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Difference between revisions of "2002 AMC 12B Problems/Problem 5"

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\qquad\mathrm{(E)}\ 120</math>
 
\qquad\mathrm{(E)}\ 120</math>
  
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== Solution ==
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== Solution 1 ==
 
The sum of the degree measures of the angles of a pentagon (as a pentagon can be split into <math>5- 2 = 3</math> triangles) is <math>3 \cdot 180 = 540^{\circ}</math>. If we let <math>v = x - 2d, w = x - d, y = x + d, z = x+2d</math>, it follows that  
 
The sum of the degree measures of the angles of a pentagon (as a pentagon can be split into <math>5- 2 = 3</math> triangles) is <math>3 \cdot 180 = 540^{\circ}</math>. If we let <math>v = x - 2d, w = x - d, y = x + d, z = x+2d</math>, it follows that  
  
 
<cmath>(x-2d)+(x-d)+x+(x+d)+(x+2d) = 5x = 540 \Longrightarrow x = 108 \ \mathrm{(D)}</cmath>
 
<cmath>(x-2d)+(x-d)+x+(x+d)+(x+2d) = 5x = 540 \Longrightarrow x = 108 \ \mathrm{(D)}</cmath>
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 +
== Solution 2 ==
 +
The sum of the degrees of a pentagon is <math>540^{\circ}</math>, which is also the sum of the arithmetic series v, w, x, y, z. Since x is the middle term, it will be the average of the terms, which is <math>\frac{540}{5} = 108 \Longrightarrow \boxed{D}</math>
  
 
== See also ==
 
== See also ==

Revision as of 06:43, 22 July 2008

Problem

Let $v, w, x, y,$ and $z$ be the degree measures of the five angles of a pentagon. Suppose that $v < w < x < y < z$ and $v, w, x, y,$ and $z$ form an arithmetic sequence. Find the value of $x$.

$\mathrm{(A)}\ 72 \qquad\mathrm{(B)}\ 84 \qquad\mathrm{(C)}\ 90 \qquad\mathrm{(D)}\ 108 \qquad\mathrm{(E)}\ 120$

Solution 1

The sum of the degree measures of the angles of a pentagon (as a pentagon can be split into $5- 2 = 3$ triangles) is $3 \cdot 180 = 540^{\circ}$. If we let $v = x - 2d, w = x - d, y = x + d, z = x+2d$, it follows that

\[(x-2d)+(x-d)+x+(x+d)+(x+2d) = 5x = 540 \Longrightarrow x = 108 \ \mathrm{(D)}\]

Solution 2

The sum of the degrees of a pentagon is $540^{\circ}$, which is also the sum of the arithmetic series v, w, x, y, z. Since x is the middle term, it will be the average of the terms, which is $\frac{540}{5} = 108 \Longrightarrow \boxed{D}$

See also

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 12 Problems and Solutions
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