Q: How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression?
If someone could explain this, it would be very helpful. I have next to no idea what the question is even asking.
Ok, upon first reading it, I too have no idea what to do. So let's go through each part:
non-similar triangles: They're probably asking us to compare 2 or more triangles, but we can't assume their sides are in proportion or their angles are the same
degree measures are distinct positive integers: So here this means that all the angles are different - by using 'degree measures' what they're doing is wording it so that you find the NUMBER of the angle rather than the DEGREE [which is the same thing just that you don't have to worry about the degree symbol at the end] (obviously they're going to have to be positive because they're angles and they have to be DISTINCT)
arithmetic progression Each triangle will probably have an angle of a, a-d, a+d (this is an important tool to remember)
Also remember that the angle sum of a triangle = 180 so a+(a-d)+(a+d) = 180 .˙. a=60 degrees which would mean that one of the angles has to be 60 degrees.
I'm not entirely sure (just basing it off of the working above) but then what I'd do is see that the minimum value of d is 1 (because they must stay integers, and it can't be 0 because the angles have to change).
Then, 60 seems to be a pretty prominent number so let's do something with that. If we put in 60 for the value of d, we find that we have the value of each angle being 0, 60, 120 for the angles (we can't have that because an angle can't be 0 if we want a triangle).
So then we try 59 (because it would seem like 60 is a limit). 60, 1, 119 which works. Then since our original formulas don't really have anything tricky about them (e.g. no 0s under fractions, square roots etc.) we can conclude that the span of d can go from d=1 to d=59. Then we can see that there will be 59-1 triangles = 58 different triangles (that aren't similar).
Hopefully this is right and helps. Feel free to point out any mistakes in my working.
