7. The base of the triangular pyramid $K L M N$ with a volume of 75 is an equilateral triangle $K L M$ with a side length of 10. Point $T$ is the midpoint of edge $M N$. It is known that the radii of the spheres inscribed in pyramids $K L M T$ and $K L N T$ are equal to each other.

a) Find all possible values of the angle between the faces of the pyramid at edge $K L$.

b) Find all possible values of the length of edge $M N$, given that the faces $K L M$ and $L M N$ are perpendicular.
Answer. a) $\angle=\arccos \left( \pm \frac{4}{5}\right)$; b) $M N=\sqrt{31}$ or $M N=3 \sqrt{39}$.

Solution. a) We use the formula for the radius of the inscribed sphere $r=\frac{3 V}{S}$, where $V$ is the volume and $S$ is the surface area of the pyramid. The volumes of pyramids $K L M T$ and $K L N T$ are equal (the face $K L T$ is common, and the vertices $M$ and $N$ are equidistant from the plane $K L T$); moreover, $S_{K M T}=S_{K N T}$ and $S_{L M T}=S_{L N T}$ (the median divides the area of the triangle in half). Therefore, the equality of the spheres inscribed in pyramids $K L M T$ and $K L N T$ is equivalent to the condition $S_{K L M}=S_{K L N}$ or the equality of the heights drawn to the side $K L$ in triangles $K L M$ and $K L N$.

Let $H$ and $Q$ be the projections of point $N$ onto the plane $K L M$ and the line $K L$ respectively. The volume of the pyramid is 75, and the area of its base is $\frac{10^{2} \sqrt{3}}{4}=25 \sqrt{3}$. Therefore, the height of the pyramid is $3 \sqrt{3}$. The height of triangle $K L M$ is $5 \sqrt{3}$. Thus, $N Q$ is also $5 \sqrt{3}$. From the right triangle $N Q H$, we find $Q H=\sqrt{Q N^{2}-N H^{2}}=4 \sqrt{3}$, i.e., point $H$ is at a distance of $4 \sqrt{3}$ from the line $K L$ ( $H$ lies on one of the two lines parallel to $K L$, at a distance of $4 \sqrt{3}$ from it).

Thus, the angle between the faces along the edge $K L$ is $\arccos \left( \pm \frac{4}{5}\right)$.

b) Under the additional condition, $H$ lies on the ray $M L$, and $\frac{L H}{L M}=\frac{4}{5}$, from which $M H=\frac{1}{5} L M=2$ or $M H=\frac{9}{5} L M=18$. From triangle $M N H$, we get that $M N=\sqrt{M H^{2}+H N^{2}}$. Therefore, $M N=\sqrt{31}$ or $M N=3 \sqrt{39}$.

## TICKET 4