Arithmagons are cool, right?

(The sum of the vertices is the value of the edge.)

Are triangles special, or can you solve them for any polygon?

What determines when you can solve one of these? Why do triangles always work, but square puzzles don’t always work?

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N equations, N unknowns, and no equation is a linear combination of the others. So there should be a unique solution for an arithmagon of any size, right?

Cool problems!

For the triangle case, I called the vertices x, y, and z, and the edges a, b, and c. You can set up a simple set of 3 equations and solve for the 3 unknowns.

We can apply the same algebraic approach to the square case (a, b, c, d for the edges, which are known, and w, x, y, z for the vertices, which are not). But we wind up with the equation d = c + a – b – meaning we’ve got a condition on our edge values for this to work.

Haven’t done it out for pentagon and up, but I’m pretty sure a solution will exist for (2k + 1)-gons, and not for 2k-gons (unless they meet a pretty stringent condition on the edge values).

As for WHY that’s true… I don’t have a good intuitive explanation yet. The triangle case is equivalent to finding the intersection of three mutually perpendicular planes in a 3D coordinate space, which will necessarily have 1 solution. But obviously I can’t picture the higher-dimensional cases.

I’m not totally happy with my approach. Have you (or your students) come across a more satisfying line of attack?

There is always a unique solution when the number of sides is odd. If the number of sides is even, there are infinitely many solutions provided that the alternating sum of the edges is 0; otherwise there are no solutions. For example, the first square does not have a solution because 5 – 8 + 7 – 6 = -2, but the other two squares have infinitely many solutions.

Instead of choosing a separate variable for each vertex, I would label one vertex as x, then label the other vertices in terms of x as we move around the polygon. The coefficient of x alternates between 1 and -1 as we do this. Note that a single equation must be satisfied in order to complete the cycle.

In the case of an odd number of edges, the equation will look like x + (x + S) = A, where A is the label on the last edge, and S is an alternating sum of the other edges. This equation has the unique solution x = (A-S)/2.

If there are an even number of edges, the equation will look like x + (-x + S) = A. This equation has infinitely many solutions if S = A, but no solutions otherwise.

Followup questions: when does an arithmagon have positive solutions? When does it have integer solutions?