15. The direction of the cuts is pretty obvious.
It will be seen that the sides of the square in Fig. 14 are marked off
into six equal parts. The sides of the cross are found by ruling lines
from certain of these points to others.

[Illustration: FIG. 14.]

[Illustration: FIG. 15.]

I will now explain, as I promised, why a Greek cross may be cut into
four pieces in an infinite number of different ways to make a square.
Draw a cross, as in Fig. 16. Then draw on transparent paper the square
shown in Fig. 17, taking care that the distance c to d is exactly
the same as the distance a to b in the cross. Now place the
transparent paper over the cross and slide it about into different
positions, only be very careful always to keep the square at the same
angle to the cross as shown, where a b is parallel to c d. If
you place the point c exactly over a the lines will indicate the
solution (Figs. 10 and 11). If you place c in the very centre of the
dotted square, it will give the solution in Figs. 8 and 9. You will now
see that by sliding the square about so that the point c is always
within the dotted square you may get as many different solutions as you
like; because, since an infinite number of different points may
theoretically be placed within this square, there must be an infinite
number of different solutions. But the point c need not necessarily be
placed within the dotted square. It may be placed, for example, at point
e to give a solution in four pieces. Here the joins at a and f may
be as slender as you like. Yet if you once get over the edge at a or
f you no longer have a solution in four pieces. This proof will be
found both entertaining and instructive. If you do not happen to have
any transparent paper at hand, any thin paper will of course do if you
hold the two sheets against a pane of glass in the window.

[Illustration: FIG. 16.]

[Illustration: FIG. 17.]

It may have been noticed from the solutions of the puzzles that I have
given that the side of the square formed from the cross is always equal
to the distance a to b in Fig. 16. This must necessarily be so, and
I will presently try to make the point quite clear.

We will now go one step further. I have already said that the ideal
solution to a cutting-out puzzle is always that which requires the
fewest possible pieces. We have just seen that two crosses of the same
size may be cut out of a square in five pieces. The reader who
succeeded in solving th