6. On the edges of a trihedral angle with vertex at point $S$, there are points $M, N$, and $K$ such that $S M^{2}+S N^{2}+S K^{2} \leq 12$. Find the area of triangle $S M N$, given that the angle $M S N$ is $30^{\circ}$, and the volume of the pyramid $S M N K$ is maximally possible.
Solution. Let's introduce the notations: $S M=m, S N=n, S K=k$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6cabcf1712d23d5ceb2bg-04.jpg?height=320&width=442&top_left_y=548&top_left_x=630)

The volume of the pyramid $S M N K$ is

$$
V=\frac{1}{3} S_{S M N} \cdot h=\frac{1}{6} m n \sin \alpha \cdot h=\frac{1}{6} m n \sin \alpha \cdot k \sin \beta=\frac{\sin \alpha \sin \beta}{6} m n k .
$$

The values of the angles $\alpha, \beta$ depend on the trihedral angle but not on the positions of the points $M, N, K$ on its edges. Therefore, the maximum volume of the pyramid $S M N K$ corresponds to the maximum value of the product $m \cdot n \cdot k$, given the condition $m^{2}+n^{2}+k^{2} \leq p$. From the inequality of the geometric mean and the arithmetic mean of three numbers, it follows that $\sqrt[3]{m^{2} n^{2} k^{2}} \leq \frac{m^{2}+n^{2}+k^{2}}{3} \leq \frac{p}{3}$

Equality is achieved when $m=n=k$. Then, when $m=n=k=\sqrt{\frac{p}{3}}$, the volume of the pyramid is maximized, and the area of the triangle $S M N$ takes the value $\frac{p \sin \alpha}{6}$. In our case, $p=12, \alpha=30^{\circ}$, so $S_{S M N}=1$.

Answer: $S_{S M N}=1$.

## Answers for other variants.

## Variant 2