Difference between revisions of "1968 AHSME Problems/Problem 20"

Revision as of 00:53, 16 August 2023

Problem

The measures of the interior angles of a convex polygon of $n$ sides are in arithmetic progression. If the common difference is $5^{\circ}$ and the largest angle is $160^{\circ}$ , then $n$ equals:

$\text{(A) } 9\quad \text{(B) } 10\quad \text{(C) } 12\quad \text{(D) } 16\quad \text{(E) } 32$

Solution

The formula for the sum of the angles in any polygon is $180(n-2)$ . Because this particular polygon is convex and has its angles in an arithmetic sequence with its largest angle being $160$ , we can find the sum of the angles.

$a_{n}=160$

$a_{1}=160-5(n-1)$

Plugging this into the formula for finding the sum of an arithmetic sequence...

$n(\frac{160+160-5(n-1)}{2})=180(n-2)$ .

Simplifying, we get $n^2+7n-144$ .

Since we want the positive solution to the quadratic, we can easily factor and find the answer is $n=\boxed{9}$ .

Hence the answer is $\fbox{A}$

1968 AHSC (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 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 28: / Line 28: @@
 == See also ==
-{{AHSME box|year=1968|num-b=19|num-a=21}}
+{{AHSME 35p box|year=1968|num-b=19|num-a=21}}
 [[Category: Introductory Geometry Problems]]
 {{MAA Notice}}

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Difference between revisions of "1968 AHSME Problems/Problem 20"

Revision as of 00:53, 16 August 2023

Problem

Solution

See also