2012 AMC 10A Problem 10

Let the smallest angle be $a$ and the common difference $d \ge 0$ (both integers). The twelve angles are $a, a + d, \dots, a + 11d$, and they fill the circle:
$$12a + (0 + 1 + \cdots + 11)d = 12a + 66d = 360 \quad\Longrightarrow\quad 2a + 11d = 60.$$

To make $a$ as small as possible, make $d$ as large as possible. Since $2a$ and $60$ are even, $11d$ must be even, so $d$ is even. Also $a \ge 1$ forces $11d \le 58$, so $d \le 5$. The largest even option is $d = 4$, which gives $2a = 60 - 44 = 16$, so $a = 8$.

Check: the angles $8, 12, \dots, 52$ sum to $12 \cdot \frac{8 + 52}{2} = 360$.

The answer is $\boxed{\textbf{(C)}\ 8}$.

An arithmetic sequence with a fixed sum trades off its first term against its common difference through one linear equation. Integer constraints then enter through parity: $2a + 11d = 60$ makes $d$ even. Push the "free" parameter to its extreme and read off the answer.

• Trying d = 5 (which makes a non-integer) or stopping at d = 2 with a = 19 without maximizing the common difference.
• Using the sum formula with $11$ or $13$ terms, or writing $12a + 12 \cdot 11 d$ instead of $12a + 66d$.