MYSTERIES OF THE EQUILATERAL TRIANGLE - HIKARI Ltd

Mathematical Recreations 105
With reference to Figure 4.3(a): 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 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. The four resulting
numbered pieces may be reassembled to form a square as in the Figure. Note
that AD, DB, BE, JK are all equal to half the side of the triangle [97]. Also,
LJ=ME [66]. As shown in Figure 4.3(b), the four pieces can be hinged in such a
way that the resulting chain can be folded into either the square or the original
triangle. Dudeney himself displayed such a table made of polished mahogany
and brass hinges at the Royal Society in 1905 [85]. An incorrect version of this
dissection appeared as Steinhaus’ Mathematical Snapshot #2 where the base
is divided in the ratio 1:2:1 (the correct ratios are approximately 0.982:2:1.018)
[291]. This was corrected as Schoenberg’s Mathematical Time Exposure #1
[272] (independently of Dudeney).
Figure 4.4: Triangle and Square Puzzle [87]
Recreation 4 (H. E. Dudeney’s Triangle and Square Puzzle [87]). It
is required to cut each of two equilateral triangles into three pieces so that the
six pieces fit together to form a perfect square [87].
Cut one triangle in half and place the pieces together as in Figure 4.4(1).
Now cut along the dotted lines, making ab and cd each equal to the side of the
required square. Then, fit together the six pieces as in Figure 4.4(2), sliding
the pieces F and C upwards and to the left and bringing down the little piece
D from one corner to the other.
Recreation 5 (H. E. Dudeney’s Square and Triangle Puzzle [87]). It
is required to fold a perfectly square piece of paper so as to form the largest
possible equilateral triangle [87]. (See Property 2.21.)