In 1956, ornithologist Chandler Robbins tagged a wild female Laysan albatross at the Midway Atoll National Wildlife Refuge in the North Pacific. The bird, dubbed Wisdom, went on to a stunning career, flying more than 3 million miles, equivalent to 120 trips around the Earth. She has been seen at the atoll as recently as last December, making her, at 72, the the oldest known wild bird in the world.

In that time she’s hatched as many as 36 chicks, a significant contribution to the struggling wild albatross population. The U.S. Fish and Wildlife Service wrote, “Her health and dedication have led to the birth of other healthy offspring which will help recover albatross populations on Laysan and other islands.” Bruce Peterjohn, chief of the North American Bird Banding Program, added, “To know that she can still successfully raise young at age 60-plus, that is beyond words.”

Suppose we drop a glass rod and it breaks into three pieces. What is the probability that the pieces can form a triangle? Mathematician D.C. Johnson found this elegant geometric solution. In order to form a triangle, none of the three pieces can be longer than the other two combined. Now consider an equilateral triangle whose altitude equals the length of the glass rod (say, 1). Viviani’s theorem tells us that the sum of the lengths of the perpendiculars from any interior point to the sides of this triangle is 1, the triangle’s altitude:

And any three non-negative numbers whose sum is 1 correspond to three such perpendiculars in some prescribed order. So the points inside the triangle correspond to all the various ways in which the glass rod can break.

Now consider the first piece of the broken rod. In order to form a triangle with the other two pieces, its length must not exceed 1/2. That means it must not extend from the base of our equilateral triangle into the shaded zone here:

And if we assign the other two pieces to the other two legs, we can make the same argument and identify two more zones:

That immediately gives the answer: The chance that all three pieces will be short enough to produce a triangle is 1 in 4.

(C. Haigh, “The Glass Rod Problem,” Mathematical Gazette 65:431 [March 1981], 37-38.)

If we have two numbers a and b such that ab + 1 is square, then it’s always possible to find a number c for which ac + 1 and bc + 1 are both square. For example, 8 × 3 + 1 = 25 = 5², and 8 × 21 + 1 = 169 = 13² and 3 × 21 + 1 = 64 = 8².

Proof:

If ab + 1 = m², then set c = a + b + 2m. Now

ac + 1 = a² + ab + 2am + 1 = a² + 2am + m² = (a + m)²

bc + 1 = ab + b² + 2bm + 1 = b² + 2bm + m² = (b + m)²

Via Edward Barbeau, Power Play, 1997.

Princeton mathematician John Horton Conway memorized π to more than a thousand decimal places by marrying it to the periodic table of the elements:

3 Neutronium 1415926535 Hydrogen 8979323846 Helium 2643383279 Lithium 5028841971 Beryllium …

Between each pair of elements are sandwiched ten digits of π. (Neutronium is Andreas von Antropoff’s notional “element of atomic number zero,” an element with zero protons in its nucleus.) This approach to memorizing digits has a number of virtues:

• It’s modular. If you forget one segment you can just look it up and plug it back in to the whole. And you can name the segment you’ve forgotten.
• The element names lend some memorable color to each segment.
• The 10-digit “mouthfuls” are relatively easy to remember, and since they’re tied to numbered elements you can jump fairly readily to, say, the 216th digit.
• They give you an excuse for stopping — you’ve run out of elements!

To remember the elements themselves Conway devised a long mnemonic. It begins

Newt? Hy! He Likes Beryl’s Boring Car for Nites Out in Florid Neon

for

Nn H He Li Be B C N O F Ne.

See the paper below for the whole package — by including unconfirmed hypothetical elements, it encodes 120 mouthfuls, or 1,200 digits.

(John Conway, “Chemical π,” Mathematical Intelligencer 38:4 [December 2016], 7-10.)

Suppose you’re hiking in the woods and become lost. What’s the best path to follow to find the boundary? You know the forest’s shape and dimensions, but you don’t know where you are within it, nor which direction you’re facing.

This has remained an open problem ever since mathematician Richard Bellman posed it in the Bulletin of the American Mathematical Society in 1956. The best strategy will cover the shortest distance in the worst case; in forests of certain simple shapes this might be as straightforward as walking in a straight line or in a spiral, but other shapes are more troubling. We know how to escape squares and circles efficiently, but not equilateral triangles.

Mathematician Scott W. Williams classed this as a “million-buck problem” because solving it is expected to cultivate techniques of particular value to mathematics. It’s known as Bellman’s lost-in-a-forest problem.

It appears that there was a club and the president decided that it would be nice to hold a dinner for all the members. In order not to give any one member prominence, the president felt that they should be seated at a round table. But at this stage he ran into some problems. It seems that the club was not all that amicable a little group. In fact each member only had a few friends within the club and positively detested all the rest. So the president thought it necessary to make sure that each member had a friend sitting on either side of him at the dinner. Unfortunately, try as he might, he could not come up with such an arrangement. In desperation he turned to a mathematician. Not long afterwards, the mathematician came back with the following reply. ‘It’s absolutely impossible! However, if one member of the club can be persuaded not to turn up, then everyone can be seated next to a friend.’ ‘Which member must I ask to stay away?’ the president queried. ‘It doesn’t matter,’ replied the mathematician. ‘Anyone will do.’

This problem, dubbed “Le Cercle Des Irascibles,” was posed by René Sousselier in Revue Française de Recherche Opérationelle in 1963. The remarkable solution was given the following year by J.C. Herz. In this figure, it’s possible to visit all 10 nodes while traveling on line segments alone, but there’s no way to close the loop and return to the starting node at the end of the trip (and thus to seat all the guests at a round table). But if we remove any node (and its associated segments), the task becomes possible. In the language of graph theory, the “Petersen graph” is the smallest hypohamiltonian graph — it has no Hamiltonian cycle, but deleting any vertex makes it Hamiltonian.

(Translation by D.A. Holton and J. Sheehan.)