# A follow-up to the regular hexagon question

This is a follow-up to the regular hexagon question.

The problem statement was:

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.

As far as I understand the solution suggested by Lopsy:

Let $\Omega :=$ the set of all regular hexagon inscribed in the sphere. Together with some appropriate $\sigma$-algebra $\mathcal{F}$ (by the way, would that be simply $2^\Omega$, or not?) and $P$, they make the probability space $(\Omega, \mathcal{F}, P)$.

For $k=1,…, 6$, we define $M_k := \{ \omega \in \Omega \ \ | \ \$the $k^\text{th}$ vertex of $\omega$ is red$\}$, and the following random variables: $X_k := \chi_{M_k}$ and $X:=X_1+…+X_6$. Then:

\begin{align} E(X) &= E(X_1+…+X_6) \\ &= E(X_1)+…+E(X_6) \\ &= P(M_1)+…+P(M_6) \\ &>3 \end{align}

And so there exists at least one regular hexagon on this sphere with at least four vertices of it placed on red surface.

My question is:

Where did we use the properties of the objects (hexagon, surface)? What is we replace sphere with circle in the setup? What if we replace hexagon with some other figure which touches the sphere in six points? Go $3D$ with an octahedron? Or just with any six points for that matter? What would it change in our proof?

Or to put it differently, I feel uneasy about $\Omega$. Since we could define it in so many ways, how do we know that we aren't counting any irrelevant objects?

That's the first time I see a proof which makes use of the probabilistic method, and I don't feel convinced which makes me think that I'm missing out some very important insights. Any help is hugely appreciated.

-
The whole difficulty in the proof is putting a probability measure on $\Omega$ and proving that $P(\{\omega \in \Omega \mid \text{the$k$-th vertex of$\omega$is red}\}) = P(\text{a random point on the sphere is red})$ – mercio Nov 24 '13 at 14:47
@mercio: Oh, I didn't even pay attention to this aspect and just assumed it tacitly. What measure can we put on $\Omega$? – Leo Nov 24 '13 at 14:59
My guess would be to pick $\Omega$ to be in fact the group of rotations of the sphere, and use its Haar measure $\mu$. Then if $P \in S^2$, you can define for $A$ a measurable subset of $S^2$, $\mu_P(A) = \mu(\{\rho \mid \rho(P) \in A\})$, which gives you a measure $\mu_P$ on $S^2$. Then you essentially have to show that $\mu_P$ doesn't depend on $P$ (easy because the action of $\Omega$ on the sphere is transitive) and that $\mu_P$ is secretly the standard Lebesgue measure $\lambda$ on $S^2$ (I don't think this is easy !) – mercio Nov 24 '13 at 15:16
@mercio: Unfortunately, that's beyond my current knowledge. I have seen a few proofs to similar problems (inscribing various things into a painted sphere) yet none of them involved the rigor you're talking about. Do you think that my confusion is rooted in this aspect of the proof? I mean, can't we still replace the hexagon with some other six-point-on-the-sphere figure and go through the same reasoning (including the remarks in your two comments)? – Leo Nov 24 '13 at 15:29