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 1 of 4: Relate the side sums and square sums to the three vertex numbers

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In plain words

Whenever configurations overlap at corners, summing over all pieces isolates the corner contributions modulo the number of pieces, here $3$ .

3s = a + b + c + \sum_{k=1}^{9} k = a + b + c + 45, \qquad 3t = a^2 + b^2 + c^2 + \sum_{k=1}^{9} k^2 = a^2 + b^2 + c^2 + 285

Detailed analysis

Let $a, b, c$ be the numbers placed in the three vertex circles, $s$ be the common sum of the four numbers on each side, and $t$ be the common sum of their squares. Adding the three sides together counts every number in $1, 2, \ldots, 9$ once and each vertex number an extra time, giving $3s = a + b + c + 45$ and $3t = a^2 + b^2 + c^2 + 285$ . Since $45$ and $285$ are multiples of $3$ , both $a + b + c$ and $a^2 + b^2 + c^2$ must be divisible by $3$ .