Nearly the most means nearly round

The most area a fence can hold proved that $4\pi A \le L^2$ for every closed curve, with equality only for the circle, and ended on the question that gives the subject its modern form. A shape that comes close to equality — whose area is within one per cent of the circle’s — must it be close to a circle?

The question matters because nothing in practice is exactly optimal. A proof that the circle is the unique best shape says nothing, on its own, about a shape that is very nearly best. It could in principle look like anything. The answer is that it cannot, and the most direct version of that answer comes with a number attached.

A ring around every shape #

To say how far a shape is from a circle, a measurement is needed, and one of the simplest comes from two circles. Every bounded convex shape has a largest circle that fits inside it, of radius $r$, and a smallest circle that contains it, of radius $R$. The shape lies in the ring between them. For a circle the two coincide and the ring has width zero; for anything else $R - r$ is positive, and the wider the ring, the further the shape strays from round.

The four shapes in the figure all have a perimeter of 300. The square’s ring is 15.5 wide, between an inner radius of 37.5 and an outer one of 53.0. The Reuleaux triangle, which measures the same across in every direction, has a ring of 14.8 — narrower than the square’s, in keeping with its higher area ratio. The ellipse twice as long as it is wide and the stadium have rings of about 30.

The ring width is a different measurement from the area ratio, and the question is whether one controls the other. It is not obvious that it must: a small shortfall in area might in principle coexist with a shape that pokes far out in one direction.

What a deficit allows #

The shortfall has a natural size, the isoperimetric deficit $L^2 - 4\pi A$, which is zero for the circle and positive for every other shape. Tommy Bonnesen proved in the 1920s that for every convex shape in the plane

$$\pi^2 (R - r)^2 \le L^2 - 4\pi A, \qquad\text{that is,}\qquad R - r \le \frac{\sqrt{L^2 - 4\pi A}}{\pi}.$$

The deficit caps the ring. For the square, whose area ratio is 0.785, the cap is 44.2 and the ring is 15.5; for the ellipse the cap is 38.1 and the ring 31.0. Every shape in the figure keeps inside its allowance, and as the deficit shrinks towards zero the allowed ring shrinks with its square root, so a shape with nearly the circle’s area is squeezed between two circles of nearly the same radius.

The square root is worth noticing. Writing $q = 4\pi A/L^2$ for the area ratio, the cap is $(L/\pi)\sqrt{1 - q}$. A shape holding 99 per cent of the circle’s area has $1 - q = 0.01$ and a cap of a tenth of $L/\pi$; a shape holding 99.99 per cent has a cap of a hundredth. Closeness in area buys closeness in shape at the rate of a square root, which is slower than one might hope and, as the next figures show, exactly as fast as the worst shapes allow.

A quadratic in the radius #

Bonnesen’s inequality comes from a stronger statement that fits in one line. For every radius $\rho$ between the inner radius and the outer one,

$$\pi\rho^2 - L\rho + A \le 0.$$

A quadratic with a positive leading coefficient is negative only between its two roots, so both $r$ and $R$ must lie between them. The roots are $\bigl(L \pm \sqrt{L^2 - 4\pi A}\bigr)/(2\pi)$, which are real by the isoperimetric inequality itself, and they are $\sqrt{L^2 - 4\pi A}/\pi$ apart. Two numbers trapped between the roots cannot be further apart than the roots are, and that is Bonnesen’s inequality.

The stadium shows the bound being touched. A stadium made of a rectangle $w$ long capped by half-circles of radius $\rho$ has perimeter $2w + 2\pi\rho$ and area $2w\rho + \pi\rho^2$, and substituting its inner radius $\rho$ into the quadratic gives $\pi\rho^2 - (2w + 2\pi\rho)\rho + 2w\rho + \pi\rho^2$, which is exactly zero. Every stadium sits on the boundary of the inequality at its inner radius. The Reuleaux triangle shows a different symmetry: its two radii add to its width, and its perimeter is $\pi$ times its width, so they sit symmetrically about the quadratic’s lowest point.

Two more bounds from the same quadratic #

The quadratic gives more than the ring. Substitute the inner radius alone. The statement $\pi r^2 - Lr + A \le 0$ rearranges to $4\pi A \le 4\pi L r - 4\pi^2 r^2$, and subtracting from $L^2$ gives

$$L^2 - 4\pi A \ge (L - 2\pi r)^2.$$

The same step at the outer radius gives $L^2 - 4\pi A \ge (2\pi R - L)^2$. So the deficit also caps how far the perimeter can exceed the inscribed circle’s circumference, and how far it can fall short of the enclosing circle’s. For the square with perimeter 300, the inscribed circle’s circumference is about 235.6, so the perimeter exceeds it by about 64.4, and the enclosing circle’s is about 333.2, which exceeds the perimeter by about 33.2. The square root of the deficit is about 139, comfortably more than either.

