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Consider a circle inscribed in a square as shown below. enter image description here

By comparing areas, it is easy to see that $\pi < 4$ from this figure.

What is the simplest proof using lengths to show that $\pi < 4$? Clearly, this is equivalent to showing that the length of the circular arc from $A$ to $B$ is less than the sum of the lengths of the straight line segments from $A$ to $C$ and from $C$ to $B$. I am interested in as elementary a proof as possible, either accessible to the Greek geometers or as close in spirit as possible to their methods.

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    $\begingroup$ If you accept that the perimeter of a convex closed shape is no larger than the perimeter of any other convex closed shape that encloses it, then $2 \pi \le 8\,$. This is not too different from the similar intuition about areas which you used. $\endgroup$ – dxiv Feb 16 '17 at 22:04
  • $\begingroup$ Use Archimedes' exhaustion method by introducing the regular circumscribed octagon to the circle with a side [A'B'] at 45° tangent to the circle in the middle of arc AB. The length of this octagon is larger than $2\pi$ $\endgroup$ – Jean Marie Feb 16 '17 at 22:05
  • $\begingroup$ I am not sure about the "methods of the greek geometers", but the problem has something of the spirit of "squaring the circle". Latter one turned out to be very hard for the greeks. $\endgroup$ – M. Winter Feb 16 '17 at 22:11
  • $\begingroup$ What exactly are the methods of the greeks concerning lengths measure of not straight line curves? $\endgroup$ – M. Winter Feb 16 '17 at 22:18
  • $\begingroup$ I start do doubt that there is any reasonable upper bound on $pi$ one can show without using "intuition"! For example, show that $pi$ < 100? $\endgroup$ – M. Winter Feb 16 '17 at 22:53
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Your idea can work, the fact that the square perimeter is larger than a circumference of an inscribed circle is so intuitive that it does not sound as a problem at the first glance.

The reason why it feels so obvious is following generalization: If P, Q are convex bodies and Q is a subset of P, then Q has smaller perimeter.

It can be proved inductively for polygons by triangle inequality and then you can generalize it by approximating the circle by polygons. But I think that Greek geometers would use another justification: Imagine Q as a solid object and P as a string. Then, if the string is shortened it get more tightly to Q up to its convex hull.

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  • $\begingroup$ Alternative thought experiment: imagine the figure in the OP made out of malleable wire. Now take two opposite corners of the square and push them towards each other until they coincide. Both the square and the circle will deform continuously while they maintain the same perimeter, neither will self-intersect because of convexity, and in the end you'll have two linear segments of wire one inside the other. $\endgroup$ – dxiv Feb 16 '17 at 22:26
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I thought about the problem for a while and sorted out, at least for me, what "elementary" geometric tools and truths can be:

  • A straight line between two points is shorter than any other path connecting these points.
  • Scaling the whole curve up, makes it longer; shrinking it makes it shorter.
  • The phytagorean theorem.
  • The definition of $\pi$ as the ratio of the circumference of a circle to its diamater.

In contrast, things that we cannot do:

  • We cannot directly measure the length of non-straight curves.
  • We cannot directly compare the length of non-straight curves with each other or with straight lines.
  • We cannot argue with limit processes, e.g. describing the circle as the limit of polygons.

The chance of using any of the first two would make the problem trivial as we can simply measure the length or compare the length using a "physical apparatus" as describes in Mirek's post (e.g. a string). Further, using limit processes seems to me not very "acient".

Now, considering our tools, one observes a bias in the direction of piecewise straight paths being provable shorter than other paths, as we have no tools which gives us any chance to say that a given non-straight curve would be shorter than any other given curve. We can of course say, that we can scale a curve down to make it arbitrary short, but we have no bounds on the scaling factor. This is also why I think, that with the tools describes in this answer, one cannot even show $\pi<100$ or any other upper bound. We need some further intuition on how to compare lengths, e.g. physical devices as simple as a string. Feel free to complete my list of "elementary" geometric tools and truths.

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What is the definition of length of a curve?

Don't have one?

I suggest we can postulate that length to be a "limit" of a decreasing monotonic sequence. Specifically for a circle, the length of a circle will be a lower bound to a sequence of perimeters of circumscribing regular polygons to the circle where the Nth term in the sequence is the perimeter of the N-sided regular circumscribing polygon for that circle.

The motivation for this definition of length of a curve is that it has to be a useful definition akin to the real world meaning of length of a curve, which is a value derived within measurement error by taking a flexible inelastic string to hug the contour of the convex curve. For a circle, this measured result is indistinguishable (within measurement error) from measuring the perimeter of a sufficiently many sided regular polygon circumscribed about the circle.

With this definition of circumference in play, we actually don't need to worry about dealing with infinite sequences to discover the limit value. We only need to show PI < 4.

If we find a regular polygon of perimeter < 4 (any regular polygon) and which circumscribes the circle, then we are done because this polygon is in the monotonic sequence and the length of the circle would be less than that perimeter by definition.

A winning polygon is the regular octagon formed by lobbing off the 4 corners of the square. [Simple geometry can calculate where to make the cuts. We can use angles or just Cartesian style coordinates depending on what the parameters of the problem are.]


We could prove the sequence is monotonic fairly easily but it is assumed. [We can calculate easily the perimeter of every such polygon. Note though that doubling the number of sides by lobbing off all vertices in a given polygon necessarily leads to a smaller perimeter by applying the triangle inequality to each such lobbing off.]

Yes, the original square is in that sequence as well, but to solve the specific question asked we need a polygon with N>4 as that implies PerimeterN < 4.

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