



 
We'll use coordinates with the center at (0,0,0,0) and the vertices at (±1/2,±1/2,±1/2,±1/2) and use Pythagoras' theorem in its 4dimensional version \[d^2 = w^2 + x^2 + y^2 + z^2\] to get distances. For a vertex of the tesseract, all four coordinates differ from the center's by 1/2: \[d = \sqrt{\left[(1/2)^2 + (1/2)^2 + (1/2)^2 + (1/2)^2\right]} = 1\] (did you get that?) For edge centers, three coordinates differ: \[d = \sqrt{\left[(1/2)^2 + (1/2)^2 + (1/2)^2 \right]} = \frac{\sqrt 3}{2}\] I think you can solve now for squareface centers and cubeface centers! Good Hunting!  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 