5 ms·
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),
by 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.
- deleted 8y ago[deleted]
- dilatedmind 8y agofor the first 8 im getting 1,3,6,14,28,60,120,248
- yantrams 8y agoJust a hunch but I think you might be skipping isosceles triangles. And possibly adding right angled triangles because of floating-point arithmetic.
- dilatedmind 8y agohow about 3,6,14,28,60,120,248,496?
- yantrams 8y agoNope :( 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 agoListed the solutions for n = 3 and 4 here. Could help you with debugging- https://imgur.com/a/fHS4tNd https://imgur.com/a/fHS4tNd