2006 AMC 10A Problem 19

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 $\boxed{\textbf{(C)}\ 59}$.

"Non-similar" means we count angle triples, not side lengths, so the whole problem is: how many integer arithmetic progressions of three distinct positive terms sum to $180$? Parametrizing an odd-length progression by its middle term and common difference makes the sum constraint trivial and leaves a single bounded parameter to count.

• Letting the smallest angle range over $1$ to $89$ (choice (D)) without using that the middle angle is forced to be $60$, or double counting $d$ and $-d$ ($178$, choice (E)).
• Including $d = 0$ (the equilateral triangle), which violates "distinct" and would give $60$, not among the choices anyway.