2D plane with a tape measure, and you measure the diameter $d$ and circumference $c$ of a circle. Now suppose that the circle is enlarged by some factor, say twice. To you, that's the same as if the circle stayed the same size, but you and the tape measure shrank by half. Since the unit on the tape measure is now half the original one, everything you measure now gives a value twice as large, since there are twice as many units fitting in the same length. So when you measure the circle again, you'll get $2 d$ for the diameter and $2 c$ for the circumference. Which means that the ratio of the two stays the same $\frac{2 c}{2 d} = \frac{c}{d} = \pi$. Of course, the same applies to any scaling factor, not just $2$.

Incidentally, this is the very reason why $\pi$ is a universal constant i.e. the ratio of the circumference to the diameter of a circle is the same whether we measure them in inches, kilometers, or parsecs.

dxiv
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