Why does the square and circle have this property in common?

Question

How can it be, that two so fundamentally different forms as the square and circle, have the thing in common, that the circumference of both are four times their average width?

Answer 1

Notice that the fact depends on how we measure "average width." In a Cartesian plane, the polygon with vertices at $(1,0),$ $(0,1),$ $(-1,0),$ and $(0,-1)$ (in that order) is a square. In which direction do we measure its height and in which direction do we measure its width?

For a square we can simply say to measure the height and width parallel to any two adjacent sides, but that does not work for an arbitrary figure. So I'll assume that when we describe a figure whose average width is to be measured, we take that figure in some specified orientation. I'll also consider only convex figures. Then if a horizontal line intersects the boundary of the figure at two points, the distance between those two points is the width of the figure at the "level" of that line. The distance between the uppermost and lowermost horizontal lines that intersect the figure is the height of the figure, and the average width is the area of the figure divided by its height.

Using these definitions, indeed the circumference of the circle is four times its average width. The perimeter of a square is four times its average width if the square is oriented with one of its sides horizontal, otherwise not. (Equivalently, the fact is true for the square if width is measured parallel to one of its sides and height is measured perpendicular to one of its sides.)

Now take any regular polygon of $2n$ sides. Orient it so that one of its sides is horizontal.

Construct a line segment from the center of the polygon to each vertex. These line segments dissect the polygon into $2n$ triangles, each of which is an isosceles triangle of height $r,$ where $r$ is the radius of the circle inscribed in the polygon.

If the length of one side of the polygon is $s,$ then that is the base of each isosceles triangle, and the area of the triangle is $\frac12 rs.$ Since there are exactly $2n$ of these triangles, the area of the entire polygon is $2n \times \frac12rs = nrs.$ The height of the polygon is $2r,$ so the polygon's average width is $\frac{nrs}{2r} = \frac12ns.$

But of course the polygon has $2n$ sides of length $s,$ so its perimeter is $2ns.$ And notice that $2ns = 4 \times \frac12ns,$ that is, the polygon's perimeter is four times its average width.

For example, consider a regular hexagon of side $s,$ oriented so that one of its sides is horizontal. Its area is $\frac32\sqrt 3 s^2$ and its height is $\sqrt3 s,$ so its average width is $\frac32 s.$ Its perimeter is $6s,$ which is four times its average width.

As another example, consider a regular octagon of side $s,$ oriented so that one of its sides is horizontal. Its area is $(2 + 2\sqrt2) s^2$ and its height is $(1+\sqrt2) s,$ so its average width is $2s.$ Its perimeter is $8s,$ which is four times its average width.

Notice that as we take larger and larger values of $n,$ shrinking the side of the polygon so that it always has the same inscribed circle of radius $r,$ the area and perimeter of the regular polygon of $2n$ sides become very close to the area and circumference of the inscribe circle. It is therefore really not surprising that the circumference of that circle (or any other circle) is four times its average width.