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. [Illustration] 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. [Illustration] 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. [Illustration] 151.--THE JOINER'S PROBLEM. [Illustration] 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. [Illustration] THE point was to find a general rule for forming a perfect square out of another square combined with a "right-angled isosceles triangle." The triangle to which geometricians give this high-sounding name is, of course, nothing more or less than half a square that has been divided from corner to corner.