Half a circle against a wall

The most area a fence can hold settled what shape a fixed length of fence should take in an open field: a circle, and nothing else. Change one thing about the field. Along one side runs a straight wall, long enough to count as endless, and the fence may start and finish against it. The wall costs nothing and encloses as well as fence does.

The question is the same — what shape holds the most — and the answer changes. It is attached to the legend of Dido, told by Virgil among others: offered as much land as an oxhide could enclose, she cut the hide into a thin strip, and the version mathematicians retell has her lay that strip in a curve against the sea.

Part of the boundary for free #

The best of the drawn fences is a half-circle, and the number it reaches is worth comparing with the open field. A circle whose whole boundary is 300 of fence holds 7,162. A half-circle whose curved edge is 300 of fence holds 14,324, exactly twice as much.

That factor of two is not a coincidence of the numbers, and it is the first clue to the proof. The half-circle’s radius is $300/\pi$, twice the radius of the circle with circumference 300, so its area is $\tfrac12 \pi (2\rho)^2 = 2\pi\rho^2$, where $\rho$ is the smaller circle’s radius. Doubling the radius quadruples the area, and keeping half of the result leaves double.

The straight-piece fences show the same pattern as in the open field: more pieces hold more. Two pieces meeting at a right angle make a triangle against the wall holding 11,250; three pieces make half a hexagon holding 12,990; four make half an octagon holding 13,580. Each fence of straight pieces falls short of the half-circle, and the shortfall shrinks as the pieces multiply.

Doubling in the wall #

The whole problem can be handed back to the open field with a mirror.

Take any fence against the wall, of length $L$, enclosing area $A$. Reflect it in the wall. The fence and its mirror image together make a closed curve, and reflection changes no lengths and no areas, so the closed curve has length $2L$ and encloses $2A$. The open-field inequality applies to every closed curve:

$$4\pi \cdot 2A \le (2L)^2, \qquad\text{so}\qquad A \le \frac{L^2}{2\pi}.$$

That is the half-circle’s area, and so no fence against a wall holds more than a half-circle with the same fence. The irregular fence in the figure holds 22,308 against a ceiling of 23,052, and its doubled curve is visibly not a circle.

The equality case comes across too. The open-field inequality is an equality only for a circle, so a fence reaches the ceiling only if its doubled curve is a circle — only if the fence is half of a circle, centred on the wall. That also says how the best fence meets the wall: at right angles. A fence that met the wall at a slant would double into a curve with a corner, and a curve with a corner is not a circle.

Why the ends meet the wall squarely #

The right angles at the wall can be understood without trusting the equality case, as a statement about corners. Suppose a fence meets the wall at an angle $\theta$, measured inside the enclosure. Doubled in the wall, that meeting becomes a point of the closed curve where it turns through an interior angle of $2\theta$, and only $\theta = 90°$ makes that a smooth point rather than a corner.

A corner sharper than a straight line can always be improved. Cut across it with a short straight segment a distance $\varepsilon$ from its tip. The cut saves length in proportion to $\varepsilon$, because a straight line is shorter than the two sides it replaces, and loses only the little triangle cut off, whose area is proportional to $\varepsilon^2$. Enlarge the whole curve slightly to use up the saved length and the area grows in proportion to $\varepsilon$, which for a small enough cut outweighs the loss. So a fence meeting the wall at less than a right angle is not the best.

A meeting at more than a right angle doubles into an inward corner, a dent, and the most area a fence can hold already removed dents: reflect the dented piece outward and the area rises at the same length. Both kinds of slant can be improved, so only a square meeting survives, which is what the half-circle does at both ends.

The argument uses nothing about the wall except that it is straight. A mirror is a straight line, and a straight wall is exactly the boundary that a mirror can make disappear.

Every fence of straight pieces #

The same mirror answers the question for fences made of a fixed number of straight pieces.

A fence of $k$ straight pieces with its ends on the wall doubles into a polygon with $2k$ sides and perimeter $2L$. Among all polygons with $2k$ sides and a given perimeter, the regular one holds the most — the polygon version of the open-field theorem, known since antiquity. A regular $2k$-gon can always be placed with two opposite corners on the wall, where it is its own mirror image, so half of it is an achievable fence. The best $k$-piece fence is half a regular $2k$-gon, and nothing else.

