A Problem

The Problem

To: geometry-puzzles@forum.swarthmore.edu
Date: 13 Jul 1995 17:21:50 GMT
From: sevgum+@pitt.edu (Sevket  Gumustekin)
Organization: University of Pittsburgh
Sender: geometry-puzzles-owner@forum.swarthmore.edu
Subject: Projective Geometry question

I have the following problem to solve:

We have an observation point p and plane P. The min distance between p & P is given : d.

A rectangle with sides a&b (given) has the orientation in 3D such that one of its corners is located at point O on P which is perpendicular to p (i.e d=|p-O|) .

The line drawn from p to the intersection of the diagonals of the rectangle is perpendicular to the rectangle.

Q: The rectangle is projected onto the plane P from point p . What is the angle between the sides of the projection of rectangle which intersect at O ?

Any comments ?
Thanks
SG

My Solution

A Commentary

Date: Fri, 14 Jul 1995 01:50:28 -0500 (CDT)
From: Gary Wang 
To: frankm@geom
Subject: Re: Projective Geometry question

I think the negative sign [before ab] is very natural; it indicates that the angle in the projection will always be greater than 90 degrees. This seems to make intuitive sense, since we started out with a rectangle, and the figure is always between our point p and the plane P.

As for the second part, I wouldn't worry about the sign in the sqrt ever being negative. Let M be the intersection of the two diagonals of the rectangle. Then OM = (1/2)sqrt(aa+bb). Since PM is perpendicular to the rectangle, we know that d (the hypotenuse) > OM. So

4dd > 4(1/4)(aa+bb) = aa+bb

This gives us 4dd-aa > 0 and 4dd-bb > 0, as desired.