More Olympic Ring Addition

Given the five olympic rings how can the digits one through nine be placed within the nine regions (five non-overlapping ring regions and four overlapping regions shared between two rings) so that each ring contains a successively incrementing total?

Or alternatively, using the labels A through I for the regions, how can the numbers one through nine be assigned to the variables such that:

A+B = B+C+D-1 = D+E+F-2 = F+G+H-3 = H+I-4

How many different totals can fit in the first circle?  Find a complete solution for each.

Source: Original. Similar to previous POTW July 25, 1997

Solutions were received from Kirk Bresniker, Joel D. Haywood, Al, Leendert Biemans, Joseph DeVincentis, Yaacov Yoseph Weiss, Adrian Atanasiu, Mechael Mendelsohn, and Claudio Baiocchi.  Joseph DeVincentis sent this logical answer, finding all seven solutions:
As in the previous problem, the first logical thing to do is add up all
the parts. The sum of all five parts of the equation gives:

(all numbers once) + B + D + F + H - 10 = 5(A+B)

And all the numbers add to 45 so we have:
B + D + F + H + 35 = 5(A+B)
B + D + F + H = 5(A+B-7)

So B+D+F+H is a multiple of 5, which could be anything from 10
(1+2+3+4) to 30 (6+7+8+9), making A+B-7 be in the range 2 to 6, or A+B in
9 to 13.

But note that H+I is 4 more than A+B, or in the range 13 to 17. This 
clearly doesn't work for 17, where H and I would need to be 8 and 9, 
but these are already used in B+D+F+H. All of this could work for 
the other cases.

So try to fill them:

A+B = 9:
I=9, H=4 is forced as the only way to reach 13.
  8 can only go in A (with B=1) or E (with D,F = 1,2 in some order).
  This leads to two solutions:

A=8     E=6     I=9
  B=1 D=2 F=3 H=4
    C=7     G=5

A=6     E=8     I=9
  B=3 D=2 F=1 H=4
    C=5     G=7

B+D+F+H=15, so any of: 1,2,3,9; 1,2,4,8; 1,2,5,7; 1,3,4,7; 1,3,5,6; 
  H+I = 14 so H is at least 5 which limits it to one or two values for
  each case above.
B,D,F,H=1,2,3,9 fails, because H is 9, but F+G+H needs to be 13 which can 
  only happen with F,G=1,3 but both are already used in B,D,F.
B,D,F,H=1,2,4,8 requires H=8, I=6. Then F=2, G=3 for the 13. So
B,D=1,4 but B+C+D=11 so C is also 6. No solution here either.
B,D,F,H=1,2,5,7 requires H=5, I=9 since H=7 would force I=7 also. Then
  F+G=8 so F=2, G=6. Then D+E=10 can only be D=7, E=3, so B is 1 and C
  must be 3 again. No solution.
B,D,F,H=1,3,4,7 fails because H=7 requires I=7 and there is no other large 
  enough H.
B,D,F,H=2,3,4,6 fails because there is nowhere to put the 9. It needs a 
  1 or 5 in a crossing circle to go on an end, or two crossings adding to 
  2, 3, or 4 to go anywhere else.
Only B,D,F,H=1,3,5,6 remains. This needs H=5, I=9 or H=6, I=8.
  Consider the cases separately. If 9 is not used at I, it can only be 
  part of 1+9=10, and if 8 is not used at I it can only be part of
  1+8+3=12. The rest of each case is forced, leading to these two

A=4     E=8     I=9
  B=6 D=3 F=1 H=5
    C=2     G=7

A=9     E=4     I=8
  B=1 D=3 F=5 H=6
    C=7     G=2

B+D+F+H=20, so many cases. 1,2,8,9; 1,3,7,9; 1,4,6,9; 1,4,7,8;
1,5,6,8; 2,3,6,9; 2,3,7,8; 2,4,5,9; 2,4,6,8; 2,5,6,7; 3,4,5,8; 3,4,6,7
  H+I=15, so is 7+8 or 6+9. This rules out the cases where both of one
  of these sets are used in B,D,F,H. In most of the other cases there are
  two ways to complete this, and by applying A+B=11 and then
  D+E+F=13 with the two unused from B,D,F,H, the cases are solved
1,2,8,9: H=8, I=7 ruled out because no A+B=11. So B=8, A=3, H=9, I=6.
  So D,F=1,2 and E must be 10!
1,3,7,9: H=7, I=8, B=9, A=2 or H=9, I=6, B=7, A=4 or H=9, I=6, B=3, A=8.
  The first two cases lead to D,F=1,3 so E=9 which is already used.
  The last case has D,F=1,7, so E=5, leading to one solution

A=8     E=5     I=6
  B=3 D=7 F=1 H=9
    C=2     G=4
1,5,6,8: H=8, I=7 ruled out because no A+B=11. So B=8, A=3, H=6, I=9.
So D,F=1,5 and E=7. But then if D=1, C is the already used 3, and if D=5,
C = -1!
2,4,5,9: H=9, I=6, nowhere to put 8. (2+8+4=14 could only go in F+G+H but H=6)
2,4,6,8: H=6, I=9 leaves nowhere to put 8, so H=8, I=7, A=9, B=2.
D,F=4,6 so E=3, but then C+D=10 but since D+F=10 there is no solution.
2,5,6,7: H=6, I=9 leaves nowhere to put 8, so H=7, I=8, A=9, B=2.
         D,F=5,6 so E=2, already used.
3,4,5,8: H=8, I=7, nowhere to put 9
3,4,6,7: H=7, I=8 leaves nowhere to put 9 so H=6, I=9. A+B=11 forces
         B=3, A=8. D,F=4,7 so E=2, leading to one solution:

A=8     E=2     I=9
  B=3 D=4 F=7 H=6
    C=5     G=1

B+D+F+H=25, so any of: 1,7,8,9; 2,6,8,9; 3,5,8,9; 3,6,7,9; 4,5,7,9; 
But H+I=16 which can only be 7 and 9, so the cases with both 7 and 9 in
       B,D,F,H are impossible. Of the remaining cases:
B,D,F,H=2,6,8,9 requires one of the rings with 3 numbers contains 2 out
  of 6,8,9. This ring has a minimum sum of 6+1+8=15, which would make it 
  F+G+H, but H is 7 or 9, contradicting this. Any other way of forming 
  this ring makes too large a sum.
B,D,F,H=3,5,8,9 requires F+G+H = 5+1+9 to avoid the pitfall of the 
  previous case. The rest is forced to a single solution:

A=4     E=6     I=7
  B=8 D=3 F=5 H=9
    C=2     G=1
B,D,F,H=4,6,7,8 requires H=7, I=9. Then F+G=8 so F=6, G=2. Then D+E=8, 
  which is impossible with D left as either 4 or 8.

So these are the seven solutions.

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