What number is written?

Question

In each of the 28 blank hexagons in the figure real numbers are written in such a way that the number in each inner hexagon is equal to the arithmetic mean of the numbers in the six adjacent ones. Find the value in the central hexagon (where ?). I solved that problem head-on by composing a system of 28 linear algebraic equations and solving it with help of Maple. My answer is $\frac {4986619541155196219}{4026303401170889720}=1.238510625\dots$ (if I am not mistaken). This is a training problem from the course of discrete complex analysis which is used to model magnetism and seepage in porous media. The comments to the problem hint that there is another solution that does not require cumbersome calculations.

Answer 1

Let the center cell value be $x$.

Mirror the honeycomb horizontally and add it cell-wise to the original. You then get nines on the bottom row, and the center cell value is $2x.$ The arithmetic mean property still applies.

Then rotate this 120 and 240 degrees and add to what you have. You now have nines on the boundary, $6x$ in the center cell value, and the arithmetic mean property still applies.

This honeycomb, with nines on the boundary has one obvious solution: nines in all cells, including the center cell. Thus $6x=9,$ i.e. $x=3/2$ is a solution.

Answer 2

I can at least confirm that solving the equations - which is not cumbersome at all - yields the unique value $x=\frac{3}{2}$.

Also some of the other values are less "cumbersome" than above. For example, $$ (x_7,x_{13},x_{20},x_{28})=\left(\frac{1}{2}, \frac{3}{2}, \frac{5}{2},\frac{7}{2}\right). $$

The advantage of using equations is that we know all values in the hexagonal pattern. There is a unique solution in positive rational numbers.

The only way to check correctness is, whether or not the equations have been written correctly. I have numbered the empty places starting from the top by $x_1,x_2,\ldots ,x_{28}$. Then the equations are

\begin{align*} n_1 & =6x_1-(x_2+x_3) \\ n_2 & =6x_2-(x_1+x_3+x_5+x_4) \\ n_3 & =6x_3-(x_1+x_2+x_5+x_6) \\ n_4 & =6x_4-(x_2+x_5+x_7+x_8) \\ n_5 & =6x_5-(x_2+x_3+x_4+x_6+x_8+x_9) \\ n_6 & =6x_6-(x_3+x_5+x_9+x_{10}) \\ n_7 & =6x_7-(x_4+x_8+x_{11}+x_{12}) \\ n_8 & =6x_8-(x_4+x_5+x_7+x_9+x_{12}+x_{13}) \\ n_9 & =6x_9-(x_5+x_6+x_8+x_{10}+x_{13}+x_{14}) \\ n_{10} &=6x_{10}-(x_6+x_9+x_{14}+x_{15}) \\ n_{11} &=6x_{11}-(x_7+x_{12}+x_{16}+x_{17}) \\ n_{12} &=6x_{12}-(x_7+x_8+x_{13}+x_{18}+x_{17}+x_{11}) \\ n_{13} &=6x_{13}-(x_8+x_9+x_{14}+x_{19}+x_{18}+x_{12}) \\ n_{14} &=6x_{14}-(x_9+x_{10}+x_{15}+x_{20}+x_{19}+x_{13}) \\ n_{15} &=6x_{15}-(x_{10}+x_{14}+x_{20}+x_{21}) \\ n_{16} &=6x_{16}-(x_{11}+x_{17}+x_{22}+x_{23}) \\ n_{17} &=6x_{17}-(x_{11}+x_{12}+x_{18}+x_{24}+x_{23}+x_{16})\\ n_{18} &=6x_{18}-(x_{12}+x_{13}+x_{19}+x_{25}+x_{24}+x_{17})\\ n_{19} &=6x_{19}-(x_{13}+x_{14}+x_{20}+x_{26}+x_{25}+x_{18})\\ n_{20} &=6x_{20}-(x_{14}+x_{15}+x_{21}+x_{27}+x_{26}+x_{19})\\ n_{21} &=6x_{21}-(x_{15}+x_{20}+x_{27}+x_{28})\\ n_{22} &=6x_{22}-(3+x_{16}+x_{23})\\ n_{23} &=6x_{23}-(5+x_{22}+x_{16}+x_{17}+x_{24})\\ n_{24} &=6x_{24}-(7+x_{23}+x_{17}+x_{18}+x_{25})\\ n_{25} &=6x_{25}-(9+x_{24}+x_{18}+x_{19}+x_{26})\\ n_{26} &=6x_{26}-(11+x_{25}+x_{19}+x_{20}+x_{27})\\ n_{27} &=6x_{27}-(13+x_{26}+x_{20}+x_{21}+x_{28})\\ n_{28} &=6x_{28}-(15+x_{27}+x_{21}) \end{align*}