Page:Amusements in mathematics.djvu/182

170 146.—AN EASY DISSECTION PUZZLE.

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.

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

148.—THE BUN PUZZLE.

secret of the bim puzzle lies in the fact that, with the relative dimensions of the circles

as given, the three diameters will form a rightangled 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 bim, 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.

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.

{{Smallcaps|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,