Before tackling this one, take a look at this one.

     +-------------D
    /|            /|
   / |           / |
  /  |          /  |
 /   |         /   |
C-------------+    |
|    |        |    |
|    B--------|----+
|   /         |   /
|  /          |  /
| /           | /
|/            |/
+-------------A

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.

this problem seems beyond my capabilities. but i've been thinking about it anyway.

there are two sets of four non-adjacent vertices on a cube.

i wonder if the average of the distances from the point to the vertices of either set is equal to the distance from the center of the cube to the vertices.

if so, then the distance from the center of the cube to a vertex is 6.5. which i believe makes the length of an edge 7.505

even if that IS correct, i can't go much further, and i have a very hard time picturing a point on the outside that shares the distance relationships with a point on the inside.

Posted by rixar on 2004-07-17 16:55:56