Page:Amusements in mathematics.djvu/187

do in Diagram 4. The circle K has one-quarter the area of the circle containing Yin and Yan, because its diameter is just one-half the length. Also L in Diagram 3 is, we know, one-quarter the area. It is therefore evident that G is exactly equal to H, and therefore half G is equal to half H. So that what F loses from L it gains from K, and F must be half of Yin or Yan.

159.—THE SQUARE OF VENEER.

Any square number may be expressed as the sum of two squares in an infinite number of different ways. The solution of the present puzzle forms a simple demonstration of this rule. It is a condition that we give actual dimensions.

In this puzzle I ignore the known dimensions of our square and work on the assumption that it is 13 by 13. The value of n we can afterwards determine. Divide the square as shown (where the dotted lines indicate the original markings) into 169 squares. As 169 is the sum of the two squares 144 and 25, we will proceed to divide the veneer into two squares, measuring respectively 12×12 and 5×5; and as we know that two squares may be formed from one square by dissection in four pieces, we seek a solution in this number. The dark lines in the diagram show where the cuts are to be made. The square 5×5 is cut out whole, and the larger square is formed from the remaining three pieces, B, C, and D, which the reader can easily fit together. Now, n is clearly 5/13 of an inch. Consequently our larger square must be 60/13 in. × 60/13 in., and our smaller square 25/13 in, × 25/13 in. The square of 60/13 added to the square of25/13 is 25. The square is thus divided into as few as four pieces that form two squares of known dimensions, and all the sixteen nails are avoided. Here is a general formula for finding two squares whose sum shall equal a given square, say $a^{2}$. In the case of the solution of our puzzle p=3, q=2, and a=5.

${\frac {2pqa}{p^{2}+q^{2}}}=x$;${\frac {{\sqrt {a^{2}(p^{2}+q^{2}}}^{2}-(2pqa)^{2}}{p^{2}+q^{2}}}=y$

Here $x^{2}+y^{2}=a^{2}$

160.—THE TWO HORSESHOES.

The puzzle was to cut the two shoes (including the hoof contained within the outlines) into four pieces, two pieces each, that would fit together and form a perfect circle. It was also stipulated that all four pieces should be different in shape. As a matter of fact, it is a puzzle based on the principle contained in that curious Chinese symbol the Monad. (See No. 158.)

The above diagrams give the correct solution to the problem. It will be noticed that 1 and 2 are cut into the required four pieces, all differ-