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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.)
 PD can't be 9. Correct Restrictions on PD. | Comment 9 of 23 |

After a lot of math, I found that if we were to keep everything the same, PD has to be between sqrt(19) and 8. I'll go through every single step, which will solve the problem too.

Let d be the length of the cube and PD=L.

A(d,d,0)
B(0,0,0)
C(d,0,d)
D(0,d,d)
P(x,y,z)

Applying the distances

(x-d)^2+(y-d)^2+z^2=9        [1]
x^2+y^2+z^2=25                 [2]
(x-d)^2+y^2+(z-d)^2=49      [3]
x^2+(y-d)^2+(z-d)^2=L^2    [4]

Take [2]-[1], [2]-[3], and [2]-[4]. Then divide by 2.

xd+yd-d^2=8                     [5]
xd+zd-d^2=-12                  [6]
yd+zd-d^2=(25-L^2)/2       [7]

Then take [6]-[7], [5]-[7], and [5]-[6].

xd-yd=-12-(25-L^2)/2        [8]
xd-zd=8-(25-L^2)/2           [9]
yd-zd=20                          [10]

Then [5]+[8], [7]+[10], and [6]-[9].

2xd-d^2=-4-(25-L^2)/2      [11]
2yd-d^2=20+(25-L^2)/2    [12]
2zd-d^2=-20+(25-L^2)/2   [13]

Solving for x, y, and z.

x=[d^2-4-(25-L^2)/2]/(2d)      [14]
y=[d^2+20+(25-L^2)/2]/(2d)   [15]
z=[d^2-20+(25-L^2)/2]/(2d)    [16]

Then plug it back into [2] And simplify to get:

3d^4-(83+L^2)d^2+3*[(25-L^2)/2]^2+916-4L^2=0 [17]

Looking at the discriminant b^2-4ac:

D=(83+L^2)^2-4*3*(3*[(25-L^2)/2]^2+916-4L^2)
=-8L^4+83L^2-1216
=-(L^2-64)(L^2-19)>=0

--> 19<=L^2<=64 (since L>0)
sqrt(19)<=L<=8.

If we're given an L in this range, we can solve for d from [17]. We can then plug d into [14], [15], and [16] to get x, y, and z.

I've checked the algebra numerous times. This should be correct.

 Posted by np_rt on 2004-07-18 18:35:48

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