State House

A quickie from Peter Winkler’s Mathematical Puzzles, 2021: Can West Virginia be inscribed in a square? That is, is it possible to draw some square each of whose four sides is tangent to this shape?

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Yes. Start by drawing an arbitrary rectangle that hugs the state. Now rotate that rectangle, adjusting its dimensions as it turns to keep each side in contact with the shape. If you turn it through 90 degrees, then the “height” and “width” will have traded places: What had been the shorter side of the rectangle is now the longer. So at some point the two must have had the same value. (This is a two-dimensional variation on a classic puzzle.)

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June 15, 2021June 14, 2021 | Puzzles

Black and White

carpenter chess problem

By George E. Carpenter. White to mate in two moves.

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1. Qh8! and the queen will mate. From Carpenter’s Chess Problems, 1886.

June 14, 2021June 12, 2021 | Puzzles

Hooper’s Paradox

https://commons.wikimedia.org/wiki/File:Hooper_paradox.svg — Image: Wikimedia Commons

William Hooper published the oddity in 1774. The rectangle at the top measures 10 units by 3, giving an area of 30. But its dissected pieces seem to produce two other rectangles, with areas 12 and 20. Where did the two extra units come from?

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Image: Wikimedia Commons The pieces don’t quite match. If the pieces in the second figure are arranged carefully back into the first, it will be seen that they overlap. The indicated parallelogram measures exactly 2 square units.

June 12, 2021 | Puzzles

“Imitative Chess”

dudeney imitative chess

A puzzle by Henry Dudeney:

A chessboard was on the table with the pieces all set up for a game. So I asked Dr. Bates to play a game with the Major on these conditions: Whatever move Bates made throughout, with the white pieces, the Major must exactly imitate with the black, and Bates must give checkmate on the fourth move. As an experiment, Bates started off with 1. e4, and Rackford replied with 1. e5. Then Bates played 2. Qh5, and the Major had to reply with 2. Qh4. This gave me a good opportunity to explain that White cannot now play 3. QxQ, because it would be impossible for Black then to imitate the move. Neither could he play 3. Qxf7+, because Black cannot do the same thing, as he would have to get out of check. White must always make a move that Black can copy, until the checkmate is actually given on the fourth move.

“This puzzle caused great interest, and it was some time before somebody (I think it was Strangways) hit on a solution.”

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1. c4 c5 2. Qa4 Qa5 3. Qc6 Qc3 4. Qxc8#: (Henry E. Dudeney, “A Budget of Christmas Puzzles,” Strand 64:12 [December 1922], 619-622.) (If you want to solve the other puzzles, the solutions are here.)

June 8, 2021June 6, 2021 | Puzzles

The Hidden Element

The name of one chemical element appears as an unbroken string in the names of four other elements. What is the element, and what are the four?

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Tin (actinium, astatine, platinum, protactinium). (Dave Morice, “Kickshaws,” Word Ways 28:3 [August 1995], 170-180.) 06/09/2021 UPDATE: In French, the name or (gold) appears as an unbroken string in the names of nine elements: bore (boron), californium, chlore (chlorine), fluor (fluorine), livermorium, phosphore (phosphorus), rutherfordium, seaborgium, and thorium. (Thanks, Jean-Claude.)

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June 7, 2021June 9, 2021 | Puzzles

Stonework

A problem from the Leningrad Mathematical Olympiad: You have 32 stones, each of a different weight. How can you find the two heaviest in 35 weighings with an equal-arm balance?

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Think of the stones as tennis players and put them through a singles tournament. There are no upsets: In each match (weighing), the better (heavier) player always wins and goes on to the next round. With 32 players, this will produce a tournament with five rounds, comprising 16, 8, 4, 2, and 1 matches, successively, enough to produce 31 losers and identify the best (heaviest) stone. Now, the second-heaviest stone must have “lost” to the heaviest at some point in the tournament (since that was the only “player” strong enough to beat it). So take the five players beaten by the heaviest stone and put them through a little tournament of their own. Four matches will be enough to find the strongest of these five, and that must be the second-heaviest stone overall. (Via Ross Honsberger’s Mathematical Delights, 2004.)

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June 4, 2021 | Puzzles

Black and White

thompson chess problem

By Theophilus A. Thompson. White to mate in two moves.

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1. Rxc6! From Chess Problems, 1873.

May 24, 2021May 23, 2021 | Puzzles

Quickie

Express 1,000,000 as the product of two numbers, neither of which contains any zeroes.

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1,000,000 = 2⁶ × 5⁶. To avoid zeroes we need to collect the 2s into one factor and the 5s into the other: 64 × 15625. (Thanks, Nick.)

May 23, 2021 | Puzzles

Match Point

A problem from the Leningrad Mathematical Olympiad: A and B take turns removing matches from a pile. The pile starts with 500 matches, A goes first, and the player who takes the last match wins. The catch is that the quantity that each player withdraws on a given turn must be a power of 2. Does either player have a winning strategy?

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If there were only 3 matches in the original pile, A would lose — he could withdraw 1 or 2 matches, but then B could immediately withdraw the remainder and win the game. If there were 4 or 5 matches in the original pile then A could saddle B with a losing 3-match pile, but if there were 6 in the original pile, then by the same principle B could saddle A with the 3-match pile. The winning strategy, then, is always to leave your opponent a quantity that’s a multiple of 3. A can accomplish this on his first turn by taking 2 matches from the pile of 500, leaving 498 = 3 × 166 matches for B, who cannot now escape his fate.

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