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 A Point and a Cube (Posted on 2004-07-16)
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: PD = 11 and not 9? - PROBLEM CORRECTION | Comment 8 of 23 |
(In reply to PD = 11 and not 9? by Bractals)

You are right.  PD=11 is too large.  PD=9 does work though.

I will petition levik to change 'PD=11' to PD='9'

I thought I had all the kinks worked out, but I still keep making change of sign errors.  Thanks for pointing that out.

 Posted by Brian Smith on 2004-07-18 12:46:40

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