# Solution

(related to Problem: The Pentagon and Square)

A regular 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 $ABCDE.$ By the cut $AC$ and the cut $FM$ ($F$ 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 $GHEA$ and form the parallelogram $GHDC.$ 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 $CK,$ and from $G$ drop the line $GL,$ perpendicular to $KC.$ The rest is easy and rather obvious. It will be seen that the six pieces will form either the pentagon or the square.

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 $1\frac 14$ 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 $1\frac{3}{10}$. So we can only hope to solve the puzzle by geometrical methods.

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### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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