Problem
ALG-B2-M02-P020 Pairing game
Peter chooses \(20\) nonnegative numbers with sum \(1\). Basil partitions them into \(10\) pairs and writes down the largest product among the pairs. Peter wants this number as large as possible, Basil as small as possible. Find the value under optimal play.
Hint 1. For the lower bound, Peter can take \(\frac12\) and nineteen equal numbers.
Hint 2. For the upper bound, Basil orders the numbers and pairs extremes.
Answer: \(\frac1{76}\). Peter takes \(\frac12,\frac1{38},\ldots,\frac1{38}\). The number \(\frac12\) must be paired with \(\frac1{38}\), so the largest product is at least \(\frac1{76}\), and all other products are no larger.
Now let the numbers be arbitrary, \(x_1\ge\cdots\ge x_{20}\). Basil pairs \(x_k\) with \(x_{21-k}\). For \(k\le10\), put \(P=x_kx_{21-k}\). From \(kx_k\le x_1+\cdots+x_k=S\) and \((21-2k)x_{21-k}\le1-S\), \[P\le\frac{S(1-S)}{k(21-2k)}\le\frac1{4k(21-2k)}\le\frac1{76}.\] Thus Basil can guarantee that the maximum is at most \(\frac1{76}\). The value is proved.
A. Source analysis. Main objects: positive quantities, a product or a sum, choosing the right terms for AM-GM, and a strict equality case. The obvious first attempt usually applies AM-GM to the visible terms, but it misses the hidden structure. The hidden observation is to normalize, order, multiply, or substitute first so that the product of the chosen terms is controlled.
B. New problem. The statement is rewritten; only the method architecture is preserved.
C-D. Hints. The hint fields contain a gentle hint and a method-level hint.
E. Full solution. Given in the solution field.
F. Difficulty justification. Regional level 7: it requires strategies for both players and the AM-GM estimate \(S(1-S)\le\frac14\).
G. Why this is not a one-step exercise. Both a construction and an upper bound for arbitrary choices are required.