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Difference between revisions of "2005 AMC 10A Problems/Problem 17"

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==Problem==
 
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In the five-sided star shown, the letters <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are replaced by the numbers <math>3</math>, <math>5</math>, <math>6</math>, <math>7</math>, and <math>9</math>, although not necessarily in this order. The sums of the numbers at the ends of the line segments <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, and <math>EA</math> form an arithmetic sequence, although not in that order necessarily. What is the middle term of the sequence?  
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In the five-sided star shown, the letters <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are replaced by the numbers <math>3</math>, <math>5</math>, <math>6</math>, <math>7</math>, and <math>9</math>, although not necessarily in this order. The sums of the numbers at the ends of the line segments <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, and <math>EA</math> form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?  
  
 
[[Image:2005amc10a17.gif]]
 
[[Image:2005amc10a17.gif]]

Revision as of 11:45, 24 September 2016

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?

2005amc10a17.gif

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

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 $\frac{60}{5}=12\Rightarrow D$

See Also

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

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