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Nine identical spheres fit exactly into a cube whose edges have length 10. Four spheres fit in the corners of the base, one sphere is in the centre of the cube, and the others are placed in the four corners above the centre sphere. What is the radius of each sphere?

Source :UK National Mathematics Contest.

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    $\begingroup$ To the close-voters: I suspect this is a nice maths puzzle rather than a simple maths problem. It's not obvious how to calculate the answer, since you need to think carefully about exactly how the spheres touch each other; there's probably an 'aha' which makes it easy. $\endgroup$ – Rand al'Thor Sep 3 '16 at 11:41
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The diagonal of the cube is $10\sqrt{3}$. The three spheres touch in a straight line along this diagonal. Let the radius of a sphere be $r$.

The distance from the centre of a corner sphere to the center of the other corner sphere is then $4r$. Place a single sphere in it's own cube, this cube has diagonal $2r\sqrt{3}$.

This is twice the distance from the centre of a corner sphere to the corner of the cube, but we have two of them to cover, so we can say $10\sqrt{3}=4r+2r\sqrt{3}$

$\therefore r=\dfrac{5\sqrt{3}}{2+\sqrt{3}}$

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    $\begingroup$ A sphere inside a cube will hardly ever touch its corner. $\endgroup$ – Sleafar Sep 3 '16 at 9:54
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    $\begingroup$ I think this answer could do with a better explanation or a diagram. It sounds like you're assuming the spheres touch the cube vertices which they don't. But if not, and there's some trick to this puzzle I would be interested to know what it is. $\endgroup$ – bg6471 Sep 3 '16 at 10:17
  • $\begingroup$ @Sleafar Hardly ever? I'd say never. $\endgroup$ – bg6471 Sep 3 '16 at 10:18
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    $\begingroup$ i'm working on a solution to this predicament as i type... $\endgroup$ – JMP Sep 3 '16 at 10:19
  • $\begingroup$ @bg6471 I translated a figure of speech which means exactly that. Maybe it doesn't work the same in English. $\endgroup$ – Sleafar Sep 3 '16 at 10:23
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unless I'm mistaken:

the radius is roughly $2.3205$

because:

consider the long diagonal from one corner of the cube to the opposite corner. There will be 3 touching spheres in a straight line along this diagonal. The spheres in the corners are touching 3 different faces of the cube. The formula for a diagonal of a cube is $\sqrt{3} \times x$, where $x$ is the side length. So the diagonal is $\sqrt{3} \times 10$. Now consider one of the spheres inscribed inside a smaller cube. The diagonal of this cube is $2r \times \sqrt{3}$. The long diagonal of the big cube will be equal to $4r + 2r \times \sqrt{3}$. Therefore $10 \times \sqrt{3} = 4r + 2r \times \sqrt{3}$, or $r = 10 \times \sqrt{3} / (4+2\times \sqrt{3})$. This is roughly $2.32050808$

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The diagonal of the cube is $\sqrt{10^2+10^2+10^2}=10\sqrt{3}$. Two of the three spheres that lie on that diagonal touch the walls and the distance from their centres to the corner of the cube is $R\sqrt{2}$. So the diagonal in therms of R will be $4R + 2R\sqrt{2}$. This leads us to the result $R=10\sqrt{3}/(4 + 2\sqrt{2})$

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Let $r$ be the radius of the spheres. The centers of the eight spheres in the corners form a smaller cube of edge $c$.

  • $r+c+r=10$ because the corner spheres touch the outside cube along an horizontal line
  • $r+2r+r=\sqrt3c$ because the inner sphere touches the corner spheres along the great diagonal of the inner cube
Thus we get $c=10-2r$ and $4r+2\sqrt3r=10\sqrt3$, and therefore $r={10\sqrt3\over4+2\sqrt3}={15\over2\sqrt3+3}$.

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