Spiraling Triangles?!?

For the bonus generalization, the area of a regular polygon with $n$ sides with length $s$ is $A = \frac{n}{4 \tan \frac{\pi}{n}} s^2$ .

Since each interior angle $x$ is $x = \frac{\pi(n - 2)}{n}$ , there will be $c = \frac{2\pi}{x} = \frac{2\pi}{\frac{\pi(n - 2)}{n}} = \frac{2n}{n - 2}$ polygons that can make up a full circle.

This means that the $k^{\text{th}}$ shape in the spiral has an area of $A_{n, k} = \frac{n}{4 \tan \frac{\pi}{n}} (k^2 - (k - c)^2)$ for $k \geq \frac{2n}{n - 2}$ , which simplifies to:

$A_{n, k} = \frac{n^2((n - 2)k - n)}{(n - 2)^2 \tan \frac{\pi}{n}}$

In this problem, $n = 3$ and $k = 18$ , so $A_{3, 18} = \frac{3^2((3 - 2)18 - 3)}{(3 - 2)^2 \tan \frac{\pi}{3}} = 45\sqrt{3} = \boxed{\frac{\sqrt{3}(18^2 - 12^2)}{4}}$ .

Since $c$ must be an integer and $c = \frac{2n}{n - 2} = 2 + \frac{4}{n - 2}$ , $n - 2$ must be a factor of $4$ , so $n - 2 = 1$ or $n - 2 = 2$ or $n - 2 = 4$ , which solves to $n = 3$ , $n = 4$ , or $n = 6$ . Therefore, only triangles, squares, and hexagons can be used. For each shape, the following general equations can be derived:

triangle: $A_{3, k} = \frac{3^2((3 - 2)k - 3)}{(3 - 2)^2 \tan \frac{\pi}{3}} = 3\sqrt{3}(k - 3)$ for $k \geq 6$

square: $A_{4, k} = \frac{4^2((4 - 2)k - 4)}{(4 - 2)^2 \tan \frac{\pi}{4}} = 8(k - 2)$ for $k \geq 4$

hexagon: $A_{6, k} = \frac{6^2((6 - 2)k - 6)}{(6 - 2)^2 \tan \frac{\pi}{6}} = \frac{9\sqrt{3}}{2}(2k - 3)$ for $k \geq 3$