Art of Problem Solving

1973 IMO Problems/Problem 6: Difference between revisions

Pf02 (talk | contribs)
mNo edit summary
Line 1: Line 1:
==Problem==
==Problem==
Let <math>a_1, a_2,\cdots, a_n</math> be <math>n</math> positive numbers, and let <math>q</math> be a given real number such that <math>0<q<1.</math> Find <math>n</math> numbers <math>b_1, b_2, \cdots, b_n</math> for which
Let <math>a_1, a_2,\cdots, a_n</math> be <math>n</math> positive numbers, and let <math>q</math> be a given real number such that <math>0<q<1.</math> Find <math>n</math> numbers <math>b_1, b_2, \cdots, b_n</math> for which


Line 7: Line 8:


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


==Solution==
==Solution==
Line 13: Line 15:


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.
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==


{{IMO box|year=1973|num-b=5|after=Last Question}}
{{IMO box|year=1973|num-b=5|after=Last Question}}
[[Category:Olympiad Geometry Problems]]
[[Category:Olympiad Agebra Problems]]
[[Category:Inequalities Problems]]

Revision as of 16:53, 18 September 2025

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
Preceded by
Problem 5
1 2 3 4 5 6 Followed by
Last Question
All IMO Problems and Solutions