Difference between revisions of "2013 AMC 10B Problems/Problem 22"
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The regular octagon <math>ABCDEFGH</math> has its center at <math>J</math>. Each of the vertices and the center are to be associated with one of the digits <math>1</math> through <math>9</math>, with each digit used once, in such a way that the sums of the numbers on the lines <math>AJE</math>, <math>BJF</math>, <math>CJG</math>, and <math>DJH</math> are all equal. In how many ways can this be done? | The regular octagon <math>ABCDEFGH</math> has its center at <math>J</math>. Each of the vertices and the center are to be associated with one of the digits <math>1</math> through <math>9</math>, with each digit used once, in such a way that the sums of the numbers on the lines <math>AJE</math>, <math>BJF</math>, <math>CJG</math>, and <math>DJH</math> are all equal. In how many ways can this be done? | ||
| − | <math> \textbf{(A)}\ | + | <math> \textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456 </math> |
==Solution== | ==Solution== | ||
Revision as of 15:50, 21 February 2013
Problem
The regular octagon 
has its center at 
. Each of the vertices and the center are to be associated with one of the digits 
through 
, with each digit used once, in such a way that the sums of the numbers on the lines 
, 
, 
, and 
are all equal. In how many ways can this be done?
Solution
First of all, note that must be , 
, or to preserve symmetry. We also notice that 
.
WLOG assume that 
. Thus the pairs of vertices must be and 
, 
and 
, 
and 
, and 
and . There are 
ways to assign these to the vertices. Furthermore, there are 
ways to switch them (i.e. do instead of ).
Thus, there are 
ways for each possible J value. There are possible J values that still preserve symmetry: 