The shares in the figure are therefore numbers already met in the open field. Two pieces reach 0.785 of the half-circle, which is the square’s ratio $4\pi A/L^2 = \pi/4$; three reach 0.907, the hexagon’s; four reach 0.948, the octagon’s; eight reach 0.987. The regular polygons of the polygon an equation forces appear here halved, and only the even ones: a wall can double a fence into a hexagon but never into a pentagon.

Bent once #

The smallest case can be settled without any theorem at all, and it is a useful check that the mirror is telling the truth.

A fence bent once has two pieces, of lengths $a$ and $300 - a$, meeting at an angle $\theta$, and the wall closes the triangle. Its area is $\tfrac12 a(300 - a)\sin\theta$. The sine is largest at a right angle, whatever the split, and the product $a(300 - a)$ is largest at an even split, whatever the angle — the same flat-topped curve that made the square the best rectangle in the open field. So the best is 150 and 150 at a right angle, holding $\tfrac12 \cdot 150 \cdot 150 = 11{,}250$.

The sweep confirms it by trying every split in steps of a 240th of the fence, at every whole degree. The curves show how forgiving the problem is near the top: at 60 degrees the best split still holds 9,743, and the curve for each angle is flat around its peak, as every maximum is. Reflected in the wall, the winning triangle — an isosceles right triangle standing on its hypotenuse — becomes a square, which is exactly what the mirror argument said it would be.

Corners #

Put two walls together at a corner and the fence has less to do.

In a corner of angle $\alpha$, measured in radians, the best fence is an arc of a circle centred at the corner, cutting off a slice. An arc of length $L$ on a circle of radius $\rho$ spans an angle $L/\rho$, so a slice filling the corner has $\rho = L/\alpha$ and area $\tfrac12 \alpha\rho^2$, which is

$$A = \frac{L^2}{2\alpha}.$$

At a right angle that is 28,648, at 60 degrees 42,972, and at 180 degrees — a single straight wall — it is the half-circle’s 14,324 again. Measured against the open field’s 7,162, the right-angled corner holds four times as much and the 60-degree corner six times: a corner of angle $\pi/k$ multiplies what the fence can hold by $2k$, exactly the number of reflected copies that close up around it. A circle unrolled gives the slice’s area in the same way it gives the disc’s: a slice is a triangle of base $L$ and height $\rho$ with its base bent round.

For a right angle the mirror proof extends directly. Reflect the fence in one wall, then both pieces in the other, and four copies surround the corner as a closed curve of length $4L$ holding $4A$. The open-field inequality gives $4\pi \cdot 4A \le 16L^2$, so $A \le L^2/\pi$, which is the quarter-circle’s area. At 60 degrees six copies close up, and at any angle $\pi/k$ exactly $2k$ copies fill the full turn around the corner, each reflection undoing the last.

Where the mirrors stop closing #

The corner of 120 degrees is the first case where that argument breaks, and the reason is about the mirrors rather than the fence. Two mirrors at angle $\alpha$ generate a family of reflections and rotations, and when $\alpha = \pi/k$ that family is the dihedral group of order $2k$: the reflected copies of the corner fit around the point exactly once, like the eight symmetries of a square in eight ways to leave a square alone, which come from two mirrors at 45 degrees.

Two mirrors at 120 degrees produce the same lines as two mirrors at 60 degrees, since a line through the corner at 120 degrees is also a line at 60 degrees the other way. The group is the same six-element one, and six copies of a 120-degree corner cover the full turn twice, overlapping. The doubled curve is no longer a simple closed curve, and the open-field inequality has nothing to say about it.

The slice is still the best shape at 120 degrees, and in every corner up to a straight half-turn. Lions and Pacella proved in 1990 that the sector, and its higher-dimensional analogue in any convex cone, is optimal. But the proof uses tools the mirror trick does not need, which is a fair measure of how much work the reflections were doing.

Past 180 degrees, where the corner bends outward, the formula stops describing the answer at all. A slice in a 270-degree corner would hold $L^2/(3\pi)$, about 9,549 — less than the half-circle’s 14,324, which fits against either wall on its own. The wider corner offers no advantage the straight wall did not already give.

Fixed posts turn the half-circle into an arc #

A variation changes the answer’s shape without changing the method much. Suppose the fence must start and finish at two fixed posts on the wall, a set distance apart, rather than wherever it likes. The half-circle is no longer available unless the posts happen to be its diameter apart, so what is best?

The answer is an arc of a circle through the two posts, and the proof uses the open-field theorem in a different way. Take the best arc, and complete its circle with the missing arc below the wall. Now take any other fence of the same length between the same posts and attach the same missing arc beneath it. Both closed curves have the same perimeter, the fence’s length plus the missing arc, and the first is a circle, so it holds at least as much as the second. Subtract the common piece below the wall and the arc holds at least as much as any other fence between the posts.

