We can distribute six consecutive integers, such as 0 to 5, on the vertices and edges of a triangle in such a way that the sum along each edge is the same. In fact, there are four essentially distinct such arrangements, shown below. 2 4 5 5 3 4 1 3 0 2 0 1 1 5 0 2 5 0 3 4 1 4 2 3 The common sums of these four triangular arrangements are 6, 7, 8, and 9 respectively. Of course, each of these arrangements has six versions if we count rotations and reflections. Similarly, arrangements of any 8 consecutive numbers can be constructed on the vertices and edges of a square. In this case there are six essentially distinct arrangements, illustrated below. 5 3 1 4 1 5 3 6 2 7 0 5 4 6 6 2 1 5 3 1 0 7 2 0 7 3 7 0 4 2 4 6 7 2 1 7 1 3 3 5 4 2 0 6 4 0 5 6 The common sums of the squares in each of these four columns are 9, 10, 11, and 12 respectively. It's easy to show algebraically that, for each square, the sums of opposite pairs of edges must be equal. If we let V and E denote the sums of the values on the vertices and edges respectively, then the common sum is S = (E+2V)/n for an n-sided polygon. We also know that there are a total of 2n vertices and edges, so we are distributing the integers 0 to 2n-1, which implies that the sum of all the numbers is E+V = n(2n-1). Solving these equations for E and V, we have E = n[2(2n-1) - S] V = n[S - (2n-1)] This shows that both E and V must be divisible by n, and also that the common sum S must be greater than (2n-1) and less than 2(2n-1). In fact, since there are n edges, each containing n distinct non- negative integers, the least possible value of E is n(n-1)/2, so the quantity 2(2n-1) - S must not only be positive, it must be greater than or equal to (n-1)/2, and thus S must be less than or equal to 2(2n-1) - (n-1)/2 = (7n-3)/2. Likewise V must be at least n(n-1)/2, so S cannot be less than (2n-1) + (n-1)/2 = (5n-3)/2. For n=4 this means S cannot be less than 8.5 and cannot exceed 12.5, consistent with the fact that the smallest common sum is 9 and the largest is 12. We can place 10 consecutive numbers on the vertices and edges of a pentagon such that the three numbers along any edge have the same sum. This can be done in six essentially distinct ways. Rather than drawing the figures, we can just list the strings of numbers around the perimeter of the figure, beginning with a vertex. The six arrangements are listed below, along with their common sum. common sum ------ 11 1738092546 13 0765283194 6523814907 14 3476185092 9234716805 16 6283547091 As in the case of the square arrangements, the alternative patterns for a single common sum differ by just two transpositions of consecutive numbers, and these two transpositions are separated by a single position. As always, the common sum must be no less than (5n-3)/2 and no greater than (7n-3)/2, which gives the limits of 11 and 16 for n=5. For hexagonal arrangements we have n=6, so the only possible common sums are in the range from 13.5 to 19.5, i.e., the set 14, 15, 16, 17, 18, and 19. The essentially distinct solutions for each of these common sums are listed below, along with the corresponding values of V and E, and (in the right-most columns) the sums of the squares of the vertex and edge numbers. S=14, V=18, E=48 6 3 5 7 2 11 1 9 4 10 0 8 82 424 5 3 6 7 1 11 2 8 4 10 0 9 82 424 8 5 1 11 2 9 3 7 4 10 0 6 94 412 S=15, V=24, E=42 0 6 9 2 4 10 1 11 3 5 7 8 156 350 S=16, V=30, E=36 6 0 10 1 5 7 4 9 3 11 2 8 190 316 7 3 6 8 2 4 10 1 5 11 0 9 214 292 8 5 3 11 2 4 10 0 6 9 1 7 214 292 2 10 4 1 11 5 0 9 7 3 6 8 226 280 0 9 7 5 4 1 11 3 2 8 6 10 226 280 7 9 0 6 10 4 2 11 3 5 8 1 226 280 S=17, V=36, E=30 9 3 5 1 11 2 4 6 7 10 0 8 292 214 11 2 4 8 5 3 9 1 7 10 0 6 292 214 11 2 4 10 3 6 8 0 9 7 1 5 292 214 9 0 8 6 3 4 10 2 5 11 1 7 280 226 9 3 5 8 4 2 11 0 6 10 1 7 280 226 6 4 7 2 8 0 9 3 5 11 1 10 256 250 S=18, V=42, E=24 2 9 7 1 10 0 8 6 4 3 11 5 354 152 S=19, V=48, E=18 11 1 7 2 10 0 9 4 6 8 5 3 412 94 10 0 9 3 7 1 11 2 6 8 5 4 412 94 3 5 11 1 7 4 8 2 9 0 10 6 424 82 These results show some interesting dualities. For example, the first solution with S=14 is the dual of the last solution with S=19, because they have the same split of the numbers into vertex and edge numbers, except that the types are reversed. In other words, the numbers that are assigned to vertices in the first solution are assigned to edges in the last solution, and vice versa. These two solutions are shown pictorially below. 6 3 5 3 5 11 8 7 6 1 0 2 10 7 10 11 0 4 4 9 1 9 2 8 We can generalize the problem by allowing k numbers to be assigned to each edge, where k is greater than 1. For example, we can construct triangles such as the two shown below 1 1 6 2 6 10 8 9 7 3 4 5 3 7 8 12 9 2 4 11 5 In general, to construct a triangle with k numbers on each edge (not counting the vertices), let N = k+2 denote how many numbers are on each edge counting the vertices, and then assign the numbers 1, N, and 2N-1 to the three vertices. Then, to fill in the edge numbers, arrange the remaining numbers into three columns, ascending on each row alternately from the left and the right. For example, to construct a triangle with N=6 numbers on each side, we first assign the numbers 1, 6, and 11 to the vertices, and then we arrange the remaining numbers in three columns as shown below 2 3 4 8 7 5 9 10 12 15 14 13 Each column sums to 34, but we actually need columns that sum to 34-5, 34, and 34+5 in order to place these numbers on the edges [6,11], [1,11], and [1,6] of the triangle, respectively. We can do this by leaving the middle column untouched, and making the transpositions (8,5) and (15,13) in the outer columns. This decreases the left-hand column by 5, and increases the right hand column by 5. As an alternate solution, we could make the transpositions (2,4) and (9,12), which would increase the left column by 5 and decrease the right column by 5. Notice that this must always lead to a solution, because the two outer columns will always differ by 2 or 3, so we can form any required adjustment (greater than 1) by transposing pairs.

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