The present King of France is bald.
Is this statement true or false? Well, it’s not true — France has no king presently. But if it’s false then its negation ought to seem true: The present King of France is not bald. That’s no better. Yet it’s not gibberish — the sentence seems to have a clear meaning that we can understand.
Bertrand Russell and Alfred North Whitehead spent much of their time at Cambridge debating this point. “It is astonishing what intricate and remote considerations can be brought to bear on this interesting question,” Russell wrote to his wife. “We finally decided that he isn’t, altho’ he has no hair of his own. Experienced people will infer that he wears a wig, but this would be a mistake.”
In 2001, USCD psychologist Vilayanur Ramachandran presented these shapes to American undergraduates and to Tamil speakers in India. He asked, “Which of these shapes is bouba and which is kiki?” Around 98% of the respondents assigned the name kiki to the spiky shape and bouba to the curvy one.
Psychologist Wolfgang Köhler had found a similar effect in 1929 using the words baluba and takete. “This result suggests that the human brain is somehow able to to extract abstract properties from the shapes and sounds,” Ramachandran wrote, “for example, the property of jaggedness embodied in both the pointy drawing and the harsh sound of kiki.”
When you’re a princess you have to kiss a lot of frogs. But you can see only one frog at a time, and once you reject a frog you can’t return to it. How can you know when to stop hoping for a prince and settle down with the frog you have?
“Surprisingly, … there is a method which enables us to select the best candidate with a probability of nearly 30% even if n is a large number,” writes Gabor Szekely in Paradoxes in Probability Theory and Mathematical Statistics (2001). “Let the first 37% (more precisely, 100/e%) of the candidates go and then select the first one better than any previous candidate (if none are better, select the last). In this case the chance of selecting the best is approximately 1/e, i.e. ≈37% however great n is.”
So if there are 100 frogs in the forest, reject the first 37 and then choose the first one that seems to beat all the others. There’s about a 37 percent chance that it’s the best one.
What is directly on the opposite side of the world from you? This map answers that question by superimposing each point on earth with its opposite. Some notable sisters:
- Beijing, China, is nearly opposite Buenos Aires, Argentina
- Madrid, Spain, is nearly opposite Wellington, New Zealand
- Bogotá, Colombia, is nearly opposite Jakarta, Indonesia
- Bangkok, Thailand, is nearly opposite Lima, Peru
- Quito, Ecuador, is nearly opposite Singapore
- Seoul, South Korea, is nearly opposite Montevideo, Uruguay
- Perth, Australia, is nearly opposite Bermuda
- Charmingly, Cherbourg, France, is opposite the Antipodes Islands south of New Zealand
W.V.O. Quine explains how an enterprising traveler can arrange to visit two precise antipodes: “Note to begin with that any route from New York to Los Angeles, if not excessively devious, is bound to intersect any route from Winnipeg to New Orleans. Now let someone travel from New York to Los Angeles, and also travel from roughly the antipodes of Winnipeg to roughly the antipodes of New Orleans. These two routes do not intersect — far from it; but one of them intersects a route that is antipodal to the other. So our traveler is assured of having touched a pair of mutually antipodal points precisely, though he will know only approximately where.”
In 2007, New Scientist announced that the best strategy in a game of rock paper scissors is to choose scissors.
Research has shown that rock is the most popular of the three moves. If your opponent expects you to choose it, he’ll choose paper in order to beat it — in which case scissors will win.
In 2005 a Japanese art collector asked Christie’s and Sotheby’s to play a match, saying the winner could sell his impressionist paintings. The 11-year-old daughter of a Christie’s director recommended scissors, saying, “Everybody expects you to choose rock.”
Sure enough, Sotheby’s chose paper, and Christie’s won the £10 million deal.
If there was a time when nothing existed, then there must have been a time before that — when even nothing did not exist. Suddenly, when nothing came into existence, could one really say whether it belonged to the category of existence or of non-existence?
A hairy ball can’t be combed flat — it must always have a cowlick.
This result arose originally in algebraic topology, but it has intriguing applications elsewhere. For example, it can’t be windy everywhere at once on Earth’s surface — at any given moment, the horizontal wind speed somewhere must be zero.
