6. The base of the pyramid $TABC$ is the triangle $ABC$, all sides of which are equal to $\sqrt{3}$, and the height of the pyramid coincides with the lateral edge $TA$. Find the area of the section of the pyramid by a plane passing through the center of the sphere circumscribed around the pyramid, forming an angle of $60^{\circ}$ with the base plane, intersecting the edge $AB$ at point $M$ such that $MB=2AM$, and intersecting the edge $BC$. It is known that the distance from point $A$ to the section plane is 0.25.
Solution: The center of the sphere $O$ lies on the perpendicular to the base plane, drawn through the center of the base $E$; $O E=A T / 2$. The distance from point $A$ to the line of intersection $M P(P \in B C)$ of the cutting plane with the base plane $A S=A R / \sin \angle A S N$, where $N \in A T, A R$ is the height in triangle $N A S$, which is equal to the distance $d=0.25$ from point $A$ to the cutting plane. $A S=\sqrt{3} / 6=A B / 6$. Then, in the right triangle $A S M$, the angle $\angle S A M=60^{\circ}$, and then the angle $\angle S A D=90^{\circ}$, and $M P \| A D, P D=A B / 6$. Draw $O N \| A D, N \in T A, T N=A N$.

Since $O N \| M P, O N$ lies in the cutting plane. Extend $M P$ to intersect line $A C$ at point $F$, then extend $F N$ to intersect edge $T C$ at point $Q$. The quadrilateral $M N Q P$ is the desired section. To determine the position of point $Q$, draw $A G \| N Q, G \in \mathrm{TC}$; from $T N=A N$ it follows that $T Q=Q G$. Since $A F=A M=1 / 3 A C, \quad Q G=1 / 3 C G$. Therefore, $T Q=1 / 5 T C$ and $A H=1 / 5 A C$, where $H$ is the projection of $Q$ on the base plane.

Let $A B=a, T A=h$. From the similarity $F N: F Q=A N: H Q$ we get $F N: F Q=(1 / 2 h):(4 / 5 h)=5: 8$. Let $A S \perp F M$, then $N S \perp F M$. From

![](https://cdn.mathpix.com/cropped/2024_05_06_d960854a9c9c598cc600g-11.jpg?height=365&width=463&top_left_y=2056&top_left_x=1456)
$F S=S M=1 / 2 M P$ it follows that $F M=1 / 2 F P$. The area of triangle $F N M \quad S_{\triangle F N M}$ is $(5 / 8) \cdot(1 / 2)=5 / 16$ of the area of triangle $F Q P \quad S_{\triangle F Q P}$, therefore, the area of the section $S_{M N Q P}=11 / 5 S_{\triangle F N M}$. Since $A S=P D=1 / 6 a, \quad F M=2 \sqrt{3} A S=a \sqrt{3} / 3$ and
$N S=\sqrt{A S^{2}+A N^{2}}=\sqrt{(a / 6)^{2}+(h / 2)^{2}}=\sqrt{a^{2}+9 h^{2}} / 6, \quad S_{\triangle F N M}=\frac{1}{2} \cdot \frac{a}{\sqrt{3}} \cdot \frac{\sqrt{a^{2}+9 h^{2}}}{6}=\frac{a \sqrt{a^{2}+9 h^{2}}}{12 \sqrt{3}} \quad$ and $S_{M N Q P}=\frac{11 a \sqrt{a^{2}+9 h^{2}}}{60 \sqrt{3}}$. Since $h=\frac{a}{3} \operatorname{tg} \varphi=\frac{a \sqrt{3}}{3}$, then $\frac{11 \sqrt{3} a^{2}}{90}=\frac{11 \sqrt{3}}{30}$.

The area of the section can be found through the area of the projection of the section on the base plane, using the formula $S_{c e 4}=\frac{S_{n p}}{\cos \varphi}$, where $\varphi$ is the angle between the section plane and the base plane. Project the section onto the base of the pyramid. The area of the projection of the section $S_{M A H P}=S_{\triangle A B C}-S_{\triangle B M P}-S_{\triangle C H P}=\left(1-\frac{2}{3} \cdot \frac{1}{3}-\frac{4}{5} \cdot \frac{2}{3}\right) S_{\triangle A B C}=\frac{11}{45} S_{\triangle A B C}=\frac{11 \sqrt{3} a^{2}}{180}$, $S_{M N Q P}=S_{M A H P} / \cos \angle N S A=S_{M A H P} / \cos 60^{\circ}=\frac{11 \sqrt{3} a^{2}}{90}=\frac{11 \sqrt{3}}{30}$.

Answer: $\frac{11 \sqrt{3}}{30}$.

## Solution for Variant №6 (11th grade)