12.251. The center of the sphere inscribed in a regular quadrilateral pyramid divides the height of the pyramid in the ratio $m: n$, measured from the vertex of the pyramid. Find the angle between two adjacent lateral faces.
Solution.

The center of the sphere $M$, inscribed in the regular pyramid $SABCD$, is on the height $SO$ (Fig. 12.117) $SM: MO = m: n$.

$OC \perp BD, OC$ is the projection of $SC$ on the base of the pyramid. Then $SC \perp BD$.

In the plane $SDC$, draw $DK \perp SC$.

Since $DK \perp SC, BD \perp SC$, the plane $BKD$ is perpendicular to $SC$ and $\angle BDK$ is the required angle between the adjacent lateral faces $BSC$ and $DSC$.

Let $\angle BKD = \alpha, \angle DCK = \beta$.

Since the pyramid is regular, $\triangle DKC = \triangle BKC$ - by hypotenuse and acute angle, and $DK = BK$, while $KO$ is the median, bisector, and altitude of triangle $BKD$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1002.jpg?height=775&width=738&top_left_y=30&top_left_x=285)

Fig. 12.117

Let $L$ be the midpoint of $CD$, then $LM$ is the bisector of $\angle SLO$ and $OL: SL = OM: SM = n: m$.

Since $CL = OL$, then $\operatorname{tg} \beta = \frac{CL}{SL} = \frac{OL}{SL} = \frac{n}{m}$.

$\sin \beta = \frac{DK}{DC}; \sin \frac{\alpha}{2} = \frac{OD}{DK}$. Then $\sin \frac{\alpha}{2} \sin \beta = \frac{OD}{DC} = \frac{1}{\sqrt{2}}$;

$\sin^2 \frac{\alpha}{2} \sin^2 \beta = \frac{1}{2}; (1 - \cos \alpha) \sin^2 \beta = 1; \cos \alpha = 1 - \frac{1}{\sin^2 \beta} = -\operatorname{ctg}^2 \beta = -\frac{n^2}{m^2} \Rightarrow \alpha = \arccos \left(-\frac{n^2}{m^2}\right) = \pi - \arccos \frac{n^2}{m^2}$.

Answer: $\pi - \arccos \frac{n^2}{m^2}$.