Difference between revisions of "1991 AHSME Problems/Problem 12"
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+ | == Problem == | ||
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The measures (in degrees) of the interior angles of a convex hexagon form an arithmetic sequence of integers. Let <math>m</math> be the measure of the largest interior angle of the hexagon. The largest possible value of <math>m</math>, in degrees, is | The measures (in degrees) of the interior angles of a convex hexagon form an arithmetic sequence of integers. Let <math>m</math> be the measure of the largest interior angle of the hexagon. The largest possible value of <math>m</math>, in degrees, is | ||
(A) 165 (B) 167 (C) 170 (D) 175 (E) 179 | (A) 165 (B) 167 (C) 170 (D) 175 (E) 179 | ||
+ | == Solution == | ||
+ | <math>\fbox{D}</math> The angles must add to <math>180(6-2) = 720</math>. Now, let the arithmetic progression have first term <math>a</math> and difference <math>d</math>, so by the sum formula, we get <math>\frac{6}{2}(2a+5d) = 720 \implies 2a+5d=240.</math> Now, <math>240</math> and <math>5d</math> are both divisible by <math>5</math> (as <math>d</math> is an integer), so <math>2a</math> is divisible by <math>5</math>, and thus <math>a</math> is divisible by 5, so <math>m = a+5d</math> is divisible by <math>5</math>. As the hexagon is convex, <math>m</math> must be less than <math>180</math>, so as it is a multiple of <math>5</math>, it can be at most <math>175</math>, and indeed this is possible with <math>a = 65, d = 22.</math> | ||
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+ | == See also == | ||
+ | {{AHSME box|year=1991|num-b=11|num-a=13}} | ||
+ | |||
+ | [[Category: Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:35, 23 February 2018
Problem
The measures (in degrees) of the interior angles of a convex hexagon form an arithmetic sequence of integers. Let be the measure of the largest interior angle of the hexagon. The largest possible value of , in degrees, is
(A) 165 (B) 167 (C) 170 (D) 175 (E) 179
Solution
The angles must add to . Now, let the arithmetic progression have first term and difference , so by the sum formula, we get Now, and are both divisible by (as is an integer), so is divisible by , and thus is divisible by 5, so is divisible by . As the hexagon is convex, must be less than , so as it is a multiple of , it can be at most , and indeed this is possible with
See also
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 | ||
All AHSME Problems and Solutions |
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