290.--BOARDS WITH AN ODD NUMBER OF SQUARES.

There are fifteen different ways of cutting the 5 × 5 board (with the

central square removed) into two pieces of the same size and shape.

Limitations of space will not allow me to give diagrams of all these,

but I will enable the reader to draw them all out for himself without

the slightest difficulty. At whatever point on the edge your cut enters,

it must always end at a point on the edge, exactly opposite in a line

through the centre of the square. Thus, if you enter at point 1 (see

Fig. 1) at the top, you must leave at point 1 at the bottom. Now, 1 and

2 are the only two really different points of entry; if we use any

others they will simply produce similar solutions. The directions of the

cuts in the following fifteen

[Illustration: Fig. 1. Fig. 2.]

solutions are indicated by the numbers on the diagram. The duplication

of the numbers can lead to no confusion, since every successive number

is contiguous to the previous one. But whichever direction you take from

the top downwards you must repeat from the bottom upwards, one direction

being an exact reflection of the other.

1, 4, 8.

1, 4, 3, 7, 8.

1, 4, 3, 7, 10, 9.

1, 4, 3, 7, 10, 6, 5, 9.

1, 4, 5, 9.

1, 4, 5, 6, 10, 9.

1, 4, 5, 6, 10, 7, 8.

2, 3, 4, 8.

2, 3, 4, 5, 9.

2, 3, 4, 5, 6, 10, 9.

2, 3, 4, 5, 6, 10, 7, 8.

2, 3, 7, 8.

2, 3, 7, 10, 9.

2, 3, 7, 10, 6, 5, 9.

2, 3, 7, 10, 6, 5, 4, 8.

It will be seen that the fourth direction (1, 4, 3, 7, 10, 6, 5, 9)

produces the solution shown in Fig. 2. The thirteenth produces the

solution given in propounding the puzzle, where the cut entered at the

side instead of at the top. The pieces, however, will be of the same

shape if turned over, which, as it was stated in the conditions, would

not constitute a different solution.

Amusements in Mathematics · 第 813 / 970 题