Let four circles (blue) pass through a single point, M. Each pair of these circles intersect at a second point (pink). Each three of the four blue circles will have three pink points among them; these trios of pink points define four new circles (brown), which intersect in a single point, P.

If we start with five circles passing through a single point M, then we can apply the procedure above to each subset of four of them. This will produce five points P that all lie on a single circle.

If we start with six circles that all pass through a single point M, then each subset of five of them defines a new circle, as we’ve just seen. These six new circles all pass through a single point.

Remarkably, this pattern continues forever. It was discovered by the English geometer William Kingdon Clifford.

Arrange cards with values ace through 9 in a row, in counting order, with the ace on the left.

Take up a card from one end of the row — left or right, your choice.

Do this twice more, each time taking up either the leftmost or the rightmost card in the remaining row.

When you have three cards, add their values, divide the total by six, and call the result n. Count the cards that remain on the table from left to right.

The card in the nth position will be the 4.

The square root of 2 is 1.41421356237 … Multiply this successively by 1, by 2, by 3, and so on, writing down each result without its fractional part:

Beneath this, make a list of the numbers that are missing from the first sequence:

The difference between the upper and lower numbers in these pairs is 2, 4, 6, 8 …

From Roland Sprague, Recreations in Mathematics, 1963.

“Fifteen young ladies in a school walk out three abreast for seven days in succession: it is required to arrange them daily, so that no two shall walk twice abreast.”

The Rev. Thomas Penyngton Kirkman posed this query innocently in The Lady’s and Gentleman’s Diary in 1850. It’s trickier than it looks — though seven solutions are possible, they’re difficult to discover by trial and error. Here’s one:

Each letter appears once in each row, but no two letters share a cell more than once. The problem’s simplicity made it popular among Victorian amateur mathematicians, and Kirkman later complained that it had eclipsed his more serious work — though he took pains to dispute James Joseph Sylvester’s claim to have invented the problem himself.

Do such problems generally have solutions? Surprisingly, the answer remained unknown until just last January, when Oxford mathematician Peter Keevash showed that the answer is yes if certain basic requirements are met. Keevash’s result was “a bit of an earthquake as far as design theory is concerned,” said Cambridge mathematician Timothy Gowers.

Now that the schoolgirl problem is under control, a related challenge has taken its place. If 20 golfers want to arrange themselves into different foursomes on five successive days, is it possible to plan the groups so that each golfer plays no more than once with any other golfer? Formally the “social golfer problem” remains unsolved — though tournament organizers work out individual solutions every day.

(Thanks, Martha.)

Divide the number 999,999,999,999,999,999,999,998,999,999,999,999,999,999,999,999 into 1 and express the result as a decimal expansion, and you’ll find the Fibonacci sequence presented in tidy 24-digit strings:

(Thanks, David.)

Suppose we put eight white and two black balls into a bag and then draw forth the balls one at a time. If we repeat this experiment many times, which draw is most likely to produce the first black ball?

Most people answer 4, but in fact the first black ball is most likely to appear on the very first draw:

By symmetry, the second black ball is most likely to appear on the final draw.

(A.E. Lawrence, “Playing With Probability,” Mathematical Gazette, vol. 53 [December 1969], 347-354.)

If every pair of people in a group have exactly one friend in common, then there’s always one person who is a friend to everyone.

This rather heartwarming fact was proven by Paul Erdős, Alfréd Rényi, and Vera T. Sós in 1966. It’s sometimes more cynically known as the paradox of the politician.

The Erdős proof uses combinatorics and linear algebra, but in 1972 Judith Longyear published a proof using elementary graph theory.

See The Elevator Problem.