10. A plane is drawn through the diagonal of a rectangular parallelepiped and a point lying on a lateral edge that does not intersect this diagonal, such that the area of the section of the parallelepiped by this plane is the smallest. Find the lengths of the sides of the base of the parallelepiped, given that the diagonals of the section are 20 and 8, and the angle between them is \(60^{\circ}\).

![](https://cdn.mathpix.com/cropped/2024_05_06_331dee27c7cee369cafag-37.jpg?height=1114&width=1447&top_left_y=214&top_left_x=313)
# Solution:

Draw $D L \perp A C, L K\left\|C C_{1}\left(K \in A C_{1}\right), P K\right\| D L$. By laying off the segment $B Q=P D_{1}$ on the lateral edge $B B_{1}$, we obtain the parallelogram $P A Q C_{1}$, which will be the section of the smallest area; in this case, $A C_{1}$ is its larger diagonal, and $P Q$ is the smaller diagonal, given in the condition.

$A C_{1}=20, P Q=8, \varphi=60^{\circ}, \cos \varphi=1 / 2, \sin \varphi=\sqrt{3} / 2$.

$D L=P K=\frac{1}{2} \cdot P Q \cdot \sin \varphi=\frac{1}{2} \cdot 8 \cdot \frac{\sqrt{3}}{2}=2 \sqrt{3} . \quad O K=\frac{1}{2} \cdot P Q \cdot \cos \varphi=\frac{1}{2} \cdot 8 \cdot \frac{1}{2}=2$.

$\frac{C L}{A L}=\frac{C_{1} K}{A K}=\frac{C_{1} O-O K}{C_{1} O+O K}=\frac{10-2}{10+2}=\frac{2}{3}$. Let $C L=2 x$, then $A L=3 x$.

$D L^{2}=C L \cdot A L$, i.e., $12=6 x^{2}, \quad x=\sqrt{2}, \quad C L=\sqrt{2}, A L=3 \sqrt{2}$.

Then $C D=\sqrt{D L^{2}+C L^{2}}=\sqrt{12+8}=2 \sqrt{5}, A D=\sqrt{D L^{2}+A L^{2}}=\sqrt{12+18}=\sqrt{30}$.

Answer: $2 \sqrt{5} ; \sqrt{30}$.