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Difference between revisions of "1991 AHSME Problems/Problem 12"

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(A) 165  (B) 167 (C) 170 (D) 175 (E) 179
 
(A) 165  (B) 167 (C) 170 (D) 175 (E) 179
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== Solution ==
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== Solution ==  
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<math>\fbox{D}</math>
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<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>
  
 
== See also ==
 
== See also ==

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 $m$ 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 $a = 65, d = 22.$

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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