2013 AMC 10B Problems/Problem 22
Contents
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 1
First of all, note that must be , , or to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have:
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:
Solution 2
As in solution 1, must be , , or giving us 3 choices. Additionally . This means once we choose there are remaining choices. Going clockwise from we count, possibilities for . Choosing also determines which leaves choices for , once is chosen it also determines leaving choices for . Once is chosen it determines leaving choices for . Choosing determines , exhausting the numbers. Additionally, there are three possible values for . To get the answer we multiply .
Side Note to Solution 1 & 2
Solution 1 and 2 states that without rigor analysis. Here it is:
Let
Because , so , but , so ,
When When When
~isabelchen
Video Solution by TheBeautyofMath
~IceMatrix
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
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Followed by Problem 23 | |
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