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.
There are 8 vertices and 7 edges. Let's suppose the magic constant
is $h$ 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
$a,b,c,d,e,f,g$ are each counted twice so
So $a+b+c+d+e+f+g$ is a multiple of 8 and it is at least
$1+2+...+7=28$ so it must be $32$ or more. Hence $8h\geq 152$ and
It is easy to check, in a similar way, for the largest value that
$h$ can take. Now it is a matter of systematically checking the
possibilities for magic labellings with $h=19$ etc.
How do we discover what edge magic graphs there are? Here the three
order-3 vertices $a, b,c$ are counted three times and the sum of
all the numbers is again $120$.
Hence $k$ is even and since $a+b+c\geq 6$ we have $7k\geq
120+12=132$ and so $k\geq 20$.
If $k=20$ then $a+b+c=10$ so this triple is $1+2+7$ or $1+3+6$ or
$1+4+5$ or $2+3+5$.
The possible edge sums for $k=20$ 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 $16-P$ 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
total of $28$.
It is left for you to find the edge magic labels for $k=20$ and for
$k=22$ (and correspondingly $k=26$) andfor $k=24$. Remember $k$
must be even and at least $20$ so these are the only