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 22:07, 3 November 2013

Problem

The regular 5-point star $ABCDE$ is drawn and in each vertex, there is a number. Each $A,B,C,D,$ and $E$ are chosen such that all 5 of them came from set $\{3,5,6,7,9\}$. Each letter is a different number (so one possible way is $A = 3, B = 5, C = 6, D = 7, E = 9$). Let $AB$ be the sum of the numbers on $A$ and $B$, and so forth. If $AB, BC, CD, DE,$ and $EA$ form an arithmetic sequence (not necessarily in increasing order), find the value of $CD$.

$(\mathrm {A}) \ 9 \qquad (\mathrm {B}) \ 10 \qquad (\mathrm {C})\ 11 \qquad (\mathrm {D}) \ 12 \qquad (\mathrm {E})\ 13$

Solution


$AB + BC + CD + DE + EA = 2(A+B+C+D+E)$. The sum $A + B + C + D + E$ will always be $3 + 5 + 6 + 7 + 9 = 30$, so the arithmetic sequence has a sum of $2 \cdot 30 = 60$. Since $CD$ is the middle term, it must be the average of the five numbers, of $\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}$.

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
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

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