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Difference between revisions of "1962 AHSME Problems/Problem 31"

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Latest revision as of 17:19, 25 March 2021

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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