Magic Tetrahedron

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Finding $(x_1 + x_2 + x_3 + x_4)$

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Let $y_j$ 's denote the white circles in the graph and $x_i$ 's denote $\{v_1, v_2, v_3, v_4\}$ .

The sum of all integer values in the puzzle is $\begin{array}{rl} \sum_{i=1}^4 x_i + \sum_{j=1}^{6k} y_j &= \dfrac{1}{2}\left(1 + 4 + 6k\right)\left(4 + 6k\right)\\ &= \left(2 + 3k\right)\left(5 + 6k\right) \end{array}$ Since the value in each of the vertices occur three times in the sum, and there are $6$ edges in the graph (representing the sum), the total of all sums is $3\sum_{i=1}^4 x_i + \sum_{j=1}^{6k} y_j = 6\sum_{i=1}^4 x_i$ where $6\sum_{i=1}^4 x_i$ denote $6$ times the edge sum that equals $\left(x_1 + x_2 + x_3 + x_4\right)$ . Elementary algebra shows that since $\sum_{j=1}^{6k} y_j = 3\sum_{i=1}^4 x_i$ and $\sum_{j=1}^{6k} y_j = \left(2 + 3k\right)\left(5 + 6k\right) - \sum_{i=1}^4 x_i$ the magic sum is $\dfrac{1}{4} (2 + 3k)(5 + 6k)$

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Solving for $k$

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There are several restrictions of the value of the magic sum that occurs in the puzzle:

• It depends on the minimum and the maximum of $\sum_{i=1}^4 x_i$ .
• Since all circles contain integers, the sum of the circles must be an integer. Otherwise, this can't happen.

From the first remark, the minimum is $10$ (which is easily found by adding all integers from $1$ to $4$ ), whereas the maximum is $\dfrac{1}{2}\left(1 + 6k + 4 + 6k\right)\left(4 + 6k - \left(1 + 6k\right) + 1\right) = 2(5 + 12k)$ Evaluating the following inequality: $10 \leq \dfrac{1}{4} (2 + 3k)(5 + 6k) \leq 2(5 + 12k)$ we see that the possible integer candidates of $k$ is between $1$ and $4$ . Checking by substitution shows that $\boxed{k = 2}$ is the only solution that works since $(2 + 3 \cdot 2)(5 + 6 \cdot 2) \equiv 0 \bmod 4$

Note : There exist such tetrahedral graphs that work! See my fun challenge under my response.