A $100$-gon $P_1$ is drawn in the Cartesian plane.  The sum of the $x$-coordinates of the $100$ vertices equals 2009.  The midpoints of the sides of $P_1$ form a second $100$-gon, $P_2$.  Finally, the midpoints of the sides of $P_2$ form a third $100$-gon, $P_3$.  Find the sum of the $x$-coordinates of the vertices of $P_3$.
This problem involves polygons and coordinates, so I will try to use some geometry and algebra to solve it. I notice that the sum of the $x$-coordinates of the vertices of a polygon is related to the centroid of the polygon, which is the average of the coordinates of the vertices. In fact, the $x$-coordinate of the centroid is the sum of the $x$-coordinates of the vertices divided by the number of vertices. So, the $x$-coordinate of the centroid of $P_1$ is $2009/100 = 20.09$. Now, I recall that the centroid of a polygon is also the centroid of the midpoints of its sides, so the centroid of $P_2$ is the same as the centroid of $P_1$. Therefore, the sum of the $x$-coordinates of the vertices of $P_2$ is also $20.09 \times 100 = 2009$. Similarly, the centroid of $P_3$ is the same as the centroid of $P_2$, so the sum of the $x$-coordinates of the vertices of $P_3$ is also $2009$. 