Difference between revisions of "2013 AMC 10B Problems/Problem 22"
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+ | ==Problem== | ||
+ | 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)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | First of all, note that <math>J</math> must be <math>1</math>, <math>5</math>, or <math>9</math> to preserve symmetry. We also notice that <math>A+E = B+F = C+G = D+H</math>. | ||
+ | |||
+ | WLOG assume that <math>J = 1</math>. Thus the pairs of vertices must be <math>9</math> and <math>2</math>, <math>8</math> and <math>3</math>, <math>7</math> and <math>4</math>, and <math>6</math> and <math>5</math>. There are <math>4! = 24</math> ways to assign these to the vertices. Furthermore, there are <math>2^{4} = 16</math> ways to switch them (i.e. do <math>2</math> <math>9</math> instead of <math>9</math> <math>2</math>). | ||
+ | |||
+ | Thus, there are <math>16(24) = 384</math> ways for each possible J value. There are <math>3</math> possible J values that still preserve symmetry: <math>384(3) = \boxed{\textbf{(C) }1152}</math> |
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: