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 {4,5,6,…nn} of consecutive positive integers, all of whose ten-element subsets have the triangle property. What is the largest possible value of n?
Solution:
The set {4,5,9,14,23,37,60,97,157,254} is a ten-element subset of {4,5,6,…,254} that does not have the triangle property. Let N be the smallest integer for which {4,5,6,…,N} has a ten-element subset that lacks the triangle property. Let {a1​,a2​,a3​,…,a10​} be such a subset, with a1​<a2​<a3​<⋯<a10​. Because none of its three-element subsets define triangles, the following must be true:
N​≥a10​≥a9​+a8​≥(a8​+a7​)+a8​=2a8​+a7​≥2(a7​+a6​)+a7​=3a7​+2a6​≥3(a6​+a5​)+2a6​=5a6​+3a5​≥8a5​+5a4​≥13a4​+8a3​≥21a3​+13a2​≥34a2​+21a1​≥34⋅5+21⋅4=254​
Thus the largest possible value of n is N−1=253​.
The problems on this page are the property of the MAA's American Mathematics Competitions