The Epitetartic Rhombus
A rhombus whose diagonals are in the ratio 5 : 4, with angles 2·arctan(4/5) = 77.32° and 2·arctan(5/4) = 102.68°. Lluis found it where Holden's line meets an arc of the Fibonacci spiral, then found that every square holds it. Its half-angle is the angle of 0.8, the ±0.8 correction drawn as a slope. This page is an exploration beyond the papers; the results are written up, with proofs, in the note The Epitetartic Rhombus (Zenodo, doi:10.5281/zenodo.23325442).
1 · Found: where Holden's line meets the arc
On the angle of 0.8, a search that joins marked points (corners, centres, the Origin) finds arctan(4/5) nowhere special. Lluis found it where that search never looked: where one of Holden's lines crosses an arc of the spiral. In his GeoGebra frame (E at the origin of the axes, the arc of radius 1 centred at (0, −1)), Holden's green line x + 3y = −2 runs through C₂ = (1, −1) on the arc and leaves it again at one more point, Q₅.
proved Put the line into the circle: x² + (y + 1)² = 1 with y + 1 = (1 − x)/3 gives 5x² − x − 4 = 0, that is (5x + 4)(x − 1) = 0. So the line meets the arc at C₂ (x = 1) and at Q₅ = (−4/5, −2/5) exactly. Seen from the arc's centre, Q₅ is the 3-4-5 point (−4/5, 3/5): a chord of slope 1/3 through the edge of a circle doubles its angle at the centre, tan 2θ₁ = (2/3)/(1 − 1/9) = 3/4. Holden's arctan(1/3) makes the 3-4-5 triangle, and the 0.8 comes with it.
2 · The same crossing in every square
Q₅ is not alone. Lluis found that in every square of the tiling Holden's lines cross the spiral exactly once, and always at a whole number of origins units (ρ = √0.1, the side of the origins grid) from the Origin: twice the next Fibonacci number. Square 1 crosses at 2h (that is Q₅), then 4h, 6h, 10h, 16h, 26h, 42h, … for ever. And seen from the centre of its arc, every crossing is the same 3-4-5 point, scaled by the side of the square and turned a quarter turn from one square to the next: the Epitetartic Rhombus grows with the spiral.
This section uses Gardiner's coordinates, as in the note: P = (1/10, 3/10) is the Origin, the first square is centred at (0, 0), and the centre of the square of side Fk is Ck = P + ik−1Lkw with w = −(1 + 3i)/10 (Holden's centres, Lk origins units from P). The crossing is Xk = P + 2Fk+1ik−1w: on the same ray as the centre, Fk units further out (2Fk+1 = Lk + Fk). In this frame Q₅ of section 1 is X₁ = −P.
proved for every k: with the arc's centre Mk = Ck + Fkik−1(1 + i)/2, Xk − Mk = Fkik−1(w − (1 + i)/2) = Fkik−1(−3/5 − 4/5·i), of length exactly Fk and pointing into the arc's quarter. computed that it is the only crossing in each square, for squares 1 to 28. Found by Lluis Garcia Torcal; see the note for the full proof.
3 · The Epitetartic Rhombus
The rhombus of section 1 has its diagonals in the ratio 5 : 4, the ratio of the major third, which the Greeks called ἐπιτέταρτος (epitetartos, “a quarter more”). Lluis named it the Epitetartic Rhombus (Ὁ Ἐπιτέταρτος Ῥόμβος). Its angles are 2·arctan(4/5) = 77.32° and 2·arctan(5/4) = 102.68°, and its side is √41 times a tenth of the long diagonal. It is not only in the spiral: it can be drawn inside any square.
Split the square into 3 × 3. The long diagonal is a √10 line: it crosses three small squares, from a third of the way up one side to two thirds up the other (√(1² + 3²) = √10 small sides, slope 1/3, Holden's angle). The short diagonal lies on the perpendicular √10 line (slope −3) but stops one tenth of the side short of each edge, from 1/10 to 9/10 of the height. So it is 8/10 = 4/5 of the long one.
With a ruler only: join the middle of the top edge of the top-left small square, the centre of the top-middle one and the middle of the bottom edge of the top-right one. That line has slope −1/2, the paper's other angle θ₂ = arctan(1/2), and it cuts the √10 line exactly at the corner (11/30, 9/10). The opposite corner comes the same way, turned half round the centre. Both of the paper's angles build the rhombus: arctan(1/3) the long diagonal, arctan(1/2) the corners.
Every square holds four: the two √10 lines through the centre come in two perpendicular pairs, and either line of a pair can carry the long diagonal. On the tilted squares of the √0.1 grid (side 3·√0.1) the long diagonal joins two nodes of the grid: that is the half rhombus of section 1, from Q₅ to N₃₀, and the Origin sees N₃₀ across exactly one grid cell (√0.2).
proved exact fractions: long diagonal (0, 1/3) → (1, 2/3), short diagonal (11/30, 9/10) → (19/30, 1/10), centre (1/2, 1/2); the slope −1/2 line through (1/6, 1), (1/2, 5/6) and (5/6, 2/3) meets the √10 line at (11/30, 9/10). Checked on Lluis's tilted square, it gives exactly M₄ (−1, −1), F (0, −1), E₈ (−0.5, −0.6) and (−0.5, −1.4).
4 · Halving never finds the corner; threes and fives do
Lluis tried to catch the corner (11/30, 9/10) by halving the square again and again, to see if it ever becomes the centre or the corner of a small square. It never does: at every halving it falls into a different quarter, for ever. Split it by threes and fives together (15 or 30 parts) instead, and it lands exactly. Threes alone or fives alone are not enough: each catches only part of it.
proved A halving grid can only hit a point whose coordinates have a power of 2 below the line; 11/30 and 9/10 have 3 and 5 there, so the corner never lands. In base 2 both are repeating fractions, so the quarters repeat: after two steps (top-left, top-right) the pattern top-left, bottom-right, bottom-right, top-right comes back for ever, every 4 halvings, because 2⁴ = 16 is one more than 15. The point is the exact centre of a cell when the square is split into 15, 45, … (15 times an odd number) and an exact corner when it is split into 30, 60, … (multiples of 30).
The angle of 0.8 lives in the arithmetic (the correction, the doubled angle of the paper's θ₂, the family of triangles on the angle of 0.8) and, as Lluis found, in the geometry too: where Holden's line meets the arc, and in the rhombus every square holds.