In 1829 a correspondent to the Mechanic’s Magazine proposed this design for a “self-moving railway carriage.” Fill the car with passengers and cargo as shown and set it on two rails that undulate across the landscape:
In the descending sections (a, c, e) the two rails are parallel. In the ascending ones (b, d) they diverge so that the car, mounted on cones, will roll forward to settle more deeply between them, paradoxically “ascending” the slope. If the track circles the world the car will “assuredly continue to roll along in one undeviating course until time shall be no more.”
“How any one could ever imagine that such a contrivance would ever continue in motion for even a short time … must be a puzzle to every sane mechanic,” wrote John Phin in The Seven Follies of Science in 1911. But what does he know?
As a joke, Michael Collins submitted a travel voucher for his trip aboard Gemini 10. NASA reimbursed him $8 per day, a total of $24.
In his autobiography, Collins notes that he could instead have claimed 7 cents a mile, which would have yielded $80,000.
But one of the original Mercury astronauts had already tried this — and had received a bill for “a couple of million dollars” for the rocket he’d used.
Draw two circles of any size and bracket them with tangents, as shown.
The chords in green will always be equal.
Any group of six people must contain at least three mutual friends or three mutual strangers.
Represent the people with dots, and connect friends with blue lines and strangers with red. Will the completed diagram always contain a red or a blue triangle?
Because A has five relationships and we’re using two colors, at least three of A’s connections must be of the same color. Say they’re friends:
Already we’re perilously close to completing a triangle. We can avoid doing so only if B, C, and D are mutual strangers — in which case they themselves complete a triangle:
We can reverse the colors if B, C, and D are strangers to A, but then we’ll get the complementary result. The completed diagram must always contain at least one red or blue triangle.
I think this problem appeared originally in the William Lowell Putnam mathematics competition of 1953. Six is the smallest number that requires this result — a group of five people would form a pentagon in which the perimeter might be of one color and the internal connections of another.
(Update: In fact the more general version of this idea was adduced in 1930 by Cambridge mathematician F.P. Ramsey. It is very interesting.) (Thanks, Alex.)
Cut a notch in a stick and label the two parts p and q. Then draw the stick around the shore of a pond. The notch will describe a curve, and, remarkably, the area between the shore and this curve will be given by πpq.
“Two things immediately struck me as astonishing,” wrote British mathematician Mark Cooker in 1988. “First, the formula for the area is independent of the size of the given curve. Second, [the equation for the area] is the area of an ellipse of semi-axes p and q, but there are no ellipses in the theorem!”
For years Raymond Smullyan sought a “metaparadox,” a statement that is paradoxical if and only if it isn’t. He arrived at this:
Either this sentence is false, or (this sentence is paradoxical if and only if it isn’t).
He wrote, “I leave the proof to the reader.”
In 1976, CUNY mathematician Robert Feinerman showed that the game of dreidel is fundamentally unfair.
Each player contributes one unit to the pot, and then all take turns spinning the top. Each spin produces one of four outcomes: the player does nothing, collects the entire pot, collects half the pot, or puts one unit into the pot. When a player collects the entire pot, then each player contributes one unit to form a new pot and play continues.
Feinerman found that, if Xn is the payoff on the nth spin and p is the number of players, the expected value of Xn is
Thus if there are more than two players, “the first player has an unfair advantage over the second player, who in turn has an unfair advantage over the third player, etc.”
Dreidel is booming nonetheless. A Major League Dreidel tournament has been held in New York City every Hanukkah since 2007. The official playing surface is called the Spinagogue, and the tournament slogan is “no gelt, no glory.”
(Robert Feinerman, “An Ancient Unfair Game,” American Mathematical Monthly 83:623-625.)
The greater honeyguide of Africa eats beeswax but isn’t always able to invade a hive on its own. So it has forged a unique partnership with human beings: The bird attracts the attention of local honey hunters with a chattering call, flies toward a hive, then stops and calls again. When they arrive at the hive, the humans open it, subdue the bees with smoke, take the honey, and leave the wax for the bird.
This arrangement saves the humans an average of 5.7 hours in searching for hives, but it’s not foolproof. “We have been ‘guided’ to an abrupt precipice and to a bull elephant by greater honeyguides,” report biologists Lester Short and Jennifer Horne. “In these cases there were bee-hives below the cliff (in a valley) and beyond the elephant. Concern for the welfare of the guided person is beyond any reasonable expectation of a honeyguide.”
