Problem 8. The base of the triangular pyramid $D A B C$ is triangle $A B C$, in which angle $A$ is a right angle, angle $A C B$ is $30^{\circ}, B C=2 \sqrt{2}$. The lengths of the edges $A D, B D, C D$ are equal to each other. A sphere with radius 1 touches the edges $A D, B D$, the extension of edge $C D$ beyond point $D$, and the plane $B A C$. Construct the image of the center $O$ of this sphere and find the length of the tangent segment drawn from point $A$ to the sphere.
Solution. The base of the given pyramid is a right-angled triangle $ABC$, and its lateral edges are equal, so the height of the pyramid is the segment $DH$ (where $H$ is the midpoint of the hypotenuse $BC$ of triangle $ABC$).

The given sphere touches the edges $AD$ and $BD$, so its center $O$ lies in the plane of symmetry $\alpha$ of the rays $DB$ and $DA$. Since triangle $ABH$ is equilateral, the plane $\alpha$ passes through point $H$ and the midpoint $K$ of segment $BA$ (Fig. 10). The given sphere touches the edge $BD$ and the ray $DC_1$ (the extension of edge $CD$ beyond point $D$), so point $O$ lies in the plane of symmetry $\beta$ of the rays $DB$ and $DC_1$.

Thus, the plane $BDC$ is perpendicular to the plane $BAC$, and the plane $\alpha$ is perpendicular to the plane $BDC$ and bisects the external angle $C_1DB$ of the isosceles triangle $BDC$. Therefore, the planes $\alpha$ and $\beta$ intersect along a line $a$ passing through point $D$ and parallel to the line $HK$.

Thus, point $O$ lies on line $a$. It follows that the quadrilateral $MODH$ is a rectangle ($M$ is the point of tangency of the sphere with the plane $ABC$) and $DH = MO = 1$.

Therefore, line $OD$ is parallel to $HM$, and $HM$ is parallel to $AC$. Hence, lines $OD$ and $AC$ are parallel, i.e., point $O$ lies in the plane $C_1DA$.

Obviously, $AC = BC \sin 60^\circ = 2\sqrt{2} \cdot 0.5\sqrt{3} = \sqrt{6}$. By the converse of the Pythagorean theorem, we get that angle $ADC$ is a right angle. Since $OC_1 = OM = 1$, the quadrilateral $OC_1DA$ is a square with a side length of 1. Therefore, the desired tangent segment, drawn from point $A$ to the sphere, is $\sqrt{3} - 1$.

Now it is clear how to construct point $O$ in the figure.

Through point $D$, draw a line $a$ parallel to line $HK$. On the extension of edge $DC$ (beyond point $D$), lay off the segment $DC_1 = 1$. Through point $C_1$, draw a line parallel to line $DA$. It intersects line $a$ at the desired point $O$.