Problem 2

Given a triangular arrangement of

9

circles along the perimeter of a triangle, with

4

circles on each of the three sides (three vertex circles shared by adjacent sides and two interior circles on each side), each of the numbers

1, 2, \ldots, 9

is to be written into one of these circles, so that each circle contains exactly one of these numbers and: (i) the sums of the four numbers on each side of the triangle are equal; (ii) the sums of the squares of the four numbers on each side of the triangle are equal. Find all ways in which this can be done.

Step 4 of 4: Solve the remaining vertex case and count all symmetric configurations

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In plain words

Once $\{2, 5, 8\}$ is placed at the corners, each side's pair of interior numbers is uniquely determined as a set; the only freedom left is the $3! = 6$ corner permutations and $2^3 = 8$ interior swaps, giving $48$ solutions.

\{a,b,c\}=\{2,5,8\} \implies s=20,\ t=126; \qquad \{3,7\}\text{ on }(2,8),\ \{4,9\}\text{ on }(2,5),\ \{1,6\}\text{ on }(5,8)

Detailed analysis

For $\{a, b, c\} = \{2, 5, 8\}$ , we get $3s = 2 + 5 + 8 + 45 = 60 \implies s = 20$ and $3t = 2^2 + 5^2 + 8^2 + 285 = 378 \implies t = 126$ . The interior pairs on the sides $(2, 8)$ , $(2, 5)$ , and $(5, 8)$ must have square sums $x^2 + y^2 = 126 - 68 = 58$ , $t_1^2 + u_1^2 = 126 - 29 = 97$ , and $m^2 + n^2 = 126 - 89 = 37$ . Among the remaining numbers $\{1, 3, 4, 6, 7, 9\}$ , the unique solutions are $\{x, y\} = \{3, 7\}$ , $\{t_1, u_1\} = \{4, 9\}$ , and $\{m, n\} = \{1, 6\}$ , each of which also gives side sum $s = 20$ . Permuting the vertex set $\{2, 5, 8\}$ in $3! = 6$ ways and ordering the two interior numbers on each side in $2^3 = 8$ ways yields $3! \cdot 2^3 = 48$ solutions.