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

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== Solution ==
 
== Solution ==
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=== Solution 1 ===
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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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Note that since <math>x</math> is the middle term of an arithmetic sequence with an odd number of terms, it is simply the average of the sequence.
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You can always assume the values are the same so <math>\frac{540}{5}=108</math>
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=== Solution 2 ===
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Let <math>v</math>, <math>w</math>, <math>x</math>, <math>y</math>, <math>z</math> be <math>v</math>, <math>v + d</math>, <math>v+2d</math>, <math>v+3d</math>, <math>v+4d</math>, respectively. Then we have <cmath>v + w + x + y + z = 5v + 10d = 180^{\circ} (5 - 2) = 540^{\circ}</cmath>
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Dividing the equation by <math>5</math>, we have <cmath>v + 2d = x = 108^{\circ} \mathrm {(D)}</cmath>
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~ Nafer
  
 
== See also ==
 
== See also ==
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{{AMC12 box|year=2002|ab=B|num-b=3|num-a=5}}
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{{AMC12 box|year=2002|ab=B|num-b=4|num-a=6}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 04:58, 8 August 2023

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

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

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

Note that since $x$ is the middle term of an arithmetic sequence with an odd number of terms, it is simply the average of the sequence.

You can always assume the values are the same so $\frac{540}{5}=108$

Solution 2

Let $v$, $w$, $x$, $y$, $z$ be $v$, $v + d$, $v+2d$, $v+3d$, $v+4d$, respectively. Then we have \[v + w + x + y + z = 5v + 10d = 180^{\circ} (5 - 2) = 540^{\circ}\] Dividing the equation by $5$, we have \[v + 2d = x = 108^{\circ} \mathrm {(D)}\]

~ Nafer

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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