Russian science writer Yakov Perelman asks: How many times must an 8-toothed cogwheel rotate on its axis to circle around a 24-toothed cogwheel?
Meteors are more commonly seen between midnight and dawn than between dusk and midnight. Why?
In 1653, Blaise Pascal composed a triangular array in which the number in each cell is the sum of the two directly above it:
In 1915, Polish mathematician Waclaw Sierpinski described an equilateral triangle in which the central fourth is removed and the same procedure is applied to all the succeeding smaller triangles. Perplexingly, the resulting structure has zero area:
Interestingly, if the odd numbers in Pascal’s triangle are shaded, they produce an approximation to Sierpinski’s triangle:
And as this triangle grows toward infinity, it becomes Sierpinski’s triangle — an arrangement of numbers that takes the shape of a geometrical object.
If two squares share a common vertex, then the centers of the squares and the midpoints of the linking segments shown form another square.
Discovered by mathematicians Paul Finsler and Hugo Hadwiger.
Draw any quadrilateral and build a square on each of its sides.
Connect the centers of opposite squares, and these two line segments will be perpendicular and of equal length.
If the centers of five circles lie along the circumference of a sixth so that they overlap like the links of a chain, and if each intersects its neighbor on the sixth circle as well, then drawing lines as shown through the remaining intersections will form a pentagram whose points lie on the five circles.
A 19th-century schoolchild defined a circle as “a round straight line with a hole in the middle.”
Entomologist Alexandre Girault expressed himself in his work. Of the 500 genera he named, many honored artists, poets, and writers whom he admired: Davincia, Shakespearia, Beethovena, Mozartella, Emersonia, Raphaelana, Ovidia, Goetheana, Thoreauella, Tennysoniana, Bachiana, Keatsia, Plutarchia, Schilleria, Aeschylia, Aligheria, Thalesanna, Rubensteina, Carlyleia, Herodotia, Cowperella, and Froudeana.
To mock his supervisor Johann Illingworth, he invented a parasitic mymarid wasp, Shillingsworthia shillingsworthi, which he described as an ephemeral creature lacking a head, abdomen, or mandibles and found only in “the chasms of Jupiter” — in other words, a nonentity. He called it “an airy species whose flight cannot be followed except by the winged mind.”
He understood even women through entomology — one of his privately printed works describes a new species of human, Homo perniciosus, known only from the female sex:
Abnormal female (loveless, without offspring); heart functionless; mammae aborted; psychology novel (as supposed) but artificial; gay, high-coloured, feral, brass-cheeked, shape lovely like Woman but nature hard (selfless, thoughtless, proud, unsympathetic, irresponsible, aggressive, irritant, insensible, luxurious, pugnacious, over-active, inquisitive, mischievous, voracious and even carnivorous; antagonistic, ungentle, immodest, critical, competitive, poisonous); conduct unstable (even inclined to treachery), the lips compressed, body strong. Everywhere but rare in natural habitat.
He was prickly, but he was dedicated — he published much of his work at his own expense, and many of his type specimens are retained today by the Queensland Museum. “Research is a labour of love,” he wrote. “Strange then to find it all done nowadays as a labour of wages! Must love, too, be a matter of cash?”
Let four circles (blue) pass through a single point, M. Each pair of these circles intersect at a second point (pink). Each three of the four blue circles will have three pink points among them; these trios of pink points define four new circles (brown), which intersect in a single point, P.
If we start with five circles passing through a single point M, then we can apply the procedure above to each subset of four of them. This will produce five points P that all lie on a single circle.
If we start with six circles that all pass through a single point M, then each subset of five of them defines a new circle, as we’ve just seen. These six new circles all pass through a single point.
Remarkably, this pattern continues forever. It was discovered by the English geometer William Kingdon Clifford.
Arrange cards with values ace through 9 in a row, in counting order, with the ace on the left.
Take up a card from one end of the row — left or right, your choice.
Do this twice more, each time taking up either the leftmost or the rightmost card in the remaining row.
When you have three cards, add their values, divide the total by six, and call the result n. Count the cards that remain on the table from left to right.
The card in the nth position will be the 4.
Alexander Graham Bell kisses his daughter Daisy inside a tetrahedral kite, October 1903.
Bang’s theorem holds that the faces of a tetrahedron all have the same perimeter only if they’re congruent triangles. Also, if they all have the same area, then they’re congruent triangles.
Buckminster Fuller proposed establishing a floating tetrahedron in San Francisco Bay called Triton City (below). It would have been assembled from modules, starting with a floating “neighborhood” of 5,000 residents, with an elementary school, a supermarket and a few specialty shops. Three to six neighborhoods would form a town, and three to seven towns would form a city. At each stage the corresponding infrastructure would be added: schools, civic facilities, government offices, and industry. A full-sized city might accommodate 100,000 people in a single building. He envisioned an even larger tetrahedron, with a million citizens, for Tokyo Bay.
The moral of Fuller’s 1975 book Synergetics was “Dare to be naïve.”
