In the 1970’s, a student of Maoist inclination asked [Columbia philosopher Sidney Morgenbesser] if he disagreed with Chairman Mao’s saying that a proposition can be true or false at the same time. Dr. Morgenbesser replied, ‘I do and I don’t.’

From Morgenbesser’s 2004 New York Times obituary.

Drill a hole straight through the center of a sphere, leaving a band in the shape of a napkin ring. Suppose the height of this band is h. Now take a sphere of a different size and drill a hole through that, contriving the hole’s width so that its depth is again h. Remarkably, the two napkin rings will have the same volume.

As the sphere expands, the band must grow both wider and thinner, and it turns out that these two effects exactly cancel one another. It’s called the napkin ring problem.

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.

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.

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:

1. Get A onto B.
2. 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.

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.”

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.)