19. Draw a line through the center of a regular pentagon so that the sum of the squares of the distances from each vertex to this line is minimized.
19. Establish a complex plane, making the center of the pentagon correspond to the number 0, and take the constructed line as the real axis. If the radius of the circumscribed circle of the pentagon is $r$, then the five vertices correspond to the complex numbers $r\left[\cos \left(\theta+\frac{2 k \pi}{5}\right)+i \sin \left(\theta+\frac{2 k \pi}{5}\right)\right] (k=1,2,3,4)$, where $\theta$ is the argument of one of the vertices. Therefore, the sum of the squares of the distances from each vertex to the constructed line, which is the sum of the squares of the imaginary parts of the corresponding complex numbers, is:
$$
\begin{array}{l}
d=r^{2}\left[\frac{5}{2}-\frac{1}{2}\left(\cos 2 \theta+\cos \left(2 \theta+\frac{4 \pi}{5}\right)+\cos \left(2 \theta+\frac{8 \pi}{5}\right)+\cos \left(2 \theta+\frac{12 \pi}{5}\right)+\right.\right. \\
\left.\left.\cos \left(2 \theta+\frac{16 \pi}{5}\right)\right)\right] . \\
\text { Let } z=\cos 2 \theta+i \sin 2 \theta, \omega=\cos \frac{4 \pi}{5}+i \sin \frac{4 \pi}{5}, \text { then } \cos 2 \theta+\cos \left(2 \theta+\frac{4 \pi}{5}\right)+ \\
\cos \left(2 \theta+\frac{8 \pi}{5}\right)+\cos \left(2 \theta+\frac{12 \pi}{5}\right)+\cos \left(2 \theta+\frac{16 \pi}{5}\right) \text { is the real part of } z+z \omega+z \omega^{2}+z \omega^{3}+z \omega^{4}=0.
\end{array}
$$

Thus, $d=\frac{5}{2} r^{2}$, meaning the sum of the squares of the distances from each vertex to this line is independent of $\theta$.
Therefore, the constructed line can be any line passing through the center.