A set of points has diameter 1 if no two points in the set are more than 1 unit apart. An example is an equilateral triangle whose side has length 1. What’s the smallest shape that can cover any such set? A circle of diameter 1 won’t cover our triangle; part of the triangle projects beyond the circle:

Of course a larger circle would work, but what’s the smallest shape will always do the job? Surprisingly, no one knows. When French mathematician Henri Lebesgue posed the problem to Gyula Pál in 1914, Pál suggested a modified hexagon (in black):

Here Pál’s shape manages to surround a circle (blue), a Reuleaux triangle (red), and a square (green), each of diameter 1, and in fact it will accommodate any such set. Its own area is 0.84529946. Will a smaller shape do the job? Well, yes, but the gains get increasingly fine: In 1936 Roland Sprague whittled Pál’s shape down to 0.844137708436, and in 1992 H.C. Hansen reduced it further to 0.844137708398. At this point observers Victor Klee and Stanley Wagon wrote, “[I]t does seem safe to guess that progress on [this problem], which has been painfully slow in the past, may be even more painfully slow in the future.” But in 2015 John Baez reached 0.8441153 with an exquisite adjustment to two regions in Hansen’s shape; the smaller of these would span only a few atoms if the shape were drawn on paper.

Is that the end of the story? No: Last October Philip Gibbs claimed a further reduction to 0.8440935944, and the search goes on. In 2005 Peter Brass and Mehrbod Sharifi showed that the universal cover must have an area of at least 0.832, so there’s room, at least in theory, for still further improvements.

(Thanks, Jacob.)

During World War II, Hungarian mathematician Pál Turán was forced to work in a brick factory. His job was to push a wagonload of bricks along a track from a kiln to storage site. The factory contained several kilns and storage sites, with tracks criss-crossing the floor among them. Turán found it difficult to push the wagon across a track crossing, and in his mind he began to consider how the factory might be redesigned to minimize these crossings.

After the war, Turán mentioned the problem in talks in Poland, and mathematicians Kazimierz Zarankiewicz and Kazimierz Urbanik both took it up. They showed that it’s always possible to complete the layout as shown above, with the kilns along one axis and the storage sites along the other, each group arranged as evenly as possible around the origin, with the tracks running as straight lines between each possible pair. The number of crossings, then, is

where m and n are the number of kilns and storage sites and denotes the floor function, which just means that we take the greatest integer less than the value in brackets. In the case of 4 kilns and 7 storage sites, that gives us

which is the number of crossings in the diagram above.

Is that the best we can do? No one knows. Zarankiewicz and Urbanik thought that their formula gave the fewest possible crossings, but their proof was found to be erroneous 11 years later. Whether a factory can be designed whose layout contains fewer crossings remains an open problem.

Today it’s possible to describe a color quantitatively, but how did people make such fine distinctions in the 18th century? German geologist Abraham Gottlob Werner proposed a solution in 1774: His Von den äußerlichen Kennzeichen der Foßilien included a “color dictionary” that located each hue in the natural world. Updated by Scottish painter Patrick Syme, it describes 110 colors, telling where each might be found in animal, vegetable, and mineral form: Number 35, for example, “bluish lilac purple,” is the shade of the male of the dragonfly Libellula depressa, the blue lilac, and the mineral lepidolite. Number 82, “tile red,” may be found in the breast of the cock bullfinch, in the shrubby pimpernel, and in porcelain jasper.

This common language gave naturalists an objective way to communicate what they were seeing. Off Brazil aboard the H.M.S. Beagle in 1832, Charles Darwin wrote, “I had been struck by the beautiful color of the sea when seen through the chinks of a straw hat. It was according to Werner nomenclature ‘Indigo with a little azure blue’. The sky at the time was ‘Berlin [blue] with little Ultra marine’.”

The Internet Archive has Syme’s full text.

The angle cos^-1(-1/3) = 109.47°, familiar from soap films and tetrahedral molecular geometry, can be produced with an ordinary piece of A4 paper: Because it has a width:length ratio of , folding it corner to corner as shown yields a shape with precisely that angle.

(Nick Lord, “A ‘Maths Bite’: How to Impress a Chemist,” Mathematical Gazette 80:489 [1996], 584-584.)

How many watchmen are needed to guard the art gallery at left, so that every part of it is under surveillance? The answer in this case is 4; four guards stationed as shown will be able to watch every part of the gallery.

In 1973 University of Montreal mathematician Václav Chvátal showed that, in a gallery with n vertices, n/3 guards will always be enough to do the job. (If n/3 is not an integer, you can dispense with the fractional guard.) And Bowdoin College mathematician Steve Fisk found a beautifully simple proof of Chvátal’s result.

The figure at right shows another art gallery. Cut its floor plan into triangles, and color the vertices of each triangle with the same three colors. The full area of any triangle is visible from any of its vertices, and that means that the whole gallery can be guarded by stationing watchmen at the points indicated by any of the three colors. Choosing the color with the fewest vertices will give us n/3 guards (again discarding fractional guards).

The Chvátal and Fisk proofs both give an answer that’s sufficient but sometimes not necessary. In this case, the gallery has 12 vertices, and 12/3 guards (say, the four green ones) will certainly do the job, but here as few as two will be enough.

(Steve Fisk, “A Short Proof of Chvátal’s Watchman Theorem,” Journal of Combinatorial Theory, Series B 24:3 [1978], 374.)