8. There are three points $A$, $B$, and $C$ on the ground. A frog is located at point $P$, which is 0.27 meters away from point $C$. The frog's first jump is from $P$ to the symmetric point $P_{1}$ with respect to $A$, which we call the frog making a "symmetric jump" from $P$ about $A$; the second jump is from $P_{1}$ to the symmetric point $P_{2}$ with respect to $B$; the third jump is from $P_{2}$ to the symmetric point $P_{3}$ with respect to $C$; the fourth jump is from $P_{3}$ to the symmetric point $P_{1}$ with respect to $A$, and so on. If the frog reaches $P_{2009}$ after 2009 symmetric jumps, what is the distance between this point and the starting point $P$ in centimeters?
8. Suppose we establish a Cartesian coordinate system on the ground such that the starting point $P$ is exactly the origin, and the coordinates of points $A$, $B$, and $C$ are $\left(a_{1}, a_{2}\right)$, $\left(b_{1}, b_{2}\right)$, and $\left(c_{1}, c_{2}\right)$, respectively. Since the coordinates of point $P$ are $(0,0)$, according to the definition of symmetric jumps, the coordinates of point $P_{1}$ are $\left(2 a_{1}, 2 a_{2}\right)$. Let $P_{2}\left(x_{2}, y_{2}\right)$. Since $B$ is the midpoint of $P_{1}$ and $P_{2}$, we have $b_{1}=\frac{2 a_{1}+x_{2}}{2}$ and $b_{2}=\frac{2 a_{2}+y_{2}}{2}$.

Thus, $x_{2}=2 b_{1}-2 a_{1}$ and $y_{2}=2 b_{2}-2 a_{2}$.
Let $P,\left(x, y_{i}\right), i=1,2, \cdots, 6$.
By the problem statement, we have $x_{3}=2 c_{1}-x_{2}=2\left(c_{1}-b_{1}+a_{1}\right)$.
$$
\begin{array}{l}
x_{1}=2 a_{1}-x_{3}=2\left(b_{1}-c_{1}\right), \\
x_{3}=2 b_{1}-x_{1}=2 c_{1}, \\
x_{5}=2 b_{1}-x_{3}=0 .
\end{array}
$$

Similarly, we get $y_{6}=0$.
This indicates that point $P_{5}$ coincides with point $P$.
Therefore, after 6 symmetric jumps about $A$, $B$, and $C$, the frog returns to the origin, meaning such symmetric jumps have a period of 6.
Since $2009=3 \times 668+5$, after 2009 steps of symmetric jumps, it is equivalent to only performing 5 symmetric jumps, or being just one step away from returning to the starting point, i.e., $P_{2009}=P_{5}$.

And $P_{5}$ is the point symmetric to $P$ about $C$, so the distance from $P_{0}$ to $P$ is equal to the distance from $P_{3}$ to $P$, which is $2 \times 0.27=0.54=54$ (centimeters).