Regular icosahedron exists

It is well known that ancient Greeks proved that there can only be at most 5 different Platonic solids. The proof is geometric and could be understood by even elementary students.

However, it just occurred to me the other day that I've never seen proof that regular icosahedron does exist. Together with its dual regular dodecahedron, they are the only "hard" Platonic solids whose existence do not seem obvious to me.

Proof:

Surely we can construct an icosahedron that has all sides of length \(1\), so the main point is that everything looks identical no matter which vertex you are, or, for every vertex \(v\) its neighboring vertices forms a regular pentagon.

We construct an icosahedron as follows. Let \(ABCDE\) and \(A'B'C'D'E'\) be two regular pentagon parallel to each other such that the segment connecting their centers is perpendicular to them, and \(AB=A'B=BB'=\ldots=1\).

\(A'BDD'\) are coplanar. \(AE\) is symmetric to \(B'C'\) with respect to plane \(AB'DD'\), given all these congruent regular triangles and that \(AE=B'C'\). So we construct \(P\) symmetric to \(E'\) with respect to plane \(AB'DD'\). Then \(PB'=PA'=1\), and \(ABB'PE'\) are coplanar because \(ABB'E'\) are coplanar.

Note that \(\angle E'AB=\angle ABB'=3\pi/5\) since \(ABB'E'\) is congruent to \(ABCE\). Then \(\angle BB'P=3\pi/5=\angle B'PE'=\angle PE'A\), and \(ABB'PE'\) is a regular pentagon, establishing \(PB'=PA'=PE'=1\).

There is only one point above regular pentagon \(A'B'C'D'E'\) with distance \(1\) to \(A'\), \(B'\), and \(E'\), which happens to be on the line central and perpendicular to \(A'B'C'D'E'\). So \(PC'=PD'=1\). Finally construct \(P'\) similarly, and we have an icosahedron with all sides equal and all vertices surrounded by a regular pentagon.

Q.E.D.

Ramsey's Theorems

Finite version

Let \(R(n_1,\ldots,n_c;m)\) be the minimum number of nodes such that any coloring of its \(m\)-subset using \(c\) colors yields a subset of size \(n_i\) all of whose \(m\)-subsets are colored \(i\) for some color \(i\). If all parameters are finite, then so is \(R(n_1,\ldots,n_c;m)\).

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Proof:

If we have $$|V|=1+R\left(R\left(n_1-1,n_2,\ldots,n_c;m\right),\ldots,R\left(n_1,n_2,\ldots,n_c-1;m\right);m-1\right)$$ nodes, pick an arbitrary vertex \(v\in V\). By induction, for some color \(i\) we have \(S\subset V-v\) of size \(R(n_1,\ldots,n_{i-1},n_i-1,n_{i+1},\ldots,n_c;m-1)\) and all \(m\)-subsets of \(S\cup\{v\}\) containing \(v\) have color \(i\). By definition \(S\) either has a subset of size \(n_k\) with all \(m\)-subsets colored \(k\) for some color\(k\ne i\), or a subset of size \(n_i-1\) with all \(m\)-subsets colored \(i\), which together with \(v\) is what we want.

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Infinite version
Any $c$-coloring of $m$-subsets of $\mathbb{N}$ yields an infinite subset all of whose \(m\)-subsets have the same color.

Proof:

With similar spirit and by induction on \(m\), fix a \(x_1\in\mathbb{N}\). There is an infinite set \(V_1\subseteq\mathbb{N}-\{x_1\}\) all of whose \((m-1)\)-subsets have color \(c_{1}\), or equivalently, all \(m\)-subsets containing \(x_1\) in \(V_1\cup\{x_1\}\) have color \(c_{1}\). We proceed to \(V_1\) and repeat on an arbitrary $x_2\in V_1$ to get $c_2$ and $V_2\subseteq V_1-\{x_2\}$. Keep going indefinitely, we get infinite sequences \(x=(x_1,x_2,\dots)\) and \(c=(c_1,c_2,\dots)\). A color appears infinite times in \(c\), and its corresponding elements in \(x\) form the desired infinite subset of $\mathbb{N}$.