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AIME 2001 II · 第 5 题

AIME 2001 II — Problem 5

专题
Contest Math
难度
L4
来源
AIME

题目详情

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,,n}\{4, 5, 6, \ldots, n\} of consecutive positive integers, all of whose ten-element subsets have the triangle property. What is the largest possible value of nn?

解析

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 S\mathcal{S}. Without loss of generality, consider any a,b,cSa, b, c \,\in \mathcal{S} with a<b<ca < b < c. S\,\mathcal{S} does not possess the triangle property, so ca+bc \geq a + b. We use this property to build up S\mathcal{S} from the smallest possible aa and bb:

S={4,5,4+5,5+(4+5),}={4,5,9,14,23,37,60,97,157,254}\mathcal{S} = \{\, 4,\, 5,\, 4+5, \,5+(4+5),\, \ldots\,\} = \{4, 5, 9, 14, 23, 37, 60, 97, 157, 254\} S\mathcal{S} is the "smallest" ten-element subset without the triangle property, and since the set {4,5,6,,253}\{4, 5, 6, \ldots, 253\} 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 n=253n = \fbox{253}.

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 4,54,5 and built up an explicit function to find the nthnth term. (The latter part is generally pretty annoying).

~Dhillonr25 ~Minor edit by Yiyj1