492 + 732 = 7730
772 + 302 = 6829
682 + 292 = 5465
542 + 652 = 7141
712 + 412 = 6722
672 + 222 = 4973
(Thanks, Alon.)
492 + 732 = 7730
772 + 302 = 6829
682 + 292 = 5465
542 + 652 = 7141
712 + 412 = 6722
672 + 222 = 4973
(Thanks, Alon.)
In 2004 University of Bristol mathematicians Hinke Osinga and Bernd Krauskopf crocheted a Lorenz manifold. They had developed a computer algorithm that “grows” a manifold in steps, and realized that the resulting mesh could be interpreted as a set of crochet instructions. After 85 hours and 25,511 stitches, Osinga had created a real-life object reflecting the Lorenz equations that describe the nature of chaotic systems.
“Imagine a leaf floating in a turbulent river and consider how it passes either to the left or to the right around a rock somewhere downstream,” she told the Guardian. “Those special leaves that end up clinging to the rock must have followed a very unique path in the water. Each stitch in the crochet pattern represents a single point [a leaf] that ends up at the rock.”
They offered a bottle of champagne to the first person who would produce another crocheted model of the manifold and received three responses in two weeks (and more since).
Of their own effort, Osinga and Krauskopf wrote, “While the model is not identical to the computer-generated Lorenz manifold, all its geometrical features are truthfully represented, so that it is possible to convey the intricate structure of this surface in a ‘hands-on’ fashion. This article tries to communicate this, but for the real experience you will have to get out your own yarn and crochet hook!” Their instructions are here.
(Hinke M. Osinga and Bernd Krauskopf, “Crocheting the Lorenz Manifold,” Mathematical Intelligencer 26:4 [September 2004], 25-37.)
As an exercise at the end of his 1887 book The Game of Logic, Lewis Carroll presents pairs of premises for which conclusions are to be found:
He gives no solutions, so you’re on your own.
Cambridge mathematician Hallard T. Croft once asked whether it was possible to have a finite set of points in the plane with the property that the perpendicular bisector of any pair of them passes through at least two other points in the set.
In 1972 Leroy M. Kelly of Michigan State University offered the elegant solution above, a square with an equilateral triangle erected outward on each side (it also works if the triangles are erected inward).
“Croft is a great problemist,” Kelley said later. “He keeps putting out lists of problems and he keeps including that one. He’s trying to get the mathematical community to get a better example — one with more points in it. … Eight is the smallest number; and whether it’s the largest number is another question.”
So far as I know Croft’s question is still unanswered.
In 1894 Sir Francis Galton devised this simple machine to demonstrate the central limit theorem: Beads inserted at the top drop through successive rows of pegs, bouncing left or right as they hit each peg and landing finally in a row of bins at the bottom. Though the path of any given bead can’t be predicted, in the aggregate they form a bell curve. Delighted with this, Galton wrote:
Order in Apparent Chaos: I know of scarcely anything so apt to impress the imagination as the wonderful form of cosmic order expressed by the Law of Frequency of Error. The law would have been personified by the Greeks and deified, if they had known of it. It reigns with serenity and in complete self-effacement amidst the wildest confusion. The huger the mob, and the greater the apparent anarchy, the more perfect is its sway. It is the supreme law of Unreason. Whenever a large sample of chaotic elements are taken in hand and marshalled in the order of their magnitude, an unsuspected and most beautiful form of regularity proves to have been latent all along.
From Recreational Mathematics Magazine, 1961, a magic square of cards:
Each row, column, and main diagonal contains an ace, king, queen, and jack and all four suits. There are numerous other subsquares and symmetrical subsets of squares that have the same property, including the center 2 × 2 square and the four corner squares.
(Recreational Mathematics Magazine 34:5, 24-29, via Pi Mu Epsilon Journal, “Unusual Magic Squares,” 6:2 [Spring 1975], 54-55.)
In a 1963 essay, Isaac Asimov pointed out that paradimethylaminobenzaldehyde can be sung to the tune of “The Irish Washerwoman.” Inspired, John A. Carroll wrote this jig:
(chorus:) Paradimethylaminobenzaldehyde
Sodium citrate, ammonium cyanide
Phosphates and nitrates and chlorides galore
Just have one o’ these and you’ll never need more.
Got messed up last night on furfuryl alcohol
Followed it down with a gallon of propanol
Drank from mid-morning til late afternoon
Then spat on the floor and blew up the saloon.
(repeat chorus)
Paradimethylaminobenzaldehyde,
Powdered aluminum, nitrogen iodide
Slop it around and add some benzene
Then top off the punch with Fluorescein
(repeat chorus)
Whiskey, tequila and rum are too tame,
The stuff that I drink must explode into flame.
When I break wind it strips all the paint in the room,
And rattles the walls with an earthshaking boom.
(repeat chorus)
Paradimethylaminobenzaldehyde
Go soak your head in a jar of formaldehyde
Scrub very hard, then rinse out your mane
In dichlorodiphenyltrichloroethane!
In “The Adventure of the Final Problem,” Sherlock Holmes flees London, pursued by his archenemy, James Moriarty. Both are headed to Dover, where Holmes hopes to escape to the continent, but there’s one intermediate stop available, at Canterbury. Holmes faces a choice: Should he get off at Canterbury or go on to Dover? If Moriarty finds him at either station he’ll kill him.
In their 1944 Theory of Games and Economic Behavior, mathematician John von Neumann and economist Oskar Morgenstern address this as a problem in game theory. They set up the following payoff matrix showing Moriarty’s calculations:
Von Neumann and Morgenstern conclude that “Moriarty should go to Dover with a probability of 60%, while Sherlock Holmes should stop at the intermediate station with a probability of 60% — the remaining 40% being left in each case for the other alternative.”
As it turns out, that’s exactly what happens in the story — Holmes and Watson get out at Canterbury and watch Moriarty’s train roar past toward Dover, “beating a blast of hot air into our faces.” “There are limits, you see, to our friend’s intelligence,” Holmes tells Watson. “It would have been a coup-de-maître had he deduced what I would deduce and acted accordingly.”
(It’s not quite that simple — in a footnote, von Neumann and Morgenstern point out that Holmes has excusably replaced the 60% probability with certainty in his calculations. In fact, they say, the odds favor Moriarty — “Sherlock Holmes is as good as 48% dead when his train pulls out from Victoria Station.”)
On Sept. 2, 1945, an American Navy squadron came ashore at Sagami Bay near Yokohama to demilitarize the Japanese midget submarines in the area. They found this notice on the door of a marine biological research station there, left by embryologist Katsuma Dan.
The Americans honored his wish: On the last of 1945 he was summoned by an officer of the U.S. First Cavalry and handed a document releasing the station back to the University of Tokyo.
The notice is on display at the Woods Hole Oceanographic Institution’s Marine Biological Laboratory (here’s the full story).
A triangle can be covered by three smaller copies of itself. A square requires four smaller copies. But in general four will do: Any bounded convex set in the plane can be covered with four smaller copies of itself (and in fact the fourth copy is needed only in the case of parallelograms, like the square).
Is this true in every dimension? In 1957 Swiss mathematician Hugo Hadwiger conjectured that every n-dimensional convex body can be covered by 2n smaller copies of itself. But this remains an unsolved problem.
(Interestingly, Russian mathematician Vladimir Boltyansky showed that this problem is equivalent to one of illumination: How many floodlights does it take to illuminate an opaque convex body from the exterior? The number of floodlights turns out to equal the number of smaller copies needed to cover the body.)