Problem: Find the number of rectangles that can be formed inside a fixed regular dodecagon ( $12$ -gon) where each side of the rectangle lies on either a side or a diagonal of the dodecagon. The diagram below shows three of those rectangles.
[asy] unitsize(0.6 inch); for(int i=0; i<360; i+=30) { dot(dir(i), 4+black); draw(dir(i)--dir(i+30)); } draw(dir(120)--dir(330)); filldraw(dir(210)--dir(240)--dir(30)--dir(60)--cycle, mediumgray, linewidth(1.5)); draw((0,0.366)--(0.366,0), linewidth(1.5)); [/asy]

Solution: We put the dodecagon in the right position that there exists a side whose slope is 0.
Note that finding a rectangle is equivalent to finding two pairs of lines, such that two lines in each pair are parallel and lines from different pairs are perpendicular.
Now, we use this property to count the number of rectangles.
Because two pairs of lines that form a rectangle are perpendicular, we only need to use the slope of one pair, denoted as $k$ , to determine the direction of the rectangle.
The slope of the other pair is thus $- \frac{1}{k}$ . To avoid overcounting, we do casework analysis by defining each case in term of $0 \leq k < \infty$ only (we make a convention that if $k = 0$ , then $- \frac{1}{k} = \infty$ ).
In our counting, we will frequently quantify the distance between two vertices of the regular dodecagon.
To characterize this in a straightforward way, we simply measure the number of vertices (on the minor arc) between our measured two vertices. For instance, two vertices on a side has distance 0. Distances between two vertices that are diagonals can be 1, 2, 3, 4, 5.
Case 1: $k = 0, \tan 30^\circ, \tan 60^\circ$
We only count for $k = 0$ . The number of solutions for $k = \tan 30^\circ$ and $\tan 60^\circ$ are the same.
Consider $k = 0$ .
We need to find a pair of horizontal segments and a pair of vertical segments to form a rectangle.
For $k = 0$ , the length of each horizontal segment can only be 0, 2, 4.
Denote by $2i$ the shorter length of two parallel horizontal segments.
Given $i$ , the number of pairs of two parallel horizontal segments is $1 + 2 \left( 4 - 2 i \right)$
Given $i$ , to form a rectangle, the number of pairs of vertical segments is $\binom{2i + 2}{2}$
Therefore, for $k = 0$ , the number of rectangles is
\begin{align*}
\sum_{i=0}^2 \left( 1 + 2 \left( 4 - 2 i \right) \right)
\binom{2i + 2}{2}
& = 54 .
\end{align*}
The number of rectangles for $k = \tan 30^\circ$ and $\tan 60^\circ$ are the same.
Therefore, the total number of rectangles in this case is $54 \cdot 3 = 162$
Case 2: $k = \tan 15^\circ$ $\tan 45^\circ$ $\tan 75^\circ$
The number of rectangles under all these $k$ s are the same.
So we only count for $k = \tan 15^\circ$
For $k = \tan 15^\circ$ , the length of each segment can only be 1, 3, 5.
However, there is only one segment with length 5.
So this cannot be the shorter length of two parallel segments with slope $\tan 15^\circ$
Denote by $2i + 1$ the shorter length of two parallel segments with slope $\tan 15^\circ$ .
Given $i$ , the number of pairs of two parallel segments is $1 + 2 \left( 3 - 2 i \right)$
Given $i$ , to form a rectangle, the number of pairs of vertical segments is $\binom{2i + 3}{2}$
Therefore, for $k = \tan 15^\circ$ , the number of rectangles is
\begin{align*}
\sum_{i=0}^1 \left( 1 + 2 \left( 3 - 2 i \right) \right)
\binom{2i + 3}{2}
& = 51 .
\end{align*}
The number of rectangles for $k = \tan 45^\circ$ and $\tan 75^\circ$ are the same.
Therefore, the total number of rectangles in this case is $51 \cdot 3 = 153$
Putting all cases together, the total number of rectangles is $162 + 153 = \boxed{315}$

Answer:  315