The Banner of St. George

A “simple but pretty little puzzle” by Henry Dudeney: If this flag measures 4 feet by 3 feet and presents equal areas of red and white bunting, how wide are the cross’s arms?

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The flag measures 4 feet by 3 feet, so its diagonal is 5 feet. “All you need to do is to deduct half the length of this diagonal (2 1/2 feet) from a quarter of the distance all round the edge of the flag (3 1/2 feet) — a quarter of 14 feet. The difference (1 ft.) is the required width of the arm of the red cross.” Proof: If the flag measures 4 feet by 3 feet, the red area is given by 4x + 3x – x² = 6, so x² – 7x + 6 = 0. Solving the quadratic equation gives x – 3.5 = ± 2.5, and as x can’t be 6, the answer must be 1 foot. From Dudeney’s Amusements in Mathematics, via Erwin Brecher’s Ultimate Book of Puzzles, Mathematical Diversions, and Brainteasers (1996). 04/18/2025 UPDATE: Reader Catalin Voinescu has a simpler solution: “Consider: the banner has the same areas of red and white as a banner of the same size with a red L of the same thickness as the cross along two of the edges. (Cut the original banner in four and rearrange the fragments.) In this case, the white area is a rectangle. The problem states that it must be half of the area of the flag, that is, 6 sq ft. Its sides are shorter than the sides of the banner by the same amount each, namely, the thickness of the red L. Without any equation, one immediately notices that 3 by 2 gives 6, and that’s 1 less than 4 by 3. So the red part is 1 ft wide.” (Thanks, Catalin.)

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April 16, 2025April 18, 2025 | Puzzles

Matching Sums

A puzzle by Russian mathematician Sergei Berlov:

Here are two decagons. Suppose that a positive integer is written at each of the 20 vertices such that the sum of the numbers around each decagon is 99. Prove that it’s possible to mark some number of successive vertices on each decagon so that the two sums of marked numbers are equal. (The marked segment on a given decagon may be a single vertex, but not all 10.)

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In place of each decagon, consider a circle whose circumference is 99 units long and divided into segments whose length corresponds to the vertex integers, in order. So if the first few vertices around one decagon bear the integers 14, 9, 30, then the corresponding segments of that circle will have those lengths. Now superimpose the two circles and rotate one of them. If at any point two pairs of segment endpoints coincide simultaneously, then we have our solution — these endpoints identify two sets of successive vertices on the two decagons that produce the same sum. What if we never get two simultaneous coincidences? Well, that’s impossible: During a full rotation, each endpoint on the first circle must encounter 10 endpoints on the other circle, so there are 100 coincidences altogether. These must occur over the course of 99 discrete opportunities (the lengths of the two circles’ circumferences), so some two coincidences must occur together. From the January-February 1995 issue of Quantum.

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April 15, 2025April 14, 2025 | Puzzles

Excerpts

J. Bryan III used to challenge his friends to think of an English word that contains either of the following strings of letters:

ILIL

ILIW

“I’ve never played the game with anyone who could solve either of these combinations.” Can you?

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OILILY and BAILIWICK. See Pillow Talk.

April 10, 2025April 9, 2025 | Puzzles

C Change

How can we make sense of this passage, composed by Willard R. Espy?

The Optic Reed has a Haste Harm. It Hose, one assumes, not to be Lever; it neither Heats the Ripple, nor Hides the Razed sinner for his Rude Rime; it is not Old. Then Leave to it; ere thy grave Loses, thou mayst find here the Hart to the Rest thou Ravest to Limb.

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Insert a C before each capitalized word (except those that begin sentences): “The Coptic creed has a chaste charm. It chose, one assumes, not to be clever; it neither cheats the cripple, nor chides the crazed sinner for his crude crime; it is not cold. Then cleave to it; ere thy grave closes, thou mayst find here the chart to the crest thou cravest to climb.” From The Game of Words, 1987.

April 9, 2025April 8, 2025 | Puzzles

Round Numbers

A circle is divided into six sectors, into which are written (say, counterclockwise) these numbers:

1, 0, 1, 0, 0, 0

You can increase any two neighboring numbers by 1. By doing this repeatedly, is there a way to make all six numbers equal?

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No. Call the current six numbers a₁ … a₆. Now I = a₁ – a₂ + a₃ – a₄ + a₅ – a₆ is initially 2, and I will remain unchanged no matter what we do. So there’s no way to reach I = 0. Via Arthur Engel, Problem-Solving Strategies, 1999.

April 8, 2025April 7, 2025 | Puzzles

Yes or No

A problem by Nikolai Rozov:

“If the brainteaser you solved before you solved this one was harder than the brainteaser you solved after you solved the brainteaser that you solved before you solved this one, was the brainteaser you solved before you solved this one harder than this one?”

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Yes. (From Quantum, November-December 1994.)

April 7, 2025April 6, 2025 | Puzzles

Road Work

https://commons.wikimedia.org/wiki/File:Maze_Type_Block.png — Image: Wikimedia Commons

By Wikimedia user Efbrazil. Fill in four squares to make a continuous dark path connecting the stars. No diagonal moves are allowed.

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Image: Wikimedia Commons

April 5, 2025April 5, 2025 | Puzzles

Late Progress

Back in 2017 I wrote about the Feynman ciphers, three coded messages that had been presented as challenges to Richard Feynman in the 1950s.

Feynman couldn’t crack them, and even at the time of my post only the first of the three had been decoded — it turned out to be a transposition of the opening of Chaucer’s Canterbury Tales in Middle English. But in May 2023 David Vierra solved the other two — one turns out to be an excerpt from A.E. Housman’s 1896 poem “Terence, This is Stupid Stuff,” and the other is the start of Feynman’s 1953 paper “Atomic Theory of the λ Transition in Helium,” from The Physical Review.

Who was the “fellow scientist” who had devised these challenges? Nick Pelling thinks the most likely candidate is Paul Olum, who had been Feynman’s officemate at Los Alamos in the 1940s, but hard evidence is lacking. More at Cipher Mysteries.

(Thanks to reader Peter Dawyndt for the tip.)

March 25, 2025March 24, 2025 | Puzzles

Black and White

henle retrograde puzzle

Smith College mathematician Jim Henle published this retrograde analysis puzzle in the Mathematical Intelligencer in 2018 as part of an appreciation of Raymond Smullyan. “After many moves, the chessboard appears as above. What were the last two moves?”

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Henle’s solution: The black king is in check, and the check must have been discovered when the white king moved from f3 to e2. But how is this possible? The white king would itself have been in check in two different ways on f3. How could Black have inflicted two checks at once? The answer is that a black pawn must have stood on f2, blocking the rook’s check. This pawn then captured some white piece on e1, promoting to a knight and inflicting a double check. So, from this position: 1. … fxe1=N+ 2. Ke2+. (Jim Henle, “The Entertainer,” Mathematical Intelligencer 40:2 [June 2018], 76-80.)

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March 24, 2025March 23, 2025 | Puzzles

The Polish Acquaintanceship Problem

A Polish problem longlisted for the 19th International Mathematical Olympiad in Yugoslavia, July 1977:

A room contains nine men. Among every three of them there are two who are mutually acquainted. Prove that four of them are mutually acquainted.

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If one of the men is unacquainted with four of the others, then those four know one another and we’re done. If not, then this means that each man is mutually acquainted with at least five of the others. It can’t be exactly five in every case, because the total number of acquaintances in the room must be an even number. Hence someone is acquainted with at least six men. Among those six, three know one another, and they together with the man himself make four.

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March 21, 2025March 20, 2025 | Puzzles