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.

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