Dmitri Shostakovich’s first opera was a setting of Nikolai Gogol’s 1836 story “The Nose.”
This scene is from the Royal Ballet & Opera’s 2016 staging.
(Thanks, Randy.)
Bootstraps
The 45-letter pneumonoultramicroscopicsilicovolcanoconiosis is often cited as one of the longest words in English — it’s been recognized both by Merriam-Webster and by the Oxford English Dictionary. The OED Supplement traced it to a 1936 puzzle book by Frank Scully called Bedside Manna, defining it as “a disease caused by ultra-microscopic particles of sandy volcanic dust.” But in fact it had appeared first in a Feb. 23, 1935, story in the New York Herald Tribune:
Puzzlers Open 103rd Session Here by Recognizing 45-Letter Word
Pneumonoultramicroscopicsilicovolcanoconiosis succeeded electrophotomicrographically as the longest word in the English language recognized by the National Puzzlers’ League at the opening session of the organization’s 103d semi-annual meeting held yesterday at the Hotel New Yorker.
The puzzlers explained that the forty-five-letter word is the name of a special form of silicosis caused by ultra-microscopic particles of silica volcanic dust …
At the meeting NPL president Everett M. Smith had claimed the word was legitimate, but in fact he’d coined it himself. Distinguished by the newspaper, it found its way into Scully’s book and thence into the dictionaries, “surely one of the greatest ironies in the history of logology,” according to author Chris Cole. Today it’s recognized as long but phony — Oxford changed its definition to “an artificial long word said to mean a lung disease caused by inhaling very fine ash and sand dust.”
(Chris Cole, “The Biggest Hoax,” Word Ways 22:4 [November 1989], 205-206.)
Invisible Artworks
For his 1959 work Zone de Sensibilité Picturale Immatérielle, Yves Klein sold the ownership of empty space.
In its 1967 Air-Conditioning Show, English conceptual collaborative Art & Language presented an empty room containing two air conditioning units; the artwork was “what is felt and said about it.”
James Lee Byars’ 1969 work The Ghost of James Lee Byars consisted of the emptiness and darkness of a pitch-black room.
Robert Barry communicated his 1969 Telepathic Piece mentally to visitors; the artwork was “a series of thoughts that are not applicable to language or image.”
Andy Warhol’s 1985 Invisible Sculpture was entirely intangible.
Tom Friedman’s 1992 work Untitled (A Curse)” consisted of a region of empty space that had been cursed by a witch.
Roman Ondak’s 2006 work More Silent Than Ever was an empty exhibition room in which a covert listening device had allegedly been hidden; visitors were told they were being monitored, but no evidence was ever given that the device really existed.
Salvatore Garau’s 2021 sculpture Io Sono occupied an area 5 feet square but was otherwise imperceptible.
Ruben Gutierrez’s 2022 work This Sculpture Makes Me Cry (A Spell) was said to represent what the artist could not see but which affected him emotionally.
Warhol and Gutierrez both presented their sculptures on white pedestals. Is there any way to prove they’re not the same piece?
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.)
Unquote
“A footnote is like running downstairs to answer the doorbell during the first night of marriage.” — John Barrymore
The Desk Calendar
This is a variation on an old puzzle by Martin Gardner. This desk calendar currently displays the date Monday, April 25. If the six cubes are capable of displaying any day of the week and any date from January 01 to December 31, what characters appear on the unseen faces in the picture? Gardner wrote, “It is a bit trickier than one might expect.”
“The Last Bard of Dixie”
South Carolina poet J. Gordon Coogler (1865-1901) was widely mocked for this terrible couplet:
Alas! for the South, her books have grown fewer —
She never was much given to literature.
He complained,
Oh you critics! — If an author errs in a single line,
That line you’ll surely quote,
And will give it as a sample fair
Of all he ever wrote.
But he was bad everywhere:
On her beautiful face there are smiles of grace
That linger in beauty serene,
And there are no pimples encircling her dimples
As ever, as yet, I have seen.
His complete works are here.
Early Arrival
When his wife died in 1893, Brooklyn retiree Jonathan Reed had a tomb built in Evergreens Cemetery, where for 10 years he kept her company. “The Reed mausoleum was furnished just like a living room in a fine house,” read his New York Times obituary, “all the usual articles of furniture being provided. It was warmed with a fine oil stove manufactured specially for the tomb.”
Mr. Reed could never be made to believe that his wife was really dead, his explanation of her condition being that the warmth had simply left her body, and that if he kept the mausoleum warm she would continue to sleep peacefully in the costly metallic casket in which her remains were put. … According to his friends, he really believed that his wife could understand what he was saying to her.
From the hour the cemetery gates opened at 6 a.m. until they closed at 6 p.m., he was in the tomb. “The old man ate all of his meals in the mausoleum and was in the habit of holding imaginary conversations with his wife. … He always appeared to be very happy.”
A visitor eventually discovered him dying on the floor of the crypt in 1905, and he was laid to rest in the empty casket that he had prepared next to hers.
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.)
Half Measures
When we read type we imagine that we read the whole of the type — but that is not so; we only notice the upper half of each letter. You can easily prove this for yourself by covering up the upper half of the line with a sheet of paper (being careful to hold the paper exactly in the middle of the letters), and you will not, without great difficulty, decipher a single word. Now place the paper over the lower half of a line, and you can read it without the slightest difficulty.
— George Lindsay Johnson, “Some Curious Optical Illusions,” Strand, October 1897