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

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== Problem 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>
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== Solution ==
 
== Solution ==
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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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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(2)}{2} = \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}}

Revision as of 23:01, 8 February 2012

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

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 $\frac{60-11(2)}{2} = \boxed{\textbf{(C)}\ 8}$

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
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
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