1960 AMC 12 Problem 27

Attempt Problem 27 of the 1960 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1960 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

27.

Let SS be the sum of the interior angles of a polygon PP for which each interior angle is 7127\dfrac12 times the exterior angle at the same vertex. Then:
S=2660∘S=2660^\circ and PP may be regular
S=2660∘S=2660^\circ and PP is not regular
S=2700∘S=2700^\circ and PP is regular
S=2700∘S=2700^\circ and PP is not regular
S=2700∘S=2700^\circ and PP may or may not be regular
Answer: E
Concepts:angle sumequiangular polygonregular polygon
Small hint:
An interior angle and its corresponding exterior angle sum to 180∘180^\circ
Big hint:
The condition fixes every angle but says nothing about the side lengths
Solution:
If an exterior angle is e,e, then e+152e=180∘, e+\frac{15}{2}e=180^\circ, so e=360∘17.e=\frac{360^\circ}{17}. Since the exterior angles total 360∘,360^\circ, the polygon has 1717 vertices and S=(17−2)180∘=2700∘. S=(17-2)180^\circ=2700^\circ. All its angles are equal, but its sides need not be equal, so it may or may not be regular.
Thus, the correct answer is E.
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