But, say you, surely there is nothing easier than for me to imagine trees, for instance, in a park, or books existing in a closet, and nobody by to perceive them. I answer, you may so, there is no difficulty in it; but what is all this, I beseech you, more than framing in your mind certain ideas which you call books and trees, and the same time omitting to frame the idea of any one that may perceive them? But do not you yourself perceive or think of them all the while? This therefore is nothing to the purpose; it only shews you have the power of imagining or forming ideas in your mind: but it does not shew that you can conceive it possible the objects of your thought may exist without the mind.
— George Berkeley, A Treatise Concerning the Principles of Human Knowledge, 1710
Prince Charming tells Sleeping Beauty, “I’m going to put you to sleep with this potion, and then I’ll flip a coin. Today is Sunday. If the coin lands heads, I’ll wake you again on Monday. If it lands tails, then I’ll wake you on Monday, put you to sleep again, and wake you on Tuesday. The potion induces a mild amnesia, so you won’t remember the intermediate awakening if it happens, but otherwise it won’t hurt you.”
When Sleeping Beauty awakes, what probability should she assign that the coin landed heads?
There seem to be two contradictory answers to this. From one perspective, the coin was fair, so it would seem the chance is 1/2. But from another, Beauty finds herself in one of three equally likely situations (heads/Monday, tails/Monday, and tails/Tuesday), so the chance of heads appears to be 1/3. Which is correct?
Given three adjacent squares,
a + b = c.
“Beauty is the first test,” wrote G.H. Hardy. “There is no permanent place in the world for ugly mathematics.”
The lowly 3×3 magic square has modest pretensions — each row, column, and diagonal produces the same sum.
But perhaps it’s magicker than we suppose:
6182 + 7532 + 2942 = 8162 + 3572 + 4922 (rows)
6722 + 1592 + 8342 = 2762 + 9512 + 4382 (columns)
6542 + 1322 + 8792 = 4562 + 2312 + 9782 (diagonals)
6392 + 1742 + 8522 = 9362 + 4712 + 2582 (counter-diagonals)
6542 + 7982 + 2132 = 4562 + 8972 + 3122 (diagonals)
6932 + 7142 + 2582 = 3962 + 4172 + 8522 (counter-diagonals)
(R. Holmes, “The Magic Magic Square,” The Mathematical Gazette, December 1970)
More: Any of the equations above will still hold if you remove the middle digit or any two corresponding digits in each of the six addends.
Yet more: (6 × 1 × 8) + (7 × 5 × 3) + (2 × 9 × 4) = (6 × 7 × 2) + (1 × 5 × 9) + (8 × 3 × 4)
I think everything above will work for any rotation or reflection of the square (that is, for any normal 3×3 magic square). I haven’t checked, though.
Is a legal chess game possible in which all the pawns promote and each player has nine queens?
Yes — Freidrich Burchard of Germany and Friedrich Hariuc of Romania reached nearly identical solutions in 1980:
1. e4 f5 2. e5 Nf6 3. exf6 e5 4. g4 e4 5. Ne2 e3 6. Ng3 e2 7. h4 f4 8. h5 fxg3 9. h6 g5 10. Rh4 gxh4 11. g5 g2 12. g6 Bg7 13. hxg7 g1=Q 14. f4 h3 15. f5 h2 16. b4 a5 17. b5 a4 18. b6 a3 19. Bb2 Ra7 20. bxa7 axb2 21. a4 b5 22. a5 b4 23. a6 b3 24. c4 h1=Q 25. c5 h5 26. c6 Bb7 27. cxb7 c5 28. d4 c4 29. d5 Nc6 30. dxc6 c3 31. c7 c2 32. c8=Q c1=Q 33. b8=Q Qc7 34. a8=Q d5 35. a7 d4 36. Nc3 dxc3 37. Qa6 c2 38. Qa8b7 c1=Q 39. a8=Q Qd5 40. gxh8=Q+ Kd7 41. g7 bxa1=Q 42. g8=Q b2 43. f7 b1=Q 44. f8=Q h4 45. f6 h3 46. f7 h2 47. Qfa3 h1=Q 48. f8=Q exf1=Q+
This may be the shortest possible such game.
Only 43 numbers have names that lack the letter N.
One of them, fittingly, is forty-three.