## Common Sums in a Hexagon

 ``` A - B - C / \ / \ D E F G / \ / \ H - I - J - K - L \ / \ / M N O P \ / \ / Q - R - S ``` Place the numbers 1-19 in the hexagon, such that each of the 12 lines of three numbers have the same sum. What are the lowest and highest possible common sums? (Or is this possible?)
(To aid in solution comparisons, place the lowest valued corner in the top-left position, and the lowest adjacent side in the top-center position.)

Source: Original. Based on a puzzle from Sudipta Das.

Solutions were received from: Al Zimmermann, Claudio Baiocchi, Jozef Hanenberg, Lorenzo Justo, Denis Borris, Philippe Fondanaiche, Brendan Owen, Sandy Thompson, Algirdas Rascius.

Here is Al Zimmermann's summary:

```Here are the attainable common sums and the number of ways (disregarding
rotations and reflections) in which they can be attained:

Sum    Number of Ways
---------------------
22           4
23           2
24           6
25           6
26           4
27           11
28           5
29           8
30           0
31           8
32           5
33           11
34           4
35           6
36           6
37           2
38           4

These are the four ways for achieving a sum of 22:

1---9--12            1--13---8
/ \     / \          / \     / \
18  19   8   6       18  19  12  10
/     \ /     \      /     \ /     \
3--17---2--16---4    3--17---2--16---4
\     / \     /      \     / \     /
14  15  13  11       14  15   9   7
\ /     \ /          \ /     \ /
5--10---7            5---6--11

1--14---7            1--18---3
/ \     / \          / \     / \
18  19  13   6       19  17  15   9
/     \ /     \      /     \ /     \
3--17---2--11---9    2--16---4---8--10
\     / \     /      \     / \     /
15  16  12   5       14  12  13   7
\ /     \ /          \ /     \ /
4--10---8            6--11---5

The four solutions for a sum of 38 can be derived by replacing each number n
in the above four solutions by the number 20-n.  ```

Claudio Baiocchi wrote an HTML/Java based solver, too: Claudio's solution
Mail to Ken