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

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Line 15: 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.
 +
 +
 +
==Remarks (added by pf02, September 2025)==
 +
 +
1.  The solution above is unacceptable for several reasons:
 +
 +
It is unacceptable to write mathematical text without using
 +
parentheses, and writing indexes and powers in the line of
 +
text, rather than use subscripts and superscripts.  On top
 +
of it, the "solution" is incomplete (the author uses "we may
 +
easily verify" to justify things which are not straightforward)).
 +
It is also imprecise (the author uses the notion of `a "near"
 +
solution...' without making it precise enough).
 +
 +
However, after a good amount of scrutiny and guessing, one
 +
can see that the text contains a good idea.  Below I will
 +
rewrite the solution in a way a reader can make sense of it.
 +
 +
2.  The author of the above solution deleted a reference to
 +
the discussion page [https://aops.com/community/p357934].
 +
On the discussion page there is an idea for another solution.
 +
(In all honesty, it is an idea, or a hint, but it can not be
 +
called a solution.)
 +
 +
3. Below I will discuss the above ideas, and give a more
 +
general class of solutions to this problem.
 +
 +
 +
==Solution 1, rewritten and completed==
 +
 +
Let <math>b_k = q^{k-1} a_1 + q^{k-2} a_2 + \dots + a_k + \dots + q^{n-k-1} a_{n-1} + q^{n-k} a_n</math>.
 +
 +
For the sake of writing this down, we assumed <math>n \ge 5</math> and
 +
<math>2 \le k \le n-2</math>.  To make things easy to follow, let us
 +
write this explicitly for <math>n = 5</math>:
 +
 +
<math>b_1 = a_1 + q a_2 + q^2 a_3 + q^3 a_4 + q^4 a_5</math>
 +
 +
<math>b_2 = q a_1 + a_2 + q a_3 + q^2 a_4 + q^3 a_5</math>
 +
 +
<math>b_3 = q^2 a_1 + q a_2 + a_3 + q a_4 + q^2 a_5</math>
 +
 +
<math>b_4 = q^3 a_1 + q^2 a_2 + q a_3 + a_4 + q a_5</math>
 +
 +
<math>b_5 = q^4 a_1 + q^3 a_2 + q^2 a_3 + q a_4 + a_5</math>
 +
 +
 +
 +
 +
 +
TO BE CONTINUED.  SAVING SO I DON'T LOSE WORK DONE SO FAR.
  
  

Revision as of 02:22, 19 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.


Remarks (added by pf02, September 2025)

1. The solution above is unacceptable for several reasons:

It is unacceptable to write mathematical text without using parentheses, and writing indexes and powers in the line of text, rather than use subscripts and superscripts. On top of it, the "solution" is incomplete (the author uses "we may easily verify" to justify things which are not straightforward)). It is also imprecise (the author uses the notion of `a "near" solution...' without making it precise enough).

However, after a good amount of scrutiny and guessing, one can see that the text contains a good idea. Below I will rewrite the solution in a way a reader can make sense of it.

2. The author of the above solution deleted a reference to the discussion page [1]. On the discussion page there is an idea for another solution. (In all honesty, it is an idea, or a hint, but it can not be called a solution.)

3. Below I will discuss the above ideas, and give a more general class of solutions to this problem.


Solution 1, rewritten and completed

Let $b_k = q^{k-1} a_1 + q^{k-2} a_2 + \dots + a_k + \dots + q^{n-k-1} a_{n-1} + q^{n-k} a_n$.

For the sake of writing this down, we assumed $n \ge 5$ and $2 \le k \le n-2$. To make things easy to follow, let us write this explicitly for $n = 5$:

$b_1 = a_1 + q a_2 + q^2 a_3 + q^3 a_4 + q^4 a_5$

$b_2 = q a_1 + a_2 + q a_3 + q^2 a_4 + q^3 a_5$

$b_3 = q^2 a_1 + q a_2 + a_3 + q a_4 + q^2 a_5$

$b_4 = q^3 a_1 + q^2 a_2 + q a_3 + a_4 + q a_5$

$b_5 = q^4 a_1 + q^3 a_2 + q^2 a_3 + q a_4 + a_5$



TO BE CONTINUED. SAVING SO I DON'T LOSE WORK DONE SO FAR.


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
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