7. The base of the triangular pyramid $A B C D$ is an equilateral triangle $A B C$. The volume of the pyramid is $\frac{100}{3 \sqrt{3}}$, and its height, drawn from vertex $D$, is 4. Point $M$ is the midpoint of edge $C D$. It is known that the radii of the spheres inscribed in pyramids $A B C M$ and $A B D M$ are equal.
a) Find all possible values of the angle between the faces of the pyramid at edge $A B$.

b) Find all possible values of the length of edge $C D$, given that the faces $B C D$ and $A B C$ are mutually perpendicular.
Answer. a) $\angle=\arccos \left( \pm \frac{3}{5}\right)$; b) $C D=\frac{8}{\sqrt{3}}$ or $C D=\frac{4 \sqrt{19}}{\sqrt{3}}$.

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 $A B C M$ and $A B D M$ are equal (the face $A B M$ is common, and the vertices $C$ and $D$ are equidistant from the plane $A B M$); moreover, $S_{A C M}=S_{A D M}$ and $S_{B C M}=S_{B D M}$ (the median divides the area of the triangle in half). Therefore, the equality of the spheres inscribed in pyramids $A B C M$ and $A B D M$ is equivalent to the condition $S_{A B C}=S_{A B D}$ or the equality of the heights drawn to the side $A B$ in triangles $A B C$ and $A B D$.

Let $H$ and $K$ be the projections of point $D$ onto the plane $A B C$ and the line $A B$ respectively. The volume of the pyramid is $\frac{100}{\sqrt{3}}$, and its height is 4. Therefore, the area of the base of the pyramid is $\frac{25}{\sqrt{3}}$. Then the side of the base $A B=\frac{10}{\sqrt{3}}$, and the height of triangle $A B C$ is 5. Thus, $D K$ is also 5. From the right triangle $D H K$, we find $K H=\sqrt{D K^{2}-D H^{2}}=3$, i.e., point $H$ is at a distance of 3 from the line $A B$ ( $H$ lies on one of two lines parallel to $A B$, at a distance of 3 from it).

Thus, the angle between the faces along the edge $A B$ is $\arccos \left( \pm \frac{3}{5}\right)$.

b) With the additional condition that $H$ lies on the ray $C B$, and $\frac{B H}{B C}=\frac{3}{5}$, we get $C H=\frac{2}{5} B C=\frac{4}{\sqrt{3}}$ or $C H=\frac{8}{5} B C=\frac{16}{\sqrt{3}}$. From triangle $C D H$, we obtain that $C D=\sqrt{C H^{2}+H D^{2}}$. Therefore, $C D=\frac{8}{\sqrt{3}}$ or $C D=\frac{4 \sqrt{19}}{\sqrt{3}}$.

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