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Tricky Triangle (Posted on 2003-12-03) Difficulty: 4 of 5
Replace the letters in the diagram with a different number from 0 to 9, such that the sum of the four numbers on all edges are the same, and the sum of the three numbers on all three corners are the same.
          A
        B   C
      D   E   F
    G   H   I   J

In other words, A+C+F+J = A+B+D+G = G+H+I+J and
A+B+C = D+G+H = F+I+J.

No Solution Yet Submitted by Kelsey    
Rating: 3.2500 (4 votes)

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Solution Multiple solutions. | Comment 1 of 15
There are several relationships involved. In addition to the given equations, these three, derived from those given, are useful:
B-H=C-I
D-C=H-F
F-B=I-D

That means that the contents of A,B,C,D and G determine all the rest:
H=A+B+C-G-D
F=H+C-D
I=C+H-B
J=A+B+C-F-I
and E is whatever is left over.

Trying all possibilities of A,B,C,D and G is easier than trying all possible permutations.

The following program goes through the possibilities, and then it prints out only those solutions where A>G>J, so that reflections and rotations are not considered as different:

CLS
FOR a = 0 TO 9
 taken(a) = 1
 FOR b = 0 TO 9
  IF taken(b) = 0 THEN
   taken(b) = 1
   FOR c = 0 TO 9
    IF taken(c) = 0 THEN
     taken(c) = 1
     FOR d = 0 TO 9
      IF taken(d) = 0 THEN
       taken(d) = 1
       FOR g = 0 TO 9
        IF taken(g) = 0 THEN
         h = a + b + c - g - d
         IF h >= 0 AND h <= 9 THEN
         IF taken(h) = 0 THEN
          taken(h) = 1
          f = h + c - d
          IF f >= 0 AND f <= 9 THEN
          IF taken(f) = 0 THEN
           taken(f) = 1
           i = c + h - b
           IF i >= 0 AND i <= 9 THEN
           IF taken(i) = 0 THEN
             taken(i) = 1
             j = a + b + c - f - i
             IF j >= 0 AND j <= 9 THEN
             IF taken(j) = 0 THEN
              taken(j) = 1
              IF g + h + i = a + c + f AND c + f + j = b + d + g THEN
              IF a > g AND g > j THEN
               FOR e = 0 TO 9
                 IF taken(e) = 0 THEN EXIT FOR
               NEXT
               r0 = 5 * (ct \ 5): c0 = 10 * (ct MOD 5)
               LOCATE r0 + 1, c0 + 5: PRINT STR$(a)
               LOCATE r0 + 2, c0 + 4: PRINT STR$(b); STR$(c)
               LOCATE r0 + 3, c0 + 3: PRINT STR$(d); STR$(e); STR$(f)
               LOCATE r0 + 4, c0 + 2: PRINT STR$(g); STR$(h); STR$(i); STR$(j)

               ct = ct + 1
              END IF
              END IF
              taken(j) = 0
             END IF
             END IF
             taken(i) = 0
           END IF
           END IF
           taken(f) = 0
          END IF
          END IF
          taken(h) = 0
         END IF
         END IF
        END IF
       NEXT
       taken(d) = 0
      END IF
     NEXT
     taken(c) = 0
    END IF
   NEXT
   taken(b) = 0
  END IF
 NEXT
 taken(a) = 0
NEXT
PRINT ct

The program finds:

5 5 6 7 7
1 7 6 2 0 4 1 3 1 5
6 2 4 1 3 9 5 8 2 2 8 5 6 0 3
4 3 9 0 4 8 4 0 2 3 7 1 5 4 6 0 3 4 8 2

7 8 8 8 8
6 0 0 5 2 3 2 4 2 4
1 4 8 6 2 3 1 4 7 3 0 6 6 1 5
3 9 3 2 3 4 9 1 7 5 6 0 6 5 7 1 1 7 9 0

8 8 8 9 9
2 6 2 6 4 3 0 4 0 6
4 1 7 7 0 4 1 2 9 2 6 5 5 1 3
7 5 9 0 4 5 9 3 7 7 6 0 8 3 7 1 8 2 8 4

9 9 9 9 9
1 3 1 4 2 4 3 0 3 1
2 4 6 3 2 6 3 1 7 1 5 6 0 5 7
6 5 7 0 6 5 8 0 6 6 8 0 4 7 4 2 7 6 4 2

9 9 9 9 9
3 2 3 2 3 4 3 5 5 1
0 4 8 1 4 8 2 0 8 4 0 7 0 2 8
8 6 5 1 6 7 6 0 8 6 7 1 7 6 8 2 8 7 3 4

--------
Thus there are 25 basic solutions, not counting rotations and reflections.
  Posted by Charlie on 2003-12-03 10:27:56
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