Just a striking image: German astronomer Johann Friedrich Julius Schmidt prepared a plaster moon for the Field Columbian Museum in 1898. From the Field Museum Library.
Science & Math
A Pi Diet
Students beginning with the compass learn to draw this rosette, sometimes called the Flower of Life. If the arcs and the circle have the same radius, a, then the area of one petal is
and the unshaded area of the circle is
Remarkably, though the area we sought is bounded entirely by the arcs of circles, the final expression is independent of π.
Related: When two cylinders of radius r meet at right angles, the volume of their intersection is 16r3/3 — again, no sign of π.
(J.V. Narlikar, “A Pi-Less Area,” Mathematical Gazette 65:431 [March 1981], 32-33.)
10/01/2024 UPDATE: Some deft rearranging shows that the unshaded area of the circle is just the area of an inscribed regular hexagon:
So the absence of π isn’t that surprising. Thanks to readers Catalin Voinescu (who sent this diagram) and Gareth McCaughan for pointing this out.
A Belt Font
Suppose you have a collection of gears pinned to a wall (disks in the plane). When is it possible to wrap a conveyor belt around them so that the belt touches every gear, is taut, and does not touch itself? This problem was first posed by Manuel Abellanas in 2001. When all the gears are the same size, it appears that it’s always possible to find a suitable path for the belt, but the question remains open.
Erik Demaine, Martin Demaine, and Belén Palop have designed a font to illustrate the problem — each letter is a collection of equal-sized gears around which exactly one conveyor-belt wrapping outlines an English letter:
Apart from its mathematical interest, the font makes for intriguing puzzles — when the belts are removed, the letters are surprisingly hard to discern. What does this say?
The Sussman Anomaly
MIT computer scientist Gerald Sussman offered this example to show the importance of sophisticated planning algorithms in artificial intelligence. Suppose an agent is told to stack these three blocks into a tower, with A at the top and C at the bottom, moving one block at a time:
It might proceed by separating the goal into two subgoals:
- Get A onto B.
- Get B onto C.
But this leads immediately to trouble. If the agent starts with subgoal 1, it will move C off of A and then put A onto B:
But that’s a dead end. Because it can move only one block at a time, the agent can’t now undertake subgoal 2 without first undoing subgoal 1.
If the agent starts with subgoal 2, it will move B onto C, which is another dead end:
Now we have a tower, but the blocks are in the wrong order. Again, we’ll have to undo one subgoal before we can undertake the other.
Modern algorithms can handle this challenge, but still it illustrates why planning is not a trivial undertaking. Sussman discussed it as part of his 1973 doctoral dissertation, A Computational Model of Skill Acquisition.
The Sandwheel
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.
Matriarch
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.”
Portal Painting
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.
The Glass Rod Problem
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.)
Perfect Numbers
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.)
Math Notes
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.