These forms say something the ring width does not. A convex shape trapped between two circles has a perimeter trapped between their circumferences — a convex curve inside another is never longer than it — and Bonnesen’s forms add that when the deficit is small, both circumferences are close to the shape’s own perimeter. Area, perimeter and the two radii are all pinned together by one small number.

Why the quadratic is never positive #

The shortest proof of the quadratic statement counts crossings, and it is the same kind of argument as dropping needles to find π. Take a circle of radius $\rho$ and consider every position it can occupy relative to the shape — every centre and every rotation. Two classical formulas of integral geometry describe the average behaviour.

Poincaré’s formula counts the total number of points where the circle’s boundary crosses the shape’s boundary, over all positions, and it depends only on the two lengths; with the standard normalisation it is $4L \cdot 2\pi\rho$. Blaschke’s formula measures how many positions put the circle and the shape in contact at all, and it is $2\pi(A + \pi\rho^2) + L \cdot 2\pi\rho$.

Now use the choice of $\rho$. It is larger than the inner radius, so the circle never fits inside the shape; it is smaller than the outer radius, so the circle never swallows the shape. Whenever the two touch, then, their boundaries must cross — and a circle’s boundary and a convex boundary that cross do so at least twice. Every position in contact contributes at least two crossings, so the crossing count is at least twice the contact count:

$$8\pi L\rho \ge 4\pi(A + \pi\rho^2) + 4\pi L\rho,$$

which rearranges to $\pi\rho^2 - L\rho + A \le 0$. The inequality about shapes is a comparison of two averages, the same move that counts crossings of a needle and a set of lines.

The same family of formulas explains why perimeter is the natural quantity to compare with a circle. Cauchy’s formula says that the width of a convex shape, averaged over every direction, is its perimeter divided by $\pi$ — the width function that the shape described from outside uses to rebuild a shape from its supporting lines. A circle of diameter $L/\pi$ has that width in every direction, so a shape with perimeter $L$ is compared, on average, with exactly the circle whose circumference matches. The Reuleaux triangle, whose width is the same in every direction, meets that average everywhere and still falls short in area, which is the gap between having a circle’s widths and having a circle’s shape.

Three shapes at 99 per cent #

Tune three different families to the same area ratio and the inequality becomes visible. An ellipse, a stadium and a curve with three gentle lobes are each adjusted until they hold exactly 99 per cent of the circle’s area. Their allowance is identical, 9.55, because the allowance depends only on the ratio and the perimeter.

Their rings are 7.78, 7.48 and 4.77. The ellipse and the stadium use most of the allowance, the three-lobed curve under half. All three are visibly round, and the drawings are the inequality’s content made literal: no convex shape holding 99 per cent of the circle’s area can have a ring wider than 9.55, around a circle of radius about 48.

The ellipse and the stadium come close to the bound for the same reason: they are elongated, pulled out in one direction and pinched in the other, which is the deviation a ring measures most directly. The lobed curve spreads its deviation over three directions, which costs area without widening the ring as much. What a map does to a circle produces exactly the elongated kind, stretching a circle into an ellipse along two directions.

Eight shapes and their allowances #

Ranked by area ratio, the table runs from the regular 12-gon at 0.977 to the ellipse four times as long as it is wide at 0.536. The share of the allowance used does not follow the ranking. The 12-gon uses 11 per cent, the hexagon 23, the square 35; the stadium and both ellipses use around 80 per cent, whatever their ratio.

That pattern says which shapes the inequality is really about. Elongated shapes press against the bound and symmetric ones do not. A regular polygon’s deviation from its circle is spread evenly around it, in many small corners, and each corner wastes some area while barely moving the inner or outer circle. The Reuleaux triangle, which has the least area of any shape of constant width with its width by the least area a width can hold, still uses under half.

Two hundred random shapes #

A theorem about every convex shape can be tested on shapes nobody chose. Scatter between 4 and 29 random points in a square, take the smallest convex polygon containing them, and measure. Dividing both sides of the inequality by $L^2$ makes the test independent of size, and every one of the 200 shapes lands above the diagonal.

