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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


Geometry

Circle

AP

Check the Answer


Answer: \(8\)

AMC-10A (2012) Problem 10

Pre College Mathematics

Try with Hints


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……..

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……..

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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