10. A regular triangular pyramid is inscribed in a sphere of radius $R$, with the height of the pyramid equal to $4 R / 3$. What is the minimum area that the cross-section of the pyramid by a plane passing through the median of the base can have? Find the ratio of the volumes of the parts into which the cutting plane divides the pyramid in this case.
# Solution:

Since $T K=4 R / 3, O K=R / 3, B K=\sqrt{1-(1 / 3)^{2}} R=\sqrt{8} R / 3 ; A B=A K \sqrt{3}=\sqrt{8 / 3} R$, $A T=\sqrt{(4 R / 3)^{2}+(\sqrt{8} R / 3)^{2}}=\sqrt{8 / 3} R$. Therefore, $A B=A T$, all edges of the pyramid are equal.

We will assume that the cutting plane passes through the median $T G$ and intersects the edge $B C$ at point $M$. Draw $G F \| B C, F \in A D$, then $D E \perp T F, E \in T F$, $E P \| B C, P \in T G$, and $P M \| D E, M \in B C . P M$ is the common perpendicular to $B C$ and $T G$.

\[
\begin{gathered}
\text { Let } a=A B=A T \text {. Then } F K=\left(\frac{1}{2}-\frac{1}{3}\right) \frac{a \sqrt{3}}{2}=\frac{a \sqrt{3}}{12}, T K=\sqrt{a^{2}-\frac{a^{2}}{3}}=\sqrt{\frac{2}{3}} a, \\
F T=\sqrt{K T^{2}+K F^{2}}=\sqrt{\frac{2}{3}+\frac{3}{12 \cdot 12}} a=\frac{\sqrt{11}}{4} a, E D=\frac{F D \cdot T K}{F T}=\frac{a \sqrt{3} \cdot \sqrt{2} \cdot 4}{4 \sqrt{3} \cdot \sqrt{11}}=\frac{a \sqrt{2}}{\sqrt{11}}=\frac{4 R}{\sqrt{33}} . \\
S_{\Delta T G M}=\frac{1}{2} \cdot T G \cdot P M=\frac{1}{2} \cdot \frac{a \sqrt{3}}{2} \cdot \frac{a \sqrt{2}}{\sqrt{11}}=\frac{a^{2} \sqrt{6}}{4 \sqrt{11}}=\frac{8}{3} \cdot \frac{\sqrt{6}}{4 \sqrt{11}} R^{2}=\frac{2 \sqrt{2}}{\sqrt{33}} R^{2} . \\
E T=\sqrt{T D^{2}-E D^{2}}=\sqrt{\left(\frac{a \sqrt{3}}{2}\right)^{2}-\left(\frac{a \sqrt{2}}{\sqrt{11}}\right)^{2}}=\frac{5 a}{2 \sqrt{11}} ; \frac{E T}{F T}=\frac{5}{2 \sqrt{11}} \cdot \frac{4}{\sqrt{11}}=\frac{10}{11} ;
\end{gathered}
\]

\[
P E=M D=\frac{10}{11} G F=\frac{10}{11} \cdot \frac{a}{4}=\frac{5 a}{22} ; B M=\left(\frac{1}{2}-\frac{5}{22}\right) a=\frac{3}{11} a \cdot V_{T B M G}=\frac{1}{2} \cdot \frac{3}{11} V_{T A B C}=\frac{3}{22} V_{T A B C}
\]

The ratio of the volumes of the parts into which the cutting plane divides the pyramid is 3:19.

![](https://cdn.mathpix.com/cropped/2024_05_06_18575ac96ab43cb21cd9g-09.jpg?height=1454&width=1407&top_left_y=618&top_left_x=356)

The pyramid is a regular tetrahedron (all its edges are equal).

\[
\begin{aligned}
& T K=\frac{4 R}{3}, O K=\frac{R}{3}, B K=\sqrt{1-\frac{R}{9}}=\frac{\sqrt{8}}{3} R, A B=B K \sqrt{3}=\sqrt{\frac{8}{3}} R \\
& B T=\sqrt{\frac{16}{9} R^{2}+\frac{8}{9} R^{2}}=\sqrt{\frac{8}{3}} R . \Rightarrow A B=B T . \quad \text { Answer: } \frac{2 \sqrt{2}}{\sqrt{33}} R^{2} ; 3: 19 .
\end{aligned}
\]

## Solution for Variant 15