The cloud is informative about how the bound behaves in bulk. Most random outlines sit well above the line, bunched at small deficits, because the outline of many random points in a square is already fairly compact. The nearest reaches 61 per cent of the way to the line. No random shape comes near equality, which is what the stadium’s exact touching already suggested: equality needs a very particular shape, and random ones are not it.

Why polygons have so much room #

For regular polygons the two sides of the inequality shrink at different rates. The ring of a regular $n$-gon has width $R(1 - \cos(\pi/n))$, which is about $R\pi^2/(2n^2)$, so its square falls like $1/n^4$. The deficit falls only like $1/n^2$. On the logarithmic axes the two lines have slopes $-2$ and $-4$, measured from the polygons themselves, and the gap between them opens without limit.

The Fourier proof in the first essay explains the difference. It writes a curve as a sum of harmonics and shows that the deficit prices each harmonic in proportion to the square of its frequency times the square of its size. A regular $n$-gon’s corners are harmonics of frequency about $n$ with size about $1/n^2$, so the deficit sees $n^2 \cdot n^{-4} = n^{-2}$, while the ring sees only the size, $n^{-2}$, and its square $n^{-4}$. Fine wiggles are expensive in perimeter and cheap in ring width; a single stretch in one direction is the opposite, and that is why the ellipse and stadium sit near the bound while the polygons drift away from it. It is also why the square root in Bonnesen’s inequality cannot be improved: a slightly squashed ellipse has a ring proportional to its squash and a deficit proportional to its square, the flat-topped behaviour every maximum has.

The version without convexity #

Bonnesen’s inequality is stated for convex shapes, and the ring is the wrong measurement without convexity: a round shape with one long, hair-thin spike breaks it, because the spike widens the ring by its whole length while adding only twice that length to the perimeter, and for a long enough spike the square of the first outgrows what the second adds to the deficit. Convexity is what rules the spike out: a convex shape contains the segment between any two of its points, so it cannot reach far in one direction without filling in the space behind the reach, and a line under every point is the same property seen from the outside. The modern form of the result replaces the ring with a measurement that ignores such spikes, the Fraenkel asymmetry — the smallest fraction of a shape’s area that would have to be moved to turn it into a disc of the same area.

Fusco, Maggi and Pratelli proved in 2008 that in every dimension the deficit, suitably scaled, is at least a constant times the square of the asymmetry, for every shape with a finite perimeter, convex or not. The square is sharp for the same reason as in the plane. Nearly optimal implies nearly the optimum, in any dimension and for any shape, with a square-root rate that cannot be beaten.

Searched radii, convex shapes and no proof by picture #

The radii are found by search. The inner and outer circles of each polygon are located by nested searches that converge to many decimal places and are checked against exact values for the shapes that have them; the curved shapes are drawn as polygons of 240 sides, whose radii differ from the true curves’ in the fourth decimal place.

Every shape drawn is convex. The figures cannot show the spiky counterexamples that make convexity necessary, and they do not attempt the asymmetry version of the inequality, which needs the best-placed disc rather than two circles.

And no figure proves the inequality. Two hundred random shapes and a dozen chosen ones are evidence, not a proof; the proof is the counting argument, which the figures illustrate only through its conclusion, the quadratic.

Still open: the lowest note of a polygon #

The same “roundest is best” pattern governs vibration. Among all drums of a given area, the circular one has the lowest note — the smallest first eigenvalue of its vibration problem — a theorem of Faber and Krahn from the 1920s, proved by the same kind of rearrangement that proves the isoperimetric inequality. The polygon version asks which drum with $n$ straight sides and a given area has the lowest note, and the natural guess is the regular $n$-gon, just as the regular $n$-gon holds the most area for its perimeter.

Pólya and Szegő proved that guess for triangles and for quadrilaterals in 1951, using a symmetrisation that preserves the number of sides only when there are three or four. For drums with five or more sides the conjecture is open, although computations strongly support it, and the area version of the same question was known more than two thousand years ago.

Measure the distance, not just the optimum #

A proof that the circle is uniquely best was the end of one question and the start of a more useful one. Bonnesen’s inequality converts a shortfall in area into a bound on how far a convex shape can stray from round, through a quadratic that a crossing count proves is never positive between the shape’s two radii. The drawings bear it out on chosen shapes, random shapes and shapes tuned to 99 per cent.

When an optimum is unique, ask how quickly a shape must approach it as its value approaches the best. The answer is usually a power law, the power is usually a square root, and the shapes that attain it are the ones that deviate in the single cheapest way.