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# Regular octagon abcdefgh has its center at z . Each of the vertices and the center are to be associated with one of the digits 1 through

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Regular octagon  abcdefgh has its center at z . Each of the vertices and the center are to be associated with one of the digits 1 through 9, with each digit used once, in such a way that the sums of the numbers on the lines aze, bzf, czg, and djh are all equal. In how many ways can this be done?

Mar 5, 2020

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Regular octagon  abcdefgh has its center at z .
Each of the vertices and the center are to be associated with one of the digits 1 through 9,
with each digit used once, in such a way that the sums of the numbers

on the lines aze, bzf, czg, and dzh are all equal.
In how many ways can this be done?

$$\begin{array}{|r|c|c|c|c|c|c|c|c|c|} \hline \text{sum} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 1 & - & 3 & 4 & 5 & 6 & 7 & 8 & \color{red} 9 & \color{red}10 \\ \hline 2 & & - & 5 & 6 & 7 & 8 & \color{red}9 & \color{red}10 & \color{red}11 \\ \hline 3 & & & - & 7 & 8 & \color{red}9 & \color{red}10 & \color{red}11 & 12 \\ \hline 4 & & & & - &\color{red} 9 & \color{red}10 & \color{red}11 & 12 & 13 \\ \hline 5 & & & & & - & \color{red}11 & 12 & 13 & 14 \\ \hline 6 & & & & & & - & 13 & 14 & 15 \\ \hline 7 & & & & & & & - & 15 & 16 \\ \hline 8 & & & & & & & & - & 17 \\ \hline 9 & & & & & & & & & - \\ \hline \end{array}$$

We need equal sums in four different ways: There are three possible sums: 9, 10, and 11.

$$\begin{array}{|rcll|} \hline \text{sum }~=9 \\ \hline 9 &=& 1+8 \\ 9 &=& 2+7 \\ 9 &=& 3+6 \\ 9 &=& 4+5 \\ \hline \text{There are }\\ ~ 4! = 24~ \text{ ways} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{sum }~=10 \\ \hline 10 &=& 1+9 \\ 10 &=& 2+8 \\ 10 &=& 3+7 \\ 10 &=& 4+6 \\ \hline \text{There are }\\ 4! = 24~ \text{ ways} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{sum }~=11 \\ \hline 11 &=& 2+9 \\ 11 &=& 3+8 \\ 11 &=& 4+7 \\ 11 &=& 5+6 \\ \hline \text{There are }\\ ~ 4! = 24~ \text{ ways} \\ \hline \end{array}$$

In total so far these are $$3\times 4!=3\times 24 = \mathbf{72} \text{ ways}$$.

Swapping the two elements of a pair of numbers:

$$\begin{array}{|rcll|} \hline \text{One of the \mathbf{72} ways} \\ \text{swapping the two elements of a pair of numbers} \\ \hline \mathbf{ab} & \mathbf{cd} & \mathbf{ef} & \mathbf{gh} \\ ab & cd & ef & hg \\ ab & cd & fe & gh \\ ab & cd & fe & hg \\ \hline ab & dc & ef & gh \\ ab & dc & ef & hg \\ ab & dc & fe & gh \\ ab & dc & fe & hg \\ \hline ba & cd & ef & gh \\ ba & cd & ef & hg \\ ba & cd & fe & gh \\ ba & cd & fe & hg \\ \hline ba & dc & ef & gh \\ ba & dc & ef & hg \\ ba & dc & fe & gh \\ ba & dc & fe & hg \\ \hline \text{There are }\\ ~ \mathbf{16}~ \text{ ways} \\ \hline \end{array}$$

This can be done  in $$16\times 72 = \mathbf{1152}$$ ways.

Mar 5, 2020