10. The base of the pyramid is a rectangle with sides $AB=24$ and $BC=30$, and the lateral edge of the pyramid $TA=16$ is perpendicular to the plane of the base. What is the minimum area that the section of the pyramid by a plane passing through the center of symmetry of the base $O$, the vertex of the pyramid, and a point $M$ lying on the side $BC$ can have? In this case, into what parts does point $M$ divide the edge $BC$?
# Solution:

Regardless of the 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 inclined line 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 $T A B \quad T K \perp T F$ 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 $M, \triangle N T M$ is the desired section.

![](https://cdn.mathpix.com/cropped/2024_05_06_331dee27c7cee369cafag-15.jpg?height=925&width=1471&top_left_y=174&top_left_x=401)

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

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

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

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

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

| $A B$ | $B C$ | $T A$ | $T F$ | $T O$ | $\cos \alpha$ | $S_{N T M}$ | $B M: M C$ | $B M$ | $M C$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 24 | 30 | 16 | 20 | 25 | $4 / 5$ | 240 | $17: 8$ | $102 / 5$ | $48 / 5$ |

First (qualifying) stage of the academic competition

"Step into the Future" School Olympiad in the subject of "Mathematics", autumn 2016