2005 AMC 10A Problems/Problem 17

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 this order. The sums of the numbers at the ends of the line segments $AB$ , $BC$ , $CD$ , $DE$ , and $EA$ form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?

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

Solution

Each corner $(A,B,C,D,E)$ goes to two sides/numbers. ( $A$ goes to $AE$ and $AB$ , $D$ goes to $DC$ and $DE$ ). The sum of every term is equal to $2(3+5+6+7+9)=60$

Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is $\frac{60}{5}=12\Rightarrow D$

Video Solution

https://youtu.be/tKsYSBdeVuw?t=544

~ pi_is_3.14

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2005 AMC 10A Problems/Problem 17

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Problem

Solution

Video Solution

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