MYSTERIES OF THE EQUILATERAL TRIANGLE - HIKARI Ltd

122 Mathematical Recreations
Figure 4.26: Barrel Sharing (N = 5) [284]
among three persons so that each receives the same amount of contents and the
same number of barrels. Using equilateral triangular coordinates, D. Singmaster
[284] has shown that the solutions of this problem correspond to triangles
with integer sides and perimeter N.
Suppose that there are N barrels of each type (full, half-full and empty)
and let fi, hi, ei be the nonnegative integer number of these that the ith
person receives (i = 1, 2, 3). Then, a fair sharing is defined as one satisfying
the following conditions:
fi + hi + ei = N, fi + hi
2
+ N
2
(i = 1, 2, 3);
3�
fi =
In turn, these conditions lead to the equivalent conditions:
ei = fi, hi = N − 2fi, fi ≤ N
2
i=1
(i = 1, 2, 3);
3�
hi =
i=1
3�
ei = N.
i=1
3�
fi = N.
However, three nonnegative lengths x, y, z can form a triangle if and only
if the three triangle inequalities hold:
x + y ≥ z, y + z ≥ x, z + x ≥ y.
Setting x + y + z = p, this is equivalent to
x ≤ p p p
, y ≤ , z ≤
2 2 2 .
i=1