10. In a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, the diagonal $C A_{1}$, equal to $d$, is inclined to the base plane at an angle of $60^{\circ}$ and forms an angle of $45^{\circ}$ with the plane passing through the diagonal $A C_{1}$ and the midpoint of the lateral edge $B B_{1}$. Find the area of the base of the parallelepiped.
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_797a639488ab43e0ad60g-10.jpg?height=1131&width=1036&top_left_y=682&top_left_x=584)

If $P$ and $Q$ are the midpoints of the lateral edges $D D_{1}$ and $B B_{1}$, then $P Q \| B D$ and $\left(A Q C_{1}\right) \|(B D)$.

Draw $(M N) \| B D, A \in(M N) ; C K \perp(M N), C F \perp K C_{1}$. Since $C F \perp\left(A K C_{1}\right)$, then $\angle C O F=\gamma$ is the given angle between the diagonal $C A_{1}$ and the plane $\left(A Q C_{1}\right)$. The given angle between the diagonal $C A_{1}$ and the base plane is $\angle C A C_{1}=\alpha$.

In the right triangle $C O F \quad C F=C O \cdot \sin \angle C O F$, i.e., $C F=\frac{d}{2} \cdot \sin \gamma$.

In the triangle $С K C_{1} \quad \frac{C K}{C C_{1}}=\frac{C F}{F C_{1}}$, hence $C K=\frac{C C_{1} \cdot C F}{\sqrt{C C_{1}^{2}-C F^{2}}}$.

Since $C C_{1}=d \cdot \sin \alpha$, then $C K=\frac{d \cdot \sin \alpha \cdot d \cdot \sin \gamma}{2 \sqrt{d^{2} \sin ^{2} \alpha-\frac{d^{2} \sin ^{2} \gamma}{4}}}=\frac{d \cdot \sin \alpha \cdot \sin \gamma}{\sqrt{4 \sin ^{2} \alpha-\sin ^{2} \gamma}}$.

The area of the base of the parallelepiped is $S=\frac{1}{2} \cdot B D \cdot C K$. Since

$$
\begin{aligned}
& B D=A C=d \cdot \cos \alpha, \text { then } S=\frac{1}{2} \cdot d \cdot \cos \alpha \cdot \frac{d \cdot \sin \alpha \cdot \sin \gamma}{\sqrt{4 \sin ^{2} \alpha-\sin ^{2} \gamma}}=\frac{\sin \alpha \cdot \cos \alpha \cdot \sin \gamma}{2 \sqrt{4 \sin ^{2} \alpha-\sin ^{2} \gamma}} \cdot d^{2} . \\
& \alpha=60^{\circ}, \gamma=45^{\circ} \cdot S=\frac{\sqrt{3} \cdot d^{2}}{2 \cdot 2 \cdot 2 \cdot \sqrt{2} \cdot \sqrt{4 \cdot \frac{3}{4}-\frac{1}{2}}}=\frac{\sqrt{3} d^{2}}{8 \sqrt{5}} .
\end{aligned}
$$

Answer: $\frac{\sqrt{3} d^{2}}{8 \sqrt{5}}$.

## Solution for Variant #1

1 Two pumps, working together, fill a tank with fuel in two hours. If 80% of the tank's volume is filled by one pump, and then the remaining part by the other, the entire process will take 3 hours and 36 minutes. How many hours does it take to fill the tank with each pump separately?

## Solution:

$$
\left\{\begin{array}{c}
\frac{1}{x}+\frac{1}{y}=\frac{1}{2}, \\
\frac{4}{5} x+\frac{1}{5} y=\frac{18}{5}
\end{array} \quad \frac{1}{x}+\frac{1}{18-x}=\frac{1}{2}, x^{2}-6 x+9=0, x=3, y=6 . \quad \text { Answer: } 3 \text { hours, } 6\right. \text { hours. }
$$