Hyper Area
Geometry
Level
pending
A five dimensional cube is inscribed inside a 5 dimensional sphere such that each corner of the hyper-cube is in contact with the surface of the hyper sphere. The sphere has a radius of 1 m. Find the total surface area of the hyper-cube in meters squared.
The answer is 64.
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1 solution
First things first: we need to find the side lengths of the hyper cube. We know that the distance from the centre of the cube to the surface of the circle is 1, we need to express this as 5 translations in perpendicular space, each of magnitude x. i.e |(x,x,x,x,x)| =1. If, for some reason, you can't imagine 5 dimensions at once, then we can eliminate them one by one. |(x,x)| by simple pythagoras is root(x^2+x^2)=root(2)x. Eliminating another dimension gives root(2x^2+x^2)=root(3)x. once all dimensions are eliminated we are left with root(5)x=1 or x=1/root(5). As (x,x,x,x,x) was the vector from the centre from the surface, the edge length of the hyper-cube is 2/root(5) making the area of each face 4/5=0.8
Now all that is left is to calculate the number of faces. We can do this by considering first the number of vertices, then the number of lines/vertex, then faces/line and the product of these divided by 8 (2 because each line has 2 ends and 4 because each face has 4 lines) is our number of faces.
Vertices =2^5 (as the number of points doubles each time)
Lines/Vertex=5 (one for each dimension (as a hube in any dimension has all perpendicular lines at a point in every dimension of the cube))
Faces/line=4 (one less than the number of dimensions)
Therefore the total number of faces is 32 5 4/8=80 and the total area =80*0.8=64