Thomas Jefferson, already absurdly accomplished by 1795, somehow found time to delve into cryptography, where he devised this cipher system. The letters of the alphabet are printed along the rim of each of 36 disks, which are stacked on an axle. One party rotates the disks until his message can be read along one of the 26 rows of letters, somewhat like a modern cylindrical bike lock. Now he can record the letters in any one of the other 25 rows and send that string safely to another party, who decodes it by reversing this procedure. If the message is intercepted, it’s useless even to someone who has the disks, because he must also know the order in which to stack them, and 36 disks can be stacked in 371,993,326,789,901,217,467,999,448,150,835, 200,000,000 different ways.
This is pretty robust. The cipher below, created in 1915 by U.S. Army cryptographer Joseph Mauborgne, has never been solved. “The known systems from this year (or earlier) shouldn’t be too hard to crack with modern attacks and technology,” writes NSA cryptologist Craig P. Bauer. “So, why don’t we have a plaintext yet? My best guess is that it used a cipher wheel” like Jefferson’s.
(L. Kruh, “A 77-year-old challenge cipher,” Cryptologia, 17(2), 172-174, 1993, quoted in Bauer’s Secret History: The Story of Cryptology, 2013.)
Squeeze six circles into a larger circle so that each is tangent to its two neighbors. Now the three lines drawn through opposite points of tangency will pass through the same point.
Remarkably, this wasn’t discovered until 1974.
If a quadrilateral circumscribes a circle, then the sums of its opposite sides are equal.
Above, a + c = b + d.
Two travelers are transporting identical antiques. Unfortunately, the airline smashes both of them. The airline manager proposes that each traveler write down the cost of his antique, any value from $2 to $100. If both write the same number, the airline will pay this amount to both travelers. If they write different numbers, the airline will assume that the lower number is the accurate price; the low bidder will receive this amount plus $2, and the high bidder will receive this amount minus $2. If they can’t confer, what strategy should the travelers take in deciding how to bid?
At first Traveler A might like to bid $100, the maximum allowed. If his opponent does the same then they’ll both net $100. But A can do better than this: If B bids $100 and A bids $99 then A will come away with $101.
Unfortunately if B is rational then he’ll have the same insight and also bid $99. So A had better undercut him again and bid $98.
This chain leads all the way down to $2. If both travelers are perfectly rational then they’ll both bid (and make) $2, the minimum price.
But this seems very unlikely to happen in actual practice — in real life both travelers would likely make high bids and get high (though perhaps unequal) payoffs.
“All intuition seems to militate against all formal reasoning in the traveler’s dilemma,” wrote economist Kaushik Basu in propounding the problem in 1994. “There is something very rational about rejecting (2, 2) and expecting your opponent to do the same. … The aim is to explain why, despite rationality being common knowledge, players would reject (2, 2), as intuitively seems to be the case.”
A Japanese geometry theorem from the Edo period: If the blue circles are equal, the green circles will be equal too.
This can be extended: Circles spanning three of these triangles will also be equal, and so on.
Jump into the sea and look up. The surface above you is dark except for a bright circle that follows you around like a portable skylight. This is Snell’s window: Because light is refracted as it enters the water, the 180-degree world above you is compressed into a tight 97 degrees.
Physicist Robert W. Wood was thinking of this effect when he created a new wide-angle lens in 1906. Fittingly, he called it the fisheye.
In 1990, Spanish philosopher Jon Perez Laraudogoitia submitted an article to Mind entitled “This Article Should Not Be Rejected by Mind.” In it, he argued:
- If statement 1 in this argument is trivially true, then this article should be accepted.
- If statement 1 were false, then its antecedent (“statement 1 in this argument is trivially true”) would be true, which means that statement 1 itself would be true, a contradiction. So statement 1 must be true.
- But that seems wrong, since Mind is a serious journal and shouldn’t publish trivial truths.
- That means statement 1 must be either false or a non-trivial truth. We know it can’t be false (#2), so it must be a non-trivial truth, and its antecedent (“statement 1 in this argument is trivially true”) is false.
- What then is the truth value of its consequent, “this article should be accepted”? If this were false then Mind shouldn’t publish the article; that can’t be right, since the article consists of a non-trivial truth and its justification.
- So the consequent must be true, and Mind should publish the article.
They published it. “This is, I believe, the first article in the whole history of philosophy the content of which is concerned exclusively with its own self, or, in other words, which is totally self-referential,” Laraudogoitia wrote. “The reason why it is published is because in it there is a proof that it should not be rejected and that is all.”
In 1964 Canadian writer Graeme Gibson bought a parrot in Mexico. The bird, which Gibson named Harold Wilson, was bright and affectionate at first, but he seemed to grow lonely in the dark Canadian winter, so in the spring Gibson made arrangements to donate him to the Toronto Zoo. At the aviary Gibson carried Harold into the cage that had been prepared for him, placed him on a perch, said his goodbyes, and turned to go.
“Then Harold did something that astonished me. For the very first time, and in exactly the voice my kids might have used, he called out ‘Daddy!’ When I turned to look at him he was leaning towards me expectantly. ‘Daddy’, he repeated.
