1960 AHSME Problems/Problem 27

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Problem

Let $S$ be the sum of the interior angles of a polygon $P$ for which each interior angle is $7\frac{1}{2}$ times the exterior angle at the same vertex. Then

$\textbf{(A)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may be regular}\qquad \\ \textbf{(B)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad \\ \textbf{(C)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is regular}\qquad \\ \textbf{(D)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad \\ \textbf{(E)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may or may not be regular}$

Solution

Let $a_n$ be the interior angle of the nth vertex, and let $b_n$ be the exterior angle of the nth vertex. From the conditions in the problem, \[a_n = 7.5b_n\] That means \[a_1 + a_2 \cdots a_n = 7.5(b_1 + b_2 \cdots b_n)\] Since the sum of the exterior angles of a polygon is $360^{\circ}$, the equation can be simplified as \[a_1 + a_2 \cdots a_n = 7.5 \cdot 360\] \[a_1 + a_2 \cdots a_n = 2700\] The sum of the interior angles is $2700$ degrees. However, there is no other constraint on what angle the interior angles can be, so we can not for sure claim that the polygon is regular or not. Thus, the answer is $\boxed{\textbf{(E)}}$.


See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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