Triangle Packing

This was based off of a geometry problem I saw previously and failed miserably, because is that problem I had thought that the square had side length 9 instead of 8, and the triangles had side lengths 2, 4, & sqrt(20). So that failure gave me the idea to make I slightly more difficult geometry problem in which I could use my failed attempt as an explanation.

in, not is

and a, not I

How do we prove that it is maximum? And how can it fill the whole square as triangle's area(2) isn't a factor of square's area(81)?

(I will be talking about the scaled up version which carries over seamlessly to the actual problem, the numbers are just easier to deal with) First, it doesn’t cover the whole square. It is visible along x = 9 that it isn’t covered and although less visible, there is a sliver at y = 4.5 that isn’t covered either. However it does cover an area of 80. So we know 20 is the maximum because each triangle covers an area of 4 and if you could fit 21 or more triangles in the square, that would imply that the square has an area greater than or equal to 84, which it clearly does not. Good question!

Also in the scaled up version, I believe that the triangles have area 4, since 1/2 x 2 x 4 = 4 If you are wondering what method I used to come up with the answer, i just kind of noted the upper bound (20) and the easiest ‘solution’ of just making many rectangles (16) and attempted to look for other ways using graph paper and a ruler. So it was partially a outside of the box type question which was intended to get people to think more abstractly in the way they would fill a box instead of just accepting the obvious ‘solution’ of 16 triangles. To be honest, I probably should have listed I hint that it was more than 16.