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

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==Problem==
 
==Problem==
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?  
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In the five-sided star shown, the letters <math>A, B, C, D,</math> and <math>E</math> are replaced by the numbers <math>3, 5, 6, 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]]
  
<math> \mathrm{(A) \ } 9\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 11\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ } 13 </math>
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<math> \textbf{(A) } 9\qquad \textbf{(B) } 10\qquad \textbf{(C) } 11\qquad \textbf{(D) } 12\qquad \textbf{(E) } 13 </math>
  
==Solution==
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==Solution 1==
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 <math>2(3+5+6+7+9)=60</math>
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Each corner <math>(A,B,C,D,E)</math> goes to two sides/numbers. (<math>A</math> goes to <math>AE</math> and <math>AB</math>, <math>D</math> goes to <math>DC</math> and <math>DE</math>). The sum of every term is equal to <math>2(3+5+6+7+9)=60</math>
  
Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is <math>\frac{60}{5}=12\Rightarrow D</math>
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Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is <math>\frac{60}{5}=\boxed{\textbf{(D) }12}</math>
  
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==Solution 2==
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We know that the smallest number in this sequence must be <math>3 + 5 = 8</math>, and the biggest number must be <math>7 + 9 = 16</math>. Since there are <math>5</math> terms in this sequence, we know that <math>8 + 4d = 16</math>, or that <math>d = 2</math>. Thus, we know that the middle term must be <math>8 + 2 \cdot 2 = \boxed{12}.</math>
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~yk2007 (Daniel K.)
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 22:58, 9 May 2022

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

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

Solution 1

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 $\frac{60}{5}=\boxed{\textbf{(D) }12}$

Solution 2

We know that the smallest number in this sequence must be $3 + 5 = 8$, and the biggest number must be $7 + 9 = 16$. Since there are $5$ terms in this sequence, we know that $8 + 4d = 16$, or that $d = 2$. Thus, we know that the middle term must be $8 + 2 \cdot 2 = \boxed{12}.$ ~yk2007 (Daniel K.)

Video Solution

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

~ pi_is_3.14

See Also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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