A Magic Pentagon 

      A
   J-- --B
 --       --
I           C
 \         /
 H\       /D
   G--F--E
  1. Place the numbers 1 thru 10 at the corners and sides of a pentagon, such that the sum of each side is the same.
  2. Find all solutions.
To distiguish among rotations and reflections, let the top (A) be the minimum corner, then let the right corner be greater than the left (C>I).

Source: Reader Chuck Johnson, citing his grandson.

Solutions were received from Jozef Hanenberg, Hidefumi Takahashi, Nick McGrath, Denis Borris, Claudio Baiocchi, Philippe Fondanaiche, Paul Botham, Adrian Atanasiu, and Sudipta Das. Here is a good summary from Sudipta Das: 
Let the common sum of each side be SUM .  Then,
(A+B+C)+(C+D+E)+(E+F+G)+(G+H+I)+(I+J+A) = 5 * SUM
=>  A+C+E+G+I+55 = 5 * SUM ( as the sum of the first 10 integers = 55 )
=>  A+C+E+G+I    = 5 * ( SUM - 11 )
Now,
1+2+3+4+5 < = A+C+E+G+I < = 6+7+8+9+10
=> 14 < = SUM < = 19
 
The only possible solutions are :
 
SUM = 14 
 
       1
   10      9
3             4
  6         8
    5  7  2
 
SUM = 16
            
       1                     1
   10      8             8       5
5             7       7             10
  2         6           6         2
    9  4  3               3  9  4
 
SUM = 17
 
       2                     1
   9       7             9       6
6             8       7             10
  1         5           2         3
    10 3  4               8  5  4
 
SUM = 19
 
       6
   5       4
8             9
  1         3
    10 2  7
 
The last 3 solutions can be obtained from the first 3 on
replacing each number by ( 11 - NUMBER ) .
Mail to Ken