I've been wondering this for a while now.
If the base and height of a triangle is constant, what variables affect how many same squares I can fit inside a triangle without them overlapping each other?
I think the highest amount of squares can be fit if the triangle is a right triangle and the least if the triangle is obtuse.
But if the triangle is acute, does the amount of same squares I can fit inside it get affected by the angles of the triangle or subsequently its shape?
I have tried illustrating this but I couldn't draw precise lines, and I don't really need a formal proof if there even is any, so an illustration alone would be more than adequate and appreciated.
You can put just as many squares in an acute triangle as in a right triangle. The width of the triangle a given distance above the base is a constant once you have the base and altitude. In the triangles below, the base is $3$ and the width one unit above the base is $\frac {12}5$. In the bottom row you can fit the width of the triangle one unit above the base rounded down, so you can fit $\lfloor \frac {12}5 \rfloor=2$ triangles. In the next row you can fit the width of the triangle two units above the base rounded down. Here that is $\frac 95$, so you can fit $1$. In the third row you can fit $\frac 65$ rounded down, which is again $1$. This works for all acute and right triangles. The reason it fails for obtuse triangles is that the width one unit above is not over part of the base. You have to take the overlap and round it down, which may give a smaller result.