From an appreciation of Ernest Rutherford by C.P. Snow in the November 1958 issue of The Atlantic:
Worldly success? He loved every minute of it: flattery, titles, the company of the high official world. He said in a speech: ‘As I was standing in the drawing room at Trinity, a clergyman came in. And I said to him: “I’m Lord Rutherford.” And he said to me: “I’m the Archbishop of York.” And I don’t suppose either of us believed the other.’
In the 1950s, mathematician C. Dudley Langford was watching his son play with blocks, two of each color, when he noticed that they formed a curious pattern: one block lay between the red blocks, two between the blue blocks, and three between the yellow blocks. Langford found that by rearranging the blocks he could add a green pair with four blocks between them.
This presented a clear challenge. He found solutions for as many as 15 pairs of blocks but came to believe that some smaller groupings (14 pairs, for example) could not produce a solution. He asked for a general investigation.
Today we know that a solution exists if and only if the number of blocks is 4k or 4k + 3, so Langford was right that no solution can be arranged with 14 pairs of blocks. The number of solutions for each quantity of pairs is listed here, and a few proofs are given here.
(C. Dudley Langford, “Problem,” Mathematical Gazette 42:341 [October 1958], 228.)
From Lee Sallows:
Thanks, Lee!
If a sphere is inscribed in a tetrahedron, and segments are drawn on each face connecting its vertices to the point of tangency, then the same three angles will appear on each face.
And the 12 triangles produced will form congruent pairs across each edge of the tetrahedron.
Into an empty vase drop balls numbered 1 to 10. Remove ball 1. Add balls numbered 11 to 20. Remove ball 2. Continue in this way, spending half an hour on the first transaction, 15 minutes on the next, and so on. After one hour all the transactions will be finished.
Obviously, in the end the vase will contain infinitely many balls, since with each step more balls have been added than removed.
But, equally obviously, after an hour the vase will be empty — since the time of each ball’s removal is known.
He, then, who says that something true exists either only asserts that something true exists or proves it. And if he merely asserts it, he will be told the opposite of his mere assertion, namely, that nothing is true. But if he proves that something is true, he proves it either by a true proof or by one that is not true. But he will not say that it is by one not true, for such a proof is not to be trusted. And if it is by a true proof, whence comes it that the proof which proves that something is true is itself true? If it is true of itself, it will be possible also to state as true of itself that truth does not exist; while if it is derived from proof, the question will again be asked ‘How is it that this proof is true?’ and so on ad infinitum. Since, then, in order to learn that there is something true, an infinite series must first be grasped, and it is not possible for an infinite series to be grasped, it is not possible to know for a surety that something true exists.
— Sextus Empiricus, Against the Logicians
In the classic Indian rope trick, a rope rises into the sky, its end lost to view. A boy disappears up the rope, and when he fails to return the angry magician climbs up after him. Body parts fall to the ground, the magician descends and places the parts in a basket, and the boy reappears uninjured.
This is all thought to be a legend, but in 1979 mathematician J.L.G. Pinhey of The Perse Boys’ School worked out that levitating a rope is possible, at least in principle. If the top of the fakir’s rope is 1.5 × 108 meters above Earth’s surface, it will simply stand erect, its position sustained by the motion of the planet.
“Since the rope between its ends is in tension the configuration is stable, and the faqir and his boy-victim can climb it in safety. However, in order to drop the bits to earth, the pair must not climb even a quarter of the way to the top.”
(J.L.G. Pinhey, “63.12 The Indian Rope Trick,” Mathematical Gazette 63:424 [June 1979], 110-111.)
From Wikimedia user Cmglee, two digital arithmetic techniques:
Above: To multiply a positive single-digit integer by 9, hold up your hands palm up, imagine the fingers numbered consecutively 1 to 10, and fold down the finger corresponding to the number to be multiplied (here, 8). The product is the two-digit number represented by the remaining two groups of fingers — here there are seven fingers to the left of the folded finger and 2 to the right, so 9 × 8 = 72.
Below: To multiply two integers between 6 and 10, imagine each hand’s fingers numbered from 6 (pinky) to 10 (thumb), as shown. Fold down the two fingers corresponding to the factors, as well as all fingers between these two (in this example we’ll calculate 6 × 7, so fold down finger 7 on the left hand, finger 6 on the right, and the finger that lies between them, the left pinky). Count the remaining upraised fingers on the left hand (3), multiply that by the remaining upraised fingers on the right hand (4), and add 10 times the number of folded fingers (30). 3 × 4 + 30 = 42.
A square can be divided into an even number of triangles of equal area — but not into an odd number.
