Optimal constructions — the series
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The largest triangle in a square
The biggest equilateral triangle a square sheet holds is tilted by exactly fifteen degrees and uses 46.4% of the paper. Both numbers come out of a quadratic — which means a compass reaches this optimum too, and folding's advantage is not needed here at all.
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The biggest one that can also be folded
Which regular polygon uses a square sheet best, and which of them a fold can actually construct, are two questions with completely different pedigrees. Answered side by side over sixteen polygons, they turn out to agree — and the reason is that both are questions about the arithmetic of the same number.
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The square is in the answer
The largest regular polygon a square sheet holds is not increasing in the number of sides, and the octagon's win is the striking part: it uses 82.8% of the paper against the twelve-gon's 80.4% and the hexagon's 69.6%. Run the same census over rectangles and the octagon's advantage is gone — on every proportion tried the hexagon leads, and the order among the even-sided polygons reverses outright.
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Every even polygon beats every odd one
Crossed with what each tool can build, the census of the largest regular polygon a sheet holds gives the same verdict on every proportion from a square to three to one: the best polygon is one a compass already builds, and the best polygon only a fold can build places fourth at best. The ranking itself stops moving at a proportion of 1.1284, where the hexagon overtakes the octagon. On every longer sheet only the short side holds a polygon, each polygon's share is a fixed constant divided by the length, and the constant — its area over the square of its least width — comes down to the circle's π⁄4 for even polygons and climbs up to it for odd ones. So every even polygon beats every odd one.
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The crossing is as hard as the polygon
Lengthen a square sheet and the largest hexagon it holds turns, pressed against all four edges, until it overtakes the polygons held by the short side alone. Every one of those overtakings happens at a proportion with a closed form, √3⁄2 + ½√(8K⁄3√3 − 1), and the number that comes out is exactly as hard to mark as the polygon being overtaken is to build. The octagon's 1.1284 is a compass number of degree eight. The heptagon's 1.0696 has degree twenty-four and needs a fold. The hendecagon's 1.0853 has degree forty and needs two folds at once.
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The sheet a polygon fits exactly
A regular polygon with 4k + 2 sides has flat edges along one axis and corners along the other, so there is one sheet, 1⁄cos(π⁄n) long, that it touches on all four edges at once. On that sheet it is beaten only by the multiples of four with fewer sides, and so it ranks exactly (n − 2)⁄4. That puts the fourteen-gon, which only a fold builds, third rather than fourth; the twenty-two-gon, which needs two folds at once, fifth rather than ninth; and the forty-six-gon, beyond two folds, eleventh. Seven sheets from a square to three to one had missed all three.
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Turning is uphill all the way
A regular polygon of 4k + 2 sides on a sheet a little longer than a square cannot lie flat: it turns, pressed against all four edges, until the sheet is exactly its own. Its share on the way has a closed form, and the closed form's slope is proportional to h² − 1 for every such polygon — flat on the square, rising all the way to the own sheet, and falling after it. So the own sheet is exactly the peak, the gain from the square to it is the average of one and the sheet's length, and the rank the census measured for polygons of this kind, (n − 2)⁄4, is now a theorem.
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An odd polygon fits like its double
A regular polygon with an odd number of sides has a corner opposite every flat, so it looked as if the closed form for the turning polygons of 4k + 2 sides would need redoing with a different offset. It does not. An odd polygon is exactly as wide, in every direction, as the regular polygon with twice its sides — that polygon is the odd one averaged with its half-turn — and a box sees only widths. So the triangle fits wherever a hexagon fits, the heptagon wherever a fourteen-gon does, and on every sheet the odd polygon's share is its double's times 2 − sec²(π/2n). It peaks on its double's own sheet with the same gain; on a square it is tilted π/4n, the triangle's fifteen degrees; and on its own sheet it beats nothing but smaller odd polygons, never once passing the circle.
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A sheet rewards the multiples of its sides
On a rectangle every odd polygon fits exactly as its double does, because a box measures widths and a width cannot see a half-turn. A triangle of paper has no half-turn, and the ranking it gives is a different kind of thing. On a regular sheet of m sides, a regular polygon beats the largest circle exactly when its sides are a multiple of m — it can then face every side of the sheet with one of its own and share the sheet's circle, taking n·tan(π/n) over m·tan(π/m) of it — and every other polygon to forty sides falls short. On a triangle that puts the triangle first, the hexagon second at two thirds, and the nonagon third: the first sheet on which a polygon only a fold can build takes a place on the podium.
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The rule has a width
On an equilateral triangle of paper the regular polygons that beat the largest circle are exactly the multiples of three. Tilt the apex a degree and every one of them loses share in proportion, while the circle loses almost nothing — a multiple's advantage is a corner and the circle's is a hilltop. So the rule holds only in a window, 44° to 79° of apex for the hexagon and 56° to 65° for the 24-gon, and past it the second place goes to polygons like the hendecagon that no single fold can build.
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The worst angle sets the price
On a triangle of paper that is nearly equilateral, the largest regular hexagon loses 0.448 percentage points of the sheet for every degree the triangle is out of true, the nonagon 0.267 and the dodecagon 0.193. Those rates were measured. They have a closed form, 2n tan²(π/n) / 9√3 a radian, and it depends only on the worst of the triangle's three angles, whatever its shape. Rate times sides falls to 2π² / 9√3, and the window in which each polygon beats the circle shrinks as ninety degrees over its number of sides.
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The window has an exact edge
How far an isosceles triangle may stray from equilateral before its largest regular hexagon falls behind its inscribed circle was measured — 15.6 degrees closing the apex, 19.1 opening it — and estimated to first order as the polygon's lead over its rate. Neither number needs measuring. The fitting problem has three sides and three unknowns, so its bound is an equality, the polygon keeps a side on the base, and its share is one line. That line carries the same factor as the circle's, so the circle's own loss cancels out of the comparison entirely, and each edge of the window is an angle in closed form. The opening half is wider because opening the apex lengthens the perimeter. And the window does not shrink as ninety degrees over the sides, as the first-order estimate says, but as (2 − √2)·180° over them — about 105 — because the corner's reach bends back as fast as it rises.
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Two sides at sixty are enough
On a nearly equilateral triangle every multiple of three beats the inscribed circle, and to first order how long it keeps doing so depends only on the worst of the triangle's angle errors, which makes the window a hexagon in the plane of two errors. Fitted exactly, the window is a scalloped star. It reaches furthest — 21.8 degrees for the hexagon, against 15.6 and 19.1 on isosceles triangles — where one angle opens and another closes with the third left at sixty, because two sides still meet at sixty and the polygon can lie flush against both. Far outside it, wherever all three sides lie along the polygon's own directions, the polygon wins again.