1962 AHSME Problems/Problem 31

Problem

The ratio of the interior angles of two regular polygons with sides of unit length is $3: 2$ . How many such pairs are there?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{infinitely many}$

Solution

The formula for the measure of the interior angle of a regular polygon with $n$ -sides is $180 - \frac{360}{n}$ . Letting our two polygons have side length $r$ and $k$ , we have that the ratio of the interior angles is $\frac{180 - \frac{360}{r}}{180 - \frac{360}{k}} = \frac{(r-2) \cdot k}{(k-2) \cdot r} = \frac{3}{2}$ . Cross multiplying both sides, we have $2rk-4k = 3kr-6r \Rightarrow -rk-4k+6r = 0$ . Using [Simon's Favorite Factoring Trick], we have $-k(r+4)+6r+24 = 24 \Rightarrow (r+4)(6-k)=24$ . Because $k$ and $r$ are both more than $2$ , we know that $6-k < r+4$ . Now, we just set these factors equal to the factors of 24. We can set $6-k$ to $1$ , $2$ , $3$ , or $4$ and $r+4$ to $24$ , $12$ , $8$ , or $6$ respectively to get the following pairs for $(k, r)$ : $(5, 20)$ , $(4, 8)$ , $(3, 4)$ , and $(2, 2)$ . However, we have to take out the solution with $k = 2$ , because $k$ and $r$ are both more than $2$ , leaving us with $\boxed{\textbf{(C)}\ 3}$ as the correct answer.

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1962 AHSME Problems/Problem 31

Problem

Solution