This is a variation on a perpetual motion machine proposed by the Indian mathematician Bhāskara II around 1150. Each of the wheel’s tilted spokes is filled with a quantity of sand. As the tubes descend on the right, the sand within them shifts outward, exerting greater torque in the clockwise direction and thus keeping the wheel turning forever.
Unfortunately the same design ensures that there’s always a greater quantity of sand on the left, so nothing happens.
In 1956, ornithologist Chandler Robbins tagged a wild female Laysan albatross at the Midway Atoll National Wildlife Refuge in the North Pacific. The bird, dubbed Wisdom, went on to a stunning career, flying more than 3 million miles, equivalent to 120 trips around the Earth. She has been seen at the atoll as recently as last December, making her, at 72, the the oldest known wild bird in the world.
In that time she’s hatched as many as 36 chicks, a significant contribution to the struggling wild albatross population. The U.S. Fish and Wildlife Service wrote, “Her health and dedication have led to the birth of other healthy offspring which will help recover albatross populations on Laysan and other islands.” Bruce Peterjohn, chief of the North American Bird Banding Program, added, “To know that she can still successfully raise young at age 60-plus, that is beyond words.”
Steven Novak uses a laser to transfer a scene to the interior of a hemisphere. The completed painting then presents an “accurate” perspective to a viewer looking about from the center.
Dick Termes creates the same impression (via a perceptual illusion) by painting on the convex exterior of a full sphere.
Suppose we drop a glass rod and it breaks into three pieces. What is the probability that the pieces can form a triangle? Mathematician D.C. Johnson found this elegant geometric solution. In order to form a triangle, none of the three pieces can be longer than the other two combined. Now consider an equilateral triangle whose altitude equals the length of the glass rod (say, 1). Viviani’s theorem tells us that the sum of the lengths of the perpendiculars from any interior point to the sides of this triangle is 1, the triangle’s altitude:
And any three non-negative numbers whose sum is 1 correspond to three such perpendiculars in some prescribed order. So the points inside the triangle correspond to all the various ways in which the glass rod can break.
Now consider the first piece of the broken rod. In order to form a triangle with the other two pieces, its length must not exceed 1/2. That means it must not extend from the base of our equilateral triangle into the shaded zone here:
And if we assign the other two pieces to the other two legs, we can make the same argument and identify two more zones:
That immediately gives the answer: The chance that all three pieces will be short enough to produce a triangle is 1 in 4.
(C. Haigh, “The Glass Rod Problem,” Mathematical Gazette 65:431 [March 1981], 37-38.)
From Lee Sallows:
As the reader can check, the English number names less than “twenty” are composed using 16 different letters of the alphabet. We assign a distinct integral value to each of these as follows:
E F G H I L N O R S T U V W X Z 3 9 6 1 -4 0 5 -7 -6 -1 2 8 -3 7 11 10
The result is the following run of so called “perfect” numbers:
Z+E+R+O = 10 + 3 – 6 – 7 = 0 O+N+E = –7 + 5 + 3 = 1 T+W+O = 2 + 7 – 7 = 2 T+H+R+E+E = 2 + 1 – 6 + 3 + 3 = 3 F+O+U+R = 9 – 7 + 8 – 6 = 4 F+I+V+E = 9 – 4 – 3 + 3 = 5 S+I+X = –1 – 4 + 11 = 6 S+E+V+E+N = –1 + 3 – 3 + 3 + 5 = 7 E+I+G+H+T = 3 – 4 + 6 + 1 + 2 = 8 N+I+N+E = 5 – 4 + 5 + 3 = 9 T+E+N = 2 + 3 + 5 = 10 E+L+E+V+E+N = 3 + 0 + 3 – 3 + 3 + 5 = 11 T+W+E+L+V+E = 2 + 7 + 3 + 0 – 3 + 3 = 12
The above is due to a computer program in which nested Do-loops try out all possible values in systematically incremented steps. The above solution is one of two sets coming in second place to the minimal (lowest set of values) solution seen here:
E F G H I L N O R S T U V W X Z –2 –6 0 –7 7 9 2 1 4 3 10 5 6 –9 –4 –3
But why does the list above stop at twelve? Given that 3 + 10 = 13, and assuming that THREE, TEN and THIRTEEN are all perfect, we have T+H+I+R+T+E+E+N = T+H+R+E+E + T+E+N. But cancelling common letters on both sides of this equation yields E = I, which is to say E and I must share the same value, contrary to our requirement above that the letters be assigned distinct values. Thus, irrespective of letter values selected, if it includes THREE and TEN, no unbroken run of perfect numbers can exceed TWELVE. This might be decribed as a formal proof that THIRTEEN is unlucky.
