Three equal masses attract by Newton's law with a small softening length ε, r̈ᵢ = Σⱼ Gm(rⱼ − rᵢ)/(|rⱼ − rᵢ|² + ε²)^(3/2), starting from the figure eight of Chenciner and Montgomery with every velocity cut to 0.44 of its value. At full speed and with ε = 0 they would circle one figure eight forever, at energy −1.29 (G = m = 1). Slowed, it is −2.26 and no closed orbit remains. Two bodies move on conics. Three have no general closed form.

How do you find every prime up to n without testing any number for divisibility? Take the first unmarked number p, mark all its multiples from p² onward, then move to the next unmarked number. A composite m has a prime factor of at most √m, so only primes p ≤ √n are needed. For n = 100 that means 2, 3, 5 and 7, and the 25 survivors are the primes below 100. Every composite is struck out by its smallest prime factor, and the total work is about n log log n steps, nearly linear in n.

A free Gaussian wave packet solves iħ ∂ψ/∂t = −(ħ²/2m) ∂²ψ/∂x² with no potential, here with ħ = m = 1, k₀ = 4 and starting width σ₀ = 1.8. Re ψ and Im ψ are two waves a quarter turn apart under one envelope, so ψ winds around the x axis as a helix. The envelope moves at v_g = ħk₀/m = 4; the crests move at ω/k = ħk₀/2m = 2, half as fast. The width grows as σ(t) = σ₀√(1 + (ħt/2mσ₀²)²), reaching 2.50 by t = 6.25.

How can a shape have no area and still not be a line? Remove the central triangle from an equilateral triangle, then repeat on each remaining piece. After n steps the area is (3/4)ⁿ of the original, which tends to 0. The piece count triples while the scale halves, giving dimension log 3 / log 2 ≈ 1.585. It sits between a curve and a surface. The same set appears from a game of chance: pick a vertex at random, jump halfway toward it, and repeat. Whatever the start, the points settle onto the triangle. Sierpiński described it in 1915.

Stokes' theorem equates the circulation ∮ F · dr of a vector field around a closed curve C with the flux ∬ (∇ × F) · dS of its curl through any surface S bounded by C. The curl measures rotation at each point. Adjacent rotations cancel along every shared edge, so only the boundary survives. For a flat S and a constant curl c normal to it, the circulation is c times the area of S, whatever the shape of C.

Why does a² + b² = c² hold in every right triangle? Put four copies of the triangle inside a square of side a + b. Arranged one way they leave a square of side c uncovered. Arranged another way they leave two squares, of sides a and b. The outer square and the triangles are the same, so the leftover areas match. Hence c² = a² + b². The result extends to any three similar shapes built on the sides, such as semicircles or equilateral triangles, because area scales with the square of length. The converse holds too: if a² + b² = c², the angle opposite c is a right angle.

The state graph of the Tower of Hanoi has a vertex for each placement of n disks on three pegs, 3ⁿ in all, and an edge for each legal move. The largest disk moves only when every smaller disk sits on the third peg, so the graph is three copies of the graph for n − 1 joined by three single edges: the Sierpiński recursion. The optimal solution runs straight along one side in 2ⁿ − 1 moves. For n = 7 that is 2187 states, 3279 edges and 127 moves.

A gear train is a set of toothed wheels, each driving the next by meshing with it. The dragonfly is a train of 93 gears with one tooth size m, so N teeth give pitch radius r = mN/2. Two gears mesh at centre distance d = r₁ + r₂, and their rims roll without slipping: ω₁r₁ = −ω₂r₂. Each gear meshes only with its parent, so the train is a tree. An odd closed loop of gears would lock, which is why three mutually meshing gears cannot turn.

What does the sum of two numbers have to do with binomial expansions? Each entry is the sum of the two above it: C(n, k) = C(n−1, k−1) + C(n−1, k). That entry is the coefficient of xᵏ in (1 + x)ⁿ, so row n sums to 2ⁿ, and the shallow diagonals sum to the Fibonacci numbers. Shade the odd entries and the Sierpiński triangle appears. Read as digits, row n gives 11ⁿ until an entry reaches 10: 11⁴ = 14641 matches row 4, while 11⁵ = 161051 needs carrying in row 5, which is 1, 5, 10, 10, 5, 1.

