2002 AMC 12B Problems/Problem 5

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)
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2002 AMC 12B Problems/Problem 5

Contents

Problem

Solution

Solution 1

Solution 2

See also