Problem
ALG-B2-M01-P030 Eliminating uniform objects
A collection contains rectangular cards with positive side lengths. Initially one card has both sides greater than \(1\). Two operations are allowed: replace a card with sides \(a,b\) by a card \(\frac1a,\frac1b\), or replace it by two cards \(c,b\) and \(\frac ac,b\), where \(c>0\). Prove that after finitely many operations it is impossible to obtain a collection in which every card has one side greater than \(1\) and the other less than \(1\).
Hint 1. Count cards with two large or two small sides.
Hint 2. Show that if such a card exists, then after an operation such a card remains.
Call a card uniform if both of its sides are greater than \(1\) or both are less than \(1\). Initially there is a uniform card. If a uniform card is replaced by its reciprocal, it remains uniform: large-large becomes small-small and conversely.
Now suppose a uniform card \((a,b)\) is split by the second operation. If \(a>1,b>1\), then from \(c\cdot\frac ac=a>1\), at least one of \(c,\frac ac\) is greater than \(1\); together with \(b>1\), the corresponding new card is uniform. If \(a<1,b<1\), then at least one of \(c,\frac ac\) is less than \(1\), otherwise their product would be at least \(1\); again a uniform card appears. Thus uniform cards cannot disappear completely. The desired final collection contains none, a contradiction.
A. Source analysis. Main objects: inequalities, order, an extremal element, or an invariant. The obvious first move usually gives only a local estimate. The hidden observation is to choose the right nondecreasing quantity, or to add/multiply inequalities only after signs are controlled. The needed step is an ordering, an invariant, a product transformation, or a boundary case.
F. Difficulty justification. Final level 8: this is an invariant problem where the number of objects changes.
G. Why this is not a one-step exercise. Tracking area alone is not enough; the classification of objects is decisive.