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DERIV-regular-polygon-fit — Regular polygon fitted uniformly in a box ​

Used by: docs/domain/shapes.md §3 (Q1: uniform mode), EN-008, US-005, US-006, EN-009, US-009. Sources read in SP-002 (note 0004) and SP-003 (note 0005); EN-013 (F08, the start at the base).

Inputs (model, integers) ​

n ≥ 3 corners, width w ≥ 0, height h ≥ 0; a horizontal and a vertical alignment, each min, mid or max (F07, step 5).

Step 1 — Unit polygon ​

The nth roots of unity e^(2πik/n) form a regular polygon with n sides, every vertex on the unit circle (REF-MATHWORLD-ROOT-OF-UNITY). By Euler's formula e^(ix) = cos x + i sin x (REF-MATHWORLD-EULER-FORMULA), the vertex at central angle α is (cos α, sin α) (REF-OPENSTAX-UNIT-CIRCLE), in the mathematical frame (y upwards). Turning every vertex by the same angle α₀ keeps them on the unit circle, 2π/n apart:

pₖ = (cos αₖ, sin αₖ), αₖ = α₀ + 2πk / n.

Two consecutive vertices are a chord a = 2 sin(π/n) apart (REF-MATHWORLD-REGULAR-POLYGON (2), R = 1).

Default orientation: flat base (a horizontal bottom edge). The bottom edge joins the angles −π/2 ± π/n: symmetric about −π/2, they have the same sine, hence the same y. So α₀ = −π/2 + π/n, and

αₖ = −π/2 + (2k + 1) π / n.

Consequence: n=3 triangle pointing up, n=4 square (not a diamond), n=6 hexagon with flat bases.

Step 2 — Order and starting vertex (Q11, Q18) ​

Increasing angles turn counter-clockwise in the mathematical frame; the picture keeps its orientation when drawn in the SVG frame (step 5), so the vertices are listed by decreasing angle to run clockwise on screen.

The start is the right end of the flat base (Q18, amended in F08 so that the base keeps points 1 and 2 whatever n). The base joins the angles −π/2 ± π/n (step 1); its right end has the positive cosine, at −π/2 + π/n, i.e. k = 0. Listed by decreasing angle, the next is −π/2 − π/n, the left end of the base, then the others clockwise. The j-th vertex in drawing order (j = 0 … n−1) has the angle

βⱼ = −π/2 + (1 − 2j) π / n.

Before F08 the start was the topmost vertex, the leftmost on a tie (βⱼ = −π/2 + (2⌊n/2⌋ + 1 − 2j) π / n): the same vertices, the list turned by ⌊n/2⌋ places.

The start is chosen by its index, never by comparing floating coordinates.

The code (unitRegularPolygon, EN-008) hands the unit polygon over already flipped to the SVG frame: (cos βⱼ, −sin βⱼ). Step 5 then uses the flipped ordinates y′ = −y (below).

Step 3 — Unit bounding box ​

Wᵤ = max xₖ − min xₖ, Hᵤ = max yₖ − min yₖ.

Step 4 — Uniform scale ​

s = min(w / Wᵤ, h / Hᵤ). At least one dimension is reached (w or h, or both), exactly in theory; in floating point within a rounding error relative to the box (about 1e-16 × its size: 1e-13 for a box of 1000, 3.7e-9 for 1e7, but a few units near the largest safe integer: 2 for a square of side 2⁵³ − 1, AUD-003). Wᵤ and Hᵤ are positive for n ≥ 3; a size of 0 gives s = 0: every vertex is at the anchored place of the box, its center for mid on both axes (Q15, step 5).

Step 5 — Placement in the SVG frame ​

The scaled polygon spans s·Wᵤ × s·Hᵤ and leaves the room Δx = w − s·Wᵤ ≥ 0 and Δy = h − s·Hᵤ ≥ 0 (step 4), zero on the axis it fills. It is placed in that room by a horizontal and a vertical alignment, each min, mid or max (F07). This is SVG's preserveAspectRatio with meet (REF-SVG2-COORDS §8.2, §8.7): the scale is the smaller of the two ratios (§8.2 step 7), and the translation adds nothing for xMin, Δx / 2 for xMid (step 11), Δx for xMax (step 12), the same for YMin, YMid, YMax (steps 13–14); min aligns the smallest coordinate with the box's, max the largest (§8.7). With the y axis flipped (y downwards, REF-SVG2-COORDS), min is the left or the top, max the right or the bottom.

x = tₓ + s · (xₖ − min xₖ), y = t_y + s · (max yₖ − yₖ), with tₓ ∈ {0, Δx / 2, Δx} and t_y ∈ {0, Δy / 2, Δy}.

The highest vertex of the mathematical frame becomes the one with the smallest SVG y: the picture is unchanged, only its coordinates are.

With the flipped unit polygon of step 2 (y′ₖ = −yₖ, so max yₖ − yₖ = y′ₖ − min y′ₖ), the same placement reads y = t_y + s · (y′ₖ − min y′ₖ): no second flip.

