Difference between revisions of "2011 AIME II Problems/Problem 3"

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== Solution ==
 
== Solution ==
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===Solution 1===
 
The average angle in an 18-gon is <math>160^\circ</math>. In an arithmetic sequence the average is the same as the median, so the middle two terms of the sequence average to <math>160^\circ</math>. Thus for some positive (the sequence is increasing and thus non-constant) integer <math>d</math>, the middle two terms are <math>(160-d)^\circ</math> and <math>(160+d)^\circ</math>. Since the step is <math>2d</math> the last term of the sequence is <math>(160 + 17d)^\circ</math>, which must be less than <math>180^\circ</math>, since the polygon is convex. This gives <math>17d < 20</math>, so the only suitable positive integer <math>d</math> is 1. The first term is then <math>(160-17)^\circ = \fbox{143^\circ.}</math>
 
The average angle in an 18-gon is <math>160^\circ</math>. In an arithmetic sequence the average is the same as the median, so the middle two terms of the sequence average to <math>160^\circ</math>. Thus for some positive (the sequence is increasing and thus non-constant) integer <math>d</math>, the middle two terms are <math>(160-d)^\circ</math> and <math>(160+d)^\circ</math>. Since the step is <math>2d</math> the last term of the sequence is <math>(160 + 17d)^\circ</math>, which must be less than <math>180^\circ</math>, since the polygon is convex. This gives <math>17d < 20</math>, so the only suitable positive integer <math>d</math> is 1. The first term is then <math>(160-17)^\circ = \fbox{143^\circ.}</math>
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===Solution 2===
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You could also solve this problem with exterior angles. Exterior angles of any polygon add up to <math>360^{\circ}</math>. Since there are <math>18</math> exterior angles in an 18-gon, the average measure of an exterior angles is <math>\frac{360}{18}=20^\circ</math>. We know from the problem that since the exterior angles must be in an arithmetic sequence, the median and average of them is <math>20</math>. Since there are even number of exterior angles, the middle two must be <math>19^\circ</math> and <math>21^\circ</math>, and the difference between terms must be <math>2</math>. Check to make sure the smallest exterior angle is greater than <math>0</math>: <math>19-2(8)=19-16=3^\circ</math>. It is, so the greatest exterior angle is <math>21+2(8)=21+16=37^\circ</math> and the smallest interior angle is <math>180-37=\boxed{143}</math>.

Revision as of 12:07, 21 August 2011

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. ! Missing $ inserted.)

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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