A star polygon {n/k} joins n points on a circle, stepping k at a time: vertex j sits at angle 2πjk/n. A point runs its edges at constant speed, so its height moves in straight lines between the values sin(2πjk/n), a sine sampled n times a turn. For {25/12}, k/n = 0.48 is just under ½, and the heights become (−1)ʲ⁺¹ sin(πj/25): a fast zigzag under one slow beat per turn. That is aliasing, the effect that makes a filmed wheel spin backwards.

A circle of radius r rolls without slipping inside a ring of radius R, and a pen at distance d from its centre traces z(t) = (R − r)eⁱᵗ + d·e^(−i(R − r)t/r). Per orbit the wheel turns back (R − r)/r times. With R = 13, r = 5, d = 4 each pen draws 13 petals, and a second pen at −d draws the same curve turned by πr/R, filling 26. The curve closes after r/gcd(R, r) turns, here 5; an irrational ratio never closes.

In 2007, a 16-year-old from Guilin named Hong Wang missed the single place Peking University's maths department held for her province, so she enrolled in earth sciences instead. She transferred into maths a year later, then doubted she was good enough for research and left to spend a semester at an architecture firm in Paris. She came back to mathematics. In February 2025, she and Joshua Zahl posted a 127-page proof of the Kakeya conjecture in three dimensions, which grew out of a 1917 question about how little space a needle needs to turn and point in every direction. In July 2026, at 35, she won the Fields Medal, only the third woman to receive it. She once doubted she could do research. Her proof settled a problem that had defeated several Fields medalists.

For n = 1, 1.5, 2, … , 20 the curves y = xⁿ and y = x^(1/n) both run from (0, 0) to (1, 1) inside the unit square, one colour per n. Each pair is a single curve read two ways: y = xⁿ exactly when x = y^(1/n), so a reflection in the line y = x exchanges them. As n grows, xⁿ stays near 0 until x is close to 1, with 0.9²⁰ ≈ 0.12, while x^(1/20) is already 0.89 at x = 0.1. In the limit the pair closes onto the edges of the square.

The Buddhabrot records where orbits go rather than which points escape: every c whose orbit under z ↦ z² + c leaves |z| ≤ 2 adds all the points it visited to one histogram. Here 3,780,478 escaping orbits are binned. Raising the cap from 20 to 5,000 steps keeps only orbits that linger near the boundary of the Mandelbrot set, and the filaments sharpen. Orbits that never escape trace the anti-Buddhabrot. Melinda Green described it in 1993.

The graphs of y = 1/x and y = −1/x are hyperbolas: the first in the first and third quadrants, the second in the second and fourth. They can never meet, since 1/x − (−1/x) = 2/x is never zero. As x → 0 one shoots to +∞ and the other to −∞, and as x → ±∞ both flatten onto the x axis from opposite sides. Each is the mirror image of the other in either axis, and both axes are asymptotes of both.

A Ford circle sits on the number line at each fraction p/q in lowest terms, with diameter 1/q². For two of them the squared distance between centres minus the squared sum of radii is ((ps − qr)² − 1)/(qs)², which is never negative. So no two circles overlap, and two touch exactly when |ps − qr| = 1, the condition for neighbours in a Farey sequence. The largest circle in the gap between two touching circles sits on the mediant (p + r)/(q + s).

Newton's method replaces a curve by its tangent and takes the tangent's zero as the next guess: xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ). For f(x) = x² − 2x − 1 from x₀ = 3.5 the guesses are 2.65, 2.4311, 2.414313, 2.4142135658, and the error obeys eₙ₊₁ = eₙ²/(2(xₙ − 1)) exactly. Each error is squared, so the correct digits of 1 + √2 run 1, 2, 4, 9, 17: five steps give seventeen digits.

"A cycloid, the curve traced by a point on a rolling circle of radius r, is turned upside down to make a ramp. A bead sliding without friction reaches the lowest point in time π√(r/g) from any starting point. A higher start means a longer path, matched exactly by a steeper initial slope. Christiaan Huygens proved this tautochrone property in 1659 and built it into a pendulum clock whose period does not depend on the swing."

Two cylinders of radius r cross at right angles. Their intersection, the Steinmetz solid, is cut by every horizontal plane at height z in a square of side 2√(r² − z²). Summing the squares gives V = ∫ 4(r² − z²) dz from −r to r = 16r³/3, with no π anywhere. The sphere inscribed in the solid has circular slices covering π/4 of each square, which recovers 4πr³/3, the route Archimedes took.