5. Solution. Let's take the side of the tetrahedron as 12. We will find the angle through the cosine, which is equal to the ratio of the area \( S_{1} \) of the triangle \( K_{1} L_{1} M_{1} \) - the projection of the triangle \( K L M \) onto the base plane, to the area \( S_{2} \) of the triangle \( K L M \) - the section. The area of the projection \( S_{1} \) is easily determined, as the vertices \( K_{1}, L_{1}, M_{1} \) divide the corresponding radii of the circumscribed circle of the base (the area of the base \( S_{0}=36 \sqrt{3} \)) in the same ratios as the corresponding points \( K, L, M \) divide the lateral sides of the tetrahedron.

\[
S_{1}=\frac{1}{3} \cdot S_{0}\left(\frac{1}{4} \cdot \frac{2}{3}+\frac{1}{2} \cdot \frac{2}{3}+\frac{1}{4} \cdot \frac{1}{2}\right)=\frac{5}{24} S_{0}=\frac{15 \sqrt{3}}{2}
\]

The sides of the section will be calculated using the cosine theorem: \( K L=7, L M= \sqrt{52}, M K=3 \sqrt{3} \). Now we will calculate the area of the section. The cosine of the angle \( \alpha \), opposite the side \( K L \), is \( \cos \alpha=\frac{5}{\sqrt{156}} \). Then \( \sin \alpha=\frac{\sqrt{131}}{\sqrt{156}} \). For the area of the section, we get the following result \( S_{2}=0.5 \cdot 3 \sqrt{3} \sqrt{52} \frac{\sqrt{131}}{\sqrt{156}}= \frac{3}{2} \sqrt{131} \).

The final step:

\[
\cos \gamma=\frac{S_{1}}{S_{2}}=\frac{5 \sqrt{3}}{\sqrt{131}}
\]
Answer: $\cos \gamma=\frac{5 \sqrt{3}}{\sqrt{131}}$.

## Lomonosov Moscow State University

Olympiad "Conquer Sparrow Hills"

Variant $6-1$ (Irkutsk)