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 smallest 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.

(12 points)
# 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 $T G$.

$$
\begin{gathered}
\text { Let } a=A B=A T . \text { Then } F K=\left(\frac{1}{2}-\frac{1}{3}\right) \frac{a \sqrt{3}}{2}=\frac{a \sqrt{3}}{12}, T K=\sqrt{a^{2}-\frac{a^{2}}{3}}=\sqrt{\frac{2}{3}} a, \\
F T=\sqrt{K T^{2}+K F^{2}}=\sqrt{\frac{2}{3}+\frac{3}{12 \cdot 12}} a=\frac{\sqrt{11}}{4} a, E D=\frac{F D \cdot T K}{F T}=\frac{a \sqrt{3} \cdot \sqrt{2} \cdot 4}{4 \sqrt{3} \cdot \sqrt{11}}=\frac{a \sqrt{2}}{\sqrt{11}}=\frac{4 R}{\sqrt{33}} . \\
S_{\Delta T G M}=\frac{1}{2} \cdot T G \cdot P M=\frac{1}{2} \cdot \frac{a \sqrt{3}}{2} \cdot \frac{a \sqrt{2}}{\sqrt{11}}=\frac{a^{2} \sqrt{6}}{4 \sqrt{11}}=\frac{8}{3} \cdot \frac{\sqrt{6}}{4 \sqrt{11}} R^{2}=\frac{2 \sqrt{2}}{\sqrt{33}} R^{2} . \\
E T=\sqrt{T D^{2}-E D^{2}}=\sqrt{\left(\frac{a \sqrt{3}}{2}\right)^{2}-\left(\frac{a \sqrt{2}}{\sqrt{11}}\right)^{2}}=\frac{5 a}{2 \sqrt{11}} ; \frac{E T}{F T}=\frac{5}{2 \sqrt{11}} \cdot \frac{4}{\sqrt{11}}=\frac{10}{11} ; \\
P E=M D=\frac{10}{11} G F=\frac{10}{11} \cdot \frac{a}{4}=\frac{5 a}{22} ; B M=\left(\frac{1}{2}-\frac{5}{22}\right) a=\frac{3}{11} a \cdot V_{T B M G}=\frac{1}{2} \cdot \frac{3}{11} V_{T A B C}=\frac{3}{22} V_{T A B C} .
\end{gathered}
$$

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-18.jpg?height=1274&width=1242&top_left_y=171&top_left_x=293)

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

$\frac{K D}{T D}=\frac{1}{3}, T D=3 K D=A D, A B=T B$.

Answer: $\frac{2 \sqrt{2}}{\sqrt{33}} R^{2} ; 3: 19$.

## Statistical data on the performance of participants in the final stage of the mathematics olympiad. Every year, the scientific and methodological department conducts analytical work on the results of academic competitions at the selection and final stages of the Schoolchildren's Olympiad "Step into the Future". The charts below provide an overview of the percentage of task completion by participants in the second stage of the olympiad, held at the Bauman Moscow State Technical University in 2016 and in comparison with 2014 and 2015.

Olympiads held at regional sites in Barnaul, Zarechny (Penza region), Kyzyl (Republic of Tuva), Yakutsk, Nizhny Novgorod, and Yaroslavl have also been analyzed.