2005 AMC 12A Problems/Problem 13

Problem

In the five-sided star shown, the letters $A$ , $B$ , $C$ , $D$ and $E$ are replaced by the numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the numbers at the ends of the line segments $\overline{AB}$ , $\overline{BC}$ , $\overline{CD}$ , $\overline{CE}$ , and $\overline{EA}$ form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?

$[asy] draw((0,0)--(0.5,1.54)--(1,0)--(-0.31,0.95)--(1.31,0.95)--cycle); label("$A$",(0.5,1.54),N); label("$B$",(1,0),SE); label("$C$",(-0.31,0.95),W); label("$D$",(1.31,0.95),E); label("$E$",(0,0),SW); [/asy]$

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

Solution

$(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)$ (i.e., each number is counted twice). The sum $A + B + C + D + E$ will always be $3 + 5 + 6 + 7 + 9 = 30$ , so the arithmetic sequence has a sum of $2 \cdot 30 = 60$ . The middle term must be the average of the five numbers, which is $\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}$ .

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

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