Talk:Divide a rectangle into a number of unequal triangles

From Rosetta Code


The idea for the task I developed from a StackOverflow question that was closed. --Paddy3118 (talk) 11:04, 18 December 2021 (UTC)

Extra Algo.

I'm guessing that you could drop a point P inside the rectangle where its distances to the corners are all different. Add random points around the outside edges, then go around the circumference in one direction and make trianges of c[i], c[i+1] and P.

You would then get triangles with edges on all four sides of the rectangle.

--Paddy3118 (talk) 10:10, 18 December 2021 (UTC)

What does random and not-similar mean?

The minimum number is three. First bisect the rectangle into two equal triangles. The problem is now reduced to splitting one of these triangles into n-1 unequal triangles, which is trivial. Is this random? The triangles are all different, are they non-similar? --Nigel Galloway (talk) 12:12, 18 December 2021 (UTC)

Similar triangles have a precise mathematical definition that is probably going too far as I think it means having the same angles, allowing rotation, flipping and scaling. Just not having the lengths of each side the same would be enough I think, so scaled triangles would not be classed as similar in this task. --Paddy3118 (talk) 17:08, 18 December 2021 (UTC)