Note that if we take the numbers 1 to 19 in 9 pairs, the sums of the paired numbers come to 20:
(1,19), (2,18), (3,17), (4,16), (5,15), (6,14), (7,13), (8,12), (9,11).
Using the numbers 1 to 9 only, we can break down 4 of these pairs by splitting the larger number of each pair:
(3,8,9), (4,7,9), (5,6,9), (5,7,8). These are the only unique groups of three, but they do not include 1 and 2. But if we include 1 and 2 we know the remaining 2 numbers to make up a group of 3 must add up to 19, which can't be achieved using a pair of numbers 1 to 9. Therefore, in some cases we need 4 numbers to make up the sum. We can only use the numbers 1 to 9 and we need to use them all and we only have 3 sides of the triangle, so if we have 4 numbers on one side, one of the other sides will have only 2 numbers, because we're only left with 5 when we've taken out 4 for one side. What to do? We have to use the vertices so that the numbers at each vertex are used twice. Let's assume that all three vertices contain such numbers, then we will have 4 numbers a side. Instead of looking at groups of three numbers, we have to look for groups of 4:
(1,2,8,9), (1,3,7,9), (1,4,6,9), (2,3,6,9), (1,4,7,8), (2,3,7,8), (1,5,6,8), (2,4,5,9), (2,4,6,8), (3,4,5,8), (3,4,6,7).
Out of these we need to find where different groups contain a common number. These common numbers will form the vertices. Let's try putting the numbers 1, 2 and 3 at the vertices. This gives us a choice of 5 groups for each of the numbers 1, 2 and 3 (we can arrange the order so that the numbers occupy the first element in each set). Here are the sets:
(1,2,8,9), (1,3,7,9), (1,4,6,9), (1,4,7,8), (1,5,6,8)
(2,1,8,9), (2,3,6,9), (2,3,7,8), (2,4,5,9), (2,4,6,8)
(3,1,7,9), (3,2,6,9), (3,2,7,8), (3,4,5,8), (3,4,6,7)
Now we have to rearrange the numbers so that there are two vertices within each set, so that means we pick groups containing 1 and 2, 2 and 3, and 3 and 1. Yes, there are some, but unfortunately some contain duplicates of the two numbers that are not at the vertices. Look at the following, for example:
(1,8,9,2), (2,6,9,3), (3,7,9,1)
9 is duplicated and we have no 4 or 5. In fact, we can"t have 1 as a vertex with 2 and 3 as the other vertices, because there's always a duplicate to spoil the arrangement. The same thing happens with 2, 3 and 4, and 3, 4 and 5. Symmetry is a common occurrence in mathematics and there's nothing more pleasing to a mathematician than an "elegant" solution. So I'm going to assume that symmetry applies to this problem and go for vertices in the middle of the range 1 to 9. The middle is occupied by numbers 4, 5 and 6. Let's see what happens if we use them as vertices:
(4,2,9,5), (5,1,8,6), (6,3,7,4)
This time it works! There are no duplications in the central pair of numbers so all the numbers from 1 to 9 are there. The other combination of 4 and 5 in the same group, (4,3,8,5), leads to a duplicated 8 in (5,1,8,6). So one answer is:
The vertices are A, B and C=4, 5 and 6 respectively, with 2 and 9 along AB, 1 and 8 along BC, and 3 and 7 along AC.