Before tackling this one, take a look at
this one.
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A, B, C and D are nonadjacent vertices of a cube. There is a point P in space such that PA=3, PB=5, PC=7, and PD=6. Find the distance from P to the other four vertices and find the length of the edge of the cube.
There are two answers, one with P outside the cube and one with P inside the cube.
(In reply to
Some observations by Bob)
Also worth noting: the sum of the squares of the labeled sides equal the sum of the squares of the unlabeled sides. This can be found by giving the point coordinates in relation to A as (w, h, d). w² + h² + d² = 9. nw, nh, and nd represent the distances from other vertices to the point by defining them as the length of a side (n) minus the w, h, or d value (width, height, depth, instead of x, y, z. Just mix things up a bit) After writing all the equations and substituting repeatedly, you can find that the sum of the squares of the distances from P to the labeled points is equal to the sum of the squares of the distances from P to the unlabeled vertices. I'm just too lazy to do it again.

Posted by John
on 20040721 01:02:42 