MYSTERIES OF THE EQUILATERAL TRIANGLE - HIKARI Ltd

120 Mathematical Recreations
and n are integers. He then states and proves his main results:
Theorem 4.1. A path is cyclic if and only if θ = 90 ◦ , or if tanθ = r/ √ 3,
where r is a nonzero rational number. Moreover,
• The length of the path traveled in each cycle may be determined as follows:
Let tanθ = p/(q √ 3) where p and q are integers with no common factor,
and where p > 0, q ≥ 0. Then the length is k � 3p 2 + 9q 2 where k = 1/2
if p and q are both odd, k = 1 otherwise; except when θ = 60 ◦ and
x = 1/2 (Fagnano orbit), when the length is 3/2.
• If x �= 1/2, the shortest path length is √ 3 and it occurs when θ = 30 ◦
and θ = 90 ◦ .
• The number of bounces occurring in each cycle may be determined as follows
(when the path leads into a corner, this is counted as three bounces,
as can be justified by a limiting argument): With p, q, k as above, the
number of bounces is k[2p + min (2p, 6q)]; except when θ = 60 ◦ and
x = 1/2 (Fagnano orbit), when the number of bounces is 3.
• If x �= 1/2, the least number of bounces per cycle is 4 and it occurs when
θ = 30 ◦ and θ = 90 ◦ . With the exception of the Fagnano orbit, the
number of bounces is always even.
Finally, Knuth takes up Gardner’s gaffe where he claimed that the path
obtained by folding Figure 4.24(c) yields a periodic orbit. However, unless
θ = 60 ◦ and x = 1/2, the angle of incidence is not equal to the angle of
reflection at x. Knuth then shows that the extended path is cyclical if and only
if x is rational. Baxter and Umble [16], evidently unaware of Knuth’s earlier
pathbreaking work on equilateral triangular billiards, provide an analysis of
this problem ab initio. However, they introduce an equivalence relation on
the set of all periodic orbits where equivalent periodic orbits share the same
number of bounces, path length and incidence angles (up to permutation).
They also count the number of equivalence classes of orbits with a specified
(even) number of bounces.
Recreation 21 (Tartaglian Measuring Puzzles [62]). The following liquidpouring
puzzle is due to the Renaissance Mathematician Niccolò Fontana,
a.k.a. Tartaglia (“The Stammerer”) [118]. An eight-pint vessel is filled with
water. By means of two empty vessels that hold five and three pints respectively,
divide the eight pints evenly between the two larger vessels by pouring
water from one vessel into another. In any valid solution, you are not allowed
to estimate quantities, so that you can only stop pouring when one of the vessels
becomes either full or empty. In 1939, M. C. K. Tweedie [311] showed
how to solve this and more general pouring problems by utilizing the trajectory
of a bouncing ball upon an equilateral triangular lattice (Figure 4.25).