Math question (should be easy)

**doc_bean** · 05-27-2007, 19:14

Hey, I'm looking to prove a few things which seem almost trivial, but I'd like a formal proof anyway (so people don't spend too much time trying to find a way where it doesn't work).

All these questions apply to a plane.

1) If you're looking from one point at a square, you can see at most two sides.

2) when looking at a square and seeing two sides, the two invisible sides will converge at the point furthest away from you, or the two sides that you do see converge at the point closest to you.

3) Polar coordinates: two objects in a plane, one is closer then the other (r is stricly lower for each pont of the ontour) , if the maximum and minimum angular coordinates of the objects furthest away fall between the maximum and minimum angular coordinates of the closest object, you can't see it.

**Rodion Romanovich** · 05-27-2007, 19:56

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.

**SwordsMaster** · 05-28-2007, 11:57

Why don't you just go: "It can be seen by observation... that blah, blah" Scientists use it all the time for stuff that is obvious.

**The_Mark** · 05-28-2007, 12:05

Proof by intimidation: "It's trivial."

**doc_bean** · 05-28-2007, 12:10

Originally Posted by SwordsMaster

Why don't you just go: "It can be seen by observation... that blah, blah" Scientists use it all the time for stuff that is obvious.

Originally Posted by The_Mark

Proof by intimidation: "It's trivial."

However obvious these things migt seem, it seems people always have the tendency to try and imagine a counter example, unfortunately.

**macsen rufus** · 05-28-2007, 14:34

I assume from what you're saying you are looking at the square from the outside?

1 & 2 are definitely true, but how do you prove it mathematically....

I'd start with the fact that the square has rotational symmetry, so your viewpoint has two possible extremes: ie with a side facing you, or with a corner facing you. Any other position falls between these two. If a side faces you, perspective dictates that the opposite side will appear shorter and therefore be fully obscured, and the two other sides will be receding behind the "front" side, and also fully obscured. Therefore in this position you can see only one side of the square. In fact this face doesn't have to be exactly perpendicular to see only one face - if it is tilted either way slightly, so that the "side" is aligned such that the front and rear corner of it are in line from your perspective, it would still be invisible - ie there is a small range of angles from which only one side is visible.

With a corner facing you, by definition the corner closest to you is the point at which the two visible sides converge, and the opposite corner is the point where the invisible sides converge. You can only see the two sides on the same principle of perspective as in the first instance.

Now you can rotate the square through 45 degrees, at which one side becomes progressively foreshortened until the side is (nearly) perpendicular to you. The foreshortened side has now been obscured, and the opposite side has not emerged, which is the position of seeing only one side. Due to the rotational symmetry, there is no other viewpoint that is not represented in this 45 degree swing, therefore no other will exist at which any more than two sides will be visible. If you need to express it numerically, subtend an arc that defines the limits of the subjective "width" of the square, the further side always has a smaller arc than the nearer side (ie perspective in numbers....) when face-on.

When rotating the square, the point of the corner is always closest to you except in the ~~one~~ few circumstance(s) where the plane is (nearly) perpendicular to your line-of-sight. The "sides" of the square from your viewpoint will disappear from view at a range of angles corresponding to the arc of your angle of vision, so when you can see only one face, that face is positioned tangential to the radius from your position to the face. As soon as the corner becomes closer than the other parts of the face (ie the corner breaks into that radius), then a second side will become visible, confirming that no 2 is true.

Not sure what exactly you mean in 3 - it looks like a statement of the obvious, but if you want some other way to express it, I guess you could easily prove that the closer object intersects all radii from you to the further object.

Hope you're less confused than I am now