Shape Court

1 · The address triangle

Shape. n = ⌊√(2i − ¾) + 3/2⌋ lists the chambers; with m = i − (n−1)(n−2)/2 the sites (−m/n, 1/n) fill the isosceles right triangle (0,0), (−1,0), (−½,½): the M₂ triangle at half scale. Its sides are T(x) − T(y−1) = 0 and its circle is T(x) + T(y−1) = 0, centred on the apex.

Key. The Desmos lines keep the sheet's own letters. In the papers' notation the sheet's i is the address N, its n is m + 1 for the chamber m, and its m is the row r.

i=[1...2000]
n=floor(sqrt(2i-3/4)+1.5)
m=i-(n-1)(n-2)/2
(-m/n,1/n)
x(x+1)/2-y(y-1)/2=0
x(x+1)/2+y(y-1)/2=0

2 · The shadow of an equilateral

Shape. Tilt an equilateral triangle of side 2 about its base by θ: its shadow has base 2 and height √3·cos θ. At cos θ = 1/√3 (θ = 54.74°, a cube's diagonal against its edge) the shadow is M₂'s right isosceles triangle. Up the dimensions, the regular simplex x₀+…+xₙ = 1 casts the corner simplex at cos θₙ = 1/√(n+1), and its shadow on a line is the B-spline Mₙ.

t=0.9553
polygon((-1,0),(1,0),(0,sqrt(3)cos(t)))
polygon((-1,0),(1,0),(0,1))
arccos(1/sqrt(3))

3 · The fold and the bit

Shape. A rectangle t × s with share p = t/(t+s) = (1+J)/2. Its fold is q = ts/(t+s)² = (1 − J²)/4. The binary entropy of the share has slope −artanh J (minus the rapidity) and curvature −1/(4q): the bit's bend is the fold. One full bit sits at the square, the fold's vertex q = ¼.

B(x)=-((1+x)/2)log_2((1+x)/2)-((1-x)/2)log_2((1-x)/2)
y=B(x)\{-1<x<1\}
y=1-x^2\{-1<x<1\}
n=[2...10]
(1/n,B(1/n))

4 · The staircase to ½

Shape. The line y = (k−1)/k − x meets the diagonal at x = (k−1)/(2k) = T(k−1)/k², climbing to ½. Because k² = T(k) + T(k−1), these are the two halves of the square: the mirror of the staircase s = T(n)/n², where the fold reads J = 1/n and q = ¼ − 1/(4n²). Spin the diagonal about the vertical axis and it sweeps the 45° cone of card 5.

k=[3...12]
y=(k-1)/k-x
y=x
((k-1)/(2k),(k-1)/(2k))

5 · The 45° cone and π/√2

Shape. Take a cone of height H, half-angle α and rim radius R = H tan α. The sphere centred on the rim plane that touches the walls has radius R_s = H sin α and meets them along one circle. Its upper hemisphere has area 2πH² sin²α and the base has πH² tan²α. They agree only when 2cos²α = 1, at α = 45°, and only there is the pole-to-equator chord √2·R_s equal to H. Nothing depends on H: the construction fixes its own scale, so H = π only names the radius π/√2, the early-2025 scalar. The staircase's diagonal x = y, spun about the vertical axis, sweeps this 45° cone. Triangle joins the apex, the pole and the end of the chord. At 45° its sides are √3·R_s, √2·R_s and (1+√2)·R_s. Below the rim plane it is a cube's right triangle (edge R_s, face diagonal √2·R_s = H, body diagonal √3·R_s), so the slant meets the rim plane at card 2's 54.74° and the axis at card 6's 35.26°. Above the rim it is half of the hemisphere's square. The axis crosses the sphere at (√2 − 1)·R_s and (√2 + 1)·R_s from the apex, the silver pair whose product is R_s², and the slant enters the sphere a third of the way up.

H=3.14159
a=0.785398
z=cot(a)sqrt(x^2+y^2)\{z<=H\}
x^2+y^2+(z-H)^2=(H sin(a))^2
z=H\{x^2+y^2<=(H tan(a))^2\}

6 · Eight triangles make the odd square

Shape. (2n+1)² = 8T(n) + 1: eight triangular blocks of T(n) cells around one centre. Lift raises each cell to the height of its ring, so ring k holds 8k cells at height k; the count does not change. Stack goes up a dimension. Pile the odd squares 1, 9, 25, … on one centre and each sector climbs into a triangular staircase of Te(n) = n(n+1)(n+2)/6 cubes, a tetrahedral number. So Σ(2j+1)² = 8·Te(n) + (n+1): eight tetrahedral stacks around a central column, the triangle going up to the tetrahedron as in the simplex shadow. Read flat as a product over the odd squares, 4·∏ 8T(k)/(8T(k)+1) → π (Wallis); the factor at T(15) = 120 = 5! is 960/961 = 1 − 1/31². Iso looks along a cube diagonal, at card 2's angle cos θ = 1/√3. The square becomes a 60°–120° rhombus with diagonals 1 : √3. The eight sectors keep equal areas but change shape: the four on the near–far diagonal turn equilateral and the four on the cross diagonal turn 30°–30°–120°. Every layer of the stack lands inside the bottom slab's outline, so the whole pyramid reads as one flat shape. The same holds for every view within 45° of vertical, the slope of the stack's own steps: each smaller layer's shadow lies inside the bottom slab's. So the largest shadow, (2n+1)√((2n+1)² + 2), is the slab's own, and the smallest, (n+1)², is the side view along an axis. Beyond 45° the smaller layers stick out in general; the iso view, 54.74° from vertical along the diagonal, is the touching case where their corners land exactly on the outline.

T(n)=n(n+1)/2
Te(n)=n(n+1)(n+2)/6
P(N)=4prod_{k=1}^{N}8T(k)/(8T(k)+1)
P(10)
P(1000)
sum_{j=0}^{4}(2j+1)^2-8Te(4)