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

[msmith12]

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

 

[quark]

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.

 

[tacman]

Hi Matt,

Yes, I can definitely help you with this problem. First, let's start by visualizing the sphere and its colors. Since 10% of the surface is white, we can imagine that the white color is spread evenly across the surface of the sphere, leaving the remaining 90% of the surface to be black.

Now, let's consider a cube inscribed in the sphere. This means that all of the cube's vertices are touching the surface of the sphere. Since the cube is inscribed in the sphere, it must be contained within the sphere, meaning that all of its points lie within the sphere's surface. This also means that the cube must have a smaller surface area than the sphere.

Since we know that the sphere's surface is made up of 10% white and 90% black, and the cube's surface area is smaller than the sphere's, it follows that there must be a larger proportion of black on the cube's surface than white. In other words, the majority of the cube's surface must be black.

Now, let's consider the cube's vertices. Since all of the vertices lie on the sphere's surface, and the majority of the cube's surface is black, it follows that all of the cube's vertices must also be black. This is because if any of the vertices were white, it would mean that the cube's surface is not predominantly black, which contradicts our initial assumption.

Therefore, we have proven that there must be a cube inscribed in the sphere such that all of its vertices are black. I hope this helps! Let me know if you have any further questions or need clarification on any of the steps. Good luck with your problem!