I was solving the following problem

Suppose I have a sphere of radius 1 metre. The sphere is colored with red and blue such that it has disconnected regions of red and blue colors.

Now I have to make a cube that fits in the sphere such that Each vertex of the cube touches a red-colored region.

This is possible if one of the following option is true

a) The aggregate area of red part is $11 m^2$ .

b) The aggregate area of red part is $10 m^2$ .

c) The aggregate area of red part is NOT $11 m^2$ .

Answer given was (a)

Where aggregate area is the sum of areas of all regions. Well I approached by calculating total surface area of sphere = $12.56 m^2$

So subtracting from $11m^2$ gives me the minimum residue and so the answer follows.

Is my reasoning correct or any other explanation for this the answer??

  • $\begingroup$ sorry, but I don't quite get what you mean by 'disconnected regions of red and blue' - do you mean that there is one blue and one red region that don't overlap, or that there are lots of blue and red regions, or what? I'm sure what you've written makes sense, but it's just not going in for some reason! $\endgroup$ – Tim Sep 20 '13 at 19:40
  • $\begingroup$ @Tim there are multiple regions of red and blue $\endgroup$ – Amartya Sep 20 '13 at 20:02
  • $\begingroup$ I'm not sure I understand why you can't simply have a completely blue sphere with tiny red dots where the vertices of a cube would be? sorry to hijack your question with my simple inability to understand! $\endgroup$ – Tim Sep 22 '13 at 18:18
  • $\begingroup$ @Tim- The question indirectly asks for the minimum aggregate area for the red regions such that it is guaranteed the cube touches only red regions.You can imagine it as a blue sphere with eight red dots at the vertices,but eight red dot simply does not guarantee that everytime the condition will be met. $\endgroup$ – Amartya Sep 23 '13 at 13:10
  • $\begingroup$ but surely if there is any blue at all then the condition won't be met every time, because we could always place the cube with one vertex on a blue region? and in your question, you introduce the sphere, then say 'now I have to make a cube that fits in the sphere such that...'. I'm not trying to be difficult, it's just that this looks like a problem that would really interest me, if I got it $\endgroup$ – Tim Sep 23 '13 at 18:31

I do not see any reasoning that you put forth, esp if you were not given any options.

(I will avoid any concerns about the validity of area, and assume that all your regions are closed / measurable. I do not think you are concerned about it either.)

You will need $> \frac{7}{8} $ of the surface to be red, in order to get a cube. This can be shown using the probabilistic method.

Consider all possible arrangements of the cube. What is the expected number of vertices that are red?

For a given vertex, as we move it across all arrangements, then the probability that the vertex is red is just the area of the red surface, which is $ > \frac{7}{8} $.

Applying the linearity of expectation, for all 8 given vertices, the expected number of vertices that are red is $> \frac{7}{8} \times 8 = 7 $.

Since the expected number of vertices that are red is more than 7, and the number of vertices that are red is an integer, there must be some figuration where there are strictly more than 7 red vertices, which means that this cube has 8 red vertices. Hence we are done.

(This part might not be true, though I think it is.)

Conversely, it is easy to find a shading of the cube sphere with less that $\frac{7}{8}$ covered, and having no cube which can fit into it.

Take a sphere where $x, y, z, \geq 0 $ is not colored red. One of the vertices of the cube must be all non-negative, and hence is not red.

Since $4 \pi = 12.56$ (and not 12.86 like you listed), $\frac{7}{8} * 4 \pi < 11$ (just barely), so 11 suffices.

  • 2
    $\begingroup$ It would help if you could give some hints why your claims are true. $\endgroup$ – Christian Blatter Sep 20 '13 at 19:56
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    $\begingroup$ Sorry this is so late, but unfortunately your construction isn't right. It's quite easy to inscribe a cube in such a way that it avoids the octant you describe (the only hard part is imagining this!). Here are two verbal descriptions to help visualize this fact: $\endgroup$ – Pat Devlin Nov 10 '16 at 22:27
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    $\begingroup$ (1) imagine the cube is in such a way that its faces are perpendicular to the axes. Rotate it around the z-axis until two corners are touching the octant you say and none of the other corners are nearby. Then rotate the thing so those two corners are both outside that octant. $\endgroup$ – Pat Devlin Nov 10 '16 at 22:30
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    $\begingroup$ (2) it's clearly possible to get TWO vertices of the cube in the interior of that octant. So there must be one octant that's completely empty. (qed) $\endgroup$ – Pat Devlin Nov 10 '16 at 22:31
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    $\begingroup$ Also, in the construction, the maps $\omega_i$ do not make sense. There can be multiple choices of these maps, because there can be multiple cubes and a given vertex that fit inside the cube with the given vertex touching the sohere at a given point. To see this, just rotate a cube along it's 3-fold axis (the main diagonal) that fixes 2 points $\endgroup$ – Breakfastisready Nov 8 '17 at 16:49

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