Small World

Three flying saucers land simultaneously on a spherical planet.

What’s the probability that all land in or on the same hemisphere?

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It’s certain. Point A, Point B, and the planet’s center determine a plane that cuts the planet in half; Point C must lie on one side or the other of this plane (or directly on the great circle).

April 18, 2012February 1, 2013 | Puzzles

Black and White

By Joseph Kling, from Chess Euclid, 1849. White to mate with the pawn in three moves.

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1. Qa2 Ke8 2. Qa8+ Kd7 3. e6#

April 13, 2012February 1, 2013 | Puzzles

Elbow Room

Five points are located in an equilateral triangle with 10-inch sides (or on its perimeter). What’s the maximum distance between the two closest points?

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Connect the midpoints of the triangle’s sides to make four smaller triangles. Because there are five points, two of them must fall within one of these triangles. The maximum distance between these two is 5 inches.

April 12, 2012February 1, 2013 | Puzzles

Hocus Pocus

A magician invites me to do the following:

Open a brand-new deck of playing cards and place it face down on the table.
Count off the top 20 cards, turn them face up, and insert them at random into the remainder of the deck.
Shuffle the deck thoroughly.
Count off the top 20 cards and hand them to him under the table.

Now, after a period of mysterious activity under the table, he places his pack of 20 cards next to mine of 32. Using the dark arts, he says, he has arranged that both packs now contain the same number of face-up cards. We count, and they do. How has he accomplished this?

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Simply by turning over the whole pack of 20 cards. This guarantees that both packs will contain the same number of face-up cards. See Lincoln Seeks Equality.

April 6, 2012February 1, 2013 | Puzzles

Black and White

From Benjamin Glover Laws, The Two-Move Chess Problem, 1890. White to mate in two moves.

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1. Ra1 Kxg6 2. Qb1#

April 5, 2012February 1, 2013 | Puzzles

Foot Paths

I have just moved to an island on which every intersection is the meeting of three two-way streets. On a lark I decide to go running, turning right at the first intersection, left at the next, and alternating in this way to decide my route. Prove that eventually I’ll return home.

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On any given street my state can be described by the street name, the direction I’m traveling on it, and the direction of my last turn (left or right). That means that each step determines that one that precedes it: If I’m running west on Primrose Lane having just turned right from Willow Avenue, then on the preceding step I was running on Willow Avenue toward Primrose Lane having just turned left, and so on backward. Since the island contains a finite number of such triples, I must eventually hit a triple for the second time. In most cases this implies a contradiction: Primrose Lane can’t be the first repeated triple, because I could only reach that triple via Willow Avenue, which would make Willow the first repeated triple. The only triple that doesn’t produce such a contradiction is the very first one … where my house is.

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April 3, 2012February 1, 2013 | Puzzles

Moving Violation

I want to take my one-piece bass flute on the train, but it’s 1.7 meters long and the baggage man won’t take any item whose greatest dimension exceeds 1 meter. What can I do?

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It will fit diagonally in a cubical box 1 meter on a side.

March 30, 2012February 1, 2013 | Puzzles

Black and White

From Joseph Kling, The Chess Euclid, 1849. White to mate in two moves.

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1. Rg6 and the queen will mate.

March 29, 2012February 1, 2013 | Puzzles

Uptown Girl

A man has two girlfriends, one who lives uptown and the other downtown. He likes them equally, so he lets the trains decide which he will visit: He arrives at the train station at random times and takes whichever train arrives first.

Over time, he finds that he’s visiting the uptown girlfriend much more often than the downtown girlfriend, even though uptown and downtown trains arrive at the station equally often. Why?

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Unless the uptown and downtown trains arrive at the station at equal intervals, the man will find himself favoring one or the other. For example, if the the uptown train arrives at 1:00, 1:10, 1:20, etc., and the downtown train arrives at 1:01, 1:11, 1:21, etc., then he’ll have a 9 times greater chance of taking an uptown train. See The Elevator Paradox.

March 27, 2012February 1, 2013 | Puzzles

Water Battle

An odd number of people armed with water guns are standing in a field so that all the pairwise distances are distinct. At a signal, each shoots at his nearest neighbor and hits him. Prove that one person doesn’t get wet.

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Consider the two people separated by the shortest distance. At the signal, these two will shoot one another. Now: If another player has also shot at one of these two, then we’ve established that one player was shot twice, and hence that some other player (in the whole group) lacked an antagonist and stayed dry. If no other player shot at these two, then we can remove them from consideration and apply the same reasoning to the remaining group: The two closest neighbors fired on one another; if another player fired on them then we have our proof; if not then remove the pair and continue. Because we’re removing players in pairs, the group under consideration always contains an odd number of players. But it can’t drop all the way to a single player, as he would be firing on no one, and we’ve said that every player fires on someone. Hence at some point we must discover a player who is fired on twice, and thus prove that some other player was not fired on at all. From the 1987 Canadian Mathematical Olympiad, based on a problem from the 6th All Soviet Union Mathematical Competition in Voronezh, 1966.

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March 26, 2012May 6, 2013 | Puzzles