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.
Are you sure that PD = 11 and not 9? With it equal to 9 I can find the distances from P to the other vertices ( still working on the edge length). With it equal to 11 I don't think there is a solution. Here is my reasoning:
Let s be the edge length and O be the vertex adjacent to vertices A, B, and C. Then we can give the five vertices the 3d coordinates O(0,0,0), A(s,0,0), B(0,s,0), C(0,0,s), and D(s,s,s). With P(x,y,z) we have
PA = 3 => (xs)^2 + y^2 + z^2 = 9
PB = 5 => x^2 + (ys)^2 + z^2 = 25
PC = 7 => x^2 + y^2 + (zs)^2 = 49
PD = 11 => (xs)^2 + (ys)^2 + (zs)^2 = 121
rearranging the first three
(xs)^2 = 9  y^2  z^2
(ys)^2 = 25  x^2  z^2
(zs)^2 = 49  x^2  y^2
summing
121 = 83  2*(x^2 + y^2 + z^2)
but, this is impossible. Do you see anything wrong with my reasoning?

Posted by Jerry
on 20040718 03:09:48 