2006 AIME II Problem 2

Attempt Problem 2 of the 2006 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AIME II solutions, or check the answer key.

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2.

The lengths of the sides of a triangle with positive area are log1012,\log_{10} 12, log1075,\log_{10} 75, and log10n,\log_{10} n, where nn is a positive integer. Find the number of possible values for n.n.

Answer: 893
Concepts:triangle inequalitylogarithm
Difficulty rating: 1890
Solution:

The triangle inequality requires logn<log12+log75=log900\log n \lt \log 12 + \log 75 = \log 900 and log12+logn>log75,\log 12 + \log n \gt \log 75, that is logn>log75log12=log254.\log n \gt \log 75 - \log 12 = \log \frac{25}{4}. The remaining inequality, log75+logn>log12,\log 75 + \log n \gt \log 12, is automatic because n1n \ge 1 and 75>12.75 \gt 12.

So 254<n<900,\frac{25}{4} \lt n \lt 900, which for integers means 7n899.7 \le n \le 899. That gives 8997+1=893899 - 7 + 1 = 893 possible values of n.n.

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