2006 AIME II Problem 2

Below is the professionally curated solution for Problem 2 of the 2006 AIME II, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:triangle inequalitylogarithm

Difficulty rating: 1890

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.

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 third inequality is the first one again.)

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.

← Problem 1Full ExamProblem 3

Problem 2 in Other Years