What about orienting a tesseract with edge length sqrt(2) in the following manner. Let one vertex sit in the origin, and let (1,0,1,0), (-1,0,1,0), (0,1,0,1), (0,-1,0,1) be the coordinates of the four vertices that are (edge-)adjacent to the origin. Since the corresponding vectors are orthogonal and have the same length sqrt(2), this indeed gives rise to a regular tesseract. Now, consider the projection (x,y,z,w) -> (x,y). The vertex set of the tesseract is mapped to the set {-1,0,1} x {-1,0,1}, and the projection of each edge has direction (1,0) or (0,1). Thus the projection of the 1D-skeleton of the tesseract is a 2x2 grid. I hope I got things right. Just let me know if I made a mistake somewhere. Jakob Jonsson ________________________________________ Från: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] för David Wilson [davidwwilson@comcast.net] Skickat: den 22 juli 2011 14:32 Till: Math Fun Ämne: [math-fun] Geometry question Can a regular tessertact can be oriented in R^4 so that the orthogonal projection of its 1D edges onto a 2D plane have the shape of a 2 x 2 grid? +-----------+------------+ | | | | | | +-----------+------------+ | | | | | | +-----------+------------+ _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun