[Illustration:  Fig. 2.]

143.—­TWO CROSSES FROM ONE.

It will be seen that one cross is cut out entire, as A in Fig. 1, while the four pieces marked B, C, D and E form the second cross, as in Fig. 2, which will be of exactly the same size as the other.  I will leave the reader the pleasant task of discovering for himself the best way of finding the direction of the cuts.  Note that the Swastika again appears.

The difficult question now presents itself:  How are we to cut three Greek crosses from one in the fewest possible pieces?  As a matter of fact, this problem may be solved in as few as thirteen pieces; but as I know many of my readers, advanced geometricians, will be glad to have something to work on of which they are not shown the solution, I leave the mystery for the present undisclosed.

144.—­THE CROSS AND THE TRIANGLE.

The line A B in the following diagram represents the side of a square having the same area as the cross.  I have shown elsewhere, as stated, how to make a square and equilateral triangle of equal area.  I need not go, therefore, into the preliminary question of finding the dimensions of the triangle that is to equal our cross.  We will assume that we have already found this, and the question then becomes, How are we to cut up one of these into pieces that will form the other?

First draw the line A B where A and B are midway between the extremities of the two side arms.  Next make the lines D C and E F equal in length to half the side of the triangle.  Now from E and F describe with the same radius the intersecting arcs at G and draw F G. Finally make I K equal to H C and L B equal to A D. If we now draw I L, it should be parallel to F G, and all the six pieces are marked out.  These fit together and form a perfect equilateral triangle, as shown in the second diagram.  Or we might have first found the direction of the line M N in our triangle, then placed the point O over the point E in the cross and turned round the triangle over the cross until the line M N was parallel to A B. The piece 5 can then be marked off and the other pieces in succession.

[Illustration]

I have seen many attempts at a solution involving the assumption that the height of the triangle is exactly the same as the height of the cross.  This is a fallacy:  the cross will always be higher than the triangle of equal area.

145.—­THE FOLDED CROSS.

[Illustration:  FIG. 1., FIG 2.]

First fold the cross along the dotted line A B in Fig. 1.  You then have it in the form shown in Fig. 2.  Next fold it along the dotted line C D (where D is, of course, the centre of the cross), and you get the form shown in Fig. 3.  Now take your scissors and cut from G to F, and the four pieces, all of the same size and shape, will fit together and form a square, as shown in Fig. 4.

[Illustration:  FIG. 3., FIG. 4.]

146.—­AN EASY DISSECTION PUZZLE.