Method to solve missing numbers The numbers from $1$ to $8$ are entered into the eight circles in this diagram, with the number $3$ placed as shown. In each triangle, the sum of the three numbers is the same. The sum of the four numbers which are at the corners of the central square is $20$. What is $x + y$?
$(A)\, 10\qquad (B) \,11\qquad (C) 12\\
(D)\, 13\qquad (E)\, 14$ 

Is there any method apart from trial and error to solve these types of problems? Ideally using methods a $13$ to $14$ year old high school students can do.
 A: The sum of all the blobs is 36; the sum of all the blobs plus the central square is equal to the sum of all the triangles. Therefore 56 is the sum of all the triangles, so each triangle sums to 14.
That means we must have 11 as the sum of the two remaining numbers in each of the triangles to which 3 is adjoined; that means it's 6/5 and 7/4, and so the remaining numbers are 1/2/8.
Note that we can't have 1/2 together in a triangle (because 11 isn't a number we can use to complete the sum to 14), so 1 and 2 must be the triangle non-central-square vertices we don't know; therefore the sum of all the blobs is 36, minus the 1 and 2 is 33, minus the 20 from the middle square is 13.
No need to calculate $x$ and $y$; we only had to place the $8$ at the bottom left square corner, and $1$ and $2$ in some order on the bottom/left triangle vertices.
A: Adding all the triangles and then subtracting the four central square corners gives you each circle once, which is 36. So all the triangles must add to 56, meaning each triangle adds to 14.
Thus the two square corners next to 3 are $14-3-x=11-x$ and $11-y$. And the final square corner is $$20-(11-x)-(11-y)-3=x+y-5$$With this, we can find the remaining outer corners as
$$
14-(11-x)-(x+y-5)=8-y
$$and $8-x$.
From here, we have to make use of the fact that all integers from 1 to 8 are used exactly once, as well as the given answer options. Which does mean some trial and error, as this isn't something that is readily encodable as an equation. For instance, $x$ can't be neither 3,4, 5 nor 8, as $x, 8-x$ and $3$ must all be distinct and strictly positive. The same goes for $y$.
That actually only leaves $6+7=13$ as a possible option. No need to check what happens with things like $x=2,y=7$, as $9$ isn't a given answer.
