2001 AIME II Problems/Problem 5
Contents
Problem
A set of positive numbers has the triangle property if it has three distinct elements that are the lengths of the sides of a triangle whose area is positive. Consider sets 
of consecutive positive integers, all of whose ten-element subsets have the triangle property. What is the largest possible value of 
?
Solution
Out of all ten-element subsets with distinct elements that do not possess the triangle property, we want to find the one with the smallest maximum element. Call this subset 
. Without loss of generality, consider any 
with 
. 
does not possess the triangle property, so 
. We use this property to build up from the smallest possible 
and 
:
![\[\mathcal{S} = \{\, 4,\, 5,\, 4+5, \,5+(4+5),\, \ldots\,\} = \{4, 5, 9, 14, 23, 37, 60, 97, 157, 254\}\]](http://latex.artofproblemsolving.com/3/e/6/3e6880c0177a3fee7a6008da1235e8d1451d5b32.png)
is the "smallest" ten-element subset without the triangle property, and since the set 
is the largest set of consecutive integers that does not contain this subset, it is also the largest set of consecutive integers in which all ten-element subsets possess the triangle property. Thus, our answer is 
.
Note
If we wanted to find this for a much larger number (say 2001), we could have noted that this is a "quasi-Fibonacci" sequence with initial terms 
and built up an explicit function to find the 
term. (The latter part is generally pretty annoying).
~Dhillonr25 ~Minor edit by Yiyj1
See also
| 2001 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 6 | |
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