1961 AMC 12 Problem 27

Attempt Problem 27 of the 1961 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 1961 AMC 12 solutions, or check the answer key.

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

27.

Given two equiangular polygons P1P_1 and P2P_2 with different numbers of sides; each angle of P1P_1 is xx degrees and each angle of P2P_2 is kxkx degrees, where kk is an integer greater than 1.1. The number of possibilities for the pair (x,k)(x,k) is:
infinite
finite, but greater than two
two
one
zero
Answer: D
Concepts:equiangular polygonangle suminequality
Small hint:
An equiangular nn-gon has angle 180∘−360∘n180^\circ-\frac{360^\circ}{n}
Big hint:
Use kx<180∘kx<180^\circ and k≥2k\ge2 to constrain the smaller angle xx
Solution:
Every polygon angle is at least 60∘,60^\circ, and every convex polygon angle is less than 180∘.180^\circ. Since k≥2k\ge2 and kx<180∘,kx<180^\circ, we need x<90∘.x<90^\circ. The only possible equiangular polygon angle in [60∘,90∘)[60^\circ,90^\circ) is the triangle angle x=60∘.x=60^\circ. Then k=2k=2 gives kx=120∘,kx=120^\circ, the angle of a regular hexagon; larger kk is impossible. Hence there is one pair.
Thus, the correct answer is D.
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