Different Number of Terms in All Equal-Valued Sums?

Let $x_1,$ $x_2,$ $x_3,$ and $x_4$ be the values in the corners. Let $y_1,$ $y_2,$ $y_3,$ and $y_4$ be the values in the sides. $y_1$ and $y_4$ are opposite each other, and $y_2$ and $y_3$ are opposite each other.

The sum of all integers is $x_1 + x_2 + x_3 + x_4 + y_1 + y_2 + y_3 + y_4 = 36$

Suppose that $y_1 + y_4 = y_2 + y_3 = x_1 + x_2 + x_3 + x_4$

Since each of the circles is intersected by two lines, then the integer in the circle occurs twice in the sum. In this case, $2\left(x_1 + x_2 + x_3 + x_4 + y_1 + y_2 + y_3 + y_4\right) = 72$

Set ${\color{#D61F06}S = x_1 + x_2 + x_3 + x_4}$ , so $\begin{array}{rl} 2\left(({\color{#D61F06}x_1 + x_2 + x_3 + x_4}) + ({\color{#D61F06}y_1 + y_4}) + ({\color{#D61F06}y_2 + y_3})\right) &= 72\\ 2 \cdot 3{\color{#D61F06}S} &= 72\\ 6{\color{#D61F06}S} &= 72\\ {\color{#D61F06}S} &= 12 \end{array}$

So since ${\color{#D61F06}S}$ is an integer, we can place distinct integers in each circle, such that $\boxed{y_1 + y_4 = y_2 + y_3 = 12}$

Since there are several ways to present the sum of $2$ terms as $12$ out of the given integers, the possibilities for $y_1,y_2,y_3,y_4$ are $(4,8)$ and $(5,7)$ . The candidates for $x_1,x_2,x_3,x_4$ are $1,2,3$ and $6$ .

Here is the corrected diagram as an example:

The sum of the 8 numbers is 36.

The circles can be divided up into 3 disjoint sets: Green Circle, Red Line, Dark Blue Line.

These 3 sets must total 36, so each set must total 12.

Since the sum is even, there must always be 2 odds or 0 odds in each row.

It becomes fairly simple to place them now, especially since 12 can only be formed as 7+5 or 8+4.

A deductive logic solution in 4 steps that should be familiar to avid players of Kakuro:

-Given that there is one sum with 4 addends, the minimum common result for all sums has to be 10. (1+2+3+4=10). Similarly, given that there are two sums with 2 addends, the maximum possible common result is 13, as 14 and 15 can only be obtained with one sum each. (8+6 and 8+7). This gives us a possible range of results of 10-13.

-It is immediately evident that the hypothetical results of 10 and 11 can only be obtained in one exact way in the four-addends sum (1+2+3+4 and 1+2+3+5), and since there are only 8 possible numbers, it is also readily evident that there's no way of making two two-addends sums with a result of 10 or 11 with the four leftover numbers for each case. Thus, we're looking at a potential result of 12 or 13.

-The fact that our minimum possible result is now 12 means that 1, 2, and 3 cannot possibly appear in two-addend sums, meaning that they have to be placed in three of the four corners AND that 1 and 2 cannot be connected by a line, as there's no one number big enough to add to their sum that results in 12 or 13. Thus, they have to be in opposite corners and we have two side sums that read (1+/ /+3 and 3+/ /+2).

-This quickly leads to the realisation that the only possible number on the 1+/ /+3 sum is 8 and that the common result is 12. The rest of the numbers, then, can be filled in easily.

Solve it like solving a sodoku :)

• Because each row (and each column) must have same sum, there are 3 rows => sum of each row must be $\frac{1 + 2 + 3 + \ldots + 8}{3} = \frac{36}{3} = 12$
• The second row or column gets only 2 elements, 1, 2, 3 must require 11, 10, 9 in addition to be 12. Hence 1, 2, 3 cannot be placed in the mid of row or column. In other word, they are placed at corners.
• The number 6 must be placed at a corner too since it need another 6 to be 12.
• 1, 2 cannot be placed in a same row or column because they need more 9.
• So we have 4 digits at the corners with their right position. Now just calculate and fill the missing number, double check the sum for each color path, verify if they are all 12.

If we write no. In each box as a variable(a,b,c,d,e,f,g,h) in clockwise direction we can solve this easily.

See, a+b+c=x as it is given .

Also,

c+d+e=x,

e+f+g=x,

g+h+a=x and also

e+h=x,b+f=x.

So, 2(a+b+c+d+e+f+g+h)=6x ,I hav added all the equations,

sum of all boxes is 36 as we know1+2+3+4+5+6+7+8=36.

so,36=3x and x=12.so a+b+c=12.

And the clockwise order of numbers is 1,8,3,7,2,4,6,5

First, notice that the eight numbers can be partitioned into three disjoint sets, the outer circle and the central horizontal and vertical lines, each of which needs to add up to the same number. Since the sum of the eight numbers is 36, each row/column/circle sum must be 12.
The central segments only have two numbers, and there are only two pairs of numbers that sum to 12: 5+7 and 4+8. Those must be placed in the four non-corner spots, with 4 opposite 8 and 5 opposite 7. 1 and 3 must be in the corners next to 8. There are two choices at this point: next to 5 or next to 7. Putting the 1 next to the 7 would require the '4' to be in two places. OTOH, if the 1 is next to the 5, the 6 slots in on the other side, with the 2 in the opposite corner.
Solution:

6 - 4 - 2

5 - * - 7

1 - 8 - 3

Any solution is a reflection/rotation (or combination) of this pattern.