2001 AIME II Problems/Problem 5

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

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

$\mathcal{S} = \{\, 4,\, 5,\, 4+5, \,5+(4+5),\, \ldots\,\} = \{4, 5, 9, 14, 23, 37, 60, 97, 157, 254\}$

$\mathcal{S}$ is the "smallest" ten-element subset without the triangle property, and since the set $\{4, 5, 6, \ldots, 253\}$ 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 $n = \fbox{253}$ .

Note

If we wanted to find this for a much larger number (say 2001), we could have noted that this is actually a "quasi-Fibonacci" sequence with initial terms $4,5$ and built up an explicit function to find the $nth$ term. (The latter part is generally pretty annoying).

~Dhillonr25

2001 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2001 AIME II Problems/Problem 5

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Problem

Solution

Note

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