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THE SQUARE OF VENEER.

(Various Dissection Puzzles)
The following represents a piece of wood in my possession, 5 in. square.
By markings on the surface it is divided into twenty-five square inches.
I want to discover a way of cutting this piece of wood into the fewest
possible pieces that will fit together and form two perfect squares of
different sizes and of known dimensions. But, unfortunately, at every
one of the sixteen intersections of the cross lines a small nail has
been driven in at some time or other, and my fret-saw will be injured if
it comes in contact with any of these. I have therefore to find a method
of doing the work that will not necessitate my cutting through any of
those sixteen points. How is it to be done? Remember, the exact
dimensions of the two squares must be given.


Answer:

[Illustration:
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|---|---|--:|---|---|:--|---|--:|---|--||:--|---|---|
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| | | :| | |: | | :| | ||: | | |
| | | :| | |: | B | :| | ||: | C | |
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| | | :| | |: | | :| | ||: | | |
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| | | :| | |: | | :|| | |: | | |
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| | | :| | |: | | :|| | |: | | |
| | | :| | |: | | :|| | |: | A | |
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| | | :| | D |: | | :|| | |: | | |
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]
Any square number may be expressed as the sum of two squares in an
infinite number of different ways. The solution of the present puzzle
forms a simple demonstration of this rule. It is a condition that we
give actual dimensions.
In this puzzle I ignore the known dimensions of our square and work on
the assumption that it is 13n by 13n. The value of n we can afterwards
determine. Divide the square as shown (where the dotted lines indicate
the original markings) into 169 squares. As 169 is the sum of the two
squares 144 and 25, we will proceed to divide the veneer into two
squares, measuring respectively 12 x 12 and 5 x 5; and as we know that
two squares may be formed from one square by dissection in four pieces,
we seek a solution in this number. The dark lines in the diagram show
where the cuts are to be made. The square 5 x 5 is cut out whole, and
the larger square is formed from the remaining three pieces, B, C, and
D, which the reader can easily fit together.
Now, n is clearly 5/13 of an inch. Consequently our larger square must
be 60/13 in. x 60/13 in., and our smaller square 25/13 in. x 25/13 in.
The square of 60/13 added to the square of 25/13 is 25. The square is
thus divided into as few as four pieces that form two squares of known
dimensions, and all the sixteen nails are avoided.
Here is a general formula for finding two squares whose sum shall equal
a given square, say a squared. In the case of the solution of our puzzle p = 3,
q = 2, and a = 5.
________________________
2pqa / a squared( p squared + q squared) squared - (2pqa) squared
--------- = x; --------------------------- = y
p squared + q squared p squared + q squared
Here x squared + y squared = a squared.










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