Filling in the blanks on a magic triangle I am struggling with this problem. I have tried out a few numbers with no luck, and I have a feeling there is a slick method to this that I am missing, rather than just guesswork. Any hints, would be appreciated. 

 A: Hint: if you add the three sides (double counting corners) the sum has to be perfectly divisible by three, so that each side sums to a whole number. The numbers sum to 45 and the two known corners make it 52. So the top circle is

 8

A: You have to put in the higher circle a number,x, such that $3|2+5+x$. So $x$ could be $2,5,8$ but you have already use the $2$ and the $5$ so the only one is $8$. Because the sum of all the number from $1$ to $9$ is $45$ and this sum with the three number on the vertices must be divisible by $3$. And now it's easy to find the rest on number : from $2$ in clockwise $2,4,9,5,6,1,8,3,7$
Or also $2,6,7,5,3,4,8,1,9$
A: 
Let the sum of the four circles which are in a line be $L$. Then, $3L=\left(S+a+b+5\right)+\left(S+e+f+2\right)+\left(S+a+b+5\right)=S+5+2+\left(S+5+2 +a+b+c+d+e+f\right)=S+7+\left(1+2+\cdots+9\right)=S+52$
Observe that $1\le S\le 9 \rightarrow 53 \le S+52=3L \le 61 \rightarrow 3L=54, 57, 60$
After some trials, we get $L=20$, when $S=8$.
Possible solution
$a=1, b=6, c=4, d=9, e=3, f=7, S=8$
