The ratio of the interior angles of two regular polygons is $3:2$. How many such pairs are there?
My original idea was to incorporate elementary modular arithmetic so either eliminate or zone in particular pairs.
Just as a background, there is no calculus required (as the book has not talked about) no too complex math (the highest level maths is probably Algebra 2, 11th grade).
Also, I'm not out here looking for a solution but hint or idea to help me, so please do not give a full solution.
On
Suppose the polygons have $m$ and $n$ sides respectively.
The interior angle of a regular $x$-gon is $180 - \frac{360}{x}$
Hence
$\frac{180 - \frac{360}{m}}{180 - \frac{360}{m}} = \frac{3}{2}$
Simplifying,
$6m - 4n = mn$
The above equation has $4$ solutions:
$(2, 2), (4, 3), (8, 4), (20, 5)$
In this case, we discard the solution $(2,2)$
Hence the final solutions are $(4, 3), (8, 4), (20, 5)$
Let the number of sides of the polygons be $n$ and $k$. We have:
$$\frac{180(n-2)/n}{180(k-2)/k} = \frac{3}{2}$$ $$2(180 \cdot\frac{n-2}{n}) = 3(180\cdot\frac{k-2}{k})$$ $$\frac{2n-4}{n} = \frac{3k-6}{k}$$ $$2nk-4k = 3nk-6n$$ $$nk + 4k - 6n = 0$$
Now if we subtract $24$ from both sides:
$$nk + 4k -6n - 24 = -24$$ $$(n+4)(k-6) = -24$$
Can you continue?