Why are there more real numbers than natural numbers? Suppose all reals in [0, 1] sit in an infinite list of decimal expansions. Build a number whose nth digit differs from the nth digit of the nth entry. It disagrees with every entry somewhere, so it is not on the list. No list can be complete: |ℝ| = 2^ℵ₀ > ℵ₀ = |ℕ|. The same move shows that any set is strictly smaller than its power set, so there is no largest infinity. Whether any size sits between ℵ₀ and 2^ℵ₀ is the continuum hypothesis, which ZFC can neither prove nor refute.

An ellipse with semi-axes a = 0.40 and b ≈ 0.259 rolls without slipping inside a unit circle. Its largest radius of curvature, a²/b ≈ 0.62, is below 1, so it touches at one point, and the arc run along its rim equals the arc run along the circle. Each rim point traces an elliptical hypocycloid, cusping on the circle at contact. The circle is exactly three rims long, so every track closes after one turn, with three cusps per point.

Why can dropping a needle on a floor estimate π? For a needle of length l on lines spaced d apart (l ≤ d), it crosses a line when the distance from its centre to the nearest line is under (l/2)sin θ. Averaging over position and angle gives P = 2l/(πd). So π ≈ 2lN/(d·C) after N drops with C crossings. When l = d the chance of a crossing is 2/π ≈ 0.6366, so about 64 drops in 100 touch a line. Buffon posed the problem in 1733 and published the solution in 1777.

A point starts at 1 and moves on the unit circle as z(t) = e^(iωt) = cos ωt + i sin ωt. In the complex plane it returns to 1 whenever t is a multiple of the period T = 2π/ω. With time as a third coordinate the path is the helix (cos ωt, sin ωt, t), and the k-th return lies kT above the start. Its projections onto the two vertical planes are cos ωt and sin ωt. The circle is the helix seen end on: the same point every turn, never the same moment.

The square wave's epicycles rebuilt with squares: square k has half-side 1/(2k + 1), a point circuits it at constant speed 2k + 1 times per period, and the next square is centred on that point. One square gives a trapezoid wave. The last point's height is the sum of all N. With every point in its corner it is 1 + 1/3 + … + 1/(2N − 1), which is 2.081 at N = 9 and grows like ½ ln N without bound.

Why are there exactly five regular convex solids? At each vertex the face angles must sum to less than 360°, and only five combinations of identical regular faces satisfy that: tetrahedron, cube, octahedron, dodecahedron, icosahedron. Each obeys V − E + F = 2, from (4, 6, 4) for the tetrahedron to (20, 30, 12) for the icosahedron. Cube and octahedron are duals, as are dodecahedron and icosahedron, and the tetrahedron is its own dual. Euclid ended the Elements with this proof, and Plato assigned the solids to fire, earth, air, water and the heavens.

What rule draws the Mandelbrot set? Take z₀ = 0 and iterate zₙ₊₁ = zₙ² + c. The set is every complex c for which the orbit stays bounded. At c = −1 it cycles 0, −1, 0, −1, … and at c = 1 it runs 0, 1, 2, 5, 26, … to infinity. Once |zₙ| exceeds 2 the orbit always escapes, so the whole set lies inside |c| ≤ 2. The main cardioid has area 3π/8 ≈ 1.178 out of a total of about 1.507, and the boundary has Hausdorff dimension 2, so its detail never smooths out under magnification.

How many petals does r = cos(kθ) have? For odd integer k the curve closes after θ = π and has k petals. For even integer k it needs θ = 2π and has 2k petals. For k = n/d in lowest terms it closes after dπ when nd is odd and 2dπ when nd is even. Every petal reaches the same length, 1, so the curve stays inside the unit circle. k = 1 is a circle through the origin, k = 3 gives three petals and k = 2 gives four. For irrational k the curve never closes and keeps adding petals.

Each gear is one turn of the golden spiral r(θ) = a φ^(2θ/π), closed by a straight step, on axles D = a(1 + φ⁴) apart. Rolling keeps the contact on the line between the axles, so r₁ + r₂ = D and ω₂/ω₁ = r₁/r₂. Those two conditions turn the partner of a logarithmic spiral into the same spiral, so one turn drives exactly one turn, while the speed ratio swings from 1/φ⁴ to φ⁴ ≈ 6.85.

The curve z(θ) = e^(iθ) + e^(iπθ) is two unit arms, the second turning π times as fast as the first. If π were a fraction p/q, the pen would be back at its start after θ = 2πq and the curve would close. π is irrational, so it never closes. Kronecker's theorem says θ and πθ, taken mod 2π, come arbitrarily close to every pair of angles, so the curve comes arbitrarily close to every point of the disc |z| ≤ 2.