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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.
    /|            /|
   / |           / |
  /  |          /  |
 /   |         /   |
C-------------+    |
|    |        |    |
|    B--------|----+
|   /         |   /
|  /          |  /
| /           | /
|/            |/
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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My numbers were wrong. | Comment 21 of 23 |

When I did this problem previously, I failed to notice the change in PD, so here we go again.  Contrary to popular believe, one does not need to know the exact point in space we're looking at.  The question asks for distances and the length of a side.

My points in space:  P(x, y, z); A(k, -k, k); B(-k, -k, -k); C(-k, k, k); D(k, k, -k); E(-k, k, -k); F(k, k, k); G(-k, -k, k); H(k, -k, -k)

From P(x, y, z) to A(k, -k, k), the distance formula is (x - k) + (y + k) + (z - k) = x + y + z + 3k - 2xk + 2yk - 2zk = 9.  No matter which point we're talking about, the first four terms remain the same while only the signs change for 2xk, 2yk, and 2zk.  Thus, all we need to do is solve for [x + y + z + 3k], 2xk, 2yk, and 2zk and add them together differently to get the remaining distances.  Respectively, I shall refer to these variables as T, A, B, and C.

Writing out equations for the distances to each point in the form of T - A + B - C = 9, etc, we end up with four equations and four variables.  Solve the system of equations any way you want.  I found that T = 119 / 4, A = 29 / 4,  B = -51 / 4 and C = 3 / 4.  Add these together as determined by the sign of k (if finding the distance to (-k, k, -k), it would be T + A - B + C.) to find the missing distances.

To find the length of a side, use the equation T = x + y + z + 3k as your basis.  x = A / 2k, y = B / 2k, z = C / 2k.  Thus, the equation is 3k + (29 + 51 + 3)/(4 * 2k) = 119 / 4, which reduces to 3k + 3251 / 64k = 119 / 4.  Multiply by 64k and you get a quadratic.  Solve for k, then take the square root of that answer and double it to get the length of a side (k is the distance from 0 to the vertex along a given edge...this is half the length of an edge, so double it). 

I found the length of an edge to be equal to (deep breath) ((119 4(238))/6).  Pay careful attention to the parenthesis.

The remaining distances are [(101/2)], [(69/2)], [(47/2)], and [(21/2)].  Check my answers for correctness.  This was all done by hand with a bunch of screaming children around, so there are no guarantees on accuracy.  But if I'm right, those answers are exact and not approximations.

  Posted by John on 2004-07-26 16:06:24
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