10. The base of the pyramid $T A B C D$ is a rectangle $A B C D$. The height of the pyramid coincides with the lateral edge $T A$, and the lateral edge $T C$ is inclined to the base plane at an angle of $45^{\circ}$. The plane passing through the edge $T C$ and parallel to the diagonal of the base $B D$ forms an angle of $60^{\circ}$ with the base plane, and the distance between this plane and the diagonal $B D$ is $a$. What is the smallest area that the section of the pyramid by a plane passing through the diagonal of the base $A C$ can have?
Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_797a639488ab43e0ad60g-28.jpg?height=1179&width=1483&top_left_y=560&top_left_x=229)

Draw $K C \| B D, A K \perp K C, L=A K \cap B D, L S \perp T K$, then $L S=a$. Let $\beta=\angle T K A$. Since $A K=2 L K$, then $A K=2 a / \sin \beta$ and $A T=A K \operatorname{tg} \beta=2 a / \cos \beta$. Let $\alpha=\angle T C A$, then $A C=A T \operatorname{ctg} \alpha=2 a \operatorname{ctg} \alpha / \cos \beta$.

Draw $A_{1} C_{1}\left\|A C, B \in\left(A_{1} C_{1}\right) ; A F \perp\left(A_{1} C_{1}\right), A Q \perp T F, Q M\right\|\left(A_{1} C_{1}\right)$ and $M N \perp A C$; then $M N$ is the height of the section $A M C$, having the shortest length, and $M N=A Q$.

Find the area of the section: $A F=L K=a / \sin \beta$,

$$
\begin{aligned}
& T F=\sqrt{A T^{2}+A F^{2}}=\sqrt{4 a^{2} / \cos ^{2} \beta+a^{2} / \sin ^{2} \beta}=a \sqrt{3 \sin ^{2} \beta+1} / \sin \beta \cos \beta \\
& A Q=A T \cdot A F / T F=2 a / \sqrt{3 \sin ^{2} \beta+1}, S_{\triangle A M C}=0.5 A C \cdot A Q=2 a^{2} \operatorname{ctg} \alpha / \sqrt{3 \sin ^{2} \beta+1} \cos \beta
\end{aligned}
$$

$\alpha=45^{\circ}, \beta=60^{\circ}$.

$A F=L K=\frac{a}{\sin 60^{\circ}}=\frac{2 a}{\sqrt{3}} ; A T=\frac{2 a}{\cos 60^{\circ}}=4 a ; A C=A T \operatorname{ctg} 45^{\circ}=4 a$.

$$
A Q=\frac{A T \cdot A F}{T F}=\frac{4 a \cdot 2 a}{\sqrt{3} \sqrt{16 a^{2}+\left(\frac{2}{\sqrt{3}} a\right)^{2}}}=\frac{4 a}{\sqrt{13}} \cdot S=\frac{1}{2} a \cdot \frac{4}{\sqrt{13}} \cdot 4 a=\frac{8}{\sqrt{13}} a^{2}
$$

Answer: $S=8 a^{2} / \sqrt{13}$.