This is a long and detailed hint which provides ideas, not just for this
particular problem, but also for other magic graph problems where similar
reasoning can be used.
Vertex Magic Investigation
There are 8 vertices and 7 edges. Let's suppose
the magic constant is
and this is the same at
each vertex. The total of all the numbers 1 + 2 +
... + 15 = 120 but the numbers on the edges, say
are each counted twice so
So
is a multiple of 8 and it is
at least
so it must be 32 or more.
Hence
and
. It is easy to check, in a similar way, for the largest value that
can take.
Now it is a matter of systematically checking the possibilities for magic
labellings with
etc.
Edge Magic Investigation
How do we discover what edge magic graphs there
are? Here the three order-3 vertices
are
counted three times and the sum of all the
numbers is again 120.
Hence
is even and since
we have
and so
.
If
then
so this triple is 1+2+7
or 1+3+6 or 1+4+5 or 2+3+5.
The possible edge sums for
are:
1+4+15, 1+5+14, 1+6+13, 1+7+12, 1+8+11, 1+9+10,
2+3+15, 2+4+14, 2+5+13, 2+6+12, 2+7+11, 2+8+10,
3+4+13, 3+5+12, 3+6+11, 3+7+10, 3+8+9, 4+5+11,
4+6+10, 4+7+9, 5+6+9, 5+7+8
It is easy to find the edge magic labelling for
k=20.
Moreover, substituting
for each of the
P-labels giving a magic total of k will give the
same numbers 1 to 15 and a magic total of 48-k.
So for all the magic graphs you find with a magic total of 20
there are corresponding graphs with a magic totl of 28.
It is left for you to find the edge magic labels for
and for
(and
correspondingly
) andfor
. Remember
must be even and at least 20 so these are the
only possibilities.