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.)

Pick any point in the interior of an equilateral triangle and draw a perpendicular to each of the three sides. The sum of these perpendiculars is the height of the triangle.

That’s Viviani’s theorem. This visual proof is by CMG Lee:

1. Choose point P and draw the three perpendiculars.
2. Now draw three lines through P, each parallel to a side of the main triangle. This creates three small similar triangles.
3. Because these smaller triangles are equilateral, we can rotate each so that its altitude is vertical.
4. Because PGCH is a parallelogram, we can slide triangle PHE to the top, and now the heights of the three constituent triangles sum to that of triangle ABC.

The converse of the theorem is also true: If the sum of the perpendiculars from a point inside a triangle to its sides is independent of the point’s location, then the triangle is equilateral.

From Lee Sallows:

(Thanks, Lee!)

The so-called four-field approach in anthropology divides the discipline into four subfields: archaeology, linguistics, physical anthropology, and cultural anthropology.

Students call these “stones, tones, bones, and thrones.”

A centered hexagonal number is a number that can be represented by a hexagonal lattice with a dot in the center, like so:

Starting at the center, successive hexagons contain 1, 7, 19, and 37 dots. The sequence goes on forever.

The sum of the first n centered hexagonal numbers is n³, and there’s a pretty “proof without words” to show that this is so:

Instead of regarding each figure as a hexagon, think of it as a perspective view of a cube, looking down along a space diagonal. The first cube here contains a single dot. How many dots must we add to produce the next larger cube? Seven, and from our bird’s-eye perspective this pattern of 7 added dots matches the 7-dot hexagon shown above. The same thing happens when we advance to a 3×3×3 cube: This requires surrounding the 2×2×2 cube with 19 additional dots, and from our imagined vantage point these again take the form of a hexagonal lattice. In the last image our 3³ cube must accrete another 37 dots to become a 4³ cube … and the pattern continues.

Reader Derek Christie sent in this surprising curiosity after Wednesday’s post about Borromean rings:

Both the ring and the karabiner clip are attached to the cord and can’t be removed.

Now complicate matters by clipping the karabiner onto the ring:

Quite unexpectedly, the cord can now be just pulled away:

(Thanks, Derek.) (A related perplexity: The Prisoners’ Release Puzzle.)