ALG-B2-M01-P016 Circle of positive numbers

Hint 1. The largest number cannot satisfy the rule.

Hint 2. Consider the two arcs from a minimum to the maximum.

Let \(M\) be the largest number. It cannot equal the absolute difference of its neighbours, because both neighbours are positive and at most \(M\). Hence \(M\) is the unique exception.

Choose a minimum \(d\). On one arc from \(d\) to \(M\), write \(d=a_0,a_1,\ldots,a_k=M\). By induction, \(a_{i+1}\ge a_i\), and since \(a_i=|a_{i+1}-a_{i-1}|\), in fact \(a_{i+1}=a_i+a_{i-1}\) for \(1\le i

The number \(d\) is not the exception, so \(d=|a_1-b_1|\). Suppose \(b_1>a_1\). Then \(b_1=a_1+d=a_2\). Continuing both recurrences gives \(a_0=b_0\le a_1

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. Regional level 7: it needs an extremum, two arc recurrences, and a final parity check.

G. Why this is not a one-step exercise. Knowing that the maximum is the exception is not enough; the whole cyclic structure must be used.