Another Six-Pointed Star 

        / \
       / A \
 -----/-----\-----
 \ B / \ D / \ F /
  \ / C \ / E \ /
   X-----X-----X
  / \ H / \ J / \
 / G \ / I \ / K \
 -----\-----/-----
       \ L /
        \ /
 Draw a regular hexagon divided into six equilateral triangles. On each side, place another equilateral triangle, to create a six-pointed star.

Can you place the numbers 1-12 into the 12 triangles such that the sum of the five triangles in each of the six lines is the same? (For example, in the figure, BCDEF and GHIJK are two lines of five.)

If it is possible, what are the largest and smallest common sums possible? If it is not possible, how many numbers would you have to change, keeping all 12 unique, to achieve the smallest possible common sum?

Source: Original.

Solutions were received from Samantha Levin, Claudio Baiocchi, David Greenspan, Philippe Fondanaiche, Stephane Higueret, and Rick (?). It turns out there is one solution for a common sum of 32 (thanks to Claudio for including the diagram): 
Sum 32; true values 1..12:
 
        / \
       / 2 \
 -----/-----\-----
 \ 3 / \ 8 / \ 5 /
  \ / 6 \ / 10\ /
   X-----X-----X
  / \ 7 / \ 1 / \
 / 9 \ / 4 \ / 11\
 -----\-----/-----
       \ 12/
        \ /
Another solution with a sum of 33 can be obtained by replacing each element with the same number subtracted from 13.
Several people sent good analysis of the relationships found in the diagram, too:
From Phillipe:
We have the following identities:
( 1) A+B+C+D+.....+K+L = 78
(2) S = A+D+C+H+G = B+C+D+E+F = G+H+I+J+K = A+D+E+J+K = B+C+H+I+L = F+E+J+I+L

Let s = A+B+G+L+K+F
Therefore (1) and (2) ==> S = 39 - s/6
So s is a multiple of 6 with the possible values s=24,30,36,42,48,54 and S=35,34,33,32,31,30.
From David:
the sums of opposite pairs of points is constant at 78 - 2*(common sum)
One person sent a very thorough proof showing the solution was impossible. (I never did find the error in the logic.)
Claudio also found some solutions with other sums by replacing various elements with the value of 13 or 14. Below are his other solutions which he says make an exhaustive list for these sums: 
Sum=30; value 12 replaced by 13:
 
        / \
       / 6 \
 -----/-----\-----
 \ 8 / \ 7 / \ 10/
  \ / 3 \ / 2 \ /
   X-----X-----X
  / \ 5 / \ 4 / \
 / 9 \ / 1 \ / 11\
 -----\-----/-----
       \ 13/
        \ /
 
Sum=31; value 11 replaced by 13:
 
        / \
       / 5 \
 -----/-----\-----
 \ 6 / \ 9 / \ 10/
  \ / 2 \ / 4 \ /
   X-----X-----X
  / \ 7 / \ 1 / \
 / 8 \ / 3 \ / 12\
 -----\-----/-----
       \ 13/
        \ /
 
Sum=32; value 9 replaced by 13:
 
        / \
       / 5 \
 -----/-----\-----
 \ 6 / \ 11/ \ 8 /
  \ / 4 \ / 3 \ /
   X-----X-----X
  / \ 2 / \ 1 / \
 / 10\ / 7 \ / 12\
 -----\-----/-----
       \ 13/
        \ /
 
Sum=32; value 9 replaced by 14:
 
        / \
       / 5 \
 -----/-----\-----
 \ 7 / \ 10/ \ 11/
  \ / 3 \ / 1 \ /
   X-----X-----X
  / \ 6 / \ 4 / \
 / 8 \ / 2 \ / 12\
 -----\-----/-----
       \ 14/
        \ /
 
Sum=32; value 11 replaced by 14:
 
        / \
       / 3 \
 -----/-----\-----
 \ 5 / \ 10/ \ 9 /
  \ / 7 \ / 1 \ /
   X-----X-----X
  / \ 4 / \ 6 / \
 / 8 \ / 2 \ / 12\
 -----\-----/-----
       \ 14/
        \ /
 
Sum=32; value 12 replaced by 14:
 
        / \
       / 2 \
 -----/-----\-----
 \ 5 / \ 8 / \ 6 /
  \ / 9 \ / 4 \ /
   X-----X-----X
  / \ 3 / \ 7 / \
 / 10\ / 1 \ / 11\
 -----\-----/-----
       \ 14/
        \ /
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