AMC 10 Math Olympiad USA Math Olympiad

Sectors in Circle | AMC-10A, 2012 | Problem 10

Try this beautiful problem from Geometry: Sectors in Circle from AMC-10A, 2012. You may use sequential hints to solve the problem

Try this beautiful problem from Geometry based on Sectors in Circle.

Sectors in Circle – AMC-10A, 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?

  • \(6\)
  • \(12\)
  • \(14\)
  • \(8\)
  • \(16\)

Key Concepts




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But try the problem first…

Answer: \(8\)

Suggested Reading

AMC-10A (2012) Problem 10

Pre College Mathematics

Try with Hints

First hint

We have to find out  the degree measure of the smallest possible sector angle.Let $x$ be the smallest sector angle and $r$ be the difference between consecutive sector angles,

Therefore the angles are $x, x+r, a+2r, \cdots. x+11r$. Now we know that sum of the angles of all sectors of a circle is \(360^{\circ}\).Can you find out the values of \(x\) and \(r\)?

can you finish the problem……..

Second Hint

Therefore using the AP formula we will get ,

\(\frac{x+x+11r}{2} . 12=360\)

\(\Rightarrow x=\frac{60-11r}{2}\)

can you finish the problem……..

Final Step

Since all sector angles are integers so $r$ must be a multiple of 2. Now an even integers for $r$ starting from 2 to minimize $x.$ We find this value to be 4 and the minimum value of $x$ to be \(\frac{60-11(4)}{2}=8\)

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