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A Point and a Cube (Posted on 2004-07-16) Difficulty: 4 of 5
Before tackling this one, take a look at this one.
    /|            /|
   / |           / |
  /  |          /  |
 /   |         /   |
C-------------+    |
|    |        |    |
|    B--------|----+
|   /         |   /
|  /          |  /
| /           | /
|/            |/
A, B, C and D are non-adjacent 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.

No Solution Yet Submitted by Brian Smith    
Rating: 4.0000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Some observations | Comment 18 of 23 |
(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).  w + h + d = 9.  n-w, n-h, and n-d 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 2004-07-21 01:02:42

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