# A convex curve inside the unit circle

Has this theorem a specific name; and I need some references for general form.

Suppose ${\gamma}$ is a convex polygon contained in the interior of the unit circle. then the length of $\gamma$ is not more than the length of the unit circle.

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You can also see this by using radial projection to the circle from a point inside the convex curve. – Misha Jul 5 '13 at 0:55
@Misha: this might work for nearest point projection, but it's not clear to me that radial projection is a 1-Lipschitz map. – Ian Agol Jul 6 '13 at 2:18
@IanAgol: Ian, yes, of course. I was thinking about nearest-point projection $S^1\to \gamma$, which is well-known to be 1-Lipschitz, but, somehow, ended up writing about the radial projection (from a convex curve to the circle, which can contract some of the distances). – Misha Jul 8 '13 at 5:15
The inequality between the length of a plane convex curve and another contained inside it is called the Archimedes axiom (was it already known to Eudoxus? BTW, to call Eudoxus to be a student of Plato is so offensive to Eudoxus). – Włodzimierz Holsztyński Aug 22 '13 at 0:08

## 6 Answers

This follows easily from the fact that the perimeter of a convex curve is in direct proportion with the mean width, and the width of the curve is in every direction smaller than the width of the circle.

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As for the name, this is an application of Crofton's Formula, and in fact Wikipedia lists the fact that of two nested convex curves the inner has a smaller perimeter as one of the notable applications of Crofron's Formula: en.wikipedia.org/wiki/Crofton_formula – Yoav Kallus Jul 4 '13 at 23:30

The drawing below illustrates a proof for the theorem as stated in your question; almost no words of explanation are necessary. To make things easy, I write a few lines. The whole proof is based on the fact that the shortest path connecting a pair of parallel lines is the (straight) line segment perpendicular to them.

From the two vertices of each side of the polygon (green) draw a pair of rays perpendicular to the side, outward of the polygon, till they meet the boundary of the circle. Together with each side, the corresponding two rays bound a strip (yellow). Since the polygon is convex, no strip overlaps another. The arc of the circle corresponding to a side of the polygon is longer than the side. The inequality follows. The arcs in the grey sectors only make the inequality deeper.

The same proof works for a more general convex shape instead of the circle, and a non-convex shape can be replaced with its convex hull, whereby perimeter decreases. If the polygon inside is replaced with an arbitrary convex shape (not necessarily polygonal), use an arbitrarily close approximation from inside by a convex polygon.

The same idea works in every dimension $n\ge3$ as well, the perimeter being replaced with the $(n-1)$-dimensional (hyper)surface area. However, the case of the outside shape being non-convex requires a more essential modification, since passing to the convex hull may increase the surface area.

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Never mind taking the convex hull: under perpendicular projection to a (hyper)plane the (hyper)surface area does not increase! Works in all dimensions. – Wlodek Kuperberg Aug 14 '13 at 0:48

This seems obvious if you make the circle out of a rubber band, and let it relax. I think one could make this model mathematically rigorous: take a supporting line to the inner convex body, and use it to cut a chord off of the circle (or any convex body containing it), which is length decreasing. Iterate this operation, getting shorter and shorter outer convex curves, which should limit to the inner convex curve as one varies the support lines densely around the inner curve.

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This is "Archimedes' axiom" (Archimedes could not prove it, so introduced it as an axiom while computing the length of the circle). As pointed out by @Yoav Kallus, Crofton's formula is a good way to prove it, but there are plenty of other ways.

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I wouldn't say Archimedes couldn't prove it, but that he needed an extra axiom to define the length of a (non-straight) curve, the ordinary Euclidean axioms not being strong enough to accomplish this. The article arxiv.org/abs/1303.0904 has more on this. – René Jun 3 '15 at 20:44

This has nothing to do with circles (or polygons). If one convex curve is inside another, then the length of the inner curve is smaller.

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In general (say in $\mathbb R^n$) the perimeter of a set $E$ containing a convex set $C$ is not smaller than the perimeter of $C$.

This is a consequence of the following facts about the projection $\pi$ on the convex set $C$:

1. $\pi(\partial E) \supset \partial C$
2. $\pi$ is a non-expanding map (i.e. $1$-lipschitz)
3. the surface area $\mathcal H^{n-1}$ of a non-expanding image of a set is not larger than the surface area of the set.
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