Is There a Cube Inscribed in a Sphere with All Black Vertices?

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The discussion centers on the mathematical problem of determining whether a cube can be inscribed in a sphere with 10% of its surface area painted white and the remaining 90% black. The key argument presented is that if the diameter of the circle formed by the projection of the white-painted segment is less than the side length of the cube, then it is possible to inscribe the cube such that all its vertices are black. The method involves calculating the arc length corresponding to the white-painted area and deriving the angle at the center of the sphere to find the diameter of the projected circle.

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msmith12
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Can anyone help me with this problem?

Suppose a sphere is colored in two colors: 10% of the surface is white, and the remaining part is black. Prove that there is a cube inscribed in the sphere such that all vertices are black

thanks
~matt
 
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This may have many solutions, provided the white paint is continuous, but I came up with this rather crude method. What you have to prove is that the diameter of the circle formed by the projection of white painted segment is lesser than the side of the cube.

If the surface area painted in white is 10% then length of the arc with the same radius that of the circle is also 10% of the circle circumference. Now you can get the angle intended by the arc at center. Now calculate the diameter of the projected circle.
 

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