Dodecahedral K3? In pondering
this
MO question and in particularly its 1st answer,  and answers to
this one recently posed, I realized there ought to be a dodecahedral K3 surface $X$.
This $X$ would fiber as an elliptic surface over   $CP^1$
with  12 singular fibers, each of type $I_2$. The corresponding singular points on $CP^1$
would form  the vertices of the icosahedron (centers of the dodecahedron). The automorphism group of
this $K3$ will then project onto the symmetry group of the dodecahedron.  Do you  know this
$K3$?   Do you have a reference? 
 A: There's a pretty K3 with icosahedral symmetry and 12 singular fibers each of type II (so a double root of the discriminant but with additive reduction); would that do?  It's $y^2 = x^3 + P(t)$ where $P$ has icosahedral symmetry.  Explicitly one can take $P(t) = t^{11} - 11 t^6 - t$.  This surface is isotrivial (i.e. with constant $j$-invariant, here $j=0$), and is closely related with the isotrivial surface of Chahal, Meijer, and Top, which also has $j=0$ but with $a(t) = t^{12} - 11 t^6 - 1$.  Both surfaces attain the maximum of $18$ for the Mordell-Weil rank of an elliptic K3 surface over ${\bf C}(t)$.  The Chahal-Meijer-Top paper is on the arXiv (9911274), and is published in Comment. Math. Univ. St. Pauli 49 (2000), 79–89.  My recent student G.Zaytman studied the icosahedral surface and its Mordell-Weil lattice in the course of his thesis research.
EDIT Come to think of it, this is up to isomorphism the unique elliptic K3 surface over ${\bf C}(t)$ with icosahedral symmetries acting on the base: for the narrow Weierstrass form $y^2 = x^3 + a(t) x + b(t)$ to have icosahedral symmetry, the same must be true of the coefficients $a$ and $b$; but $a$ has degree $8$, so must vanish identically, while $b$ has degree 12 and is thus determined up to a scalar factor.  
