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DISSECTING A MITRE.

(Various Dissection Puzzles)
The figure that is perplexing the carpenter in the illustration
represents a mitre. It will be seen that its proportions are those of a
square with one quarter removed. The puzzle is to cut it into five
pieces that will fit together and form a perfect square. I show an
attempt, published in America, to perform the feat in four pieces, based
on what is known as the "step principle," but it is a fallacy.
We are told first to cut oft the pieces 1 and 2 and pack them into the
triangular space marked off by the dotted line, and so form a rectangle.
So far, so good. Now, we are directed to apply the old step principle,
as shown, and, by moving down the piece 4 one step, form the required
square. But, unfortunately, it does _not_ produce a square: only an
oblong. Call the three long sides of the mitre 84 in. each. Then, before
cutting the steps, our rectangle in three pieces will be 84 x 63. The
steps must be 101/2 in. in height and 12 in. in breadth. Therefore, by
moving down a step we reduce by 12 in. the side 84 in. and increase by
101/2 in. the side 63 in. Hence our final rectangle must be 72 in. x 731/2
in., which certainly is not a square! The fact is, the step principle
can only be applied to rectangles with sides of particular relative
lengths. For example, if the shorter side in this case were 61+5/7
(instead of 63), then the step method would apply. For the steps would
then be 10+2/7 in. in height and 12 in. in breadth. Note that 61+5/7 x
84 = the square of 72. At present no solution has been found in four
pieces, and I do not believe one possible.


Answer:

The diagram on the next page shows how to cut into five pieces to form a
square. The dotted lines are intended to show how to find the points C
and F--the only difficulty. A B is half B D, and A E is parallel to B H.
With the point of the compasses at B describe the arc H E, and A E will
be the distance of C from B. Then F G equals B C less A B.
This puzzle--with the added condition that it shall be cut into four
parts of the same size and shape--I have not been able to trace to an
earlier date than 1835. Strictly speaking, 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.










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