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Amusements in Mathematics
Henry Ernest Dudeney
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144.—THE CROSS AND THE TRIANGLE.
Cut a Greek cross into six pieces that will form an equilateral triangle. This is another
hard problem, and I will state here that a solution is practically impossible without a
previous knowledge of my method of transforming an equilateral triangle into a square
(see No. 26, "Canterbury Puzzles").
145.—THE FOLDED CROSS.
Cut out of paper a Greek cross; then so fold it that with a single straight cut of the
scissors the four pieces produced will form a square.
VARIOUS DISSECTION PUZZLES.
We will now consider a small miscellaneous selection of cutting-out puzzles, varying in
degrees of difficulty.
146.—AN EASY DISSECTION PUZZLE.
First, cut out a piece of paper or cardboard of the shape shown in the illustration. It will
be seen at once that the proportions are simply those of a square attached to half of
another similar square, divided diagonally. The puzzle is to cut it into four pieces all of
precisely the same size and shape.
147.—AN EASY SQUARE PUZZLE.
If you take a rectangular piece of cardboard, twice as long as it is broad, and cut it in half
diagonally, you will get two of the pieces shown in the illustration. The puzzle is with
five such pieces of equal size to form a square. One of the pieces may be cut in two, but
the others must be used intact.
148.—THE BUN PUZZLE.
The three circles represent three buns, and it is simply required to show how these may
be equally divided among four boys. The buns must be regarded as of equal thickness
throughout and of equal thickness to each other. Of course, they must be cut into as few
pieces as possible. To simplify it I will state the rather surprising fact that only five pieces