Clearance

In the 1928 film Steamboat Bill, Jr., a falling facade threatens to flatten Buster Keaton, but he’s spared by the fortunate placement of an open attic window. “As he stood in the studio street waiting for a building to crash on him, he noticed that some of the electricians and extras were praying,” writes Marion Meade in Cut to the Chase, her biography of Keaton. “Afterward, he would call the stunt one of his greatest thrills.”

It’s often said that the falling wall missed Keaton by inches. Is that true? James Metz studied the problem in Mathematics Teacher in 2019. Keaton was 5 feet 5 inches tall; if that the “hinge” of the facade is 5 inches above the surface of the ground, the attic window is 12 feet above that, and the window is 3 feet high, he finds that the top of the window came only within about 1.5 feet of Keaton’s head.

“The window was tall enough to allow an ample margin of safety, so the legend about barely missing his head cannot be true,” Metz writes. “Apparently, Keaton had more headroom than was previously suspected.”

(James Metz, “The Right Place at the Right Time,” Mathematics Teacher 112:4 [January/February 2019], 247-249.)

All for One

In 1988, Florida International University mathematician T.I. Ramsamujh offered a proof that all positive integers are equal. “The proof is of course fallacious but the error is so nicely hidden that the task of locating it becomes an interesting exercise.”

Let p(n) be the proposition, ‘If the maximum of two positive integers is n then the integers are equal.’ We will first show that p(n) is true for each positive integer. Observe that p(1) is true, because if the maximum of two positive integers is 1 then both integers must be 1, and so they are equal. Now assume that p(n) is true and let u and v be positive integers with maximum n + 1. Then the maximum of u – 1 and v – 1 is n. Since p(n) is true it follows that u – 1 = v – 1. Thus u = v and so p(n + 1) is true. Hence p(n) implies p(n+ 1) for each positive integer n. By the principle of mathematical induction it now follows that p(n) is true for each positive integer n.

Now let x and y be any two positive integers. Take n to be the maximum of x and y. Since p(n) is true it follows that x = y.

“We have thus shown that any two positive integers are equal. Where is the error?”

(T.I. Ramsamujh, “72.14 A Paradox: (1) All Positive Integers Are Equal,” Mathematical Gazette 72:460 [June 1988], 113.)

(The error is explained here and here.)

Mirror Therapy

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When a limb is paralyzed and then amputated, the patient may perceive a “phantom limb” in its place that is itself paralyzed — the brain has “learned” that the limb is paralyzed and has not received any feedback to the contrary.

University of California neuroscientist V.S. Ramachandran found a simple solution: The patient holds the intact limb next to a mirror, looks at the reflected image, and makes symmetric movements with both the good and the phantom limb. In the reflected image, the brain is now able to “see” the phantom limb moving. The impression of paralysis lifts, and the patient can now move the phantom limb out of painful positions.

A 2018 review called the technique “a valid, simple, and inexpensive treatment for [phantom-limb pain].”

A Perfect Bore

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If we assume the existence of an omniscient and omnipotent being, one that knows and can do absolutely everything, then to my own very limited self, it would seem that existence for it would be unbearable. Nothing to wonder about? Nothing to ponder over? Nothing to discover? Eternity in such a heaven would surely be indistinguishable from hell.

— Isaac Asimov, “X” Stands for Unknown, 1984

An Even Dozen

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The surface of a standard soccer ball is covered with 20 hexagons and 12 pentagons. Interestingly, while we might vary the number of hexagons, the number of pentagons must always be 12.

That’s because the Euler characteristic of a sphere is 2, so VE + F = 2, where V is the number of vertices, or corners, E is the number of edges, and F is the number of faces. If P is the number of pentagons and H is the number of hexagons, then the total number of faces is F = P + H; the total number of vertices is V = (5P + 6H) / 3 (we divide by 3 because three faces meet at each vertex); and the total number of edges is E = (5P + 6H) / 2 (dividing by 2 because two faces meet at each edge). Putting those together gives

\displaystyle V-E+F={\frac {5P+6H}{3}}-{\frac {5P+6H}{2}}+P+H={\frac {P}{6}},

and since the Euler characteristic is 2, this means P must always be 12.

A Late Arrival

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Image: Wikimedia Commons

There’s only one normal magic hexagon — only one hexagonal arrangement of cells like the one shown here that can be filled with the consecutive integers 1 to n such that every row, in all three directions, sums to the same total. (This excludes the trivial example of a single cell on its own, as well as rotations and reflections of the hexagon shown here.)

Amazingly, it appears that this lonely example may not have been discovered until 1888! In a 1988 letter to the Mathematical Gazette, Martin Gardner reported that the magic hexagon was given as a problem in the German magazine Zeitschrift für mathematischen und naturwissenschaftlichen Unterricht (Volume 19, page 429) in 1888. The proposer was identified only as von Haselberg, of Stralsund; his solution was published in the next volume.

Gardner writes, “The structure could easily have been discovered by mathematicians in ancient times, but as of now, this is the earliest known publication.”

(Martin Gardner, “The History of the Magic Hexagon,” Mathematical Gazette 72:460 [June 1988], 133.)