Suppose we have a group of n men and n women. Each of the women can find some subset of the men whom she would be happy to marry. And each of the men would be happy with any woman who will have him. Is it always possible to pair everyone off into happy marriages?

Clearly this won’t work if, for example, two of the women have their hearts set on the same man and won’t be happy with anyone else. In general, for any subset of the women, we need to be sure that they can reconcile their preferences so that each of them finds a mate.

Surprisingly, though, that’s all that’s required. So long as every subset of women can collectively express interest in a group of men at least as numerous as their own, it will always be possible to marry off the whole group into happy couples.

The theorem was proved by English mathematician Philip Hall in 1935. Another application of the same principle: Shuffle an ordinary deck of 52 playing cards and deal it into 13 piles of 4 cards each. Now it’s always possible to assemble a run of 13 cards, ace through king, by drawing one card from each pile.

As early as the 1st century B.C., the Chinese text Zhou Bi Suan Jing reflected the reasoning of the Pythagorean theorem, showing how to find the hypotenuse of the 3-4-5 triangle. Arrange four 3×4 rectangles around a unit square, as shown, producing a 7×7 square. The diagonals of the four rectangles produce a tilted square. Now, the area of the 7×7 square is 49, and the area of one right triangle with legs 3 and 4 is 6. So the area of the tilted square is 49 – (4 × 6), or 25. This shows that the hypotenuse of each of the right triangles is 5.

In Mathematics and the Aesthetic (2007), Nathalie Sinclair writes, “The Chinese diagram … is the same as one given by the twelfth-century Indian scholar Bhaskara, whose one-word injunction Behold! recorded his sense of awe.”

Drape a chain of evenly spaced weights over a pair of (frictionless) inclined planes like this. What will happen? There’s more mass on the left side, but the slope on the right side is steeper. Simon Stevin (1548-1620) realized that in fact the chain won’t move at all — if it did, we could link the ends as shown and produce a perpetual motion machine.

This is remembered as the “Epitaph of Stevinus.” Richard Feynman wrote, “If you get an epitaph like that on your gravestone, you are doing fine.”

The denominations of U.S. coins make intuitive sense, but they can be unwieldy: It can take up to eight coins to assemble an amount up through 99¢. Indeed, producing 99¢ takes (1 × 50¢) + (1 × 25¢) + (2 × 10¢) + (4 × 1¢). What five denominations would minimize the number of coins ever needed to make change?

In The Math Chat Book, Frank Morgan reports that with coins of 1¢, 3¢, 11¢, 27¢, and 34¢, you never need more than 5 coins to make change. For example, now 99¢ = (2 × 34¢) + (1 × 27¢) + (1 × 3¢) + (1 × 1¢). Of the 1,129 possible solutions, this one requires the fewest coins on average (3.343).

Unfortunately, this system is a bit tricky too — to assemble some totals, it’s more efficient to use a few middle-size coins rather than starting with the largest value possible. For example, if you assemble 54¢ by starting with a 34¢ coin, it takes four additional coins to gather the remaining 20¢: (1 × 11¢) + (3 × 3¢). It would have been simpler to choose 2 × 27¢, but that’s not immediately evident.

This pretty proof of the Pythagorean theorem is attributed to Leonardo da Vinci. Draw a right triangle and construct a square on each side, and make a copy of the original triangle and add it to the bottom of the hypotenuse square as shown. Now the shaded hexagon in the first figure can be rotated 90 degrees clockwise around the indicated point to occupy the position shown in the second figure. The orange and green quadrilaterals in the second figure are seen to be congruent to those in the first figure: The three shortest sides of the orange quadrilateral in the second figure correspond to their counterparts in the first, and the angles between them are assembled from the same constituents. The same is true of the green quadrilaterals. In each figure the shaded hexagon contains two instances of the original right triangle; remove these and we can see that the two squares in the first figure equal the large square in the second figure, proving Pythagoras.

10/10/2021 UPDATE: A number of readers point out that only the orange quadrilateral here can properly be said to turn; in the second diagram the green quadrilateral has been reflected as well. (Thanks, Mark and Bill.)

In a December 1985 letter to the Mathematical Gazette, Middlesex Polytechnic mathematician Ivor Grattan-Guinness writes that Astronomer Royal George Biddell Airy “would sometimes go around the Observatory, and on finding an empty box, insert a piece of paper saying ‘Empty box’ and thereby falsify its description! This last achievement deserves, in my proposal, the name of ‘Airy’s paradox’.”