An Eight-Pointed Star 

    a   b
      c
d   e   f   g
  h       i
j   k   l   m
      n
    o   p
  1. Place the numbers 1 to 16 in the 16 locations on the eight-pointed star, such that each of the eight lines of numbers has the same sum and the sum of the eight points of the star has the same sum as the lines. Or prove it can't be done.

  2. Without the condition that the sum of the points equal that of the lines, find a solution such that each of the eight lines has a common sum. Can solutions for different common sums be found?

Source: Original. Based on a puzzle sent from a reader.

Solutions were received from Joseph DeVincentis and Philippe Fondanaiche. Philippe's answers are shown here. 
Let consider the 10 identities:
 (1) a+b+c+d+e+f+g+h+i+j+k+l+m+n+o+p = 136
 (2) a+e+k+o = S
 (3) b+f+l+p = S
 (4) d+e+f+g = S
 (5) j+k+l+m = S
 (6) a+c+f+i+m = S
 (7) b+c+e+h+j = S
 (8) d+h+k+n+p = S
 (9) g+i+l+n+o = S
 (10) a+b+d+g+j+m+o+p = S

By adding the identities (6) to (9), we find out c+h+i+n = 4*S - 136 
By adding the identities (2) to (10), we obtain c+h+i+n = 3*136 - 9*S= 408 - 9*S
Therefore 13*S = 544. As S is an integer, there is a contradiction.

Question 2

First we have the relation c+h+i+n = 4*S-136 and by adding the above 
identities (2) to (5), we have a second interesting relation e+f+k+l = 
c+h+i+n + 4*S-136 = 8*S-272.
As c+h+i+n>=10, then S>36.
As e+f+k+l<=13+14+15+16=58, then S<42.
The possible values of S are 37,38,39,40 and 41 and for each value of S, the 
sums c+h+i+n and e+f+k+l are known.So it is easy to determine the complete 
set of solutions with a simple computer program. 
Hereafter,a possible solution for each value of S:
S     a   b   c   d   e   f   g   h   i   j   k   l   m   n   o   p 
37   14  15   3  16   4   5  12   2   6  13   8   7   9   1  11  10
38   14  15   8  16   2   7  13   4   3   9  12  11   6   1  10   5
39    9  16   1  13   4  14   8   6  10  12  15   7   5   3  11   2
40   11  10   8   7   4  15  14  12   1   6  16  13   5   3   9   2
41   13   7   9   4  11  14  12   6   3   8  16  15   2   10  1   5

For S=37, there is a unique solution (excluding the symmetries). At the 
opposite, for S=40 the number of different solutions is maximum.
I translated the above table into the stars - KCD.
37
    14  15
       3
16   4   5  12
   2       6
13   8   7   9
       1
    11  10
38
    14  15
       8
16   2   7  13
   4       3
 9  12  11   6
       1
    10   5
39
     9  16
       1
13   4  14   8
   6      10
12  15   7   5
       3
    11   2
40
    11  10
       8
 7   4  15  14
  12       1
 6  16  13   5
       3
     9   2
41
    13   7
       9
 4  11  14  12
   6       3
 8  16  15   2
      10
     1   5
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