SOLUTIONS
length of the side of the required square ; from E with distance
E H, describe the arc H J, and make J K equal to BE; now,
from the points D and K drop perpendiculars on E J at L and M.
If you have done this accurately you will now have the required
directions for the cuts.
I exhibited this problem before the Royal Society, at Burlington
House, on 17th May, 1905, and also at the Royal Institution in the
following month, in the more
general form :—" A New
Problem on Superposition :
a demonstration that an equi-
lateral triangle can be cut /
into four pieces that may be j
reassembled to form a square,
\
with some examples of a \
general method for trans- A
forming all rectilinear tri-
angles into squares by dis-
section." It was also issued as a challenge to the readers of the
Daily Mail
(see issues of 1st and 8th February, 1905), but though
many hundreds of attempts were sent in there was not a single
solver.
27.—
The Dyers Puzzle.
The correct answer is 18,816 different ways. The general
formula for six fleurs-de-lys for all squares greater than 2
2
is simply
this : Six times the square of the number of combinations of
n
things,
taken three at a time, where
n
represents the number of fleurs-de-lys
in the side of the square. Of course where
n
is even the remainders
in rows and columns will be even, and where
n
is odd the remainders
will be odd.
28.—
The Great Dispute between the Friar and the Sompnour.
In this little problem we attempted to show how, by sophistical
reasoning, it may apparently be proved that the diagonal of a square
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