On the anchored side. For min, the extreme vertex is at x = 0 + s · 0 = 0; for max, at Δx + s · Wᵤ = w. In floating point these hold within a rounding error relative to the box, like step 4: the unit vertices of one edge come from cos and sin and may differ in their last bit (the base of the unit triangle at y = 0.5 and 0.5000000000000003), so a base anchored at the bottom of a 100 box may be computed at 99.99999999999999. The output writes 5 decimals (Q10): every vertex of that edge is written 100, the integer edge of the box. For mid, the placement of F02 is kept unchanged: x = Δx / 2 + s · (xₖ − min xₖ).

On the axis the polygon fills, Δ = 0 in theory and the three alignments give the same points; in floating point they differ by a rounding error relative to the box (about 1e-16 × its size, step 4), far below what the output writes (5 decimals, Q10): the written drawing is the same, unless a coordinate falls on a rounding boundary of the 5th decimal (none in boxes from 0 to 1000 for n = 3 … 12, review of EN-009).

The default is mid on both axes: F02's centering. In the code, steps 3–5 are boundingBox and fitInBox (its share of the room: alignmentFactor, EN-009), and the whole fit is regularPolygonContour (US-005, US-009).

Step 6 — Maximal rounding: the incircle ​

At the maximal radius, every edge carries two equal fillets, clamped to meet at its middle (DERIV-local-radius-clamp): setback a/2. With the interior angle θ = π − 2π/n, the radius is ρ = (a/2) · tan(θ/2) (DERIV-fillet-setback step 4), and tan(π/2 − π/n) = cot(π/n):

ρ = ½ a cot(π/n), the inradius r of the polygon (REF-MATHWORLD-REGULAR-POLYGON (3)).

The fillet's center D is the point of the bisector at distance ρ from both edges of the corner B (DERIV-fillet-setback step 1). The incircle is tangent to every side (REF-MATHWORLD-INCIRCLE): its center O is at distance r = ρ from both edges, the feet E and F of its perpendiculars being the points of contact (Euclid III.18, REF-EUCLID-III18). The right triangles BEO and BFO share the hypotenuse BO and have OE = OF, so BE = BF (Euclid I.47, REF-EUCLID-I47); their three sides being equal, their angles at B are equal (Euclid I.8, REF-EUCLID-I8): O lies on the bisector. On the bisector, the distance to the edges grows with the distance to B, so one point only is at distance ρ: O = D. Every fillet is therefore an arc of the incircle, and the fillets meet at the middles of the edges: the rounded polygon is its incircle, of radius s · cos(π/n) once scaled (REF-MATHWORLD-REGULAR-POLYGON (4), R = s). In the code nothing is specific to polygons: regularPolygonCorners gives every vertex the requested radius and the clamp of F01 finds the incircle (US-006).

Checks ​

nBoxWᵤ × HᵤResultVertices in drawing order (5 decimals)Maximal radius
3100 × 100√3 × 1.5100 × 86.60254 (width)(100, 93.30127), (0, 93.30127), (50, 6.69873)28.86751
4100 × 50√2 × √250 × 50 (height)(75, 50), (25, 50), (25, 0), (75, 0)25
5100 × 1001.90211 × 1.80902100 × 95.10565 (width)(80.9017, 97.55283), (19.0983, 97.55283), (0, 38.7743), (50, 2.44717), (100, 38.7743)42.53254
6100 × 1002 × √3100 × 86.60254 (width)(75, 93.30127), (25, 93.30127), (0, 50), (25, 6.69873), (75, 6.69873), (100, 50)43.30127
8100 × 1001.84776 × 1.84776100 × 100 (both)(70.71068, 100), (29.28932, 100), …, (100, 70.71068)50

Anchored (F07): the room is all before (max) or all after (min) the shape instead of split in two.

nBoxAlignment (horizontal, vertical)Vertices in drawing order (5 decimals)
3100 × 100any, max (bottom)(100, 100), (0, 100), (50, 13.39746)
3100 × 100any, min (top)(100, 86.60254), (0, 86.60254), (50, 0)
4100 × 50min (left), any(50, 50), (0, 50), (0, 0), (50, 0)
4100 × 50max (right), any(100, 50), (50, 50), (50, 0), (100, 0)
6100 × 100any, max (bottom)(75, 100), (25, 100), (0, 56.69873), (25, 13.39746), (75, 13.39746), (100, 56.69873)

By hand: the triangle's room is Δy = 100 − 86.60254 = 13.39746; at the bottom the apex is at y = 13.39746 and the base at 100; at the top the apex is at 0 and the base at 86.60254. The square of side 50 leaves Δx = 50: left from x = 0, right from x = 50. The hexagon is shifted down by 13.39746 − 6.69873 = 6.69873 from its centered place.

By hand: n = 3, s = 100/√3 = 57.73503, height 1.5 s = 86.60254, top margin (100 − 86.60254)/2 = 6.69873; maximal radius s cos(π/3) = 28.86751. n = 6, s = 50, maximal radius 50 cos(π/6) = 43.30127, a circle of diameter 86.60254.

Note ​

The fit is computed on the sharp-cornered polygon. A corner radius then rounds inwards: a vertex that touched the box no longer does, a flat edge that touched it still does; at the maximal radius only the incircle remains (accepted by the Product Owner, 2026-10-09).