10. The base of the pyramid $TABC$ is the triangle $ABC$, all sides of which are equal to $a$, and the height of the pyramid coincides with the lateral edge $TA$. Find the area of the section of the pyramid by a plane that passes through the midpoint of the base side $AC$, is parallel to the median $AM$ of the lateral face $ATB$, and intersects the edge $AT$ at point $N$ such that $TN = 3AN$, and the distance from $AM$ to the cutting plane is $d$.

| Option | 1 |
| :---: | :---: |
| $a$ | $4 \sqrt{2}$ |
| $d$ | $\sqrt{4 / 7}$ |
Solution: Draw

$P N \| A M, S=(P N) \cap(A B)$. Let $H-$ be the midpoint

$A B, \quad M H=1 / 2 \cdot A T$,

$A N=1 / 4 \cdot A T=1 / 2 \cdot M H$.

$\Rightarrow \quad S A=1 / 2 \cdot A H=1 / 4 \cdot A B$.

Since $P N=3 / 4 \cdot A M$, $S N=1 / 2 A M \Rightarrow S N=2 / 3 \cdot P N$, or $S N=2 / 5 \cdot S P$.

Draw $S D, \quad F=(S D) \cap(B C), C E \| S F$. Since $A D=D C, \quad S D=1 / 2 \cdot E C$ and $E C=6 / 5 \cdot S F$, $S D=3 / 5 \cdot S F$.

We have $\quad S_{N S D}=\frac{1}{2} \cdot S N \cdot S D \sin \angle N S D=\frac{1}{2} \cdot \frac{2}{5} S P \cdot \frac{3}{5} S F \sin \angle P S F=\frac{6}{25} S_{P S F}$. Then the area of the section $S_{N D F P}=\frac{19}{25} S_{P S F}=\frac{19}{6} S_{N S D}$.

Draw $A K \perp S D, A L \perp K N$, the length of $A L$ is equal to the given distance $d$ between $A M$ and the cutting plane. In triangle $S A D$ we have
$S D=\sqrt{A S^{2}+S D^{2}-2 A S \cdot S D \cos 120^{\circ}}=\sqrt{\frac{a^{2}}{16}+\frac{a^{2}}{4}-2 \frac{a}{4} \cdot \frac{a}{2}\left(-\frac{1}{2}\right)}=\frac{a \sqrt{7}}{4}, \quad S_{S A D}=\frac{1}{2} S A \cdot A D \sin 120^{\circ}=\frac{a^{2} \sqrt{3}}{32}$.

But $S_{S A D}=\frac{1}{2} A K \cdot S D$, hence $\frac{1}{2} A K \cdot \frac{a \sqrt{7}}{4}=\frac{a^{2} \sqrt{3}}{32}$ and $A K=\frac{a \sqrt{3}}{4 \sqrt{7}}$. In triangle $A K N$ we have

$K L=\sqrt{A K^{2}-A L^{2}}=\sqrt{\frac{3 a^{2}}{16 \cdot 7}-d^{2}}=\frac{\sqrt{3 a^{2}-112 d^{2}}}{4 \sqrt{7}}, \quad K N=\frac{A K^{2}}{K L}=\frac{3 a^{2}}{4 \sqrt{7} \sqrt{3 a^{2}-112 d^{2}}} . \quad$ The area

of triangle $N S D: S_{N S D}=\frac{1}{2} \cdot S D \cdot K N=\frac{3 a^{3}}{32 \sqrt{3 a^{2}-112 d^{2}}}$. Then $S_{N D F P}=\frac{19 a^{3}}{64 \sqrt{3 a^{2}-112 d^{2}}}$.

Answer: $19 / 2$

## Diagrams for the problem

![](https://cdn.mathpix.com/cropped/2024_05_06_159c92d956ef263142f0g-06.jpg?height=1506&width=942&top_left_y=1142&top_left_x=520)

## Solution for Variant #4

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