https://www.youtube.com/watch?v=aJhiY70KY5o

Stand a bicycle so that one pedal is in its lowest position and one in its highest. Now if we pull backward on the low pedal, will the bicycle move forward or backward?

Intuition suggests it will move forward: We’re turning the pedals in the same direction that a rider would, and normally this motion drives the rear wheel to propel the bike forward.

Surprisingly, though, in the experiment the bike moves backward. That’s because (in most bikes, in most gears) each pedal is constantly moving forward with respect to the ground. So pulling backward on the pedal doesn’t produce the intuitive result — it moves the pedal backward relative to the ground, and so produces the opposite result to the one we expect.

An exception: If the bike is in a sufficiently low gear, pulling backward on the pedal will drive the bike forward. But the sprocket ratio must be so low that really we’re betraying intuition rather than the reverse (see the video).

In the 1970s, dolphin trainer Jim Mullen sought to encourage the dolphins at Marine World in Redwood City, California, to tidy up their pool at the end of the day. Each dolphin received a reward of fish for each piece of litter it brought to him.

“It worked very well,” Mullen told psychologist Diana Reiss. “The pool was kept neat and clean, and the dolphins seemed to enjoy the game.”

One day in the summer of 1978, a dolphin named Spock seemed unusually diligent, bringing one piece after another of brown paper to Mullen and receiving a reward each time. Eventually Mullen grew suspicious and asked an assistant to go below and look through the pool windows.

“It turned out that there was a brown paper bag lodged behind an inlet pipe,” Mullen said. “Spock went to the paper bag, tore a piece off, and brought it to me. I then gave Spock a fish, as per our arrangement, and back he went. The second time my assistant saw Spock go to the paper bag, Spock pulled at it to remove a piece, but the whole bag came out. Spock promptly shoved the bag back into place, tore a small piece off, and brought it to me. He knew what he was doing, I’m sure. He completely had me.”

Spock hadn’t been trained to tear debris to pieces, and in doing so he was certainly maximizing his reward, Reiss writes. “And when he pushed the bag back behind the pipe when it came out in one piece, that certainly had the ring of deliberate action. Whether you can call it deliberate deception is a tough call.”

(Diana Reiss, The Dolphin in the Mirror, 2011.)

Native to Chile and Argentina, Boquila trifoliolata has a remarkable ability: Once it’s wrapped its vines around a host plant, it can alter its leaves to mimic those of the host, a phenomenon called mimetic polymorphism.

“It modifies its size, shape, color, orientation, and even the pattern of its veins in such a way that it fuses perfectly with the foliage of the tree that bears it,” writes botanist Francis Hallé in his 2018 Atlas of Poetic Botany. “If, in the course of growing, it changes its support, the same stalk can even display leaves that are completely different, corresponding to the new tree — even if these leaves are much bigger.”

This helps it to avoid predators. If the plant grew along the forest floor it would be eaten by weevils, snails, and leaf beetles, but these tend to leave it alone when it disguises itself with “tree leaves.” But how it accomplishes the mimicry remains unclear.

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.