The completed-circle argument is a close cousin of the mirror. Both extend a fence into a closed curve by adding something whose contribution is the same for every competitor, and both then let the closed-curve inequality decide. The mirror adds a copy of the fence itself, which is why it doubles; the completed circle adds a fixed arc, which is why it needs the answer guessed in advance.

The same mirror in space #

Nothing in the reflection depends on the plane. In space the question becomes a dome: a surface of fixed area standing on a flat floor, and the most volume it can cover. Reflect the dome in the floor and it becomes a closed surface with twice the area and twice the volume, to which the three-dimensional isoperimetric inequality — the sphere holds the most volume for its surface — applies.

So the best dome is a hemisphere, meeting the floor at right angles, exactly as the half-circle met the wall. The gain is smaller than in the plane. A half-circle holds twice what a full circle of the same fence holds; a hemisphere holds $\sqrt 2$ times what a full sphere of the same surface holds, because volume grows as the area to the power three halves rather than as the square of the length. In a corner where three walls meet at right angles, three reflections produce eight copies, and the best shape is an eighth of a ball.

The mirror is a statement about symmetry, not about dimension. A flat boundary that the problem’s answer can be reflected across turns a constrained problem into a free one, in the plane, in space, and in every dimension beyond.

The same move elsewhere #

Replacing a boundary by a mirror image of everything on the other side is one of the most portable tricks in geometry. A bounce is a fold of the table uses it on billiards: a ball bouncing off a cushion travels in a straight line through the reflected table, so questions about paths with bounces become questions about straight lines through copies of the table — and a table can tile the plane by reflection only if every one of its corners divides a half-turn, the same condition as here.

A circle can serve as a mirror too. The map that trades circles for lines reflects the plane in a circle rather than a line, and it carries problems about circles through a point into problems about straight lines, which is the same exchange of a hard configuration for an easy one.

In all these places the trick has the same shape. A constraint that looks like it changes the problem is removed by enlarging the space, and the answer in the larger space is folded back. The fence against the wall is a closed curve that has been folded in half; the bouncing ball is a straight line that has been folded at every cushion.

The mirror borrows the hard part #

Every fence drawn is made of straight pieces or circular arcs. The theorem concerns every curve from the wall back to the wall, including curves with no straight piece anywhere, and the drawings establish the ceiling only for the fences they draw.

The mirror argument borrows the hard part. Doubling reduces the wall problem to the open-field inequality, and that inequality — with its existence question and its equality case — is carried in whole from the previous essay. Nothing here proves it again.

And the fence must stay on its side of the wall. A fence that crossed the wall would not double into a simple closed curve, and the open-field inequality is a statement about curves that do not cross themselves — the kind that, as a curve that has area explains, have a well-defined inside at all. It is also allowed to meet the wall only at its two ends; letting it touch in between splits it into separate enclosures, each of which is its own wall problem.

Still open: the best way to share walls #

A wall is boundary supplied for free. The next question lets many enclosures share their walls, so that every piece of fence separates two regions and does double duty. In the plane, what is the least perimeter needed to divide space into regions of equal area? The answer is the honeycomb of regular hexagons, the pattern that nearest neighbours dividing the plane produce from a triangular lattice of sites. It was believed since antiquity and proved by Thomas Hales in 1999.

In three dimensions the corresponding question is Kelvin’s, from 1887: how should space be divided into cells of equal volume with the least total wall area? Kelvin proposed a packing of slightly curved truncated octahedra. Denis Weaire and Robert Phelan found a structure in 1993, with two kinds of cell, using about 0.3 per cent less area. Whether Weaire and Phelan’s structure is the best possible is unknown, and no proof that any structure is optimal has come close.

Borrow the answer from the problem without the constraint #

The wall looked like a new problem and turned out to be the old problem folded in half. One reflection doubled the fence into a closed curve, the closed-curve theorem gave the ceiling and the equality case, and the half-circle — meeting the wall squarely — fell out. The same reflection made the best polygonal fences half of regular polygons, and repeated reflections gave slices of circles in every corner whose angle divides a half-turn.

When a boundary is straight, try reflecting across it. If the reflected copies fit together without overlapping, the constrained problem is the unconstrained one in disguise; where they overlap, the disguise stops working, and that is exactly where the harder proofs begin.