The dashed disk here has radius 1. Suppose we want to cover it entirely with n smaller disks. How small can those disks be?

Pleasingly, no one has yet found a general answer to this question. If we have only a single covering disk, then obviously it will need to be fully as large as the target. But if we’re allowed six discs, they can do the job even if each has a radius of only 0.555905…, as shown here.

Similar configurations work up to n = 10. But if we’re allowed 11 disks then some creative thinking again becomes necessary to find the best solution. No one has yet found a general strategy that reliably finds the minimum successful size.

One might conjecture that there is an interesting fact concerning each of the positive integers. Here is a ‘proof by induction’ that such is the case. Certainly, 1, which is a factor of each positive integer, qualifies, as do 2, the smallest prime; 3, the smallest odd prime; 4, Bieberbach’s number; etc. Suppose the set S of positive integers concerning each of which there is no interesting fact is not vacuous, and let k be the smallest member of S. But this is a most interesting fact concerning k! Hence S has no smallest member and therefore is vacuous. Is the proof valid?

— Edwin F. Beckenbach, “Interesting Integers,” American Mathematical Monthly, April 1945

The coastline of Nova Scotia was once frequented by pirates, and people occasionally dig for buried pirate treasure. On a local radio program a few years ago I heard an interview with someone who had done a study of attempts to find pirate treasure. He claimed that in most of the cases in which treasure was actually found, it was in a place where treasure-hunters had dug before, rather than in a brand new, previously undug, location. Past diggers simply hadn’t dug deep enough. The previous digger had, in fact, often stopped just short of the treasure. If the previous digger had dug a little deeper than he did, he would have found it.

The interviewer asked him what advice he would give to treasure hunters on the basis of this study; and, producing an interesting application of induction, he lamely suggested that diggers should dig a little deeper than they in fact do. Can you see why this advice is impossible to follow?

— Robert M. Martin, There Are Two Errors in the the Title of This Book, 2002

“The telescope makes the world smaller; it is only the microscope that makes it larger.” — G.K. Chesterton

In 1999, while serving as research fellows at Cambridge University’s Cavendish Laboratory, physicists Thomas Fink and Yong Mao made a mathematical study of necktie knots. They published a summary in Nature that year and a detailed exposition in Physica A in 2000.

They found that, if knots are modeled as persistent random walks on a triangular lattice, there are exactly 85 ways to tie a tie. Of the 10 knots they scored as most aesthetic (for symmetry and balance), only four (four-in-hand, Pratt knot, half-Windsor, Windsor) are well known to Western men; interestingly, the simplest of the remainder, the unassuming small knot, above, is popular in the communist youth organization in China.

Here’s a list of the most aesthetic knots in their list.

Western Illinois University mathematician Iraj Kalantari published an unusual puzzle in Math Horizons in February 2019. A sphere B of radius 150 is centered at (150, 150, 0). A sphere M of radius 144, centered on the z-axis, lies entirely below the (x, y)-plane so that the volume of its intersection with B is 1/2. “Can we find a sphere S of radius 73 that has its center on the circle (x – 73)² + (y – 73)² = 150² in the plane z = 73 so that the volume of B minus its intersections with M and S equals the volume of M minus its intersection with B plus the volume of S minus its intersection with B?”

The answer is no, because Vol(B – (M ∪ S)) = Vol((M – B) ∪ (S – B)) if and only if Vol(B) = Vol(M) + Vol(S), and that’s the case if and only if , where r_B, r_M, and r_S are the radii of the three spheres. “[A]nd because the radii are integers, this equality is impossible by Fermat’s last theorem!”

The placement of the spheres and the fact that the values differ by 1 are red herrings.

(Iraj Kalantari, “The Three Spheres,” Math Horizons 26:3 [February 2019]: 13, 25.)

02/28/2026 UPDATE: In my original statement of the problem I left out a vital phrase in Kalantari’s presentation: The sphere S should have its center on the circle (x – 73)² + (y – 73)² = 150² in the plane z = 73.

I’d omitted the last phrase, a condition that guarantees that S lies above the xy plane and so does not intersect sphere M, which is required to deduce the equation involving the volumes. Many thanks to reader Francesco Veneziano for pointing this out.