Maths Olympiad Prep

Library / /71 of 520

Combinatorics Difficulty 6.4 National olympiad Find the answer

A strip of width 11 is to be divided by rectangular panels of common width 11 and denominations long a1a_1, a2a_2, a3a_3, .... . . be paved without gaps (a11a_1 \ne 1). From the second panel on, each panel is similar but not congruent to the already paved part of the strip. When the first nn slabs are laid, the length of the paved part of the strip is snsn. Given a1a_1, is there a number that is not surpassed by any sns_n? The accuracy answer has to be proven.

A number or a short expression. Spacing and $ signs are ignored.

Solution

1. Understanding the Problem:
- We have a strip of width 1.
- The strip is divided into rectangular panels of width 1 and lengths a1,a2,a3,a_1, a_2, a_3, \ldots.
- The panels are laid without gaps.
- From the second panel onwards, each panel is similar but not congruent to the already paved part of the strip.
- After laying the first nn slabs, the length of the paved part of the strip is sns_n.
- We need to determine if there is a number that is not surpassed by any sns_n.

2. Analyzing the Panels:
- The first panel has length a1a_1.
- The second panel is similar but not congruent to the first panel, meaning it is a scaled version of the first panel.
- Let the scaling factor be kk. Then the length of the second panel is ka1k \cdot a_1.
- The third panel is similar to the first two panels, so its length is k2a1k^2 \cdot a_1, and so on.

3. Generalizing the Lengths:
- The length of the nn-th panel is kn1a1k^{n-1} \cdot a_1.
- The total length of the paved part of the strip after nn panels is:
sn=a1+ka1+k2a1++kn1a1 s_n = a_1 + k \cdot a_1 + k^2 \cdot a_1 + \cdots + k^{n-1} \cdot a_1

4. Summing the Series:
- The series a1+ka1+k2a1++kn1a1a_1 + k \cdot a_1 + k^2 \cdot a_1 + \cdots + k^{n-1} \cdot a_1 is a geometric series with the first term a1a_1 and common ratio kk.
- The sum of the first nn terms of a geometric series is given by:
sn=a11kn1kfork1 s_n = a_1 \frac{1 - k^n}{1 - k} \quad \text{for} \quad k \neq 1

5. Behavior of the Series:
- If 0<k<10 < k < 1, as nn approaches infinity, knk^n approaches 0.
- Therefore, the sum sns_n approaches:
sna111k s_n \to a_1 \frac{1}{1 - k}
- This means that there is a limit to the length of the paved part of the strip, which is a111ka_1 \frac{1}{1 - k}.

6. Conclusion:
- Given a1a_1, there is indeed a number that is not surpassed by any sns_n, which is a111ka_1 \frac{1}{1 - k}.

The final answer is a111k \boxed{ a_1 \frac{1}{1 - k} } .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.