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Source: 1. Henry Ernest Dudeny, 536 Curious Problems & Puzzles, #394. 2. Original.
1) Since each letter is used twice and 1+2+...+12 = 78, all six lines total 2*278 => each line = 26 We can assume A is the smallest of the values of any of the points of the star and that B < E (otherwise rotate and flip if needed). With the restriction on the sum of the 6 points = 26 there are 6 different solutions: (1 4 10 5 7 12 11 3 8 6 9 2) (1 3 9 6 8 11 12 5 10 4 7 2) (1 3 7 12 4 8 11 10 9 5 2 6) (1 3 11 8 4 12 7 2 5 9 10 6) (1 2 10 9 5 7 12 8 11 3 4 6) (1 2 12 9 3 6 11 7 10 4 5 8) 2) Let the consecutive numbers be: x3, x2, x1, x, x+1, x+2, x+3. One of these is the sum of the points. Depending on which one that is, the other 6 add up to some number between 6x3 and 6x+3. But they still must add to 2*278 which is a multiple of 6; so x = 26 and the 6 lines total 23, 24, 25, 27, 28 and 29. There are 499 possible solutions! (1 5 8 7 9 10 12 6 11 4 3 2) (1 5 7 8 9 11 12 6 10 4 3 2) [... 495 other solutions removed  KD ...] (1 2 10 12 3 9 7 4 6 8 5 11) (1 2 10 12 3 8 6 4 7 9 5 11)