2012 AMC 10A Problems/Problem 10

The following problem is from both the 2012 AMC 12A #7 and 2012 AMC 10A #10, so both problems redirect to this page.

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 1

Let $a_1$ be the first term of the arithmetic progression and $a_{12}$ be the last term of the arithmetic progression. From the formula of the sum of an arithmetic progression (or arithmetic series), we have $a_1 + a_{12} = 60$ . $a_{12}$ , the largest term of the progression, can also be expressed as $a_1+11d$ , where $d$ is the common difference. Since each angle measure must be an integer, d must also be an integer. We can isolate $d$ by subtracting $a_1$ from $a_{12}$ like so: $a_{12}-a_1=a_1+11d-a_1=11d$ . Since d is an integer, the difference between the first and last terms, $11d$ , must be divisible by $11.$ In conjunction with the knowledge that the first and last terms add up to $60$ , we can start checking the options for the first term, $a_1$ , that we are given, starting with the smallest term. A: $a_{12}=60-5=55$ . $a_{12}-a_1=55-5=50$ . $50$ is not divisible by $11$ . B: $a_{12}=60-6=54$ . $a_{12}-a_1=54-6=48$ . $48$ is also not divisible by $11$ . $a_{12}=60-8=52$ . $a_{12}-a_1=52-8=44$ . $44$ is divisible by $11$ , so the answer is $\boxed{\textbf{(C)}\ 8}$

-Solution by Rhiju

Solution 2

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 $\frac{60-11(4)}{2} = \boxed{\textbf{(C)}\ 8}$

Solution 3

Starting with the smallest term, $a - 5x \cdots a, a + x \cdots a + 6x$ where $a$ is the sixth term and $x$ is the difference. The sum becomes $12a + 6x = 360$ since there are $360$ degrees in the central angle of the circle. The only condition left is that the smallest term in greater than zero. Therefore, $a - 5x > 0$ . $2a + x = 60$ $x = 60 - 2a$ $a - 5(60 - 2a) > 0$ $11a > 300$ Since $a$ is an integer, it must be $28$ , and therefore, $x$ is $4$ . $a - 5x$ is $\boxed{\textbf{(C)}\ 8}$

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

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2012 AMC 10A Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also