Difference between revisions of "1973 IMO Problems/Problem 6"

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==Solution==
 
==Solution==
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We notice that the constraints are linear, in the sense that if bi is a solution for ai, q, and bi' is a solution for ai', q, then for any k, k' > 0 a solution for kai + k'ai', q is kbi + k'bi'. Also a "near" solution for ah = 1, other ai = 0 is b1 = qh-1, b2 = qh-2, ... , bh-1 = q, bh = 1, bh+1 = q, ... , bn = qn-h. "Near" because the inequalities in (a) and (b) are not strict.
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However, we might reasonably hope that the inequalities would become strict in the linear combination, and indeed that is true. Define br = qr-1a1 + qr-2a2 + ... + qar-1 + ar + qar+1 + ... + qn-ran. Then we may easily verify that (a) - (c) hold.
  
 
==See Also==
 
==See Also==

Latest revision as of 09:58, 30 September 2022

Problem

Let $a_1, a_2,\cdots, a_n$ be $n$ positive numbers, and let $q$ be a given real number such that $0<q<1.$ Find $n$ numbers $b_1, b_2, \cdots, b_n$ for which

(a) $a_k<b_k$ for $k=1,2,\cdots, n,$

(b) $q<\dfrac{b_{k+1}}{b_k}<\dfrac{1}{q}$ for $k=1,2,\cdots,n-1,$

(c) $b_1+b_2+\cdots+b_n<\dfrac{1+q}{1-q}(a_1+a_2+\cdots+a_n).$

Solution

We notice that the constraints are linear, in the sense that if bi is a solution for ai, q, and bi' is a solution for ai', q, then for any k, k' > 0 a solution for kai + k'ai', q is kbi + k'bi'. Also a "near" solution for ah = 1, other ai = 0 is b1 = qh-1, b2 = qh-2, ... , bh-1 = q, bh = 1, bh+1 = q, ... , bn = qn-h. "Near" because the inequalities in (a) and (b) are not strict.

However, we might reasonably hope that the inequalities would become strict in the linear combination, and indeed that is true. Define br = qr-1a1 + qr-2a2 + ... + qar-1 + ar + qar+1 + ... + qn-ran. Then we may easily verify that (a) - (c) hold.

See Also

1973 IMO (Problems) • Resources
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