## Haberdasher's PuzzleWeekly Dispatch in 1902 and then as problem no. 26 in The Canterbury Puzzles (1907).^{1} The accompanying diagram
shows the solution, which Dudeney describes as follows:
Bisect AB in D and BC in E; produce the line AE to F making EF equalA remarkable feature of the solution is that the each of the pieces can be hinged at one vertex, forming a chain that can be folded into the square or the original triangle. Two of the hinges bisect sides of the triangle, while the third hinge and the corner of the large piece on the base cut the base in the approximate ratio 0.982: 2: 1.018. Dudeney showed just such a model of the solution, made of polished mahogany with brass hinges, at a meeting of the Royal Society on May 17, 1905. ## Reference- Dudeney, H. E.
*The Canterbury Puzzles*. London: Nelson, 1907. Reprinted Mineola, NY: Dover, 1958.
## Related category• GAMES AND PUZZLES | ||||||

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