6. The base of the pyramid $TABC$ is the triangle $ABC$, all sides of which are equal to 3, and the height of the pyramid, equal to $\sqrt{3}$, coincides with the lateral edge $TA$. Find the area of the section of the pyramid by a plane that passes through the center of the sphere circumscribed around the pyramid, is parallel to the median $AD$ of the base, and forms an angle of $60^{\circ}$ with the base plane.
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 $E$ to the line of intersection of the cutting plane with the base plane $E L=O E \cdot \operatorname{ctg} \angle O L E$. In all cases, the conditions are chosen such that $E L=A B / 6$, i.e., the length $E L$ is equal to the distance from point $E$ to the line $M P$, and then the line of intersection of the cutting plane with the base plane is $M P$, with $M P \| A D$. 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 the line $A C$ at point $F$, then $F N$ - to intersect the edge $T C$ at point $Q$. The quadrilateral $M N Q P$ is the desired section. To determine the position of point $Q$, draw $A G \| N Q, G \in T C$; from $T N=A N$ it follows that $T Q=Q G$. Since $A F=A M=1 / 3 A C, Q G=1 / 3 C G$. 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 $A K \perp F M$, then $N K \perp F M$. From $F K=K M=1 / 2 M P$ 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 $A K=P D=1 / 6 a, F M=2 \sqrt{3} A K=a \sqrt{3} / 3$ and $N K=\sqrt{A K^{2}+A N^{2}}=\sqrt{(a / 6)^{2}+(h / 2)^{2}}=\sqrt{a^{2}+9 h^{2}} / 6, S_{\triangle F N M}=\frac{1}{2} \cdot \frac{a}{\sqrt{3}} \cdot \frac{\sqrt{a^{2}+9 h^{2}}}{6}=\frac{a \sqrt{a^{2}+9 h^{2}}}{12 \sqrt{3}}$ and $S_{M N Q P}=\frac{11 a \sqrt{a^{2}+9 h^{2}}}{60 \sqrt{3}}$.

Project the section onto the base of the pyramid. The area of the projection of the section $S_{M A H P}=S_{\triangle A B C}-S_{\triangle B M P}-S_{\triangle C H P}=\left(1-\frac{2}{3} \cdot \frac{1}{3}-\frac{4}{5} \cdot \frac{2}{3}\right) S_{\triangle A B C}=\frac{11}{45} S_{\triangle A B C}=\frac{11 \sqrt{3} a^{2}}{180}$, $\cos \angle N K A=\frac{A K}{N K}=\frac{a}{\sqrt{a^{2}+9 h^{2}}}, S_{\text {MNQP }}=S_{\text {MAHP }} / \cos \angle N K A=S_{\text {MAHP }} / \cos 60^{\circ}=\frac{11 \sqrt{3} a^{2}}{90}=\frac{11 \sqrt{3}}{10}$. Answer: $\frac{11 \sqrt{3}}{10}$.

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## Solution for Variant 11 (11th Grade)