Page:Amusements in mathematics.djvu/183

Rh 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.

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.—AN0THER JOINER'S PROBLEM.

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.

The precise relative proportions of the square and triangle are of no consequence whatever.