Suppose we have a sphere and more than a half of its surface is red. Prove or disprove that we can place all vertices of a regular hexagon on this sphere and at least four vertices of it will be placed on red surface.
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It depends on what you mean by "half". Probably the simplest definition is that P(A random point turns out to be red) > 50%. Consider the expected number of red vertices on a random hexagon on the sphere. By properties of expectation - in particular, expected value is additive, even for nonindependent events - the expected number of red vertices is P(vertex one is red) + P(vertex two is red) + ... + P(vertex six is red) > 50% + 50% + ... + 50% > 3. If every hexagon had three or less red vertices, the expected value would be less than three. However, it's more than three, so there must indeed exist a hexagon with four red vertices. |
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