The square root of 2 is 1.41421356237 … Multiply this successively by 1, by 2, by 3, and so on, writing down each result without its fractional part:
Beneath this, make a list of the numbers that are missing from the first sequence:
The difference between the upper and lower numbers in these pairs is 2, 4, 6, 8 …
From Roland Sprague, Recreations in Mathematics, 1963.
“Fifteen young ladies in a school walk out three abreast for seven days in succession: it is required to arrange them daily, so that no two shall walk twice abreast.”
The Rev. Thomas Penyngton Kirkman posed this query innocently in The Lady’s and Gentleman’s Diary in 1850. It’s trickier than it looks — though seven solutions are possible, they’re difficult to discover by trial and error. Here’s one:
Each letter appears once in each row, but no two letters share a cell more than once. The problem’s simplicity made it popular among Victorian amateur mathematicians, and Kirkman later complained that it had eclipsed his more serious work — though he took pains to dispute James Joseph Sylvester’s claim to have invented the problem himself.
Do such problems generally have solutions? Surprisingly, the answer remained unknown until just last January, when Oxford mathematician Peter Keevash showed that the answer is yes if certain basic requirements are met. Keevash’s result was “a bit of an earthquake as far as design theory is concerned,” said Cambridge mathematician Timothy Gowers.
Now that the schoolgirl problem is under control, a related challenge has taken its place. If 20 golfers want to arrange themselves into different foursomes on five successive days, is it possible to plan the groups so that each golfer plays no more than once with any other golfer? Formally the “social golfer problem” remains unsolved — though tournament organizers work out individual solutions every day.
Divide the number 999,999,999,999,999,999,999,998,999,999,999,999,999,999,999,999 into 1 and express the result as a decimal expansion, and you’ll find the Fibonacci sequence presented in tidy 24-digit strings:
Amateur magician Oscar Weigle invented this surprising effect in 1949. Assemble a deck of 20 playing cards, 10 red and 10 black, in strictly alternating colors. Hold this deck under a table. Now turn over the top two cards as one, place them on top, and cut the deck. Repeat this procedure as many times as you like — turn two, cut, turn two, cut. When you’ve finished, the deck will contain an unknown number of reversed cards, distributed randomly.
Now, still holding the deck under the table, shift the top card to the bottom, then turn over the next card and place it on the table. Do this repeatedly — shift a card to the bottom, then reverse the next card and put it on the table — continuing until you’ve put 10 cards on the table. Surprisingly, these cards are sorted by color — the face-up cards are of one color, and the face-down ones are of the other.
You’re still holding 10 cards under the table. Divide these into two stacks and weave them together under the table randomly. Do this as many times as you like — divide the 10 cards into two groups and merge them together however you like, so long as no card is turned upside down. Turn over the packet and shuffle it in the same way a few more times. Give it a final cut if you like.
Now deal these cards out as before: Shift the top card to the bottom, reverse the next card and put it on the table. Like the first group, this one will sort itself by color, with one color face up and the other face down.
Suppose we put eight white and two black balls into a bag and then draw forth the balls one at a time. If we repeat this experiment many times, which draw is most likely to produce the first black ball?
Most people answer 4, but in fact the first black ball is most likely to appear on the very first draw:
By symmetry, the second black ball is most likely to appear on the final draw.
(A.E. Lawrence, “Playing With Probability,” Mathematical Gazette, vol. 53 [December 1969], 347-354.)
If every pair of people in a group have exactly one friend in common, then there’s always one person who is a friend to everyone.
This rather heartwarming fact was proven by Paul Erdös, Alfréd Rényi, and Vera T. Sós in 1966. It’s sometimes more cynically known as the paradox of the politician.
The Erdös proof uses combinatorics and linear algebra, but in 1972 Judith Longyear published a proof using elementary graph theory.
See The Elevator Problem.
“The origin of science is in the desire to know causes; and the origin of all false science and imposture is in the desire to accept false causes rather than none; or, which is the same thing, in the unwillingness to acknowledge our own ignorance.” — William Hazlitt
Three coins are lying on a table: a quarter, a half dollar, and a silver dollar. You claim one coin, I’ll claim the other two, and then we’ll toss all three. A coin that lands tails counts zero, and a coin that lands heads wins its value (in cents, 25, 50, or 100) for its owner. Whichever of us has the larger score wins all three coins. If all three coins land tails then we toss again.
Which coin should you claim to make the game fair — that is, so that each of us has an expected win of zero?
Arrange three congruent equilateral triangles so that their corners meet at a point, like the red triangles above. The arrangement doesn’t have to be symmetric; the triangles can even overlap. Now draw lines BC, DE, and FA to complete a hexagon inscribed in a circle. The midpoints of these three lines will form the vertices of an equilateral triangle.
