I have received what purported to be a solution in five pieces, but the method was based on the rather subtle fallacy that half the diagonal plus half the side of a pentagon equals the side of a square of the same area. I say subtle, because it is an extremely close approximation that will deceive the eye, and is quite difficult to prove inexact. I am not aware that attention has before been drawn to this curious approximation.
Another correspondent made the side of his square 1¼ of the side of the pentagon. As a matter of fact, the ratio is irrational. I calculate that if the side of the pentagon is 1—inch, foot, or anything else—the side of the square of equal area is 1.3117 nearly, or say roughly 13/10. So we can only hope to solve the puzzle by geometrical methods.
[156.—THE DISSECTED TRIANGLE.—solution]
Diagram A is our original triangle. We will say it measures 5 inches (or 5 feet) on each side. If we take off a slice at the bottom of any equilateral triangle by a cut parallel with the base, the portion that remains will always be an equilateral triangle; so we first cut off piece 1 and get a triangle 3 inches on every side. The manner of finding directions of the other cuts in A is obvious from the diagram.
Now, if we want two triangles, 1 will be one of them, and 2, 3, 4, and 5 will fit together, as in B, to form the other. If we want three equilateral triangles, 1 will be one, 4 and 5 will form the second, as in C, and 2 and 3 will form the third, as in D. In B and C the piece 5 is turned over; but there can be no objection to this, as it is not forbidden, and is in no way opposed to the nature of the puzzle.