“I don’t remember what I said to him. Something about him being happier there, that he’d soon make friends. The kind of things you say to kids when you abandon them at camp. But outside the aviary I could still hear him calling ‘Daddy! Daddy!’ as we walked away. I was shattered to discover that Harold knew my name, and that he did so because he’d identified himself with my children.
“I now believe he’d known it all along, but was using it — for the first time — out of desperation. Both Konrad Lorenz and Bernd Heinrich mention instances of birds calling out the private names of intimates when threatened by serious danger. I am no longer surprised by such information. We think of our captive birds as our pets, but perhaps we are theirs as well.”
(From Gibson’s Perpetual Motion, 1982.)
Inscribe a hexagon in a unit circle such that AB = CD = EF = 1.
Now the midpoints of BC, DE, and FA form an equilateral triangle.
See A Better Nature.
Arthur W.J.G. Ord-Hume calls this “the most graceful and simple perpetual motion machine of all time.” It was offered by American inventor F.G. Woodward in the 19th century. A heavy wheel is mounted between two rollers so that the wheel’s weight causes it to roll continuously in the direction of the arrow.
Or so Woodward hoped. Ord-Hume notes that the principle required the left half of the wheel always to be heavier than the right half. “Sadly, it wasn’t.”
How do I know that I’m not just a fictional character in some imagined story? What could I learn about myself that would prove that I’m real? “I am human, male, brunette, etc., but none of that helps,” writes UCLA philosopher Terence Parsons. “I see people, talk to them, etc., but so did Sherlock Holmes.”
Descartes would say that the very fact that I’m thinking about this shows that I exist: cogito ergo sum. But a fictional character could make the same argument. “Hamlet did think a great many things,” writes Jaakko Hintikka. “Does it follow that he existed?” Robert Nozick adds, “Could not any proof be written into a work of fiction and be presented by one of the characters, perhaps one named ‘Descartes’?”
Tweedledee tells Alice that she’s only a figment of the Red King’s dream. “If that there King was to wake,” adds Tweedledum, “you’d go out — bang! — just like a candle!”
Alice says, “Hush! You’ll be waking him, I’m afraid, if you make so much noise.”
“Well, it’s no use YOUR talking about waking him,” replies Tweedledum, “when you’re only one of the things in his dream. You know very well you’re not real.”
“It seems to me that this is a philosophical problem that deserves to be treated seriously on a par with issues like the reality of the external world and the existence of other minds,” Parsons writes. “I don’t know how to solve it.”
(Terence Parsons, Nonexistent Objects, 1980; Charles Crittenden, Unreality, 1991; Robert Nozick, “Fiction,” Ploughshares 6:3 (1980), pp. 74-78; Jaakko Hintikka, “Cogito, Ergo Sum: Inference or Performance?”, The Philosophical Review, 71:1 (January 1962), pp. 3-32.)
If a flock of birds disperses gradually, at what point does it cease to be a flock?
“There is at the moment a pipe on my desk,” wrote MIT philosopher Richard Cartwright in 1987. “Its stem has been removed, but it remains a pipe for all that; otherwise no pipe could survive a thorough cleaning.”
But he also owned a two-volume set of John McTaggart’s The Nature of Existence, one volume of which was in Cambridge and the other in Boston. Do those two volumes still make one thing? If so, is there a “thing” composed of the Eiffel Tower and the Old North Church? Why not?
(From Cartwright’s Philosophical Essays.)
Draw a circle whose circumference is the golden mean. Choose a point and label it 1, then move clockwise around the circle in steps of arc length 1, labeling the points 2, 3, and so on. At each step, the difference between each pair of adjacent numbers on the circle is a Fibonacci number.
- What time is it at the North Pole?
- The shortest three-syllable word in English is W.
- After the revolution, the French frigate Carmagnole used a guillotine as its figurehead.
- 823502 + 381252 = 8235038125
- PRICES: CRIPES!
- “Conceal a flaw, and the world will imagine the worst.” — Martial
When Montenegro declared independence from Yugoslavia, its top-level domain changed from .yu to .me.
If I buy two toothbrushes in a “buy one, get one free” offer … which one did I buy, and which was free?
(From philosopher Roy Sorensen.)
How long is a coastline? If we measure with a long yardstick, we get one answer, but as we shorten the scale the total length goes up. For certain mathematical shapes, indeed, it goes up without limit.
English mathematician Lewis Fry Richardson discovered this perplexing result in the early 20th century while examining the relationship between the lengths of national boundaries and the likelihood of war. If the Spanish claim that the length of their border with Portugal is 987 km, and the Portuguese say it’s 1,214 km, who’s right? The ambiguity arises because a wiggly boundary occupies a fractional dimension — it’s something between a line and a surface.
“At one extreme, D = 1.00 for a frontier that looks straight on the map,” Richardson wrote. “For the other extreme, the west coast of Britain was selected because it looks like one of the most irregular in the world; it was found to give D = 1.25.”
