A sphere of radius $R$ divides each of the edges $S A, S C, A B$ and $B C$ of the triangular pyramid $S A B C$ into three equal parts and passes through the midpoints of the edges $A C$ and $S B$. Find the height of the pyramid dropped from the vertex $S$.
Let the specified sphere pass through points $P, Q$ of edge $A S$ (Fig.1), points $E, F$ of edge $A B$, through point $M$ of edge $A C$ and point $N$ of edge $B S$, and

$$
A P=P Q=Q S, A E=E F=B F, A M=C M, B N=N S
$$

Let $A P=P Q=Q S=x$. Consider the section of the sphere by the plane of face $A S B$ (Fig.2). We get a circle passing through points $P, Q, E, F$ and $N$. From the equality $A P \cdot A Q=A E \cdot A F$ it follows that

$$
A E=E F=B F=x
$$

Therefore, $A B=A S=3 x$. Let $N 1-$ be the other common point of the sphere and the line $B S$. Then $S N 1 \cdot S N=S Q \cdot S P=$ $2 \times 2$ and $B N 1 \cdot B N=B F \cdot B E=2 \times 2$, and since $S N=B N$, then $S N 1=B N 1$, i.e., $N 1$ is also the midpoint of segment $B S$, which is impossible. Therefore, the line $B S$ has a unique common point $N$ with the sphere, i.e., it is tangent to the sphere at point $N$. Similarly, $B C=3 x, S C=3 x$ and the line $A C$ is tangent to the sphere at point $M$. From the equalities $B N 2=B F \cdot B E=2 \times 2$ and $A M 2$ $=A P \cdot A Q=2 \times 2$ it follows that $B S=A C=2 x \sqrt{2}$. Let $G-$ be a point on edge $B C$, such that $B G=\frac{1}{3} B C$. Since the sphere passes through points $F$ and $G$, its center lies in the plane passing through the midpoint of segment $F G$ and perpendicular to $F G$, and since triangles $A B C$ and $A S C$ are isosceles, this plane passes through points $B, S$ and $M$ (Fig.3). Planes $B S M$ and $A B C$ are perpendicular, since plane $A B C$ passes through line $F G$, perpendicular to plane $B S M$. Therefore, the height $S H$ of the pyramid $S A B C$ lies in plane $B S M$. The section of the sphere by plane $B S M$ is a circle of radius $R$, tangent to the base $B S$ of isosceles triangle $B S M$ at point $N$ - the midpoint of $B S$, and passing through vertex $M$. Therefore, the height $M N$ of triangle $B S M$ is $2 R$, and the height of this triangle, drawn from vertex $S$, coincides with the height $S H$ of pyramid $S A B C D$. From triangle $A B C$ we find that

$$
B M=\sqrt{A B^{2}-A M^{2}}=\sqrt{9 x^{2}-2 x^{2}}={ }_{x} \sqrt{7}
$$

Thus, in triangle $B S M$ it is known that

$$
B M=M S=x \sqrt{7}, B N=S N=x \sqrt{2}, M N=2 R
$$

Since $B S \cdot M N=B M \cdot S H$, then

$$
S H=\frac{B S \cdot M N}{B M M}=\frac{2 x \sqrt{2} \cdot 2 R}{x \sqrt{7}}=\frac{4 R \sqrt{2}}{\sqrt{7}}=\frac{4 R \sqrt{14}}{7}
$$

## Answer

$\frac{4[\sqrt{14}}{7}$

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