Before tackling this one, take a look at
this one.
+D
/ /
/  / 
/  / 
/  / 
C+ 
   
 B+
 /  /
 /  /
 /  /
/ /
+A
A, B, C and D are nonadjacent 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.
This was done on napkins and Winnie the Pooh paper in the Golden Horseshoe Saloon at Disneyland.
First of all, I imagined the cube on a threedimensional coordinate system with the center of the circle at (0,0,0). Since we do not know the length of an edge, each point is based on a variable k which will be half the length of a side. A, therefore is located at (k, k, k) (let's assume that a positive z is toward us), B is at (k, k, k), C is at (k, k, k), and D is at (k, k, k). Now, we do the nasty, notmuchlookedforwardto approach of creating equations.
As far as A goes, we have (xk)©÷ + (y+k)©÷ + (zk)©÷ = 9, correct? This becomes x©÷ + y©÷ + z©÷ + 3k©÷ 2kx + 2ky  2kz = 9, right? Looking at what the other equations would be, I saved some time and said T would be equal to x©÷ + y©÷ + z©÷ + 3k©÷, since that shows up in EVERY EQUATION.
Now, the equations are as follows:
2kx + 2yk  2zk + T = 9
2xk + 2yk + 2zk + T = 25
2xk  2yk  2zk + T = 49
2xk  2yk + 2zk + T = 121
Now, use the first three to solve for 2xk, 2yk, and 2zk. I found them to be 37, 17, and 29, respectively. Divide each by 2k, and you have an exact value, although still unsolved for x, y, and z.
Plug in the values for 2xk, 2yk, and 2zk into the fourth equation to get 37  17  29 + T = T  83 = 121, or T = 204 T ends up being (2499/(4k©÷)) + 3k©÷ = 204. Multiply by 4k©÷ and rearrange and you have 12k^4  816k©÷ + 2499 = 0, or a quadratic equation. k is equal to ¡î((68 ¡¾ ¡î(3791))/2). x, y, and z, respectively are 37, 17, and 29 divided by 2k or ¡î(136 ¡¾ ¡î(3791)). (Note that 2k is also the length of a side. There's part I of the answer. That gives two different values for k, thus giving different values for x, y, and z, as is explained in the problem. Use graphing software to see the relationships between the two answers)
Now, to find the distances from the other points: (k, k, k);l (k, k, k); (k, k, k); (k, k, k) **I still use k because the above number is ugly!**
To compute these distances, take the value of T (204) and either add or subtract the values of 2xk, 2yk, and 2zk, based on the OPPOSITE of the signs of the coordinates (think about writing an equation for a sphere, and you'll know why).
This creates the following distances (in the same order as above)
¡î(204 + 37  17  29) = ¡î(195)
¡î(204 37  17 + 29) = ¡î(179)
¡î(204 + 37 + 17 + 29) = ¡î(287)
¡î(204  37 + 17  29) = ¡î(155)
Those are the remaining distances. There you go. Feel free to check my numbers, though. I know the process works. Just remember to switch the signs, althoguh I'm sure that keeping them the same would yield the same result, although I'm not all that intent on trying. Oh, by the way, the second solution mentioned is inherent in the value for k, due to the plus and minus. The EXACT answers are given here. For the decimal approximations, plug it into your own calculators. I hate approximations, personally. :D

Posted by John
on 20040721 00:57:30 