Difference between revisions of "2011 AIME II Problems/Problem 3"
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== 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. | + | 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]]. |
+ | __TOC__ | ||
== Solution == | == 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 | + | 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> |
===Solution 2=== | ===Solution 2=== | ||
− | You could also solve this problem with exterior angles. Exterior | + | You could also solve this problem with exterior angles. [[Exterior angle]]s 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>. |
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+ | ==See also== | ||
+ | {{AIME box|year=2011|n=II|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 09:33, 23 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.
Contents
[hide]Solution
Solution 1
The average angle in an 18-gon is . In an arithmetic sequence the average is the same as the median, so the middle two terms of the sequence average to . Thus for some positive (the sequence is increasing and thus non-constant) integer , the middle two terms are and . Since the step is the last term of the sequence is , which must be less than , since the polygon is convex. This gives , so the only suitable positive integer is 1. The first term is then
Solution 2
You could also solve this problem with exterior angles. Exterior angles of any polygon add up to . Since there are exterior angles in an 18-gon, the average measure of an exterior angles is . We know from the problem that since the exterior angles must be in an arithmetic sequence, the median and average of them is . Since there are even number of exterior angles, the middle two must be and , and the difference between terms must be . Check to make sure the smallest exterior angle is greater than : . It is, so the greatest exterior angle is and the smallest interior angle is .
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |