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Difference between revisions of "1980 USAMO Problems/Problem 2"

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m (→‎Solution)
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<cmath>f(n) = f(n-1) + \frac{n-1}{2}</cmath>
 
<cmath>f(n) = f(n-1) + \frac{n-1}{2}</cmath>
 
<cmath>f(n) = \frac{{(n-1)}^2 - 2(n-1)}{4} +  \frac{n-1}{2}</cmath>
 
<cmath>f(n) = \frac{{(n-1)}^2 - 2(n-1)}{4} +  \frac{n-1}{2}</cmath>
−
\boxed{  f(n \text{even) = \frac{{(n-1)}^2}{4}}<cmath>
+
<cmath>\boxed{  f(n \text{even) = \frac{{(n-1)}^2}{4}}</cmath>
  
 
Note the ambiguous wording in the question! If the "arithmetic progression" is allowed to be a disordered subsequence, then every progression counts twice, both as an ascending progression and as a descending progression.
 
Note the ambiguous wording in the question! If the "arithmetic progression" is allowed to be a disordered subsequence, then every progression counts twice, both as an ascending progression and as a descending progression.
 
Double the expression to account for the descending versions of each triple, to obtain:
 
Double the expression to account for the descending versions of each triple, to obtain:
−
</cmath>\boxed{f(n \text{even}) = \frac{n^2 - 2n}{2}}<cmath>
+
<cmath>\boxed{f(n \text{even}) = \frac{n^2 - 2n}{2}}</cmath>
−
</cmath>\boxed{f(n \text{odd}) = \frac{{(n-1)}^2}{2}}<math></math>   
+
<cmath>\boxed{f(n \text{odd}) = \frac{{(n-1)}^2}{2}}</cmath>   
  
 
~Lopkiloinm
 
~Lopkiloinm

Revision as of 09:48, 26 March 2023

Problem

Find the maximum possible number of three term arithmetic progressions in a monotone sequence of $n$ distinct reals.

Solution

Consider the first few cases for $n$ with the entire $n$ numbers forming an arithmetic sequence \[(1, 2, 3, \ldots, n)\] If $n = 3$, there will be one ascending triplet (123). Let's only consider the ascending order for now. If $n = 4$, the first 3 numbers give 1 triplet, the addition of the 4 gives one more, for 2 in total. If $n = 5$, the first 4 numbers give 2 triplets, and the 5th number gives 2 more triplets (135 and 345). Repeating a few more times, we can quickly see that if $n$ is even, the nth number will give \[\frac{n}{2} - 1\] more triplets in addition to all the prior triplets from the first $n-1$ numbers. If $n$ is odd, the $n$th number will give \[\frac{n-1}{2}\] more triplets. Let $f(n)$ denote the total number of triplets for $n$ numbers. The above two statements are summarized as follows: If $n$ is even, \[f(n) = f(n-1) + \frac{n}{2} - 1\] If $n$ is odd, \[f(n) = f(n-1) + \frac{n-1}{2}\]

Let's obtain the closed form for when $n$ is even: \[f(n) = f(n-2) + n-2\] \[f(n) = f(n-4) + (n-2) + (n-4)\] \[f(n) = \sum_{i=1}^{n/2} n - 2i\] \[\boxed{f(n \text{odd}) = frac{n^2 - 2n}{4}}\]

Now obtain the closed form when $n$ is odd by using the previous result for when $n$ is even \[f(n) = f(n-1) + \frac{n-1}{2}\] \[f(n) = \frac{{(n-1)}^2 - 2(n-1)}{4} + \frac{n-1}{2}\] 

\[\boxed{  f(n \text{even) = \frac{{(n-1)}^2}{4}}\] (Error compiling LaTeX. Unknown error_msg)

Note the ambiguous wording in the question! If the "arithmetic progression" is allowed to be a disordered subsequence, then every progression counts twice, both as an ascending progression and as a descending progression. Double the expression to account for the descending versions of each triple, to obtain: \[\boxed{f(n \text{even}) = \frac{n^2 - 2n}{2}}\] \[\boxed{f(n \text{odd}) = \frac{{(n-1)}^2}{2}}\]

~Lopkiloinm

See Also

1980 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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