The Pentametric spiral
Square n sits at distance n·√0.1 (n origins units) from the Origin P (Holden's point) and has the Pentametric area 𝔓(n). In the golden rectangle arrangement every square turns around P step by step: the Lucas squares fall exactly into the Fibonacci tiling, and the steps in between are shadow squares. In the pentagon arrangement every square turns 72°, a fifth of a turn, onto the corners of a pentagon. Choose “only the Fibonacci squares” to isolate the Fibonacci spiral contained in it, growing square by square with its golden arcs. Everything starts from nothing, at n = 0.
Why the ±0.8 shows at the centre
Take a square of area n²/5. Its side is n/√5 and half its diagonal is n/√10 = n·√0.1, which is exactly its distance from O. So when its diagonal points along the ray, its inner corner lands exactly on P, for every n. That is the square Sk of the paper (a vertex on the Origin P), now for every n and turned by 72° each step.
The Pentametric square has area 𝔓(n) = n²/5 ± 0.8 instead. On the corners n ≡ 1, 4 (mod 5) it is a little larger, so its corner passes just beyond P; on the corners n ≡ 2, 3 it is a little smaller and stops just short; on the corner n ≡ 0 it touches P exactly. The gap is about 0.63/n: large for the first steps, where the magnifier shows it (n = 1 to 3), and soon too small to see.
With circles, the circle around the square (radius √(𝔓(n)/2)) would pass exactly through P for the area n²/5; the Pentametric circle's area differs from that one by exactly 2π/5, 0 or −2π/5, the circle law of the paper. In 3D the squares become cubes and the circles spheres, each centred at height h(n): the cubes all lean on the vertical axis, and the spheres all brush against it.
From the pentagon to the golden rectangle
At a Lucas number n = Lk the formula gives the area 𝔓(Lk) = Fk² (the Pentametric–Lucas Theorem), and the distance n·√0.1 = Lk/√10 is exactly Holden's distance from the Origin P to the centre of the k-th Fibonacci square. Size and distance are already right; only the direction differs. So a single turn around P, keeping the distance, carries each Lucas square to its place, and together they tile the golden rectangle: n = 1 and n = 3 give the two squares of side 1, n = 4 the square of side 2, n = 7 the 3, n = 11 the 5, n = 18 the 8, and so on. Switch the arrangement between “pentagon” and “golden rectangle” to watch every square turn around P, with Holden's two lines appearing through O.
All the other n turn too and stay on the picture, drawn dashed: they are the shadow squares between the Fibonacci squares. Their area is still 𝔓(n) and their distance still n·√0.1, but they do not tile: they overlap the golden rectangle between one Fibonacci square and the next.
Holden's geometry and the formula fix the direction only at the Lucas numbers, a quarter turn per Lucas number. For the shadows the direction is our interpolation: the quarter turn from Lk to Lk+1 is shared evenly among the steps in between, and each square tilts by the same angle as it turns, as if carried round P (for example n = 5 and 6 sit ⅓ and ⅔ of the way from the direction of L3 = 4 to that of L4 = 7). It is exact at every Lucas number; other rules are possible. n = 2 is also a Lucas number (L0), with 𝔓(2) = F0² = 0: it has no square. The pentagon arrangement itself is a way of drawing the paper's results (the pentagonal angle form gives the turn, the origins unit the distance, the formula the area), not a separate theorem; the golden-rectangle positions are Holden's (1975).