10. The base of the pyramid $T A B C D$ is a rectangle with sides $A B=12$ and $A D=4$, and the lateral edges are respectively equal to $T A=3, T D=5, T C=13$. What is the minimum area that the cross-section of the pyramid can have when cut by a plane passing through vertex $T$, the center of symmetry of the base, and a point $M$ lying on edge $B C$? In this case, into what parts does point $M$ divide edge $B C$?
# Solution:

When constructing the drawing, the following should be taken into account.

$$
\begin{aligned}
& T A^{2}+A D^{2}=T D^{2},\left(3^{2}+4^{2}=5^{2}\right) \Rightarrow \angle T A D=\pi / 2 \\
& A B=D C=12 \\
& D C^{2}+T D^{2}=T C^{2},\left(12^{2}+5^{2}=13^{2}\right) \Rightarrow \angle T D C=\pi / 2 \\
& D C \perp T A D, A B \perp T A D, \angle T A B=\pi / 2 \\
& \text { Therefore, } T A \perp A B C D .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_331dee27c7cee369cafag-22.jpg?height=928&width=1465&top_left_y=947&top_left_x=361)

For any position of point \( M \) on side \( B C \), the face \( T A B \) is the orthogonal projection of the section \( T M N \). The area of the section will be the smallest if the angle between the cutting plane and the face \( T A B \) is the smallest. Since the cutting plane passes through the center of symmetry of the base \( O \) and the vertex of the pyramid \( T \), the segment \( O T \) is the oblique to the plane of the face \( T A B \), and the smallest possible angle will be \( \angle O T F \), where \( O F \perp T A B, F \in A B \). The line of intersection of the cutting plane and the plane of the face \( T A B \) is \( T K \perp T F \) and intersects the line \( A B \) at point \( K \). If the condition \( T K \perp T F \) is not met, then \( F T_{1}\operatorname{tg} \alpha, \cos \alpha_{1}\frac{S_{\triangle A T B}}{\cos \alpha} \).

The line passing through points \( K \) and \( O \) intersects the edge \( A D \) at point \( N \) and the edge \( B C \) at point \( M \), and \( \triangle N T M \) is the desired section.

If we denote \( A B=a, B C=b, T A=c \), then \( T F=\sqrt{(a / 2)^{2}+c^{2}} \);

\( T O=\sqrt{T F^{2}+O F^{2}}=\sqrt{(a / 2)^{2}+c^{2}+(b / 2)^{2}}=\sqrt{a^{2}+b^{2}+4 c^{2}} / 2, \cos \angle O T F=\frac{T F}{T O}=\frac{\sqrt{a^{2}+4 c^{2}}}{\sqrt{a^{2}+b^{2}+4 c^{2}}} \).

\( S_{\triangle N T M}=\frac{S_{\triangle A T B}}{\cos \alpha}=\frac{a c \sqrt{a^{2}+b^{2}+4 c^{2}}}{2 \sqrt{a^{2}+4 c^{2}}}=\frac{a c \sqrt{(a / 2)^{2}+(b / 2)^{2}+c^{2}}}{2 \sqrt{(a / 2)^{2}+c^{2}}} \).

In \( \triangle K T F \) where \( \angle K T F=90^{\circ}, A K=A T^{2} / A F=2 c^{2} / a . B K=A K+A B=\frac{2 c^{2}}{a}+a \).

Point \( M \) divides the segment \( B C \) in the ratio \( \frac{B M}{M C}=\frac{B M}{A N}=\frac{B K}{A K}=\frac{a^{2}+2 c^{2}}{2 c^{2}}=1+\frac{1}{2} \frac{a^{2}}{c^{2}} \).

| \( A B \) | \( B C \) | \( T A \) | \( T F \) | \( T O \) | \( \cos \alpha \) | \( S_{N T M} \) | \( B M: M C \) | \( B M \) | \( M C \) |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 12 | 4 | 3 | \( 3 \sqrt{5} \) | 7 | \( 3 \sqrt{5} / 7 \) | \( 42 / \sqrt{5} \) | \( 9: 1 \) | \( 18 / 5 \) | \( 9 / 5 \) |

## First (preliminary) stage of the academic competition of the "Step into the Future" School Students' Olympiad in the subject of "Mathematics", autumn 2016.  Variant No. 9