2011 AIME II Problem 3

Below is the professionally curated solution for Problem 3 of the 2011 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 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:angle sumarithmetic sequenceparity

Difficulty rating: 1920

3.

The degree measures of the angles of a convex 1818-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.

Solution:

The interior angles of an 1818-gon sum to 18016=2880180 \cdot 16 = 2880 degrees. If the smallest angle is aa and the common difference is d,d, then 18a+153d=2880,18a + 153d = 2880, i.e. 2a+17d=320.2a + 17d = 320. Since aa and dd are integers, 17d17d must be even, so dd is even, and d2d \ge 2 because the sequence is increasing.

Convexity requires the largest angle a+17d=320+17d2a + 17d = \frac{320 + 17d}{2} to be less than 180,180, so 17d<4017d \lt 40 and d2.d \le 2. Thus d=2d = 2 and a=320342=143.a = \frac{320 - 34}{2} = 143.

← Problem 2Full ExamProblem 4

Problem 3 in Other Years