That’s called the asymmetric propeller theorem, and it’s been known since the 1930s. But in 1979 Beverly Hills dentist and geometry enthusiast Leon Bankoff told Martin Gardner of some further discoveries. Bankoff never found time to write them up, so after the dentist’s death in 1997 Gardner published them in the College Mathematics Journal:
- The three equilateral triangles need not be congruent. Each can be of any size and the theorem still holds.
- The triangles need not meet at a point. They can meet at the corners of any equilateral triangle.
- They need not even be equilateral! If three similar triangles of any sizes meet at a point, the midpoints of the three added lines will form a triangle similar to each of the “propellers.”
- The similar triangles need not meet at a point! If they meet at the corners of a fourth triangle (of any size) that’s similar to each propeller, then the midpoints of the added lines will form a triangle similar to each propeller, provided that the vertices of the central triangle touch the corresponding corners of the propellers.
Given all this flexibility, Gardner asked, do the propellers even have to be triangles? It turns out that the answer is yes. Still, the discoveries above form a fitting tribute to Bankoff, whom Gardner called “one of the most remarkable mathematicians I have been privileged to know.”
(Martin Gardner, “The Asymmetric Propeller,” College Mathematics Journal 30:1 [January 1999], 18-22.)
It was a good answer that was made by one who when they showed him hanging in a temple a picture of those who had paid their vows as having escaped shipwreck, and would have him say whether he did not now acknowledge the power of the gods, — ‘Aye,’ asked he again, ‘but where are they painted that were drowned after their vows?’ And such is the way of all superstition, whether in astrology, dreams, omens, divine judgments, or the like; wherein men, having a delight in such vanities, mark the events where they are fulfilled, but where they fail, though this happens much oftener, neglect and pass them by.
— Francis Bacon, Novum Organum, 1620
Imagine a staircase with 1001 stairs, numbered -500 to 500. You’re standing in the middle, on stair 0, and you want to reach the top. On each step you can play either of two coin-flipping games — if the result is heads then you move up a step; if it’s tails then you move down a step:
- In game 1 you flip coin A, which is slightly biased: It comes up heads 49.5 percent of the time and tails 50.5 percent.
- In game 2 you use two coins, B and C. Coin B produces heads 9.5 percent of the time and tails 90.5 percent. Coin C produces heads 74.5 percent of the time and tails 25.5 percent. In game 2 if the number of the stair you’re on is a multiple of 3 then you flip coin B; otherwise you flip coin C.
Both of these are losing games — if you played either game 1 or game 2 exclusively, you’d eventually find yourself at the bottom of the staircase. But in 1996 Spanish physicist Juan Parrondo found that if you play the two games in succession in random order, keeping your place on the staircase as you switch between them, you’ll rise to the top of the staircase. It’s not, properly speaking, a paradox, but it’s certainly counterintuitive.
This example is from David Darling’s Universal Book of Mathematics. (Thanks, Nick.)
We’ve mentioned before that you can estimate π by dropping needles on the floor. (Reader Steven Karp also directed me to this remarkable solution, from Daniel A. Klain and Gian-Carlo Rota’s Introduction to Geometric Probability .)
Here’s a related curiosity. If a circle of diameter L is placed at random on a pattern of circles of unit diameter, which are arranged hexagonally with centers C apart, then the probability that the placed circle will fall entirely inside one of the fixed circles is
If we put k = C/(1 – L), we get
And a frequency estimate of P will give us an estimate of π.
Remarkably, in 1933 A.L. Clarke actually tried this. In Scripta Mathematica, N.T. Gridgeman writes:
His circle was a ball-bearing, and his scissel a steel plate. Contacts between the falling ball and the plate were electrically transformed into earphone clicks, which virtually eliminated doubtful hits. With student help, a thousand man-hours went into the accumulation of N = 250,000. The k was about 8/5, and the final ‘estimate’ of π was 3.143, to which was appended a physical error of ±0.005.
“This is more or less the zenith of accuracy and precision,” Gridgeman writes. “It could not be bettered by any reasonable increase in N — even if the physical error could be reduced, hundreds of millions of falls would be needed to establish a third decimal place with confidence.”
(N.T. Gridgeman, “Geometric Probability and the Number π,” Scripta Mathematica 25:3 [November 1960], 183-195.)
(From Raymond F. Lausmann’s Fun With Figures, 1965.)
This trick seems to have been invented independently by Martin Gardner and Karl Fulves. A blindfolded magician asks a spectator to lay three pennies on a table, in any arrangement of heads and tails. The magician’s goal is to put all three coins into the same state, all heads or all tails.
If the three coins already match, then the trick is done. If not, then the magician gives three instructions: Flip the left coin, flip the middle coin, flip the left coin. After each step he asks whether the three coins now match. By the third flip, they will.
“It’s no surprise that the magician can eventually equalize all the coins,” writes MIT computer scientist Erik Demaine, “but it’s impressive that it always takes at most three moves.” The technique exploits a principle used in Gray codes, which are used to reduce errors when using analog signals to represent digital data. Demaine relates a similar trick involving four coins in the November-December 2010 issue of American Scientist.