Black and White

agnel chess problem

By H.R. Agnel. White to mate in two moves.

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1. Bc8! Kg4 2. Re4# From Benjamin Glover Laws, The Two-Move Chess Problem, 1890.

January 6, 2015January 11, 2015 | Puzzles

Progress

The Martian parliament consists of a single house. Every member has three enemies at most among the other members. Show that it’s possible to divide the parliament into two houses so that every member has one enemy at most in his house.

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Consider all the possible divisions, and in each case count up the total number of cases E in which a member has an enemy in his own house. The division in which E is minimized must have the required property. If it didn’t, that would mean that some member has at least two enemies in his own house. In that case he’d have one enemy at most in the other house, and could reduce E further by moving to the opposite house. That’s a contradiction, so the required task must be possible.

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January 2, 2015January 11, 2015 | Puzzles

Red and Black

Take two decks of cards, minus the jokers, shuffle them together, and divide them into two piles of 52 cards. What is the probability that the number of red cards in the Pile A equals the number of black cards in Pile B? How many cards would you have to view to be certain of your answer?

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The probability is 100 percent. First suppose that Pile A is all red and Pile B all black. This fulfills the condition. Now move one red card from A to B. In order to keep 52 cards in each pile, we must also move a black card from B to A, so this new arrangement also fulfills the condition. And so on: So long as there are 52 cards of each color and 52 cards in each pile, the number of red cards in one pile must equal the number of black cards in the other. So you don’t have to view any cards at all. See Alcohol Problem.

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December 27, 2014January 4, 2015 | Puzzles

Black and White

dickinson chess problem

By S.C. Dickinson. White to mate in two moves.

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1. Nd5! Kxd5 2. Rb5# From O.A. Brownson Jr., Chess Problems, 1876.

December 26, 2014January 4, 2015 | Puzzles

Square Deal

square regions puzzle

This unit square is divided into four regions by a diagonal and a line that connects a vertex to the midpoint of an opposite side. What are the areas of the four regions?

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a + b = 1/4, b + c = 1/2, and a + d = 1/2. Triangles a and c are similar, and c has twice the linear dimensions of a, so c = 4a. That’s enough to work out the areas: a = 1/12, b = 1/6, c = 1/3, and d = 5/12. From University of Toronto mathematician Ed Barbeau’s After Math (1995).

December 15, 2014December 21, 2014 | Puzzles

Black and White

pradignat chess problem

A rather baroque problem by Émile Leonard Pradignat. White to mate in two moves.

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The surprising 1. Qxh8! leaves Black with no options. If he captures on c4 or d3 with his knight, then White recaptures with his king, giving mate. Any other response allows mate along the first rank or first file. From Prager Tagblatt, Jan. 17, 1904.

December 14, 2014December 21, 2014 | Puzzles

Black and White

maximov chess problem

By Nikolai Maximov. White to mate in two moves.

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1. Qd2!, waiting. From Deutsche Schachzeitung, 1905.

November 30, 2014December 7, 2014 | Puzzles

Sizing Up

Two lines divide this equilateral triangle into four sections. The shaded sections have the same area. What is the measure of the obtuse angle between the lines?

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Triangles CAM and ABN have the same area. Rotating triangle ABC 120° takes CAM into ABM1. Since ABN and ABM1 have the same area and both N and M1 fall on BC, N = M1. Since rotating CM produces AN, the angle between them is 120°, or 180° – 120° = 60°, which is the same. (Posed by V. Proizvolov in Math Horizons, Spring 1994.)

November 25, 2014November 30, 2014 | Puzzles

Black and White

loyd chess problem

By Sam Loyd. White to mate in two moves.

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1. Re3! wins at once: If the king takes the rook then 2. Qd2# is mate; if the pawn moves then 2. Re4#; and if the knight moves then 2. Qe5#. From Alain C. White, Sam Loyd and His Chess Problems, 1913.

November 16, 2014November 23, 2014 | Puzzles

Exact Change

To pay for 1 frognab you’ll need at least four standard U.S. coins. To pay for 2, you’ll need at least six coins. But you can buy three with two coins. How much is 1 frognab?

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Seventeen cents: 17 cents = 1 dime + 1 nickel + 2 pennies 34 cents = 1 quarter + 1 nickel + 4 pennies 51 cents = 1 half dollar + 1 penny From Oswald Jacoby’s Mathematics for Pleasure (1962). See Cash and Carry.

November 13, 2014November 23, 2014 | Puzzles