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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 18: / Line 18: @@
 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.
-You can always assume the values are the same so <math>\frac{540}{5}=109</math>
+You can always assume the values are the same so <math>\frac{540}{5}=108</math>
 === Solution 2 ===

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