THE PENTAGON AND SQUARE.
(
Various Dissection Puzzles)
I wonder how many of my readers, amongst those who have not given any
close attention to the elements of geometry, could draw a regular
pentagon, or five-sided figure, if they suddenly required to do so. A
regular hexagon, or six-sided figure, is easy enough, for everybody
knows that all you have to do is to describe a circle and then, taking
the radius as the length of one of the sides, mark off the six points
round the circumference. But a pentagon is quite another matter. So, as
my puzzle has to do with the cutting up of a regular pentagon, it will
perhaps be well if I first show my less experienced readers how this
figure is to be correctly drawn. Describe a circle and draw the two
lines H B and D G, in the diagram, through the centre at right angles.
Now find the point A, midway between C and B. Next place the point of
your compasses at A and with the distance A D describe the arc cutting H
B at E. Then place the point of your compasses at D and with the
distance D E describe the arc cutting the circumference at F. Now, D F
is one of the sides of your pentagon, and you have simply to mark off
the other sides round the circle. Quite simple when you know how, but
otherwise somewhat of a poser.
Having formed your pentagon, the puzzle is to cut it into the fewest
possible pieces that will fit together and form a perfect square.
Answer:
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 11/4 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+3/10. So we can only hope to solve the puzzle by geometrical methods.
Informational
