DERIV-stretched-polygon — A polygon stretched to its box, sides equal while its corners stay sound

Used by: ADR-0037, F06 (rule C chosen by the Product Owner, research note 0011). Sources: REF-CRANE-DISCRETE-CURVES (a plane polygon is fixed by its side lengths and turning angles), REF-CARROLL-2013 (the turning angle), REF-OPENSTAX-UNIT-CIRCLE, REF-MATHWORLD-VECTOR-ADDITION, REF-MATHWORLD-AFFINE, REF-MDN-ATAN2, REF-MDN-ACOS, REF-MATHWORLD-BISECTION, REF-SVG2-COORDS; DERIV-regular-polygon-fit steps 1–2, DERIV-turning-angle, DERIV-free-shape-stretch.

Inputs

n corners, 3 ≤ n ≤ 12 (Q19); the box w × h, integers ≥ 0; its ratio r = h / w.

Notation

Symbol	Definition
m	sides of the right half after the half base: (n − 1)/2 for an odd n, n/2 − 1 for an even n
θⱼ	direction of side j of the regular polygon, 2πj/n, j = 1 … m
λ	vertical scale of those directions, λ > 0; λ = 1 is the regular polygon
φⱼ	direction of side j of the stretched polygon
τ	turning angle at a corner
A(λ)	aspect of the unit polygon: its height over its width

Step 1 — A symmetric polygon of sides 1, built from directions

Mathematical frame (y up). A plane polygon is fixed, up to its place, by its side lengths and turning angles (REF-CRANE-DISCRETE-CURVES); so a polygon of sides 1 is fixed by the directions of its sides. A side of length 1 and direction φ is the vector (cos φ, sin φ) (REF-OPENSTAX-UNIT-CIRCLE), and the vertices are the running sums of the sides (REF-MATHWORLD-VECTOR-ADDITION).

The polygon is symmetric about the vertical axis x = 0, its base on y = 0 (F06). Only its right half is built: from R₀ = (1/2, 0), the right end of the half base, Rⱼ = Rⱼ₋₁ + (cos φⱼ, sin φⱼ) for j = 1 … m, with 0 < φ₁ < … < φₘ < π, so that y grows along the half. For an odd n, Rₘ is the apex, on the axis: 1/2 + Σ cos φⱼ = 0. For an even n, Rₘ is the right end of the top side, horizontal, which goes back to the axis: Σ cos φⱼ = 0, so Rₘ = (1/2, H). The left half is the mirror image (x, y) ↦ (−x, y); the apex of an odd n is its own image.

Step 2 — The λ family: the regular directions scaled vertically

In the regular polygon of DERIV-regular-polygon-fit step 1, the vertices are 2π/n apart on the circle; turning the polygon by 2π/n maps each side onto the next, so the directions of the sides grow by 2π/n: from the base (direction 0), θⱼ = 2πj/n.

The stretched polygon keeps these directions scaled vertically by λ, the axis scaling (x, y) ↦ (x, λy), an affine map, which does not keep angles nor lengths (REF-MATHWORLD-AFFINE); the order of the directions is enforced by the corner guard of step 3, a negative turn failing it: the side of direction θⱼ, (cos θⱼ, sin θⱼ), becomes (cos θⱼ, λ sin θⱼ), of direction

φⱼ = atan2(λ sin θⱼ, cos θⱼ) (REF-MDN-ATAN2: the angle in [−π, π] of the point (x, y); here sin θⱼ > 0, so φⱼ ∈ (0, π)), and every side is given length 1 again.

• Even n: θⱼ and θ_{n/2 − j} = π − θⱼ have opposite cosines and equal sines, so φ_{n/2−j} = π − φⱼ: their cosines cancel, and θ_{n/4} = π/2 gives φ = π/2. The closure Σ cos φⱼ = 0 of step 1 holds for every λ.
• Odd n: the last side closes the half on the axis: cos φₘ = −1/2 − Σ_{j<m} cos φⱼ, φₘ = acos(…) in [0, π] (REF-MDN-ACOS); outside [−1, 1] no side closes it, and acos gives NaN: such a λ has no polygon.

