Question 199322: In a cube, 10" on a side, a sphere of maximum radius is inscribed. What is the size (diameter) of the largest sphere that can be placed in the space remaining in any of the eight corners?
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! A rough sketch will greatly aid in solving this
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diagonal of the cube is dia ^2 = 3 * 10 ^2,,,,diagonal is 17.320
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the largest sphere in the box has a dia of 10,,,sides of the box
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alone the diagonal, the distance left between sphere (r=5) and corner(diag = 17.32)
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is (17.32 - 10 ) /2 = 3.6602.
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now comes the hard part, a sphere of that dia will not fit, it goes to corner.
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to calculate a sphere that touches the sides of the cube, remember that 3.6602 is
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also the diagonal of a small box, that would fit in corner, with a small sphere inside.
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if the diagonal of the box is 3.6602, but s^2 +s^2 +s^2 = diag^2,,,s= sqrt ( diag^2 /3)
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s = sqrt 3.6602^2 /3 = 2.11,,,the diameter of the sphere
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