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 Check out our web site! http://mathforum.org/dr.math/
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