is perhaps asked himself: "Can it be done in
fewer pieces?" This is just the sort of question that the true puzzle
lover is always asking, and it is the right attitude for him to adopt.
The answer to the question is that the puzzle may be solved in four
pieces--the fewest possible. This, then, is a new puzzle. Cut a square
into four pieces that will form two Greek crosses of the same size.

[Illustration: FIG. 18.]

[Illustration: FIG. 19.]

[Illustration: FIG. 20.]

The solution is very beautiful. If you divide by points the sides of the
square into three equal parts, the directions of the lines in Fig. 18
will be quite obvious. If you cut along these lines, the pieces A and B
will form the cross in Fig. 19 and the pieces C and D the similar cross
in Fig. 20. In this square we have another form of Swastika.

The reader will here appreciate the truth of my remark to the effect
that it is easier to find the directions of the cuts when transforming a
cross to a square than when converting a square into a cross. Thus, in
Figs. 6, 8, and 10 the directions of the cuts are more obvious than in
Fig. 14, where we had first to divide the sides of the square into six
equal parts, and in Fig. 18, where we divide them into three equal
parts. Then, supposing you were required to cut two equal Greek crosses,
each into two pieces, to form a square, a glance at Figs. 19 and 20 will
show how absurdly more easy this is than the reverse puzzle of cutting
the square to make two crosses.

Referring to my remarks on "fallacies," I will now give a little example
of these "solutions" that are not solutions. Some years ago a young
correspondent sent me what he evidently thought was a brilliant new
discovery--the transforming of a square into a Greek cross in four
pieces by cuts all parallel to the sides of the square. I give his
attempt in Figs. 21 and 22, where it will be seen that the four pieces
do not form a symmetrical Greek cross, because the four arms are not
really squares but oblongs. To make it a true Greek cross we should
require the additions that I have indicated with dotted lines. Of course
his solution produces a cross, but it is not the symmetrical Greek
variety required by the conditions of the puzzle. My young friend
thought his attempt was "near enough" to be correct; but if he bought a
penny apple with a sixpence he probably would not have thought it "near
enough" if he had been given only fourpence change. As the read