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Perplexus Icon plus one (Posted on 2005-06-18) Difficulty: 2 of 5
A square table (a meter a side) has two spheres on its surface. The spheres have two special properties:

1. The larger is twice the diameter of the smaller, and
2. They are the largest size that will fit on the table without falling off. (They may extend over the edge of the table.)

I. What are the dimensions of the spheres?

II. A third sphere is added next to the other two. What is its largest possible size?

See The Solution Submitted by Leming    
Rating: 4.5000 (6 votes)

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Solution Solution | Comment 2 of 12 |
As a preliminary -
  Let two spheres of radii R and r rest on
  the table ( R > r ). Let C be the distance
  between their points of contact. By the
  Pythagorean Theorem:
     C^2 + (R-r)^2 = (R+r)^2
     C = 2*sqrt(R*r)
For the first part, R = 2*r and the maximum
distance between points of contact is the
diagonal of the square,
     sqrt(2) = 2*sqrt((2*r)*r)
     r = 1/2 and R = 1
For the second part, let the vertices of the
square be (0,0), (1,0), (1,1), and (0,1). Let
the smaller sphere be at (0,0) and the larger
sphere at (1,1). Then the point of contact of
the third sphere, of radius a, will fall on the
intersection of the two circles,
     x^2 + y^2 = [2*sqrt([1/2]*a)]^2 = 2*a         (1)
     (x-1)^2 + (y-1)^2 = [2*sqrt([1]*a)]^2 = 4*a   (2)
Combining equations (1) and (2) we get
     (x+1)^2 + (y+1)^2 = 4                         (3)
Clearly, the maximum value for the radius a will
occur when the point of contact is at the edge
of the table (x=0 or y=0). Using this with
equations (1) and (3) gives
     x^2 = 2*a  and  (x+1)^2 = 3
     a = 2 - sqrt(3).
All radii are in meters.

Edited on June 18, 2005, 4:10 pm
  Posted by Bractals on 2005-06-18 14:53:47

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