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 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.

Solution 2

Assume that the pentagon is regular, with the angles is an arithmetic sequence with common difference $0$ . Then $x$ , and any other angle for that matter, will be $108$ .

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

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2002 AMC 12B Problems/Problem 5

Contents

Problem

Solution 1

Solution 2

See also