4 ms·

n=1, A=2: h b #tri #non-congruent tri 1 4 b-1 roundup(#tri/2) 2 2 b-1 roundup(#tri/2) . . . n=4, A=16: h b

n=1, A=2: h b #tri #non-congruent tri 1 4 b-1 roundup(#tri/2) 2 2 b-1 roundup(#tri/2) . . . n=4, A=16: h b #tri #non-congruent tri 1 2A/h b-1 roundup(#tri/2) 2 2A/h b-1 roundup(#tri/2) 4 ... ... 8 ... 16 ... ... summing up #non-congruent tris for each n I see n tri 1 3 2 7 3 15 4 31 so I think that answer may be 2^(n+1) - 1. Could be wrong because I forgot what congruent really means.

tylerhou 8y ago
You're undercounting some triangles and missing the acute condition. For n = 4, there is more than one triangle with base 4 and height 4, for example: [(0, 0), (4, 0), (1, 4)]; [(0, 0), (4, 0), (2, 4)]; (and [(0, 0), (4, 0), (3, 4)] but it's congruent to the first so we don't count it). Furthermore, a triangle with base 16 and height 1 might be non-acute.
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- dilatedmind 8y ago
  for the first 8 im getting 1,3,6,14,28,60,120,248
  - yantrams 8y ago
    Just a hunch but I think you might be skipping isosceles triangles. And possibly adding right angled triangles because of floating-point arithmetic.
    - dilatedmind 8y ago
      how about 3,6,14,28,60,120,248,496?
      - yantrams 8y ago
        Nope :( The sequence starts with 1 because there is just one triangle with an area of 2 that satisfies the conditions. I ended up with a sequence similar to your previous one when I forgot to take the absolute value of area -- x1(y2-y3) + x2(y3-y1) + x3*(y1-y2))/2 . That could be a possibility in case you were using the same formula. In your first result - 1, 3, 6, 14.... , first two are right, third over-counted and rest are undercounted.
      - yantrams 8y ago
        Listed the solutions for n = 3 and 4 here. Could help you with debugging- https://imgur.com/a/fHS4tNd https://imgur.com/a/fHS4tNd
yantrams 8y ago
Close enough but wrong :) I went the same route as you did and realized my mistake after a while. Base and altitude values can be fractional/irrational too. PS : I bruteforced my way to the answer.
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tylerhou 8y ago

You're undercounting some triangles and missing the acute condition. For n = 4, there is more than one triangle with base 4 and height 4, for example: [(0, 0), (4, 0), (1, 4)]; [(0, 0), (4, 0), (2, 4)]; (and [(0, 0), (4, 0), (3, 4)] but it's congruent to the first so we don't count it). Furthermore, a triangle with base 16 and height 1 might be non-acute.

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dilatedmind 8y ago
for the first 8 im getting 1,3,6,14,28,60,120,248
- yantrams 8y ago
  Just a hunch but I think you might be skipping isosceles triangles. And possibly adding right angled triangles because of floating-point arithmetic.
  - dilatedmind 8y ago
    how about 3,6,14,28,60,120,248,496?
    - yantrams 8y ago
      Nope :( The sequence starts with 1 because there is just one triangle with an area of 2 that satisfies the conditions. I ended up with a sequence similar to your previous one when I forgot to take the absolute value of area -- x1(y2-y3) + x2(y3-y1) + x3*(y1-y2))/2 . That could be a possibility in case you were using the same formula. In your first result - 1, 3, 6, 14.... , first two are right, third over-counted and rest are undercounted.
    - yantrams 8y ago
      Listed the solutions for n = 3 and 4 here. Could help you with debugging- https://imgur.com/a/fHS4tNd https://imgur.com/a/fHS4tNd

yantrams 8y ago

Close enough but wrong :) I went the same route as you did and realized my mistake after a while. Base and altitude values can be fractional/irrational too. PS : I bruteforced my way to the answer.

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