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Difference between revisions of "2013 AMC 10B Problems/Problem 22"

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<math> \textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576  \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456 </math>
 
<math> \textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576  \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456 </math>
 +
 +
<asy>
 +
pair A,B,C,D,E,F,G,H,J;
 +
A=(20,20(2+sqrt(2)));
 +
B=(20(1+sqrt(2)),20(2+sqrt(2)));
 +
C=(20(2+sqrt(2)),20(1+sqrt(2)));
 +
D=(20(2+sqrt(2)),20);
 +
E=(20(1+sqrt(2)),0);
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F=(20,0);
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G=(0,20);
 +
H=(0,20(1+sqrt(2)));
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J=(10(2+sqrt(2)),10(2+sqrt(2)));
 +
draw(A--B);
 +
draw(B--C);
 +
draw(C--D);
 +
draw(D--E);
 +
draw(E--F);
 +
draw(F--G);
 +
draw(G--H);
 +
draw(H--A);
 +
dot(A);
 +
dot(B);
 +
dot(C);
 +
dot(D);
 +
dot(E);
 +
dot(F);
 +
dot(G);
 +
dot(H);
 +
dot(J);
 +
label("A",A,NNW);
 +
label("B",B,NNE);
 +
label("C",C,ENE);
 +
label("D",D,ESE);
 +
label("E",E,SSE);
 +
label("F",F,SSW);
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label("G",G,WSW);
 +
label("H",H,WNW);
 +
label("J",J,SE);
 +
</asy>
  
 
==Solution==
 
==Solution==

Revision as of 17:18, 21 February 2013

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?

$\textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456$

[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label("A",A,NNW); label("B",B,NNE); label("C",C,ENE); label("D",D,ESE); label("E",E,SSE); label("F",F,SSW); label("G",G,WSW); label("H",H,WNW); label("J",J,SE); [/asy]

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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