At λ = 1 every φⱼ = θⱼ (the odd closure gives θₘ back, the regular polygon closing by symmetry): the regular polygon. For n = 5 and 6 (m = 2: the closure and the aspect fix both directions) this family gives the only equal-sided polygon filling the box; from 7 on, the octagon included (m = 3, one freedom left), it picks one of several (note 0011).

Step 3 — Turning angles and the corner guard

The turning angle at a corner is the angle between the directions of the two sides meeting there (REF-CARROLL-2013; DERIV-turning-angle): φ₁ at each base corner (from the base, direction 0), φⱼ₊₁ − φⱼ at the next corners, 2(π − φₘ) at the apex of an odd n (the side after it is the mirror of side m run downwards, (cos φₘ, −sin φₘ), of direction 2π − φₘ), and π − φₘ at each top corner of an even n (towards the top side, direction π). The regular polygon turns 2π/n at every corner.

Corner guard (Product Owner, 2026-10-12): λ is sound when every corner turns at least π/n, half of its regular turn. A λ without polygon (step 2) is not sound. The sound λ form one interval [λ_lo, λ_hi] containing 1; this is checked numerically, not proved (step 5).

In the code the turns are those of the whole polygon in the SVG frame (turningAngles, DERIV-turning-angle, clockwise positive on screen): the same values, and a NaN coordinate gives a NaN turn, which fails the guard.

Step 4 — The aspect

The unit polygon spans 2W in width, W the largest x of the right half, and H in height, the largest y (the apex or the top side, y growing along the half): A(λ) = H / (2W). A grows with λ for n ≥ 5 wherever the polygon closes (checked, step 5). The triangle (m = 1, φ₁ = 2π/3 by closure) and the square (φ₁ = π/2) do not depend on λ: one shape each, A = √3/2 and 1, always sound.

Step 5 — Choosing λ (rule C)

Bisection keeps a bracket [a, b] around the solution and halves it at each step; after k steps the error is at most 2⁻ᵏ (b − a) (REF-MATHWORLD-BISECTION). It runs on ln λ, in [−3, 3], which contains every sound λ of n = 5 … 12: from λ_lo = 1/3 (hexagon) to λ_hi = 6.74 (pentagon, ln λ = 1.908), found by the sweep below and confirmed by the review of SP-009; 60 steps: an error below 3 · 2⁻⁶⁰ ≈ 2.6e-18 in exact arithmetic; in doubles the bracket stops shrinking at one unit in the last place of ln λ (about 4e-16, after some 54 steps), still far under EPSILON. A bound found this way may lie on the unsound side by that much: a clamped polygon's guard corner turns π/n within about 1e-16.

1. λ_lo: the bisection in [−3, 0] on "not sound" (true at −3, false at 0; for the triangle and the square, sound everywhere, it ends at −3, which gives their only shape).
2. λ_hi: the bisection in [0, 3] on "sound".
3. The bisection in [ln λ_lo, ln λ_hi] on A(λ) < r. If r ≤ A(λ_lo), the predicate is false everywhere and the bisection ends at λ_lo; if r ≥ A(λ_hi), true everywhere, it ends at λ_hi (the box is too wide or too tall for sound equal sides): the clamp of rule C needs no step of its own.

Each bisection needs its predicate to change once in its bracket: "sound" on each side of 1, and A(λ) < r between λ_lo and λ_hi. Neither is proved (note 0011); both are checked by the tests over ln λ from −3 to 3 in steps of 0.001 (6 001 values): "sound" changes at most twice and holds at λ = 1 for every n from 3 to 12 (never changes for 3 and 4); A never decreases where the polygon closes (n = 5 and 9 do not close below ln λ ≈ −0.58 and −0.71), constant for 3 and 4. The triangle and the square, independent of λ, give their only shape whatever the branch.

A box of size 0 gives r = 0, +∞ or NaN (0/0): every comparison with NaN is false, so the last bisection ends at λ_lo; any λ is right, the polygon being then flattened to a line or a point (step 6).

Step 6 — Placing it in the box

The full polygon in drawing order (DERIV-regular-polygon-fit step 2, clockwise on screen from the right end of the base): R₀, its mirror L₀, L₁ … Lₘ₋₁, then the apex Rₘ (odd n) or Lₘ, Rₘ (even n), then Rₘ₋₁ … R₁; n points. It goes to the SVG frame with its box at the origin: x = W + X, y = H − Y (y down, REF-SVG2-COORDS §8.4). Then each axis is scaled to the box, the top-left fixed (stretchToSize, DERIV-free-shape-stretch): every side of the box is touched, the far ones within one unit in the last place, since the span of the unit polygon is irrational (a base at y = 6.000000000000001 in a box of height 6), far below the 5 decimals written (Q10). When r lies within [A(λ_lo), A(λ_hi)], A(λ) = r and both factors are equal: the sides stay equal; past the guard, the limit polygon is stretched on each axis (rule C). The triangle becomes isosceles and the square the rectangle of the box: the plain stretch, as they have no freedom.

Checks (test cases)

Values of note 0011, recomputed by the agent's own implementation of these steps (5 decimals, SVG frame, drawing order).

Case	Expected
any n, r = the regular aspect	λ = 1: the polygon of F02 filling its box
triangle 100 × 100	(100, 100), (0, 100), (50, 0)
square 300 × 50	(300, 50), (0, 50), (0, 0), (300, 0)
pentagon 100 × 100	(81.75623, 100), (18.24377, 100), (0, 39.16418), (50, 0), (100, 39.16418); every side 63.51246
pentagon 300 × 50	(228.85967, 50), (71.14033, 50), (0, 12.85950), (150, 0), (300, 12.85950)
hexagon 100 × 100	(77.42919, 100), (22.57081, 100), (0, 50), (22.57081, 0), (77.42919, 0), (100, 50); every side 54.85838
hexagon 300 × 50	(204.90381, 50), (95.09619, 50), (0, 25), (95.09619, 0), (204.90381, 0), (300, 25)
hexagon 50 × 300	(41.47297, 300), (8.52703, 300), (0, 150), (8.52703, 0), (41.47297, 0), (50, 150)
octagon 100 × 50	(67.55767, 50), (32.44233, 50), (0, 39.16136), (0, 10.83864), (32.44233, 0), (67.55767, 0), (100, 10.83864), (100, 39.16136)
heptagon 100 × 100	(72.54712, 100), (27.45288, 100), (0, 64.22529), (9.67353, 20.18085), (50, 0), (90.32647, 20.18085), (100, 64.22529)

By hand, without the harmonic identity of note 0011: hexagon 100 × 100. With φ₂ = π − φ₁ (step 2), A = 2 sin φ₁ / (1 + 2 cos φ₁); at φ₁ = 65.7048°, 2 sin φ₁ = 1.82288 and 1 + 2 cos φ₁ = 1.82288: A = 1 = r. The half width 1/2 + cos φ₁ = 0.91144, the scale 50 / 0.91144 = 54.858 is the side, and the base ends at x = 50 ± 27.429. Hexagon 300 × 50, at its guard: φ₁ = 30° turns exactly π/6; width 1 + 2 cos 30° = 2.73205, height 2 sin 30° = 1, A = 0.36603 > 50/300: λ = λ_lo, x scale 300 / 2.73205 = 109.808 (the base), base end x = 150 + 54.904 = 204.904.

Limits

• The two monotonic properties of step 5 are checked, not proved; a proof would remove the sweeps from the tests.
• λ_lo and λ_hi depend on n only: computed at each call (three bisections of 60 steps on n points, negligible), not tabulated, so that no table needs a derivation of its own.
• In very elongated boxes, every corner of the limit polygon is stretched: sides and corners become uneven, less than with the plain stretch of the regular polygon (note 0011).