2011 AIME II Problem 3

Attempt Problem 3 of the 2011 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 2011 AIME II solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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.
Answer: 143
Concepts:angle sumarithmetic sequenceparity
Small hint:
The interior angles of an 1818-gon sum to 180⋅16=2880180 \cdot 16 = 2880 degrees
Big hint:
With smallest angle aa and common difference d,d, 2a+17d=320,2a + 17d = 320, so dd is even; convexity caps the largest angle below 180180
Solution:
The interior angles of an 1818-gon sum to 180⋅16=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 d≥2d \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 d≤2.d \le 2. Thus d=2d = 2 and a=320−342=143.a = \frac{320 - 34}{2} = 143.
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