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Singapore International Mathematical Olympiad Committee National Team Selection Test

Combinatorics Difficulty 8.0 Shortlist Prove it Singapore

Let nn be an integer ≥2\ge 2. Let SS be a set of triples of nonnegative integers (a,b,c)(a, b, c) such that a+b+c=na + b + c = n and if (a,b,c)(a, b, c) and (a′,b′,c′)(a', b', c') are distinct triples in SS, then a≠a′a \ne a', b≠b′b \ne b' and c≠c′c \ne c'. Find the maximum value of ∣S∣|S|.

Solution

Suppose ∣S∣=s|S| = s. Then ∑(a,b,c)∈Sa≥∑i=0s−1i=s(s−1)2\sum_{(a, b, c) \in S} a \ge \sum_{i=0}^{s-1} i = \frac{s(s-1)}{2}. Likewise ∑Sb≥s(s−1)2\sum_S b \ge \frac{s(s-1)}{2} and ∑Sc≥s(s−1)2\sum_S c \ge \frac{s(s-1)}{2}. Therefore
3s(s−1)2≤∑Sa+b+c=ns⇒s≤⌊2n3⌋+1. \frac{3s(s-1)}{2} \le \sum_{S} a + b + c = ns \Rightarrow s \le \left\lfloor \frac{2n}{3} \right\rfloor + 1.
The following shows the bound can be attained.

n = 3k - 1n = 3kn = 3k + 1
aabbccaabbccaabbcc
0k+1k + 12k−22k - 20kk2k2k0kk2k+12k + 1
1k+2k + 22k−42k - 41k+1k + 12k−22k - 21k+1k + 12k−12k - 1
...........................
k−1k - 12k2k0kk2k2k0kk2k2k1
kk02k−12k - 1k+1k + 102k−12k - 1k+1k + 102k2k
k+1k + 112k−32k - 3k+2k + 212k−32k - 3k+2k + 212k−22k - 2
...........................
2k−12k - 1k−1k - 112k2kk−1k - 112k2kk−1k - 12

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