Labeling edges of an icosahedron with sum constraints The question is inspired by this previous MO question. There it was shown that it's possible to label the edges of a cube by the numbers $\{1,2,\ldots,6,8,9, \ldots, 13\}$ in such a way that:


*

*Three edges meeting at a vertex sum to 21.

*Four edges constituting a face sum to 28.


The solution is unique up to isometry (numbers in red correspond to invisible edges) :

Similary, is it possible the label the edges of a regular icosahedron (or dodecahedron) with distinct integers in such a way that the two conditions above are satisfied (with different sums of course). If yes, is the solution unique up to isometry?
Note that for the regular tetrahedron, the answer is no: a computation shows that two opposite edges must be equal.
Moreover, what would be natural generalizations of this problem for polytopes in $n$-dimensional space?
 A: Alas there's no such labeling by
$\{1, 2, 3, \ldots, 15, 17, 18, 19, \ldots, 31\}$,
assuming I made no computational error somewhere along the way.
The vertex and face conditions give $32$ linear equations on the $30$ edge labels,
but with enough redundancy that there is a $6$-dimensional space of solutions.
The obvious solution is to label all the edges $1$.  A less trivial solution
is obtained by fixing two opposite vertices $v,v'$ of the icosahedron, and
using the edges that connect a neighbor of $v$ to a neighbor of $v'$: 
there are $10$ such, forming a cycle, and we can label them alternately
$+1$ and $-1$, and label the remaining $20$ edges zero.  This gives $6$
generators (in addition to the all-$1$ vector), satisfying one relation 
(they sum to zero if the signs are chosen consistently), for a total
dimension of $1 + (6-1) = 6$.  These turn out to generate the space of
solutions even over the integers; that is, every integer solution of
the linear equations is an integer linear combination of the generators.
Subtracting off $16 \cdot (1,1,1,\ldots, 1)$, we are looking for a vector in
the $5$-dimensional span of the remaining generators whose coordinates
are a permutation of $(\pm 1, \pm 2, \pm3, \ldots, \pm15)$.
There is no obvious obstruction: no two of the $30$ coordinates
are identically equal (and coordinates at opposite edges are
automatically each other's negatives, which should only help).
It would now be feasible to choose $5$ independent coordinates,
try all $30 \cdot 29 \cdot 28 \cdot 27 \cdot 26$ choices of labeling,
and check whether any of them results in a solution of the desired kind.
But first I tried the shortcut of looking at all solutions modulo $2$ and $3$.
The former failed: there are $15$ linear combinations with the requisite 
$16$ odd and $14$ even entries.  But modulo $3$, (unwelcome) success:
none of the $121 = (3^5-1) / 2$ pairs of nonzero linear combinations has
coordinates equally split among $0,1$, and $2 \bmod 3$.  (The possible
counts of $0 \bmod 3$ coordinates are $6, 8, 12, 14, 20$, occurring with
multiplicities $15, 60, 25, 15, 6$ respectively.)  Hence there is no
edge labeling satisfying all the conditions, QED.  
[added later: I should have noticed that none of $6,8,12,14,20$
is $1 \bmod 3$.  This can be seen directly from the matrix of inner products
of our five nontrivial generators, which has rank $1 \bmod 3$.
So no exhaustive search is required at all, though there's still
the integer linear algebra to find our generators.]
A: This is pretty useless, definitely not an answer, because the edge labels are not distinct. The face sum is constant
at $145$, and the vertex sum constant at $87$.

          


          

$$\textrm{Labels}=\left(
2, 2, 4, 4, 8, 8, 10, 10, 23, 27, 27, 27, 27, 29, 29, 29, 29, 31, 31, \
31, 31, 35, 48, 48, 50, 50, 54, 54, 56, 56
\right)$$


There are likely many similar solutions.
All labels distinct might be challenging.
A: If you require that your labels are the set $\{1,2,\dotsc,31\}$ with one label $k$ removed, a back-of-the-envelope calculation reveals that $k$ must be $16$. Moreover, your vertex-sums (the sums of three edges meeting at any vertex) must be 36 48, and your face-sums must be 100 80. 
