The base of the pyramid $A B C D$ is a triangle $A B C$ with sides $A C=6, B C=8, A B=10$. All lateral edges are equal to $5 \sqrt{2}$. Find a) the radius of the sphere circumscribed around the pyramid $A B C D$; b) the distance between the lines $D M$ and $A C$ and the distance between the lines $D M$ and $B C$, where $D M$ is the height of the pyramid $A B C D$.
a) Since the lateral edges of the pyramid are equal, its height $DM$ passes through the center $M$ of the circle circumscribed around the base, and since the base $ABC$ is a right triangle ($AC^2 + BC^2 = 36 + 64 = 100 = AB^2$), the point $M$ is the midpoint of its hypotenuse $AB$, and the radius of this circle is half the hypotenuse, i.e., 5. From the right triangle $DMB$, we find that

$$
MD = \sqrt{DB^2 - MB^2} = \sqrt{50 - 25} = 5
$$

Thus, $MA = MB = MC = MD = 5$. Therefore, $M$ is the center of the sphere circumscribed around the pyramid $ABCD$, and the radius of this sphere is 5. b) Let $K$ be the midpoint of $AC$. Then $MK$ is the midline of triangle $ABC$, so $MK \perp AC$. On the other hand, $MK \perp DM$, since the line $DM$ is perpendicular to the plane $ABC$, in which the line $MK$ lies. Therefore, $MK$ is the common perpendicular to the skew lines $DM$ and $AC$, and $MK = \frac{1}{2} BC = 4$. Similarly, we find that the distance between the lines $DM$ and $BC$ is 3.

## Answer

a) $5;$ b) 4 and 3.