2013 AMC 10B Problems/Problem 22

Problem

The regular octagon $ABCDEFGH$ has its center at $J$ . 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 $AJE$ , $BJF$ , $CJG$ , and $DJH$ are all equal. In how many ways can this be done?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8$

Solution

First of all, note that $J$ must be $1$ , $5$ , or $9$ to preserve symmetry. We also notice that $A+E = B+F = C+G = D+H$ .

WLOG assume that $J = 1$ . Thus the pairs of vertices must be $9$ and $2$ , $8$ and $3$ , $7$ and $4$ , and $6$ and $5$ . There are $4! = 24$ ways to assign these to the vertices. Furthermore, there are $2^{4} = 16$ ways to switch them (i.e. do $2$ $9$ instead of $9$ $2$ ).

Thus, there are $16(24) = 384$ ways for each possible J value. There are $3$ possible J values that still preserve symmetry: $384(3) = \boxed{\textbf{(C) }1152}$

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2013 AMC 10B Problems/Problem 22

Problem

Solution