Square Deal

square deal puzzle

A puzzle by Sam Loyd. The red strips are twice as long as the yellow strips. The eight can be assembled to form two squares of different sizes. How can they be rearranged (in the plane) to form three squares of equal size?

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October 23, 2013January 30, 2022 | Puzzles

Black and White

møller chess problem

By Jørgen Thorvald Møller. White to mate in two moves.

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1. Kf6! and White will mate with 2. Qh1# or 2. Kg6#. From Skakbladet, August 1919.

October 18, 2013October 25, 2013 | Puzzles

All Relative

A problem from Dick Hess’ All-Star Mathlete Puzzles (2009):

A man points to a woman and says, “That woman’s mother-in-law and my mother-in-law are mother and daughter (in some order).” Name three ways in which the two can be related.

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The woman can be the man’s daughter-in-law, because his wife is the daughter of his mother-in-law. The woman can be the man’s wife’s aunt (his mother-in-law’s brother’s wife), because his mother-in-law is the daughter of her own brother’s wife’s mother-in-law. The woman can be the wife of the man’s nephew (the wife of his wife’s sister’s son), because his wife is the daughter of his mother-in-law. Equivalently, the man can be the woman’s father-in-law, her niece’s husband (her husband’s sister’s daughter’s husband), or her aunt’s husband (her husband’s mother’s sister’s husband).

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October 12, 2013October 18, 2013 | Puzzles

Black and White

pradignat chess problem

By Émile Leonard Pradignat. White to mate in two moves.

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1. Bd5! and White will win by Qxe6#, Qd3#, or c8=N#. From La Stratégie, 1896.

October 11, 2013October 18, 2013 | Puzzles

Ballot Measure

A Russian problem from the 1999 Mathematical Olympiad:

In an election, each voter writes the names of n candidates on his ballot. Each ballot is then placed into one of n+1 boxes. After the election, it’s noted that each box contains at least one ballot, and that if one ballot is drawn from each box, these n+1 ballots will always have a name in common. Show that for at least one box, there’s a name that appears on all of its ballots.

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Suppose, by way of contradiction, that this isn’t the case — that there is no box whose ballots all show a common name. Look at an arbitrary ballot in the first box. It lists n names; call these Candidates 1 through n. Now we can find a ballot in the second box that doesn’t list Candidate 1, a ballot in the third box that doesn’t list Candidate 2, and so on. Collect these n+1 ballots together and they form a group with no name in common. This contradicts the stated fact that no such group exists. So our assumption must be wrong: All of the ballots in at least one box must have a name in common.

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October 8, 2013October 18, 2013 | Puzzles

Black and White

von holzhausen chess problem

By Walter Freiherr Von Holzhausen. White to mate in two moves.

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1. Na7! and Black must succumb: 1. … Rxa7 2. Rc8# 1. … Kxc7+ 2. bxa8=N# 1. … Kxa7+ 2. b8=N# From Akademische Schachblätter, 1901.

October 4, 2013October 11, 2013 | Puzzles

Black and White

dos santos chess problem

By Arthur Napoleão Dos Santos. White to mate in two moves.

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1.Bg3! b2 2.Bd6# From Caissana Brasileira, 1898.

September 27, 2013October 4, 2013 | Puzzles

Edge Case

A circular table stands in a corner, touching both walls. A certain point on the table’s edge is 9 inches from one wall and 8 inches from the other. What’s the diameter of the table?

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Let the table’s center be A and its radius be r. ABC is a right triangle, so (r – 8)² + (r – 9)² = r², or r² – 34r + 145 = 0, or (r – 5) × (r – 29) = 0. So r is 5 or 29, and the diameter is 10 or 58 inches — one solution is for a point on the table’s inside edge (shown), the other for a point on the outside edge. From Steven Jerome Bryant et al., Nonroutine Problems in Algebra, Geometry, and Trigonometry, 1965.

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September 25, 2013October 4, 2013 | Puzzles

Going Down

In antiquity Aristotle had taught that a heavy weight falls faster than a light one. In 1638, without any experimentation, Galileo saw that this could not be true. What had he realized?

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He imagined connecting a heavy weight to a light one and dropping them together. Now Aristotle’s theory makes two contradictory predictions: The heavy weight should fall more slowly than normal, since the light weight now hinders it. The heavy weight should fall more quickly than normal, since the two connected weights now form one very heavy object. These can’t both be true, so Aristotle can’t be right. Instead the two weights must fall at the same rate. At the end of the final Apollo 15 moon walk, astronaut David Scott dropped a geologic hammer and a feather simultaneously, to test this idea without air resistance. Both hit the surface simultaneously. Mission controller Joe Allen wrote that this was “reassuring, considering both the number of viewers that witnessed the experiment and the fact that the homeward journey was based critically on the validity of the particular theory being tested.”

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