Difference between revisions of "2013 AMC 10B Problems/Problem 22"
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Revision as of 15:50, 3 July 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?
![[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]](http://latex.artofproblemsolving.com/d/3/0/d305e9074b588800bd9056981b0e283a8488bc59.png)
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: 
See also
| 2013 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
