Magic I

Use the numbers 1 - 7.
Each row in the figure below (horizontal, vertical, and diagonal) must add up to 12.

                  X  X  X
                      4
                   X  X  X 

x x x
  x 
x x x

so that all five possible sums of three numbers on a line are equal to 12 has exatly the same solution as yours.

Indeed, each number, except for the middle one, appears in two sums, while the middle one is included into three. The sum of all sums therefore includes each number twice and the middle number three times. There 5 such sums each equal to 12. So we have

5·12 = 2·(1 + 2 + 3 + 4 + 5 + 6 + 7) + m,

where m is the middle number. In other words,

60 = 56 + m,

so that m = 4.

Also, this one problem does appear in Peggy Kaye's book Games for Math.