From Lee Sallows:

The drawing at left above shows an unusual type of 3×3 geomagic square, being one in which the set of four pieces occupying each of the square’s nine 2×2 subsquares can be assembled so as to tile a 4×8 rectangle. The full set of subsquares become more apparent when it is understood that the square is to be viewed as if drawn on a torus, in which case its left-hand and right-hand edges will coincide, as will its upper and lower edges. In an earlier attempt at producing such a square several of the the pieces used were disjoint, or broken into separated fragments. Here, however, the pieces used are nine intact octominoes.

(Thanks, Lee!)

A sure-fire way to avoid hangover, from philosopher Josh Parsons:

It’s well known that one can alleviate hangover symptoms with a “hair of the dog” — another alcoholic drink. The problem is that this incurs another later hangover.

But think about this. Suppose that a pint of beer produces an hour of drunkenness followed by an hour of hangover, and that smaller quantities produce proportionately shorter periods. Begin, then, from a sober state, and drink half a pint of beer. Wait half an hour, until you’re just about to start feeling the hangover. Then drink a quarter pint of beer and wait a quarter of an hour, then an eighth of a pint, and so on. After an hour, you will have drunk one pint of beer and experienced no hangover, as it’s manifestly true that every incipient hangover you had was prevented by a further drink.

In his paper, cited below, Parsons addresses some objections to this scheme, including the fact that most publicans won’t stock 1/64-pint glasses and that eventually you’ll be swallowing at greater than light speed, and indeed swallowing something that may no longer qualify as beer. He grants that the task might be a “medical impossibility.”

But “by the time you get to the point where you can’t continue, you will only be incurring a very very short hangover with your last sip of beer. You don’t mind suffering a millisecond’s hangover. Besides, isn’t it worth speculating about whether the cure would work, on the counterfactual supposition that you can swallow at any finite speed, and that alcoholic beverages are made out of infinitely divisible gunk? Such speculations can tell us a lot about our concepts of infinity and matter.”

(He thanks “my colleagues and students for many helpful suggestions, and Central Bar for providing a pleasant environment in which to test my theories.”)

(Josh Parsons, “The Eleatic Hangover Cure,” Analysis 64:4 [October 2004], 364-366.)

In this position, devised by Harvard mathematician Noam Elkies, White is in serious trouble. He’s been reduced to a bare king, and his opponent has a knight as well as two pawns on the verge of queening.

It’s tempting to snap up the c2 pawn — this gains material and keeps Black’s king bottled up in the corner, which stops the a-pawn from queening. But now Black can win with 1. … Nd3! This stops the white king from moving to c1, which had been his only way to keep the black king immobilized. Now he’ll have to back off (even though perhaps winning the knight), and Black will play 2. … Kb1 and queen his pawn.

Interestingly, in the original position if White plays 1. Kc1, he’s guaranteed a draw: He can capture the c2 pawn on his next move and then shuffle happily forever between c1 and c2. Because he’s “lost a move,” Black can’t execute the maneuver described above to force the white king away from the corner. That’s because as the white king shuffles between a white and a black square, the knight must move about the board, also alternating between squares of different colors — and always landing on the wrong color to achieve his goal. The knight can wander as far afield as he likes, and execute any complex maneuver; he’ll always arrive at the wrong moment to achieve his aim.

(Noam D. Elkies and Richard P. Stanley, “The Mathematical Knight,” Mathematical Intelligencer 25:1 [December 2003], 22-34.)

Can any 10 points on a plane always be covered with some number of nonoverlapping unit discs?

It’s not immediately clear that the answer is yes. The most efficient way to pack circles together on the plane is the hexagonal packing shown above; it covers about 90.69 percent of the surface. But if our 10 points are inconveniently arranged, it’s not clear that we’ll always be able to shift the array of circles around in order to get them all covered.

In this case there’s a neat proof that takes advantage of a technique called the probabilistic method — if, for a group of objects, the probability is less than 1 that a randomly chosen object does not have a certain property, then there must exist an object that has this property.

Take a hexagonal packing randomly. Then, for any point on the plane, the probability that it’s not covered by the chosen packing is about 1 – 0.9069 = 0.0931. This means that for any 10 points, the chance that one or more points are not covered is approximately 0.0931 × 10 = 0.931. And that’s less than 1.

“Therefore, we obtain from the principle that there exists some closest packing that covers all the 10 points,” writes mathematician Hirokazu Iwasawa of the Institute of Actuaries of Japan. “And, in such a packing, we actually need at most 10 discs to cover the 10 points.”

(Hirokazu Iwasawa, “Using Probability to Prove Existence,” Mathematical Intelligencer 34:3 [September 2012], 11-14. The puzzle was created by Naoki Inaba.)

I found this surprising. What’s the volume of a ball of radius 1 in various dimensions?

In one dimension it’s a line segment of length 2.

In two dimensions it’s a unit disc in the plane, with area π.

In three dimensions it’s a unit ball with volume 4π/3.

Intuitively we might expect the number to keep rising. But it doesn’t!

In fact it peaks at five dimensions, and it drops quite sharply after that. In 20 dimensions the volume is only 0.026, and the limiting value is zero. Wikipedia explains the math.

A curious physics puzzle from Mark Levi’s excellent Why Cats Land on Their Feet: Suppose two astronauts, Al and Bob, are strapped to opposite ends of a space capsule’s interior, Al on the left and Bob on the right. Al is holding a large helium balloon, and everything is at rest. If Al pushes the balloon toward Bob, which way will the capsule drift?

It would be reasonable to guess that the capsule will drift to the left. Newton’s third law says that action equals reaction, so as Al pushes the balloon to the right, the balloon pushes Al to the left, and since he’s strapped to the capsule, he and it should drift left.

In fact the capsule will drift right as well. Because there are no external forces, the center of mass of the whole system is fixed. The helium balloon has less mass than the air it displaces, so from Al’s point of view the center of mass moves left. But the center of mass of the whole system is fixed in space, so the capsule must move right from the point of view of an external observer.

One way to make this intuitive is to imagine that the capsule is full of water rather than air. The mass of water essentially stays in place while we transfer a bubble of helium from the water’s left to its right. To accommodate this, the shell (whose mass we neglect) must move to the right.

This arrangement of rubber bands is “Brunnian”: Though the bands are entangled, no two are directly linked; and though no band can be extricated from the mass, cutting any one of them will free all the others.

By following the pattern here, any number of bands can be joined in a configuration with the same properties.