8,9

The height of a regular triangular pyramid is $6 \sqrt{6}$, and the lateral edge forms an angle of $45^{\circ}$ with the base plane. Find the distance from the center of the base of the pyramid to a lateral face.
Let $M$ be the center of the base $ABC$ of a regular pyramid $ABCD$, and $P$ be the projection of point $M$ onto the height $DL$ of the face $BCD$. According to the problem, $\angle DAM = 45^{\circ}$. From the isosceles right triangle $AMD$, we find that $AM = DM = 6\sqrt{6}$. The line $MP$ lies in the plane $ADL$, which is perpendicular to the line $BC$, so $MP \perp BC$. Since $MP \perp DL$, $MP$ is the perpendicular to the plane $BCD$. Therefore, the desired distance is the length of the segment $MP$. In the right triangle $DML$, the legs $DM = 6\sqrt{6}$ and $ML = \frac{1}{2} AM = 3\sqrt{6}$ are known, hence,

$$
DL = \sqrt{DM^2 + ML^2} = \sqrt{(6\sqrt{6})^2 + (3\sqrt{6})^2} = 3\sqrt{30}
$$

From the relation $ML \cdot DM = DL \cdot MP$, we find that

$$
MP = \frac{ML \cdot DM}{DL} = \frac{3\sqrt{6} \cdot 6\sqrt{6}}{3\sqrt{30}} = \frac{36}{\sqrt{30}}
$$

## Answer

The problem