**Chessboard brain teasers**

The mutilated chessboard

A chessboard (8X8), and 32 dominoes. Each domino is of such size that it exactly covers two adjacent squared on the board. The 32 dominoes therefore can cover all 64 of the chessboard squares. If the two white corners are removed from the board then 30 white squares and 32 black squares remain to be covered by dominoes, so this is impossible. If the two black corners are removed instead, then 32 white squares and 30 black squares remain, so it is again impossible.

A table has a chessboard integrated into the center of the table. A round coin fits exactly inside of a square on a chessboard. The coin is dropped on the table and lands face-down, with at least part of the coin on the chessboard.

What is the probability that the coin covers the corner of four chessboard squares?

Answer

49Pi/(320+Pi) ~ .47638

The diameter of the coin is equal to the length of the side of a square on the chessboard. For part of the coin to land on the chessboard, the center of the coin must be no more than the radius of the coin from the chessboard. If the length of a side of a square on the chessboard is one, then this describes a 9 x 9 square with rounded corners. The area missing from the rounded corners is equal to the size of a chessboard square minus the size of an inscribed circle (the coin). So the size of the area that the center of the coin must land in is:

9 x 9 - (1 - Pi/4) = 80 + Pi/4

For part of the coin to cover a corner, the center of the coin must be within a circle centered over the corner that has a radius equal to the radius of the coin. The area of this circle is

(1/2)^2 x Pi = Pi/4

There are 7 x 7 = 49 corners that are shared by four squares, so the total area that the center of the coin can land in to cover the corner of four squares is:

49 * Pi/4

The probability is then:

(49Pi/4) / (80 + Pi/4)

= 49Pi / (4 x (80 + Pi/4))

= 49Pi / (320 - Pi) ~ .47638

Wheat and chessboard problem

If a chessboard were to have wheat placed upon each square such that one grain were placed on the first square, two on the second, four on the third, and so on (doubling the number of grains on each subsequent square), how many grains of wheat would be on the chessboard at the finish?

If a chessboard were to have wheat placed upon each square such that one grain were placed on the first square, two on the second, four on the third, and so on (doubling the number of grains on each subsequent square), how many grains of wheat would be on the chessboard at the finish?