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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.)
 Some observations | Comment 1 of 23

The maximum length of the face diagonal is 8. Spheres centered at A and B with radii 3 and 5, respectively, will not intersect if the face diagonal is larger than 8. Thus a side cannot be longer than (8/sqrt(2)), about 5.656854249.

Also, logically, the intersection of two spheres is a circle. The intersection of a circle and a sphere is (generally) two points, though it could be a circle or a single point in rare cases. The intersection of two points with a sphere is, well, you get the point. Assuming they intersect, either one or both points.

There's more math I don't feel like doing just this moment.

 Posted by Eric on 2004-07-16 13:05:01

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