# 1961 AHSME Problems/Problem 27

## Problem

Given two equiangular polygons $P_1$ and $P_2$ with different numbers of sides; each angle of $P_1$ is $x$ degrees and each angle of $P_2$ is $kx$ degrees, where $k$ is an integer greater than $1$. The number of possibilities for the pair $(x, k)$ is: $\textbf{(A)}\ \infty\qquad \textbf{(B)}\ \text{finite, but greater than 2}\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$

## Solution

Each angle in each equiangular polygon is $\frac{180(n-2)}{n} = 180 - \frac{360}{n}$, where $n$ is the number of sides. As $n$ gets larger, each angle in the equiangular polygon approaches (but does not reach) $180^{\circ}$. That means $kx < 180$, so $x < \frac{180}{k}$.

Recall that each angle in an equiangular triangle has $60^{\circ}$, each angle in an equiangular quadrilateral has $90^{\circ}$, and so on. If $k = 2$, then $x < 90$. That means the only option available is $x = 60$ because an equiangular triangle is the only equiangular polygon to have acute angles. If $k = 3$, then $x < 60$, and there are no values of $x$ that satisfy. As $k$ increases, $\frac{180}{k}$ decreases even further, so we can for sure say that there is only 1 possibility for the pair $(x, k)$, which is answer choice $\boxed{\textbf{(D)}}$.

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