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I. In order to separate the two spheres (radius = a and b) to the maximum extent, they should rest at opposite corners of the table. The distance across the table represents the maximum separation of the vertical centerlines of the spheres. For a unit length table, this places them sqrt(2) apart.
For two spheres of radius a and b (and a = 2b) the distance between vertical centerlines “s” can be calculated as follows.
s^2 = (a + b)^2 – (a – b)^2
= a^2 +2ab + b^2 – (a^2 – 2ab + b^2)
= 4ab
s = 2sqrt(ab) [equation (1)]
Since a=2b and s=sqrt(2)
sqrt(2) = 2*sqrt(2bb)
sqrt(2) = 2b*sqrt2
2b = 1
b = ½ and a = 1
II. Placing the third sphere (radius = c) on the table so that it is touching the other two spheres and is balancing on the edge of the table adjacent to the smaller of the first two spheres, will give it the most amount of room.
Find the distance between vertical centerlines.
Using equation (1): t=2sqrt(ac) and u=2sqrt(bc)
and the right triangle t^2 = 1^2 + (1-u)^2
(2sqrt(ac))^2 = 1 + (1 – 2sqrt(bc))^2
4ac = 1 + 1 – 4sqrt(bc) + 4bc
4c = 2 – 4sqrt(c/2) + 2c
2c + 2sqrt(2c) – 2 =0
c + sqrt(2c) – 1 = 0
quadratic equation gives
c = 2 – sqrt(3) = 0.2679 |
