Difference between revisions of "2011 AIME II Problems/Problem 3"
Eclipse471 (talk | contribs) m |
|||
(10 intermediate revisions by 7 users not shown) | |||
Line 1: | Line 1: | ||
− | == Problem | + | ==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. | + | 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]]. |
− | == Solution == | + | ==Solution== |
− | 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> | + | ===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> | ||
+ | |||
+ | ===Solution 2=== | ||
+ | Another way to solve this problem would be to use 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>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | The sum of the angles in a 18-gon is <math>(18-2) \cdot 180^\circ = 2880 ^\circ.</math> Because the angles are in an arithmetic sequence, we can also write the sum of the angles as <math>a+(a+d)+(a+2d)+\dots+(a+17d)=18a+153d,</math> where <math>a</math> is the smallest angle and <math>d</math> is the common difference. Since these two are equal, we know that <math>18a+153d = 2880 ^\circ,</math> or <math>2a+17d = 320^\circ.</math> The smallest value of <math>d</math> that satisfies this is <math>d=2,</math> so <math>a=143.</math> There are other values of <math>d</math> and <math>a</math> that satisfy that equation, but if we tried any of them the last angle would be greater then <math>180,</math> so the only value of <math>a</math> that works is <math>a=\boxed{143}</math>. | ||
+ | ~Eclipse471 | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2011|n=II|num-b=2|num-a=4}} | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:38, 29 May 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 . 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
Another way to solve this problem would be to use 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 .
Solution 3
The sum of the angles in a 18-gon is Because the angles are in an arithmetic sequence, we can also write the sum of the angles as where is the smallest angle and is the common difference. Since these two are equal, we know that or The smallest value of that satisfies this is so There are other values of and that satisfy that equation, but if we tried any of them the last angle would be greater then so the only value of that works is . ~Eclipse471
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.