Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with 'In triangle ABC, AB=<math>/frac{20/11}</math>')
 
m (→‎Solution 3)
 
(22 intermediate revisions by 12 users not shown)
Line 1: Line 1:
−
In triangle ABC, AB=<math>/frac{20/11}</math>
+
==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]].
 +
 
 +
==Solution==
 +
===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> Other values of <math>d</math> and <math>a</math> satisfy that equation, but if we tried any of them the last angle would be greater than <math>180,</math> so the only value of <math>a</math> that works is <math>a=\boxed{143}</math>.
 +
 
 +
Note: The equation <math>2a+17d = 320^\circ</math> can also be obtained by using the sum of an arithmetic sequence formula <math>\frac{2a_1+(n-1)d}{2} \cdot n</math>. We set <math>n = 18</math> and equate it to 2880, thereby achieving the same result.
 +
~Eclipse471
 +
~note by cxsmi
 +
 
 +
===Solution 4===
 +
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>.
 +
 
 +
Our smallest possible angle is <math>\boxed {143}</math>
 +
 
 +
~Arcticturn
 +
 
 +
==See also==
 +
{{AIME box|year=2011|n=II|num-b=2|num-a=4}}
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 00:13, 31 January 2024

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 $a=143.$ Other values of $d$ and $a$ satisfy that equation, but if we tried any of them the last angle would be greater than $180,$ so the only value of $a$ that works is $a=\boxed{143}$.

Note: The equation $2a+17d = 320^\circ$ can also be obtained by using the sum of an arithmetic sequence formula $\frac{2a_1+(n-1)d}{2} \cdot n$. We set $n = 18$ and equate it to 2880, thereby achieving the same result. ~Eclipse471 ~note by cxsmi

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 (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_3&oldid=213879"