But not all situations call for an unbroken series of perfect numbers. Sixteen distinct numbers occur in the following, eight positive, eight negative. This lends itself to display on a checkerboard:
Choose any number on the board. Call out the letters that spell its name, adding up their associated numbers when on white squares, subtracting when on black. Their sum is the number you selected.
(Thanks, Lee.)
If we have two numbers a and b such that ab + 1 is square, then it’s always possible to find a number c for which ac + 1 and bc + 1 are both square. For example, 8 × 3 + 1 = 25 = 52, and 8 × 21 + 1 = 169 = 132 and 3 × 21 + 1 = 64 = 82.
Proof:
If ab + 1 = m2, then set c = a + b + 2m. Now
ac + 1 = a2 + ab + 2am + 1 = a2 + 2am + m2 = (a + m)2
bc + 1 = ab + b2 + 2bm + 1 = b2 + 2bm + m2 = (b + m)2
Via Edward Barbeau, Power Play, 1997.
Princeton mathematician John Horton Conway memorized π to more than a thousand decimal places by marrying it to the periodic table of the elements:
3 Neutronium 1415926535 Hydrogen 8979323846 Helium 2643383279 Lithium 5028841971 Beryllium …
Between each pair of elements are sandwiched ten digits of π. (Neutronium is Andreas von Antropoff’s notional “element of atomic number zero,” an element with zero protons in its nucleus.) This approach to memorizing digits has a number of virtues:
To remember the elements themselves Conway devised a long mnemonic. It begins
Newt? Hy! He Likes Beryl’s Boring Car for Nites Out in Florid Neon
for
Nn H He Li Be B C N O F Ne.
See the paper below for the whole package — by including unconfirmed hypothetical elements, it encodes 120 mouthfuls, or 1,200 digits.
(John Conway, “Chemical π,” Mathematical Intelligencer 38:4 [December 2016], 7-10.)
In 2015 Nature published an alarming article suggesting that dragons are real and had only gone to sleep during the Little Ice Age. A medieval document discovered “under a pile of rusty candlesticks” in the Bodleian Library showed that the creatures were once common but had entered a state of brumation when temperatures dropped and their traditional diet of knights began to thin. Rising temperatures in the modern age have correlated with increasing mentions in fictional literature, which “suggests that these fire-breathing lizards are being sighted more frequently.”
It gets worse: “Sluggish action on global warming is set to compound the problem, and policies such as the restoration of knighthoods in Australia are likely to exacerbate the predicament yet further by providing a sustained and delicious food supply.” The date of the article was April 2.
(Andrew J. Hamilton, Robert M. May, and Edward K. Waters, “Here Be Dragons,” Nature 520:7545 [April 2, 2015], 42-43.)
Suppose you’re hiking in the woods and become lost. What’s the best path to follow to find the boundary? You know the forest’s shape and dimensions, but you don’t know where you are within it, nor which direction you’re facing.
This has remained an open problem ever since mathematician Richard Bellman posed it in the Bulletin of the American Mathematical Society in 1956. The best strategy will cover the shortest distance in the worst case; in forests of certain simple shapes this might be as straightforward as walking in a straight line or in a spiral, but other shapes are more troubling. We know how to escape squares and circles efficiently, but not equilateral triangles.
Mathematician Scott W. Williams classed this as a “million-buck problem” because solving it is expected to cultivate techniques of particular value to mathematics. It’s known as Bellman’s lost-in-a-forest problem.
