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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.
     +-------------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.

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

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Question PD = 11 and not 9? | Comment 7 of 23 |

Are you sure that PD = 11 and not 9? With it equal to 9 I can find the distances from P to the other vertices ( still working on the edge length). With it equal to 11 I don't think there is a solution. Here is my reasoning:

Let s be the edge length and O be the vertex adjacent to vertices A, B, and C. Then we can give the five vertices the 3-d coordinates O(0,0,0), A(s,0,0), B(0,s,0), C(0,0,s), and D(s,s,s). With P(x,y,z) we have

PA = 3  =>  (x-s)^2 + y^2 + z^2 = 9

PB = 5  =>  x^2 + (y-s)^2 + z^2 = 25

PC = 7  =>  x^2 + y^2 + (z-s)^2 = 49

PD = 11  =>  (x-s)^2 + (y-s)^2 + (z-s)^2 = 121

                     rearranging the first three

(x-s)^2 = 9 - y^2 - z^2

(y-s)^2 = 25 - x^2 - z^2

(z-s)^2 = 49 - x^2 - y^2

                      summing

121 = 83 - 2*(x^2 + y^2 + z^2)

but, this is impossible. Do you see anything wrong with my reasoning?   


  Posted by Jerry on 2004-07-18 03:09:48
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