2006 AMC 10A Problem 19

"Non-similar" means different angle sets, so we only need to count sets of three angles.

1. The middle angle. Three numbers in arithmetic progression are evenly spaced, so the middle one is their average. The angles add to $180$, so the average is $180 \div 3 = 60$. The middle angle is always $60^\circ$.
2. Name the spacing. Call the gap between neighboring angles $d$. The angles are $60 - d$, $60$, $60 + d$. For example, $d = 10$ gives $50^\circ, 60^\circ, 70^\circ$.
3. What $d$ can be. The angles must be different, so $d$ is not $0$. We can take $d$ positive (a negative $d$ just lists the same three angles in the other order). The smallest angle must be at least $1$, so $60 - d \ge 1$, which means $d \le 59$.
4. Count. $d$ can be $1, 2, \ldots, 59$, and each gives a different set of angles, so a different triangle. The last one, $d = 59$, gives $1^\circ, 60^\circ, 119^\circ$, a real triangle.

That is $59$ triangles.

The answer is $\boxed{\textbf{(C)}\ 59}$.

For an odd number of evenly spaced values, describe them by the middle value and the gap. The middle value is fixed by the total, so only the gap is left to count. "Non-similar" tells you to count angle sets, not side lengths.

Write the three angles as $60 - d$, $60$, $60 + d$: in any three-term arithmetic progression the middle term is the average, and the average of the angles is $180/3 = 60$. So the middle angle is always $60^\circ$, and the triangle is determined by $d$.

Constraints on $d$: the angles are distinct, so $d \ne 0$, and we may take $d > 0$ (negative $d$ just lists the same three angles in the other order). All angles are positive integers, so $60 - d \ge 1$, i.e. $d \le 59$.

Thus $d = 1, 2, \dots, 59$, and each $d$ gives a different angle triple, hence a different similarity class: $59$ triangles.

The answer is $\boxed{\textbf{(C)}\ 59}$.