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.
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.
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