1 to 12

Here is one possible arrangement of the numbers 1 - 12 in the twelve vertices.

The sum of all numbers 1-12 equals 78. If we select 4 random numbers from this set their sum most likely will be 78/3 = 26.

Now assign each circle a letter (any order you like) and write down the equations for the picture: - A+B+C+D=sum - E+F+G+H=sum - E+X+B+Y=sum - M+G+N+D=sum - A+X+F+M=sum - Y+C+N+H=sum

Sum last 4 equations and subtract 2 first ones to find out that - M+N+X+Y=sum also

Now we have that - A+B+C+D=sum - E+F+G+H=sum - M+N+X+Y=sum

Meaning that we have to split numbers 1-12 into 3 subsets such that sums inside each subset is equal. It is possible only if sum=26.

26 is even, and numbers 1-12 contain 6 odd and 6 even numbers, so each line should have either 2 or 4 odd numbers for the sum to be even.

Considering these 2 facts It is pretty easy to find a fitting combination.

Here's some Python 3 code I wrote that does it all for you:

import itertools # for generating all permutations
eqs = ((0, 2, 4, 6), 
       (1, 2, 3, 5),
       (0, 3, 7, 10),
       (1, 4, 8, 11),
       (10, 9, 8, 6),
       (5, 7, 9, 11)) # the sums for each line



str_to_format = \
"""    {:2}      {:2}    
        {:2}        
    {:2}      {:2}    
{:2}              {:2}
    {:2}      {:2}    
        {:2}        
    {:2}      {:2}     Sum: {}\n\n\n""" # so that it looks like the picture



for i in itertools.permutations(list(range(1, 13))): #for every possible arrangement of [1,12]
    sums = []
    for eq in eqs: #for each line on the picture
        sum = 0
        for q in eq: #take the sum of that line
            sum += i[q]
        sums.append(sum)
    if sums.count(sums[0]) == len(sums): #and if they're all the same
        print(str_to_format.format(*i, sums[0])) #show me!

Numbers 1,2,9 and 8 add up to 20. Similarly numbers 7,2,6 and 5 add up to 20.