Difference between revisions of "2012 AMC 10A Problems/Problem 10"

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== Problem 10 ==
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{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #7]] and [[2012 AMC 10A Problems|2012 AMC 10A #10]]}}
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== Problem ==
  
 
Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
 
Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
  
 
<math> \textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12 </math>
 
<math> \textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12 </math>
== Solution ==
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[[Category: Introductory Geometry Problems]]
When you say the smallest sector is A and the common difference is D then you have, adding the angles together, you get A+A+D+A+2D....+A+11D. This gives 12A+66D. 12A+66D must equal 360 degrees. 12A+66D=360 because a full circle is 360 degrees. When you divide by 6 it gives A+11D=60. To get the smallest A you must have the largest D. The largest D can be is 5 so A+55=60.This means A is 5.
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== Solution 1==
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If we let <math>a</math> be the smallest sector angle and <math>r</math> be the difference between consecutive sector angles, then we have the angles <math>a, a+r, a+2r, \cdots. a+11r</math>. Use the formula for the sum of an arithmetic sequence and set it equal to 360, the number of degrees in a circle.
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<cmath>\begin{align*} \frac{a+a+11r}{2}\cdot 12 &= 360\\
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2a+11r &= 60\\
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a &= \frac{60-11r}{2} \end{align*}</cmath>
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All sector angles are integers so <math>r</math> must be a multiple of 2. Plug in even integers for <math>r</math> starting from 2 to minimize <math>a.</math> We find this value to be 4 and the minimum value of <math>a</math> to be <math>\frac{60-11(4)}{2} = \boxed{\textbf{(C)}\ 8}</math>
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== Solution 2==
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Starting with the smallest term, <math>a - 5x \cdots a, a + x \cdots a + 6x</math> where <math>a</math> is the sixth term and <math>x</math> is the difference. The sum becomes <math>12a + 6x = 360</math> since there are <math>360</math> degrees in the central angle of the circle. The only condition left is that the smallest term in greater than zero. Therefore, <math>a - 5x > 0</math>.
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<cmath>2a + x = 60</cmath>
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<cmath>x = 60 - 2a</cmath>
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<cmath>a - 5(60 - 2a) > 0</cmath>
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<cmath>11a > 300</cmath>
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Since <math>a</math> is an integer, it must be <math>28</math>, and therefore, <math>x</math> is <math>4</math>. <math>a - 5x</math> is <math>\boxed{\textbf{(C)}\ 8}</math>
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== See Also ==
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{{AMC10 box|year=2012|ab=A|num-b=9|num-a=11}}
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{{AMC12 box|year=2012|ab=A|num-b=6|num-a=8}}
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{{MAA Notice}}

Revision as of 02:30, 10 February 2017

The following problem is from both the 2012 AMC 12A #7 and 2012 AMC 10A #10, so both problems redirect to this page.

Problem

Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$

Solution 1

If we let $a$ be the smallest sector angle and $r$ be the difference between consecutive sector angles, then we have the angles $a, a+r, a+2r, \cdots. a+11r$. Use the formula for the sum of an arithmetic sequence and set it equal to 360, the number of degrees in a circle.

\begin{align*} \frac{a+a+11r}{2}\cdot 12 &= 360\\ 2a+11r &= 60\\ a &= \frac{60-11r}{2} \end{align*}

All sector angles are integers so $r$ must be a multiple of 2. Plug in even integers for $r$ starting from 2 to minimize $a.$ We find this value to be 4 and the minimum value of $a$ to be $\frac{60-11(4)}{2} = \boxed{\textbf{(C)}\ 8}$

Solution 2

Starting with the smallest term, $a - 5x \cdots a, a + x \cdots a + 6x$ where $a$ is the sixth term and $x$ is the difference. The sum becomes $12a + 6x = 360$ since there are $360$ degrees in the central angle of the circle. The only condition left is that the smallest term in greater than zero. Therefore, $a - 5x > 0$. \[2a + x = 60\] \[x = 60 - 2a\] \[a - 5(60 - 2a) > 0\] \[11a > 300\] Since $a$ is an integer, it must be $28$, and therefore, $x$ is $4$. $a - 5x$ is $\boxed{\textbf{(C)}\ 8}$

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2012 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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