Difference between revisions of "2011 AIME II Problems/Problem 3"
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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>. | 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== | + | ===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>. | 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 | ~Eclipse471 |
Revision as of 15:38, 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.