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AIME 2011 II · 第 3 题

AIME 2011 II — Problem 3

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.

解析

Solution

Solution 1

The average angle in an 18-gon is 160160^\circ. In an arithmetic sequence the average is the same as the median, so the middle two terms of the sequence average to 160160^\circ. Thus for some positive (the sequence is increasing and thus non-constant) integer dd, the middle two terms are (160d)(160-d)^\circ and (160+d)(160+d)^\circ. Since the step is 2d2d the last term of the sequence is (160+17d)(160 + 17d)^\circ, which must be less than 180180^\circ, since the polygon is convex. This gives 17d<2017d < 20, so the only suitable positive integer dd is 1. The first term is then (16017)=143.(160-17)^\circ = \fbox{143}.

Solution 2

Another way to solve this problem would be to use exterior angles. Exterior angles of any polygon add up to 360360^{\circ}. Since there are 1818 exterior angles in an 18-gon, the average measure of an exterior angles is 36018=20\frac{360}{18}=20^\circ. We know from the problem that since the exterior angles must be in an arithmetic sequence, the median and average of them is 2020. Since there are even number of exterior angles, the middle two must be 1919^\circ and 2121^\circ, and the difference between terms must be 22. Check to make sure the smallest exterior angle is greater than 00: 192(8)=1916=319-2(8)=19-16=3^\circ. It is, so the greatest exterior angle is 21+2(8)=21+16=3721+2(8)=21+16=37^\circ and the smallest interior angle is 18037=143180-37=\boxed{143}.

Solution 3

The sum of the angles in a 18-gon is (182)180=2880.(18-2) \cdot 180^\circ = 2880 ^\circ. Because the angles are in an arithmetic sequence, we can also write the sum of the angles as a+(a+d)+(a+2d)++(a+17d)=18a+153d,a+(a+d)+(a+2d)+\dots+(a+17d)=18a+153d, where aa is the smallest angle and dd is the common difference. Since these two are equal, we know that 18a+153d=2880,18a+153d = 2880 ^\circ, or 2a+17d=320.2a+17d = 320^\circ. The smallest value of dd that satisfies this is d=2,d=2, so a=143.a=143. Other values of dd and aa satisfy that equation, but if we tried any of them the last angle would be greater than 180,180, so the only value of aa that works is a=143a=\boxed{143}.

Note: The equation 2a+17d=3202a+17d = 320^\circ can also be obtained by using the sum of an arithmetic sequence formula 2a1+(n1)d2n\frac{2a_1+(n-1)d}{2} \cdot n. We set n=18n = 18 and equate it to 2880, thereby achieving the same result. ~Eclipse471 ~note by cxsmi

Solution 4

Each individual angle in a 1818-gon is (182)18018=160\frac {(18-2) \cdot 180^\circ}{18} = 160^\circ. Since no angle in a convex polygon can be larger than 180180^\circ, the smallest angle possible is in the set 159,161,157,163,155,165,153,167,151,169,149,171,147,173,145,175,143,177159, 161, 157, 163, 155, 165, 153, 167, 151, 169, 149, 171, 147, 173, 145, 175, 143, 177.

Our smallest possible angle is 143\boxed {143}

~Arcticturn