How do we know that S = 12, 13, and 15 yield no solutions to this problem? https://i.stack.imgur.com/RlG1i.png
Question: Four overlapping triangles create eight regions, not including the shaded region. If one places the digits 1 to 8 into the diagram, one digit per region, so that the sum of the numbers inside each of the rectangles is the same, then what digit must be placed in the region marked with Z?
I've understood everything in this problem so far, and I understand that we must get the range of values for 4S in order to minimize the number of possible sums for each rectangle. However, what I do not understand is the process to finding out that 12, 13, and 15 yield no solutions.
The picture I have attached above is a solution given to me by my teacher, but he does not provide any detailed explanation for the part where 12, 13, and 15 yield no solutions. Did he just do trial and error, or is there a way where we can figure it out without solving for each case?
I'm trying to find an efficient solution because this problem will help me in my training for a math contest.
Additional Information:
1+2+3+4 is the least possible sum of A, B, C, and Z, while 5+6+7+8 is the greatest.
 A: You basically need to show that a certain number cannot go into any area given, with the assumption that $S$ sums up to a certain number.
I don't think there is no "easy" way of showing that these numbers don't work, so I did a bit of an exhaustive proof below. I've labeled the two triangles on the left $A$ and $B$, middle one $C$, and right one $D$. Here is how you show it:
S=15:
Let's suppose $S=15$. Then we know that $D=8+7$, because any lower number would not yield a solution. To get $15$ for either $A$ or $B$, we must do $6+5+4=15$, and then it's impossible for the other left rectangle to be $15$. So $S\ne15$.
S=12:
Let's suppose $S=12$. We know the $8$ is inside only 1 rectangle because to get $12$, we would need $\{1,3\}$ twice. So let's place the $8$ in $D$, on the right. If it is in the right part, then we must sum $x+y+z+4=12$ in $C$. But if $x,y,z\ne4 \Rightarrow \{x,y,z\}=\{1,2,5\}$. Thus, in the remaining areas, we need to use $\{3,6,7\}$. We know that $\{6,7\}$ aren't together, so we place them in the middle of $C$. Leaving $3$ to go the intersection. Now, $3+6+a=12$. So $a=3$, which has already been used, so the $8$ cannot be in $D$.
So we place it in $B$ (same will go for $A$, as they are symmetrical), then to get $12$ in $D$, we need $\{7,5\}$. If $7\in C$ and $7\in D$, then we need $x+y+z$ to be $5$, which is impossible. So $5\in C$. So now, $x+y+z=7$, which means $\{x,y,z\}=\{1,2,4\}$. Seeing that $8$ must be paired with $\{1,3\}$, we know that the intersections of $B$ are $3$ (with $A$), and 1 (with $C$). In $A\cap C$, must now be $\{2,4\}$. Because $3+2=5$, and $3+4=7$, and we used $7$ and $5$, there is no way to make it work. Do the same for $A$, and $S\ne 12$.
S=13: Let's suppose $S=13$. We know it's not in the middle of $A$ or $B$, because then the intersections would have to be $\{1,4\}$ and $\{2,3\}$, and they don't share a number.
If the $8$ is in $A\cap B$, then the other pairs of numbers are $\{1,4\}$ (e.g. top) and $\{2,3\}$ (e.g. bottom). $D$ must now be $\{6,7\}$. If the outside part of $D$ is $6$, the inside one is $7$, so the other members of $C$ must have $\{1,2,3\}$, which is impossible ($2,3$ are in the bottom left rectangle). So $6$ is in the middle, which yields a similar contradiction ($1,4$ are in the top intersection). So it's not in the left intersection.
So $8$ is in $D$, or in $C$ alone. It's not in $C$, otherwise $x+y+z=5$. So it's on the outside in $D$, $5$ on the inside. So now we need $x+y+z=8$ without using $5$, which means $\{x,y,z\}=\{1,3,4\}$, leaving us with $\{3,6,7\}$. $6,7$ can't be together, so they are in $\{A,B\}$ (e.g. $6\in A$, $7\in B$) alone, leaving $2$ in $A\cap B$. So now, the top rectangle we have $2+6+x=8$, so $x=5$, which we used. So $S\ne13$.
