Difference between revisions of "1968 AHSME Problems/Problem 20"
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== Solution == | == Solution == | ||
| − | <math>\fbox{}</math> | + | |
| + | The formula for the sum of the angles in any polygon is <math>180(n-2)</math>. Because this particular polygon is convex and has its angles in an arithmetic sequence with its largest angle being <math>160</math>, we can find the sum of the angles. | ||
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| + | <math>a_{n}=160</math> | ||
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| + | <math>a_{1}=160-5(n-1)</math> | ||
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| + | Plugging this into the formula for finding the sum of an arithmetic sequence... | ||
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| + | <math>n(\frac{160+160-5(n-1)}{2})=180(n-2)</math>. | ||
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| + | Simplifying, we get <math>n^2+7n-144</math>. | ||
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| + | Since we want the positive solution to the quadratic, we can easily factor and find the answer is <math>n=\boxed{9}</math>. | ||
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| + | Hence the answer is<math>\fbox{A}</math> | ||
== See also == | == See also == | ||
Revision as of 14:32, 7 June 2017
Problem
The measures of the interior angles of a convex polygon of 
sides are in arithmetic progression. If the common difference is 
and the largest angle is 
, then equals:
Solution
The formula for the sum of the angles in any polygon is 
. Because this particular polygon is convex and has its angles in an arithmetic sequence with its largest angle being 
, we can find the sum of the angles.
Plugging this into the formula for finding the sum of an arithmetic sequence...
.
Simplifying, we get 
.
Since we want the positive solution to the quadratic, we can easily factor and find the answer is 
.
Hence the answer is
See also
| 1968 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
