In the pyramid $ABCD$, the length of segment $BD$ is $\frac{5}{2}$, point $E$ is the midpoint of $AB$, and $F$ is the point of intersection of the medians of face $BCD$, with $EF=8$. A sphere of radius 5 touches the planes $ABD$ and $BCD$ at points $E$ and $F$ respectively. Find the dihedral angle between the faces $ABD$ and $BCD$, the area of face $BCD$, and the volume of the pyramid $ABCD$.
Let O be the center of the sphere (Fig.1). Since the radii OE and OF are drawn to the points of tangency of the sphere with the planes $ABD$ and $BCD$, the line OE is perpendicular to the plane $ABD$, and the line OF is perpendicular to the plane $BCD$, meaning $OE \perp BD$ and $OF \perp BD$, as well as $OE \perp DE$ and $OF \perp DF$. Let the plane passing through the intersecting lines $OE$ and $OF$ intersect the line $BD$ at point $H$. Since the line $BD$ is perpendicular to the intersecting lines $OE$ and $OF$ of this plane, the plane $OEF$ is perpendicular to the line $BD$, meaning $EHF$ is the linear angle of the dihedral angle between the faces $ABD$ and $BCD$. Let $\angle EHF = 2\alpha$. From the equality of the segments $DE$ and $DF$ (segments of tangents drawn to the sphere from one point), the equality of the right triangles $EHD$ and $FHD$ follows (by the leg and hypotenuse), meaning $EH = FH$. Thus, in the quadrilateral $OENF$ (Fig.2), points $O$ and $H$ are equidistant from the ends of the diagonal $EF$, so $HO \perp EF$ and the diagonal $OH$ passes through the midpoint $K$ of the diagonal $EF$. Then

$$
\angle OEK = \angle EHO = \frac{1}{2} \angle EHF = \alpha, \cos \alpha = \frac{EK}{OE} = \frac{4}{5}
$$

Therefore,

$$
\cos 2\alpha = 2 \cos^2 \alpha - 1 = 2 \cdot \frac{16}{25} - 1 = \frac{32}{25} - 1 = \frac{7}{25}
$$

From the right triangle $NKF$ we find that

$$
FH = \frac{KF'}{\sin \angle KF'H} = \frac{KF'}{\sin 2\alpha} = \frac{4}{\sqrt{1 - \cos^2 \alpha}} = \frac{4}{\sqrt{1 - \frac{16}{25}}} = \frac{20}{3}.
$$

Since $FH \perp BD$, the segment $FH$ is the height of the triangle $DBF$, and since $F$ is the intersection of the medians of the triangle $BCD$,

$$
S_{\triangle BDC} = 3 S_{\triangle BDF} = 3 \cdot \frac{1}{2} BD \cdot FH = \frac{3}{2} \cdot \frac{5}{2} \cdot \frac{20}{3} = 25
$$

Since $EH \perp BD$, the segment $EH$ is the height of the triangle $BED$, and since $E$ is the midpoint of the edge $AB$ and $EH = FH = \frac{20}{3}$,

$$
S_{\triangle ABD} = 2 S_{\triangle BDE} = 2 \cdot \frac{1}{2} BD \cdot EH = \frac{5}{2} \cdot \frac{20}{3} = \frac{50}{3}
$$

We will use the formula $V = \frac{2}{3} \cdot \frac{S_1 \cdot S_2 \sin \varphi}{a}$, where $V$ is the volume of the tetrahedron, $S_1$ and $S_2$ are the areas of two of its faces, $\varphi$ is the angle between these faces, and $a$ is the length of their common edge. In our case,

$$
V_{ABCD} = \frac{2}{3} \cdot \frac{S_{ABD} \cdot S_{BCD} \sin 2\alpha}{BD} = \frac{2}{3} \cdot \frac{\frac{50}{3} \cdot 25 \cdot \sin 2\alpha}{\frac{5}{2}} = \frac{2}{3} \cdot \frac{\frac{50}{3} \cdot 25 \cdot \frac{24}{25}}{\frac{5}{2}} = \frac{320}{3}
$$

## Answer

$\arccos \frac{7}{25} = \arcsin \frac{24}{25}; 25; \frac{320}{3}$.