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

(Solution)
(Solution)
Line 5: Line 5:
 
<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 ==
 
== Solution ==
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. This gives A+11D=60. To get the smallest A you must have the largest D. D=5 so A+55=60.This means A is 5.
+
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.

Revision as of 23:33, 8 February 2012

Problem 10

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

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.