Difference between revisions of "2005 AMC 12A Problems/Problem 13"
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<math>AB + BC + CD + DE + EA = 2(A+B+C+D+E)</math>. The sum <math>A + B + C + D + E</math> will always be <math>3 + 5 + 6 + 7 + 9 = 30</math>, so the arithmetic sequence has a sum of <math>2 \cdot 30 = 60</math>. Since <math>CD</math> is the middle term, it must be the average of the five numbers, of <math>\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}</math>. | <math>AB + BC + CD + DE + EA = 2(A+B+C+D+E)</math>. The sum <math>A + B + C + D + E</math> will always be <math>3 + 5 + 6 + 7 + 9 = 30</math>, so the arithmetic sequence has a sum of <math>2 \cdot 30 = 60</math>. Since <math>CD</math> is the middle term, it must be the average of the five numbers, of <math>\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}</math>. | ||
Revision as of 21:07, 3 November 2013
Problem
The regular 5-point star 
is drawn and in each vertex, there is a number. Each 
and 
are chosen such that all 5 of them came from set 
. Each letter is a different number (so one possible way is 
). Let 
be the sum of the numbers on 
and 
, and so forth. If 
and 
form an arithmetic sequence (not necessarily in increasing order), find the value of 
.
Solution
. The sum 
will always be 
, so the arithmetic sequence has a sum of 
. Since is the middle term, it must be the average of the five numbers, of 
.
See also
| 2005 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
