Part 78

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.

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.

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.

146.--AN EASY DISSECTION PUZZLE.

+===========+===========+- | . | . : \ | . | . : \ | . | . : \ | . | . : \ | . | . : \ +-----------+===========+===========+ | / : . | . : \ | / : . | . : \ | / : . | . : \ | / : . | . : \ | / : . | . : \ +===========+===========+===========+===========+

The solution to this puzzle is shown in the illustration. Divide the figure up into twelve equal triangles, and it is easy to discover the directions of the cuts, as indicated by the dark lines.

147.--AN EASY SQUARE PUZZLE.

+-----------------------------------------+ | . /| | . / | | . / | | / / | | / . / | | / . / | | / . / | | / ./ | | +--------------------+ | | / / | | / / | | / / | | / . / | | / . / | | / . / | | / . / | | / . | | / . | | / . | |/ . | +-----------------------------------------+

The diagram explains itself, one of the five pieces having been cut in two to form a square.

148.--THE BUN PUZZLE.

. . . . . . . |\ A . | \ . C | \ | | \ . | \ / . |_____________________\/ | | . . . B . . . . . -

. | . . | . . | . | | | D | E | | | . | . . | . . | .

_ . | . . -+- . . . . . - - . | G| F | |

- - . . . . . - - . . | . -

-+- . . - - . | H |

- - . . - -

The secret of the bun puzzle lies in the fact that, with the relative dimensions of the circles as given, the three diameters will form a right-angled triangle, as shown by A, B, C. It follows that the two smaller buns are exactly equal to the large bun. Therefore, if we give David and Edgar the two halves marked D and E, they will have their fair shares--one quarter of the confectionery each. Then if we place the small bun, H, on the top of the remaining one and trace its circumference in the manner shown, Fred's piece, F, will exactly equal Harry's small bun, H, with the addition of the piece marked G--half the rim of the other. Thus each boy gets an exactly equal share, and there are only five pieces necessary.

149.--THE CHOCOLATE SQUARES.

Square A is left entire; the two pieces marked B fit together and make a second square; the two pieces C make a third square; and the four pieces marked D will form the fourth square.

150.--DISSECTING A MITRE.

The diagram on the next page shows how to cut into five pieces to form a square. The dotted lines are intended to show how to find the points C and F--the only difficulty. A B is half B D, and A E is parallel to B H. With the point of the compasses at B describe the arc H E, and A E will be the distance of C from B. Then F G equals B C less A B.

This puzzle--with the added condition that it shall be cut into four parts of the same size and shape--I have not been able to trace to an earlier date than 1835. Strictly speaking, it is, in that form, impossible of solution; but I give the answer that is always presented, and that seems to satisfy most people.

We are asked to assume that the two portions containing the same letter--AA, BB, CC, DD--are joined by "a mere hair," and are, therefore, only one piece. To the geometrician this is absurd, and the four shares are not equal in area unless they consist of two pieces each. If you make them equal in area, they will not be exactly alike in shape.

151.--THE JOINER'S PROBLEM.

Nothing could be easier than the solution of this puzzle--when you know how to do it. And yet it is apt to perplex the novice a good deal if he wants to do it in the fewest possible pieces--three. All you have to do is to find the point A, midway between B and C, and then cut from A to D and from A to E. The three pieces then form a square in the manner shown. Of course, the proportions of the original figure must be correct; thus the triangle BEF is just a quarter of the square BCDF. Draw lines from B to D and from C to F and this will be clear.

152.--ANOTHER JOINER'S PROBLEM.