Sums on a Triangle and Tetrahedron

  1.       a
         / \
        b - c
       / \ / \
      d - e - f
     / \ / \ / \
    g - h - i - j
    Can you place the numbers 1-10 at the locations a-j on the triangle, such that every side (abdg, acfj, ghij) has the same sum and every internal line (bei, ceh, def) has the same sum?
  2. Can you place the numbers 1-10 on a tetrahedron, one at each corner and one on each edge, such that (1) each edge has the same sum, or (2) each face has the same sum? Is it possible to do both at once? If either is not possible, can you change one or two numbers to meet the conditions? Try to minimize the sums.
Source: Kevin King's Calculus teacher, Al Zimmermann. #1 was found later at Math Forum's MathMagic, Challenge for 10-12, Nov 28, 1999.
Solutions were received from Al Zimmermann and Sandy Thompson.
  1. There are two solutions, disregarding rotations and reflections. The first is shown. The second can be found by substituting each element X with (11-X).
          1
         / \
        6 - 10   Side Sums = 21
       / \ / \   Internal Lines Sums = 17
      5 - 4 - 8
     / \ / \ / \
    9 - 3 - 7 - 2
    
  2. Sandy Thompson sent the following solution for the tetrahedron:
    My answers for this question use the following flattened version of a
    tetrahedron.  I know it might look a little strange, but if you fold that
    thing up, you'll find that it does work out.
    
              a
             / \
            h   i
           /     \
          b - e - c
         / \     / \
        /   j   f   \
       /     \ /     \
      /-- g - d ------\
    
    Sides: ahb, aic, agd, bec, bjd, cfd
    Faces: abcehi, bcdefj, abdghj, acdfgi
    
    Phew!
    
    First, faces with the same sum.  All 14 possibilities are listed here, from
    the smallest to largest sum.  Only one of the 24 possible transformations
    for each is listed.  All lists of integers are in order from a to j.
    
    Face sums of 31:
    [1, 2, 3, 8, 6, 5, 4, 9, 10, 7]
    [1, 2, 4, 7, 9, 6, 8, 10, 5, 3]
    
    Face sums of 32:
    [1, 2, 5, 10, 8, 3, 6, 9, 7, 4]
    [1, 2, 6, 9, 8, 4, 7, 10, 5, 3]
    [1, 3, 4, 10, 8, 2, 6, 7, 9, 5]
    [1, 4, 6, 7, 2, 5, 3, 9, 10, 8]
    [2, 3, 4, 9, 8, 1, 6, 5, 10, 7]
    
    Face sums of 34:
    [1, 6, 9, 10, 3, 2, 5, 8, 7, 4]
    [1, 7, 8, 10, 3, 4, 5, 9, 6, 2]
    [2, 5, 9, 10, 3, 1, 4, 7, 8, 6]
    [2, 7, 8, 9, 3, 6, 5, 10, 4, 1]
    [4, 5, 7, 10, 9, 2, 8, 6, 3, 1]
    
    Face sums of 35:
    [3, 8, 9, 10, 5, 2, 7, 6, 4, 1]
    [4, 7, 9, 10, 2, 1, 3, 5, 8, 6]
    
    Edges with the same sum.  I didn't find any combination of the numbers that
    resulted in all 6 edges having the same sum... and my program tried all the
    possible combinations.  There aren't any!
    

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