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

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== Problem 3 ==
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==Problem==
 
The degree measures of the angles in a [[convex polygon|convex]] 18-sided polygon form an increasing [[arithmetic sequence]] with integer values. Find the degree measure of the smallest [[angle]].
 
The degree measures of the angles in a [[convex polygon|convex]] 18-sided polygon form an increasing [[arithmetic sequence]] with integer values. Find the degree measure of the smallest [[angle]].
  
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==Solution==
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== Solution ==
 
 
===Solution 1===
 
===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}.</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}.</math>
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==See also==
 
==See also==
 
{{AIME box|year=2011|n=II|num-b=2|num-a=4}}
 
{{AIME box|year=2011|n=II|num-b=2|num-a=4}}
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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:47, 9 March 2020

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.

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 $(160-17)^\circ = \fbox{143}.$

Solution 2

Another way to solve this problem would be to use 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 $180-37=\boxed{143}$.

See also

2011 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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