Warping spacetime into a moving bubble allows apparent superluminal travel. - The metric is ds² = −dt² + (dx − v_s f(r_s) dt)² + dy² + dz². - The shaping function f(r_s) = [tanh(σ(r_s + R)) − tanh(σ(r_s − R))] / [2 tanh(σ R)] creates the bubble using shell thickness σ and radius R. - York time θ = v_s (x_s/r_s) (df(r_s)/dr_s) measures spatial expansion or contraction. It models hypothetical interstellar warp propulsion in general relativity.

A pen is driven by two perpendicular oscillations, x = 3 cos(50t) and y = 3 sin(49t). Since 49t = 50t − t, each lap of the x-wave is an ellipse x = 3 cos u, y = 3 sin(u − t) whose phase lags by the slowly growing t. It starts as a circle, flattens onto the diagonal y = −x at t = π/2, reopens at π, flattens onto y = x at 3π/2 and closes at 2π. Any two frequencies one apart weave the same square: n laps against n − 1, one full turn of phase.

A closed curve is a sum of rotating circles: z(t) = ∑ cₖ e^(ikt), where the cₖ are its Fourier coefficients and circle k turns k times per lap. Keeping the N largest |cₖ| gives the best N-circle drawing in the least-squares sense, by Parseval’s identity. Doubling N from 4 to 64 turns each rough loop into the symbol, until every outline lies within 0.3 % of the glyph. A symbol with holes needs one chain per boundary, so ∞ takes three.

This supertoroid sweeps the curve X¹⁰ + Y¹⁰ = 1, a square with rounded corners, around a circle of radius 4. In polar form the section is Rⁿ(cosⁿ v + sinⁿ v) = 1, and n must be even: for odd n the bracket vanishes at v = 3π/4. The section turns through tu as u goes round, so with t = 3 each corner, at radius 2^0.4 ≈ 1.32, winds three times round the tube.

Homeschooled in Nassau, the Bahamas, Hannah Cairo learned maths mostly from Khan Academy, finished calculus by 11 and taught herself from graduate textbooks. At 14 she joined the Berkeley Math Circle, and she later took graduate courses at UC Berkeley. In one of them, a homework problem came with an optional extra tied to the Mizohata-Takeuchi conjecture, a 40-year-old question about how overlapping waves concentrate their energy. She built a counterexample and posted it on 10 February 2025, at 17. She then applied to 10 PhD programs, and six rejected her for not having a college degree. The University of Maryland took her. She had no high school diploma. She had just disproved a 40-year-old conjecture.

T(x, y) = (xy², −x²y) is a cubic map of the plane, applied to a square grid point by point and blended in as (1 − a)p + a·T(p). For c ≠ 0 every grid line lands on a parabola: the row y = c on Y = −X²/c³, the column x = c on X = Y²/c³. The axes x = 0 and y = 0 collapse to the origin, where the Jacobian 3x²y² vanishes. Each point is sent at right angles to itself, since p · T(p) = x²y² − x²y² = 0.

Five hundred birds fly at one constant speed. At each step every bird turns a tenth of the way toward the mean heading of the birds within radius R, plus a random wobble of width η. No bird sees the whole flock. For small η the local rule still settles on one shared heading, and the order φ = |(1/N) ∑ e^(iθₖ)| climbs from 0.02 to 0.99. Past a critical η the order is lost. Vicsek and coauthors described this phase transition in 1995 naming it Vicsek Model.

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.

"A ball tiled by P pentagons and H hexagons, with three faces at every corner, has E = (5P + 6H)/2 edges and V = (5P + 6H)/3 vertices. Euler's formula V − E + F = 2 then reduces to P/6 = 2, so P = 12 whatever H is. The football has H = 20; the Goldberg balls GP(2,0), GP(3,0) and GP(4,0) have 30, 80 and 150. The same count fixes the twelve pentagons in every fullerene and geodesic dome."

n identical books of length 1 are stacked so that the centre of mass of everything above each book sits exactly over its edge. The k-th book from the top then overhangs the one beneath it by 1/(2k). The total overhang is ½(1 + ½ + ⅓ + … + 1/n), half the harmonic number Hₙ. Four books give 25/24 > 1, so the top book clears the table; 31 give 2.01. Hₙ grows like ln n without bound, so the overhang has no limit.