3. A regular quadrilateral pyramid is given. The side of the base is 6, the length of the lateral edge is 5. A sphere $Q_{1}$ is inscribed in the pyramid. A sphere $Q_{2}$ touches $Q_{1}$ and all the lateral faces of the pyramid. Find the radius of the sphere $Q_{2}$.
Answer: $\frac{3 \sqrt{7}}{49}$.

Solution. Let $r_{1}$ and $r_{2}$ be the radii of the spheres $Q_{1}$ and $Q_{2}$, respectively. Let $A B C D$ be the base of the pyramid, $M$ the apex of the pyramid, $E$ the center of the base, and $F$ the midpoint of $C D$. Using the Pythagorean theorem, we find the slant height of the lateral face $M F$ and the height of the pyramid $h_{1}=M E$:

$$
M F=\sqrt{M C^{2}-F C^{2}}=4 ; \quad h=\sqrt{M F^{2}-E F^{2}}=\sqrt{7}
$$

Now we calculate the volume $V$ of the pyramid and its total surface area $S$:

$$
V=\frac{1}{3} S_{A B C D} \cdot h_{1}=12 \sqrt{7} ; \quad S=S_{A B C D}+4 S_{M C D}=36+4 \cdot 12=84
$$

Using the known formula, we determine the radius of the inscribed sphere:

$$
r_{1}=\frac{3 V}{S}=\frac{36 \sqrt{7}}{84}=\frac{3 \sqrt{7}}{7}
$$

We draw a tangent plane to the sphere $Q_{1}$, parallel to the base of the pyramid. It will cut off a smaller pyramid $T=A_{1} B_{1} C_{1} D_{1} M$ from the original pyramid. Clearly, the sphere $Q_{2}$ is inscribed in this smaller pyramid. It is easy to find the height of the cut-off pyramid: $h_{2}=h_{1}-2 r_{1}=\frac{\sqrt{7}}{7}$. The smaller pyramid $T$ is similar to the original pyramid with a similarity ratio $k=\frac{h_{2}}{h_{1}}=\frac{1}{7}$. The radii of their inscribed spheres are in the same ratio. Therefore, $r_{2}=k r_{1}=\frac{3 \sqrt{7}}{49}$.

Evaluation. Full solution: 12 points. If only the radius of the first sphere is found, 4 points.