Math question (should be easy)

[Rodion Romanovich]

Hm...

At most 3 corners can be visible from any point <=> at most 2 sides being visible from any point (this might need further proof to be completely formal). From this it follows that in no. 1, you just need to show that at most 3 corners can be visible from any point.

Perhaps a good idea could be to choose some arbitrary coordinates for square, such as (0,0) with extent (1,1). This is allowed because any change of coordinate system could be made to fix size, positioning and rotation of the square to fit any case. Then you can call the viewer coordinate (x,y), then try to prove that for all x and y in R^2, a maximum of 3 corners will be visible. Visibility could be tested by forming a projection matrix that projects points onto a one-dimensional line and a "depth coordinate" denoting distance from the point, in the matrix you would include x,y and the square coordinates you choose. Forming the projection matrix might be a bit tricky but once this is done the rest should be easy, apart from that it requires a damn lot of calculations and algebraic manipulations.

This would work, but there may be easier and less time consuming methods.

The other 2 could be solved with a somewhat similar technique.

To be honest, at first glance it doesn't seem like an easy problem Perhaps there is an easier way.