Problem 8. For the tetrahedron $K L M N$, it is known that $K L=5, N M=6, \angle L M N=\angle K N M=35^{\circ}, \angle L N M=$ $\angle K M N=55^{\circ}$. A sphere is circumscribed around the tetrahedron. Consider the set of all points on this sphere for which the sum of the spherical distances from these points to $K, L, M, N$ is no more than $6 \pi$. What is the area of this set? The spherical distance between two points on a sphere is the length of the shortest arc of a great circle connecting these points.
Answer: $18 \pi$, i.e., half the area of a sphere with radius 3.

Solution. Since $\angle L M N+\angle L N M=\angle K M N+\angle K N M=90^{\circ}$, triangles $L M N$ and $K M N$ are right triangles with a common hypotenuse $M N$. If $O$ is the midpoint of segment $M N$, then by the property of the median of a right triangle, $O K=O L=O M=O N=3$. Therefore, the radius of the circumscribed sphere is 3, and point $O$ is its center.

Let $d(X, Y)$ denote the spherical distance between points $X$ and $Y$. According to the problem, we need to find the area of the set $\omega$ on the sphere, consisting exactly of points $X$ for which

$$
d(X, K)+d(X, L)+d(X, M)+d(X, N) \leqslant 6 \pi
$$

Since $M N$ is the diameter of the sphere, points $M, X$, and $N$ lie on the same great circle; therefore, $d(M, X)+d(X, N)=d(M, N)=3 \pi$. Inequality (1) can be rewritten as

$$
d(X, K)+d(X, L) \leqslant 3 \pi
$$

Let $L_{1}$ be the point symmetric to point $L$ with respect to the center of the sphere $O$. Since $L_{1}$ and $L$ are the endpoints of the sphere's diameter, $d(L, X)+d\left(X, L_{1}\right)=d\left(L, L_{1}\right)=3 \pi$. Substituting $d(X, L)=3 \pi-d\left(X, L_{1}\right)$ into inequality (2), we get

$$
d(X, K) \leqslant d\left(X, L_{1}\right)
$$

Since $K L=5 \neq 6$, $K L$ is not a diameter, and therefore $L_{1} \neq K$. Thus, $\omega$ is the set of points on the sphere whose spherical distance from one point on the sphere does not exceed the spherical distance from another point on the sphere. By symmetry (with respect to the plane passing through the center of the sphere perpendicular to the segment $\left.L_{1} K\right), \omega$ is half the sphere, and its area is $\frac{1}{2} \cdot 4 \cdot \pi \cdot 3^{2}=18 \pi$.