2010 IMO Problems/Problem 6

Problem

Let $a_1, a_2, a_3, \ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that $a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.$ Prove there exist positive integers $\ell \leq s$ and $N$ , such that $a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.$

Author: Morteza Saghafiyan, Iran

Solution

So for solving This Problem, we need to take a assumption that,

Let

$w1 = \frac{a_{1}}{1}, w2 = \frac{a_{2}}{2}, ......, ws = \frac{a_{s}}{s}$

\text{We claim the right choice of l is the one maximizing w_{l}}

Let w1 = a1 1 , w2 = a2 2 , . . . , ws = as s . (The choice of the letter w is for “weight”.) We claim the right choice of ℓ is the one maximizing wℓ . Our plan is to view each an as a linear combination of the weights w1, . . . , ws and track their coefficients. To this end, let’s define an n-type to be a vector T = ⟨t1, . . . , ts⟩ of nonnegative integers such that • n = t1 + · · · + ts; and • ti is divisible by i for every i. We then define its valuation as v(T) = Ps i=1 witi . Now we define a n-type to be valid according to the following recursive rule. For 1 ≤ n ≤ s the only valid n-types are T1 = ⟨1, 0, 0, . . . , 0⟩ T2 = ⟨0, 2, 0, . . . , 0⟩ T3 = ⟨0, 0, 3, . . . , 0⟩ . . . Ts = ⟨0, 0, 0, . . . , s⟩ for n = 1, . . . , s, respectively. Then for any n > s, an n-type is valid if it can be written as the sum of a valid k-type and a valid (n − k)-type, componentwise. These represent the linear combinations possible in the recursion; in other words the recursion in the problem is phrased as an = max T is a valid n-type v(T). In fact, we have the following description of valid n-types: Claim — Assume n > s. Then an n-type ⟨t1, . . . , ts⟩ is valid if and only if either • there exist indices i < j with i + j > s, ti ≥ i and tj ≥ j; or • there exists an index i > s/2 with ti ≥ 2i. Proof. Immediate by forwards induction on n > s that all n-types have this property. The reverse direction is by downwards induction on n. Indeed if P i ti i > 2, then we may subtract off on of {T1, . . . , Ts} while preserving the condition; and the case P i ti i = 2

2010 IMO (Problems) • Resources
Preceded by Problem 5	1 • 2 • 3 • 4 • 5 • 6	Followed by Last Question
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2010 IMO Problems/Problem 6

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