[ Sphere inscribed in a pyramid ]  [

The distance from the center $O$ of a sphere with radius 12, circumscribed around a regular quadrilateral pyramid, to a lateral edge is $4 \sqrt{2}$. Find: 1) the height of the pyramid; 2) the distance from point $O$ to a lateral face of the pyramid; 3) the radius of the sphere inscribed in the pyramid.
1) Let $O N$ be the perpendicular dropped from the center $O$ of the circumscribed sphere to the lateral edge $S C$ of the regular quadrilateral pyramid $S A B C D$ with vertex $S$ (Fig.1), and $S H$ be the height of the pyramid. Since $S O$ is the radius of the circumscribed sphere, from the right triangle $S O N$ we find that

$$
S N=\sqrt{S O^{2}-O N^{2}}=\sqrt{144-16 \cdot 2}=4 \sqrt{9-2}=4 \sqrt{7}
$$

The perpendicular dropped from the center of the sphere to a chord bisects it, so $S C=2 S N=8 \sqrt{7}$.

The right triangles $S H C$ and $S N O$ are similar, so $\frac{S H}{S N}=\frac{S C}{S O}$, from which we get

$$
S H=S N \cdot \frac{S C}{S O}=4 \sqrt{7} \cdot \frac{5 \sqrt{7}}{12}=\frac{56}{3}
$$

2) Let $A B=B C=C D=A D=a$. Then $C H=\frac{\frac{1}{2}}{2} A C=\frac{a \sqrt{2}}{2}$. From the right triangle $S H C$ we get that $C H=\frac{16 \sqrt{2}}{3}$. From the equation $\frac{a \sqrt{2}}{2}=\frac{16 \sqrt{2}}{3}$ we find that $a=\frac{16 \sqrt{7}}{3}$. Let $O P$ be the perpendicular dropped from point $O$ to the median $S M$ of the isosceles triangle $A S D$. Since the line $A D$ is perpendicular to the intersecting lines $A D$ and $S H$ in the plane $S H M$, then $A D \perp O P$, so $O P$ is the perpendicular to the plane of the lateral face $A S D$. In the right triangle $S H M$ it is known that

$$
S H=\frac{56}{3}, H M=\frac{a}{2}=\frac{8 \sqrt{7}}{3}, S O=12 .
$$

Then

$$
S M=\sqrt{S H^{2}+H M^{2}}=\sqrt{\left(\frac{56}{3}\right)^{2}+\left(\frac{8 \sqrt{7}}{3}\right)^{2}}=\frac{16 \sqrt{14}}{3} .
$$

From the similarity of the right triangles $S P O$ and $S H M$ we find that

$$
O P=M H \cdot \frac{S O}{S M}=\frac{5 \sqrt{7}}{3} \cdot \frac{12 \sqrt{11}}{3 /}=3 \sqrt{2}
$$

3) Let $\beta$ be the angle between the lateral face of the pyramid $SABCD$ and the plane of its base. Since the pyramid is regular, the center $Q$ of its inscribed sphere lies on the height $S H$, the sphere touches the base plane at point $H$, and the planes of the lateral faces at points lying on the apothems of the pyramid. Draw a section through the apothem $S M$ and the apothem $S K$ lying in the face $B S C$ (Fig.2). We get an isosceles triangle $M S K$ and a circle inscribed in it with center $Q$ and radius $r$, equal to the radius of the sphere inscribed in the pyramid. Since $S M H$ is the linear angle of the dihedral angle between the plane of the lateral face of the pyramid and the plane of its base, $\angle S M H=\beta$. From the right triangle $S M H$ we find that

$$
\operatorname{tg} \beta=\frac{S H}{H M}=\frac{\frac{g}{g \sqrt{7}}}{3}=\sqrt{7}
$$

$\underline{2 \operatorname{tg} g^{8}}$

Applying the formula $\operatorname{tg} \beta=\overline{1-t g^{2 \frac{2}{2}}}$, we get the equation

$$
\sqrt{7}=\frac{2 \operatorname{tg} \frac{3^{2}}{2}}{1-\lg ^{2 \frac{8}{2}}}
$$

from which we find that $\operatorname{tg} \frac{\frac{9}{2}}{=} \frac{2 \sqrt{2}-1}{\sqrt{7}}$, and since $M O$ is the bisector of angle $S M H$, from the right triangle $O H M$ we find that

$$
r=O H=M H \operatorname{tg} \frac{\frac{9}{2}}{2}=\frac{\xi \sqrt{7}}{3} \cdot \frac{2 \sqrt{2}-1}{\sqrt{7}}=\frac{\frac{8}{3}}{3}(2 \sqrt{2}-1) .
$$

## Answer

$\frac{56}{3} ; 3 \sqrt{2} ; \frac{8}{3}(2 \sqrt{2}-1)$.