This is a mathematical notion, but it’s also a practical problem. On the fjord-addled panhandle of Alaska, the boundary with British Columbia was originally defined as “formed by a line parallel to the winding of the coast.” Who gets to define that? On the map below, the United States claimed the blue border, Canada wanted the red one, and British Columbia claimed the green. The yellow border was arbitrated in 1903.
Write out the positive powers of 10 in both base 2 and base 5:
Now for any integer n > 1, we’ll find exactly one number of length n somewhere on the two lists. They contain one 3-digit number, one 4-digit number, and so on forever — if n = 100 we find a 100-digit number in the 30th position on the base 2 list.
(This result first appeared in the 1994 Asian Pacific Mathematics Olympiad. I found it in Ravi Vakil’s A Mathematical Mosaic.)
Two further curious lists: If we write out the triangular numbers, those in positions 3, 33, etc. show a pattern:
T(3) = 6
T(33) = 561
T(333) = 55611
T(3,333) = 5556111
T(33,333) = 555561111
T(333,333) = 55555611111
T(6) = 21
T(66) = 2211
T(666) = 222111
T(6,666) = 22221111
T(66,666) = 2222211111
T(666,666) = 222222111111
Draw any triangle and divide each leg into three equal segments. Connect each vertex to one of the trisection points on the opposite leg, as shown, and the triangle formed in the center will have 1/7 the area of the original triangle.
A square inscribed in a semicircle has 2/5 the area of a square inscribed in a circle of the same radius.
Draw a square and connect each vertex to the midpoint of an opposite side, as shown. The square formed in the center will have 1/5 the area of the original square.
A “proof without words”:
Steven Bartlett and Peter Suber’s Self-Reference: Reflections on Reflexivity contains a bibliography of works on reflexivity.
It includes an entry for Steven Bartlett and Peter Suber’s Self-Reference: Reflections on Reflexivity.
Until 1911, the U.S. House of Representatives grew along with the country. Accordingly, when the 1880 census showed an increase in population, C.W. Seaton, chief clerk of the census office, worked out apportionments for all House sizes between 275 and 350, in order to see which states would get the new seats.
He was in for a surprise. The method was straightforward: Take the total U.S. population and divide it by the proposed number of seats in the House, rounding all fractions down. This would dispose of most of the seats; any leftover seats would be awarded to the states whose fractional remainders had been highest. But Seaton discovered an oddity:
If the House had 299 seats, Alabama would get 8 representatives (because its remainder, .646, was higher than that of Texas or Illinois). But if the House had 300 seats it would get only 7 (the extra representative would now go to Illinois, whose remainder had surpassed Alabama’s). The problem is that the “fair share” of a large state increases more quickly than that of a small state.
Seaton called this the Alabama paradox. A related problem is the population paradox: If the method above had been used in 1901 to reallocate 386 seats in the House, Virginia would have lost a seat to Maine even though the ratio of their populations had increased from 2.67 to 2.68:
Here, even though the size of the House has not changed, a fast-growing state receives fewer representatives than a slow-growing one.
In 1982 mathematicians Michel Balinski and Peyton Young showed that if each party gets one of the two numbers closest to its fair share of seats, then any system of apportionment will run into one of these paradoxes. The solution, it seems clear, is to start cutting legislators into pieces.
(These data are from Hannu Nurmi’s Voting Paradoxes and How to Deal With Them, 1999. Balinski and Young’s book is Fair Representation: Meeting the Ideal of One Man, One Vote.)
In 1981, when science journalist Marcus Chown was an undergraduate physics student, his mother watched a profile of Richard Feynman on the BBC series Horizon. She had never shown an interest in science before, and he wanted to encourage her, so when he advanced to Caltech to study astrophysics, he told Feynman of his mother’s interest and asked him to send her a birthday note. She received this:
Happy Birthday Mrs. Chown!
Tell your son to stop trying to fill your head with science — for to fill your heart with love is enough!
Richard P. Feynman (the man you watched on BBC “Horizons”)
Male bees come from unfertilized eggs, so they have mothers but no fathers. Females come from fertilized eggs, so they have parents of both sexes. This produces an interesting pattern: The number of males in a given generation equals the number of females in the succeeding generation. And the number of females in a given generation equals the number of females in the succeeding two generations:
So the total number of bees, male and female, in generation n is the Fibonacci number Fn.
W. Hope-Jones discovered the relationship in 1921; this example is from Thomas Koshy’s Fibonacci and Lucas Numbers With Applications, 2001.
The first 10 digits of the golden ratio φ can be rearranged to give the first 10 digits of 1/π:
φ = 1.618033988 …
1/π = .3183098861 …
And the first nine digits of 1/φ can be rearranged to give the first 9 digits of 1/π:
1/φ = .618033988 …
1/π = .318309886 …
In 1983 amateur mathematician George Odom discovered that if points A and B are the midpoints of sides EF and DE of an equilateral triangle, and line AB meets the circumscribing circle at C, then AB/BC = AC/AB = φ. Odom used this fact to construct a pentagon, which H.S.M. Coxeter published in the American Mathematical Monthly with the single word “Behold!”