Page:Amusements in mathematics.djvu/184

172 It is only necessary to cut the wood or material into five pieces.

Suppose our original square to be A C L F in the above diagram and our triangle to be the shaded portion C E D. Now, we first find half the length of the long side of the triangle (C D) and measure ofE this length at A B. Then we place the triangle in its present position against the square and make two cuts—one from B to F, and the other from B to E. Strange as it may seem, that is all that is necessary! If we now remove the pieces G, H, and M to their new places, as shown in the diagram, we get the perfect square B E K F.

Take any two square pieces of paper, of different sizes but perfect squares, and cut the smaller one in half from comer to comer. Now proceed in the manner shown, and you will find that the two pieces may be combined to form a larger square by making these two simple cuts, and that no piece will be required to be turned over.

The remark that the triangle might be " a little larger or a good deal smaller in proportion " was intended to bar cases where area of triangle is greater than area of square. In such cases six pieces are necessary, and if triangle and square are of equal area there is an obvious solution in three pieces, by simply cutting the square in half diagonally.

153.—A CUTTING-OUT PUZZLE.

illustration shows how to cut the four pieces and form with them a square. First find the side of the square (the mean proportional between the length and height of the rectangle), and the method is obvious. If om: strip is exactly in the proportions 9×1, or 16×1, or 25×1, we can clearly cut it in 3, 4, or 5 rectangular pieces respectively to form a square. Excluding these special cases, the general law is that for a strip in length more than $$n^2$$ times the breadth, and not more than $$(n+1)^2$$ times the breadth, it may be cut in $$n+2$$ pieces to form a square, and there will be $$n-1$$ rectangular pieces like piece 4 in the diagram. Thus, for example, with a strip 24×1, the length is more than 16 and less than 25 times the breadth. Therefore it can be done in 6 pieces (n here being 4), 3 of which will be rectangular. In the case where n equals 1, the rectangle disappears and we get a solution in three pieces. Within these limits, of course, the sides need not be rational: the solution is purely geometrical

154.—MRS. HOBSON'S HEARTHRUG.

I gave full measurements of the mutilated rug, it was quite an easy matter to find the precise dimensions for the square. The two pieces cut off would, if placed together, make an oblong piece 12×6, giving an area of 72 (inches or yards, as we please), and as the original complete rug measured 36 × 27, it had an area of 972. If, therefore, we deduct the pieces that have been cut away, we find that om: new rug will contain 972 less 72, or 900; and as 900 is the square of 30, we know that the new rug must measure 30 × 30 to be a perfect square. This is a great help towards the solution, because we may safely conclude that the two horizontal sides measuring 30 each may be left intact.

There is a very easy way of solving the puzzle in four pieces, and also a way in three pieces that can scarcely be called difficult, but the correct answer is in only two pieces.

It will be seen that if, after the cuts are made, we insert the teeth of the piece B one tooth lower down, the two portions will fit together and form a square.

155.—THE PENTAGON AND SQUARE.

pentagon may be cut into as few as six pieces that will fit together without any turning over and form a square, as I shall show below. Hitherto the best answer has been in seven pieces—the solution produced some years ago by a foreign mathematician, Paul Busschop. We first form a parallelogram, and from that the square. The process will be seen in the diagram on the next page.

The pentagon is A B C D E. By the cut A C and the cut F M (F being the middle point between A and C, and M being the same distance from A as F) we get two pieces that may be placed in position at G H E A and form the parallelogram G H D C. We then find the mean proportional between the length HD and the height of the parallelogram. This distance we mark off from C at K, then draw C K,