[ Sphere touching the edges or sides of the pyramid ]

The height of the pyramid is 5, and the base is a triangle with sides 7, 8, and 9. A sphere touches the planes of all the lateral faces of the pyramid at points lying on the sides of the base. Find the radius of the sphere.

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Let a sphere with center $O$ and radius $R$ touch the planes of the lateral faces $ABD, BCD$, and $ACD$ at points $C1, A1$, and $B1$, which lie on the sides $AB, BC$, and $AC$ of the base $ABC$, respectively, with $AB=7, BC=8, AC=9$ (Fig.1). The section of the sphere by the plane of the base $ABC$ is a circle with center at some point $O1$, inscribed in the triangle $ABC$. The lines $OC1, OA1$, and $OB1$ are perpendicular to the planes of the faces $ABD, BCD$, and $ACD$, respectively. The right triangles $OC1D, OA1D$, and $OB1D$ are equal by the leg and hypotenuse, so the lateral edges $DC1, DA1$, and $DB1$ of the triangular pyramid $A1B1C1D$ are equal. Therefore, the height $DH$ of this pyramid passes through the point $O1$ - the center of the circle circumscribed around the triangle $A1B1C1$. Since $DO1$ is also the height of the pyramid $ABCD$, then $DO1=5$. Moreover, $OO1$ is perpendicular to the plane of the base $ABC$, so the points $D, O1$, and $O$ lie on the same line, and this line is perpendicular to the plane of the base of the given pyramid. Let $r$ be the radius of the inscribed circle of the triangle $ABC, p$ be the semiperimeter of the triangle, $S$ be its area. Then

$$
\begin{gathered}
p=12, S=\sqrt{12(12-7)(12-8)(12-9)}=\sqrt{12 \cdot 5 \cdot 4 \cdot 3}=12 \sqrt{5} \\
O1A1=r=\frac{S}{p}=\frac{12 \sqrt{5}}{12}=\sqrt{5}
\end{gathered}
$$

In the right triangle $DA1O$, the height $A1O1=\sqrt{5}$, drawn from the vertex of the right angle, and the segment $DO1=5$ (Fig.2) are known. Let $\angle A1DO=\angle OA1O1=\alpha$. We further find:

$$
A1D=\sqrt{OD^2+OA^2}=\sqrt{25+5}=\sqrt{30}, \frac{DOO1}{DA1}=\cos \alpha=\frac{A1O1}{OA1}, \frac{5}{\sqrt{30}}=\frac{\sqrt{5}}{R}
$$

from which $R=\sqrt{30} \cdot \frac{\sqrt{5}}{5}=\sqrt{6}$.

## Answer

$\sqrt{6}$.