Difference between revisions of "2012 AMC 10A Problems/Problem 10"
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<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 | + | When you say the smallest sector is A and the common difference is D then you have, adding the angles together, 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. |
Revision as of 22:32, 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?
Solution
When you say the smallest sector is A and the common difference is D then you have, adding the angles together, 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.
