bob012345's latest activity

We need to subtract the "singular" triangles, i.e., "triangles" with their three vertices overlapping. This happens in the vertices 7...

Ok. You can figure the upper bound of total possible triangles and go down from there.

It can be approached as a combinatorics problem. Mark the lines, e.g., Every three lines that intersect, make a triangle. There are 12...

I would consider that to be automating a "brute force counting" method. IMO, although it is not what the OP asked for, it is still an...

I'd start with marking the vertices, e.g., Then, I'd make a 25x25 matrix with 1s for each pair of vertices connected by straight line...

The game is to find the number of triangles in this complicated figure by other than brute force counting and explain your method.

Congrats @fresh_42 @PeroK @martinbn @FactChecker @DaveC426913 @WWGD @Agent Smith @mathwonk @Charles Link @anuttarasammyak @pasmith...

The formula for the 2nd and the 3rd case is written as $$\frac{a^2+s^2-as(\sin\theta + \cos\theta)}{2 \sin\theta \cos\theta}$$ which is...

Interesting observation! It appears so if both values are within the range of the functions. Looking at ##(a,s)## being (4,5) vs...

The answers for the first case and the 4th case are symmetric for exchange of s and a. May we expect the same for the 2nd and the 3rd...

I worked out the overlap of the two squares as a function of angle. Given ##a## is the half edge length of the original square and...

This is my interpretation of the problem statement in the original post.

Agreed. Now the larger square is the 1m square. The overlap depends on the relative orientation until ##s## shrinks to...

$$s=\sqrt{2}/2$$is the minimum value satisfying it though not larger any more. And also $$s=\frac{1}{2\sqrt{2}} $$is the maximum.

The next level is if we let the length ##s## of the larger square vary, what range of values of ##s## will your statement not be true?