1991 AHSME Problems/Problem 12

Problem

The measures (in degrees) of the interior angles of a convex hexagon form an arithmetic sequence of integers. Let $m$ be the measure of the largest interior angle of the hexagon. The largest possible value of $m$ , in degrees, is

(A) 165 (B) 167 (C) 170 (D) 175 (E) 179

Solution

$\fbox{D}$ The angles must add to $180(6-2) = 720$ . Now, let the arithmetic progression have first term $a$ and difference $d$ , so by the sum formula, we get $\frac{6}{2}(2a+5d) = 720 \implies 2a+5d=240.$ Now, $240$ and $5d$ are both divisible by $5$ (as $d$ is an integer), so $2a$ is divisible by $5$ , and thus $a$ is divisible by 5, so $m = a+5d$ is divisible by $5$ . As the hexagon is convex, $m$ must be less than $180$ , so as it is a multiple of $5$ , it can be at most $175$ , and indeed this is possible with $a = 65, d = 22.$

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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 • 26 • 27 • 28 • 29 • 30
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1991 AHSME Problems/Problem 12

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