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 log⁡1012,\log_{10} 12, log⁡1075,\log_{10} 75, and log⁡10n,\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
Small Hint:

The three lengths must satisfy the triangle inequality; two of the three conditions bound log⁡10n\log_{10} n from both sides.

Big Hint:

log⁡75−log⁡12\log 75 - \log 12 <log⁡n<\lt \log n \lt log⁡75+log⁡12,\log 75 + \log 12, so 254<n<900.\frac{25}{4} \lt n \lt 900.

Solution:

The triangle inequality requires log⁡n<log⁡12+log⁡75=log⁡900\log n \lt \log 12 + \log 75 = \log 900 and log⁡12+log⁡n>log⁡75,\log 12 + \log n \gt \log 75, that is log⁡n>log⁡75−log⁡12=log⁡254.\log n \gt \log 75 - \log 12 = \log \frac{25}{4}. The remaining inequality, log⁡75+log⁡n>log⁡12,\log 75 + \log n \gt \log 12, is automatic because n≥1n \ge 1 and 75>12.75 \gt 12.

So 254<n<900,\frac{25}{4} \lt n \lt 900, which for integers means 7≤n≤899.7 \le n \le 899. That gives 899−7+1=893899 - 7 + 1 = 893 possible values of n.n.

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