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 Perplexus Icon plus one (Posted on 2005-06-18)
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 | 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`
`                or`
`     C = 2*sqrt(R*r)`
`For the first part, R = 2*r and the maximumdistance between points of contact is thediagonal of the square,`
`     sqrt(2) = 2*sqrt((2*r)*r)`
`                or`
`     r = 1/2 and R = 1`
`For the second part, let the vertices of thesquare be (0,0), (1,0), (1,1), and (0,1). Letthe smaller sphere be at (0,0) and the largersphere at (1,1). Then the point of contact ofthe third sphere, of radius a, will fall on theintersection of the two circles,`
`     x^2 + y^2 = [2*sqrt([1/2]*a)]^2 = 2*a         (1)`
`                and`
`     (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 willoccur when the point of contact is at the edgeof the table (x=0 or y=0). Using this with equations (1) and (3) gives`
`     x^2 = 2*a  and  (x+1)^2 = 3`
`                 or`
`     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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