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

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

Revision as of 13:58, 30 October 2021

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 $180-37=\boxed{143}$.

Solution 3

The sum of the angles in a 18-gon is $(18-2) \cdot 180^\circ = 2880 ^\circ.$ Because the angles are in an arithmetic sequence, we can also write the sum of the angles as $a+(a+d)+(a+2d)+\dots+(a+17d)=18a+153d,$ where $a$ is the smallest angle and $d$ is the common difference. Since these two are equal, we know that $18a+153d = 2880 ^\circ,$ or $2a+17d = 320^\circ.$ The smallest value of $d$ that satisfies this is $d=2,$ so $a=143.$ There are other values of $d$ and $a$ that satisfy that equation, but if we tried any of them the last angle would be greater then $180,$ so the only value of $a$ that works is $a=\boxed{143}$. ~Eclipse471

Solution 4

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

Our smallest possible angle is $\boxed {143}$

~Arcticturn

See also

2011 AIME II (ProblemsAnswer KeyResources)
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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