2011 AIME II Problems/Problem 3

Problem 3

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.

Solution

Solution 1

The average angle in an 18-gon is $160^\circ$ . In an arithmetic sequence the average is the same as the median, so the middle two terms of the sequence average to $160^\circ$ . Thus for some positive (the sequence is increasing and thus non-constant) integer $d$ , the middle two terms are $(160-d)^\circ$ and $(160+d)^\circ$ . Since the step is $2d$ the last term of the sequence is $(160 + 17d)^\circ$ , which must be less than $180^\circ$ , since the polygon is convex. This gives $17d < 20$ , so the only suitable positive integer $d$ is 1. The first term is then $(160-17)^\circ = \fbox{143^\circ.}$ (Error compiling LaTeX. Unknown error_msg)

Solution 2

You could also solve this problem with exterior angles. Exterior angles of any polygon add up to $360^{\circ}$ . Since there are $18$ exterior angles in an 18-gon, the average measure of an exterior angles is $\frac{360}{18}=20^\circ$ . We know from the problem that since the exterior angles must be in an arithmetic sequence, the median and average of them is $20$ . Since there are even number of exterior angles, the middle two must be $19^\circ$ and $21^\circ$ , and the difference between terms must be $2$ . Check to make sure the smallest exterior angle is greater than $0$ : $19-2(8)=19-16=3^\circ$ . It is, so the greatest exterior angle is $21+2(8)=21+16=37^\circ$ and the smallest interior angle is $180-37=\boxed{143}$ .

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2011 AIME II Problems/Problem 3

Contents

Problem 3

Solution

Solution 1

Solution 2