Tetrahedron Projected on a Plane

Date: 10/29/96 at 23:54:2
From: Lottie English
Subject: Geometry

How would you project a regular tetrahedron perpendicularly onto a 
plane to get the maximum area shadow?  

I have no idea how to do this problem.  I haven't had geometry since 
10th grade and I really don't know what I can use to figure it out.  
I know what a tetrahedron is: a platonic solid with 4 faces, 6 edges, 
and 4 vertices.  Please give me a hint to get me started!

Date: 12/06/96 at 16:12:21
From: Doctor Lorenzo
Subject: Re: Geometry

I can only suggest a few possibilities because it is a very hard 
problem:

1) (This WILL work, but might take you forever).  Pick a fixed 
*regular* tetrahedron with vertices at 0, A, B, and C (e.g. you can 
pick A = (1,0,0), B = (1/2, sqrt(3)/2,0), and I forget the formula for 
C.)  Now pick a general unit vector (say v= (x,y,sqrt(1-x^2-y^2))), 
and project the four vertices onto the plane through the origin 
perpendicular to v.  That is, 0 goes to 0, A goes to A - (A dot v) v, 
etc.  The four projected vertices either make a quadrilateral or a 
triangle containing the origin.  Either way, you can compute the area, 
which will be a (complicated!) function of x and y.  Using calculus, 
maximize this function.  There are probably several local maxima, so 
you should compare the areas for these local maxima to choose the 
global maximum.

2) The best direction is bound to have a lot of symmetry.  Two obvious
candidates are the line between a vertex and the center of the 
opposite face (i.e. projecting the whole tetrahedron onto one of its 
faces), or the line between the midpoint of one edge (say, 0A) and the 
midpoint of the opposite edge (BC).  Compute the area of these two 
projections, and see which is higher.  That's not a proof of anything, 
but it's probably the right answer.

-Doctor Lorenzo,  The Math Forum
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