h ever extended a
hundred miles!

26.--_The Haberdasher's Puzzle._

[Illustration]

The illustration will show how the triangular piece of cloth may be cut
into four pieces that will fit together and form a perfect square. Bisect
AB in D and BC in E; produce the line AE to F making EF equal to EB;
bisect AF in G and describe the arc AHF; produce EB to H, and EH is the
length of the side of the required square; from E with distance EH,
describe the arc HJ, and make JK equal to BE; now, from the points D and
K drop perpendiculars on EJ at L and M. If you have done this accurately,
you will now have the required directions for the cuts.

[Illustration]

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 equilateral triangle can be cut into four pieces
that may be reassembled to form a square, with some examples of a general
method for transforming all rectilinear triangles into squares by
dissection." 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. Credit,
however, is due to Mr. C. W. M'Elroy, who alone sent me the correct
solution when I first published the problem in the _Weekly Dispatch_ in
1902.

I add an illustration showing the puzzle in a rather curious practical
form, as it was made in polished mahogany with brass hinges for use by
certain audiences. It will be seen that the four pieces form a sort of
chain, and that when they are closed up in one direction they form the
triangle, and when closed in the other direction they form the square.

27.--_The Dyer's 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.

For further solution, see No. 358 in _A. in M._

28.--_The Great Dispute between the Friar and the Sompnour._

In this little problem we attempted to show how, by sophistical
re