AMC 10 Step by Step

2006 AMC 10A Problem 19

P16-20~3 minPrint

How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression?

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Key insight

Three evenly spaced angles adding to 180 have middle angle 60. Write them as 60 − d, 60, 60 + d, with d from 1 to 59.

Solution

"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 180180, so the average is 180÷3=60180 \div 3 = 60. The middle angle is always 60∘60^\circ.
  2. Name the spacing. Call the gap between neighboring angles dd. The angles are 60−d60 - d, 6060, 60+d60 + d. For example, d=10d = 10 gives 50∘,60∘,70∘50^\circ, 60^\circ, 70^\circ.
  3. What dd can be. The angles must be different, so dd is not 00. We can take dd positive (a negative dd just lists the same three angles in the other order). The smallest angle must be at least 11, so 60−d≥160 - d \ge 1, which means d≤59d \le 59.
  4. Count. dd can be 1,2,…,591, 2, \ldots, 59, and each gives a different set of angles, so a different triangle. The last one, d=59d = 59, gives 1∘,60∘,119∘1^\circ, 60^\circ, 119^\circ, a real triangle.

That is 5959 triangles.

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Write the three angles as 60−d60 - d, 6060, 60+d60 + d: in any three-term arithmetic progression the middle term is the average, and the average of the angles is 180/3=60180/3 = 60. So the middle angle is always 60∘60^\circ, and the triangle is determined by dd.

Constraints on dd: the angles are distinct, so d≠0d \ne 0, and we may take d>0d > 0 (negative dd just lists the same three angles in the other order). All angles are positive integers, so 60−d≥160 - d \ge 1, i.e. d≤59d \le 59.

Thus d=1,2,…,59d = 1, 2, \dots, 59, and each dd gives a different angle triple, hence a different similarity class: 5959 triangles.

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

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.