2005 AMC 12A Problems/Problem 13

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

Solution I

$AB + BC + CD + DE + EA = 2(A+B+C+D+E)$ . 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 $\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}$ .

Solution II

Assume AB is the smallest term (this assumption will drop out by the end). If AB is the smallest the sequence goes A+B,B+C,C+D,D+E,E+A. If (A+B)<(B+C) then A must be less than C. Using this method we eventually get A<C<E and B<D<A so we know B<D<A<C<E. Therefore D is 5 and C is 7 and CD is 12. If AB is the biggest it just works out to D being 7 and C being 5. D

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 12 Problems and Solutions

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2005 AMC 12A Problems/Problem 13

Contents

Problem

Solution

Solution I

Solution II

See also