A needle can turn around in as little area as you like, yet in 3D the set it sweeps can never be thin — proved by Hong Wang and Joshua Zahl in 2025
How much room do you need to turn a needle around? In the plane, as little area as you like. Yet in 2025 Hong Wang and Joshua Zahl proved that in three dimensions the set it sweeps can never be thin.
A unit needle makes a half-turn in a triangle and in a deltoid. Follow the white dot to see it end up reversed.
The white dot marks one end of the needle: after a half-turn it is at the other end. Kakeya suspected the deltoid was optimal. Besicovitch showed in 1928 that the answer is: no minimum at all.
In 1917 Sōichi Kakeya asked for the smallest region in which a unit needle can be turned through 180°. A disc of diameter 1 works, with area π/4 ≈ 0.785. Gyula Pál showed in 1920 that among convex regions the best is the equilateral triangle of height 1, with area 1/√3 ≈ 0.577. The needle pivots at a corner, slides to the next corner, and repeats.
Without convexity you can do better. The deltoid, a three-cusped curve traced by a point on a circle rolling inside another three times its size, has area π/8 ≈ 0.393, and the needle glides round inside it, always touching the curve. Kakeya suspected this was optimal.
It is not, and in fact nothing is. Abram Besicovitch had already built a set of measure zero containing a unit segment in every direction, answering a question in integration theory, and in 1928 he adapted it to Kakeya’s problem. A needle can be turned around in a region of area less than any ε you name. The region becomes a thin tangle of spikes, and the needle has to travel a long way, but the area can be made as small as you like.
Raise the stage and watch the area fall. Then drag the direction: a full-height segment is always there.
The blue segment joins a Cantor point below to a Cantor point above, so it survives every stage. The pieces multiply by four each time while the area only creeps down, yet in the limit it is zero. The fan spans about 72°, so three copies turned by 60° cover every direction.
This construction is Jean-Pierre Kahane’s, from 1969. Take the Cantor set C of numbers whose base-4 digits are all 0 or 3. Put C on the bottom edge and a half-size copy on the top edge, and join every point below to every point above. The run of a segment from (x, 0) to (y/2, 1) is y/2 − x, and its base-4 digits are 1.5 times one of −2, −1, 0, 1. Four consecutive values are enough to fill an interval, so the runs cover all of [−1, 1/2].
Every horizontal slice of the limit set is a projection of the four-corner Cantor set C × C. That set is a scattered dust of total length zero in almost every direction, which is Besicovitch’s projection theorem. So almost every slice has length zero, and the set has area zero.
Measure zero is not the same as small in every sense. Roy Davies proved in 1971 that any set in the plane containing a segment in every direction has Hausdorff dimension 2: it is as large as the plane itself in the sense of dimension. The Kakeya conjecture says the same holds in every dimension: a set in ℝⁿ containing a unit segment in every direction has dimension n. It sits underneath some of the central open problems in harmonic analysis, including the restriction conjecture, which is why so much work has gone into it.
Over a finite field every line is a row of q points. Step through the directions and compare the set’s size with Dvir’s bound.
Every one of the 57 directions has a full line inside the orange set (the blue one is highlighted). As q grows this set fills about half of the plane and a quarter of space, and Dvir’s bound says no cleverer arrangement can fall below a fixed fraction.
In 1999 Thomas Wolff proposed a model problem: replace ℝⁿ by 𝔽_qⁿ, the n-dimensional space over a finite field with q elements, and ask how few points a set can have if it contains a full line in every direction. Zeev Dvir settled it in 2008 with an argument short enough to sketch here. If a set K has fewer than C(q + n − 1, n) points, some nonzero polynomial of degree at most q − 1 vanishes on all of K. Restricted to any line in K it vanishes at q points, more than its degree, so it vanishes on the whole line. Then its top-degree part vanishes at every direction vector, hence everywhere, and a nonzero polynomial of degree below q cannot vanish on the whole space. So K has at least C(q + n − 1, n) ≈ qⁿ/n! points, a fixed fraction of the space.
The real problem is much harder, because ℝ has no analogue of the counting that makes polynomials work so well over a finite field. In three dimensions Wolff proved in 1995 that a Kakeya set has dimension at least 5/2, and for thirty years the improvements came in tiny increments. The modern approach works with thickened tubes rather than segments: for each of about δ⁻² well-separated directions, take a tube of radius δ, and show that the union cannot be much smaller than the whole region however the tubes are arranged.
In February 2025 Hong Wang and Joshua Zahl posted a proof that every Kakeya set in ℝ³ has Hausdorff dimension 3 (arXiv:2502.17655), building on their earlier work on “sticky” Kakeya sets, where tubes pointing in nearby directions stay close together. The conjecture is still open in four dimensions and higher.
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