8,9

Given a cube $A B C D A_1 B_1 C_1 D_1$ with edge $a$. Point $E$ is the midpoint of edge $A D$. Vertices $M$ and $N$ of the regular tetrahedron $M N P Q$ lie on the line $E D_1$, and vertices $P$ and $Q$ lie on the line passing through point $A_1$ and intersecting line $B C$ at point $R$. Find a) the ratio $B R: B C$; b) the distance between the midpoints of segments $M N$ and $P Q$.
Since the opposite edges $M N$ and $P Q$ of a regular tetrahedron are perpendicular, the lines $E D 1$ and $A 1 R$ containing them are also perpendicular. Drop the perpendicular $R F$ from point $R$ to the plane $A A 1 D 1 D$. Since $C D$ and $A B$ are perpendicular to the plane $A A 1 D 1 D$, $R F \| C D$, so point $F$ lies on the line $A D$. $A 1 F$ is the orthogonal projection of the inclined line $A 1 R$ onto the plane $A A 1 D 1 D$. By the theorem of three perpendiculars, $E D 1 \perp A 1 F$. Consider the plane $A A 1 D 1 D$. Let the line $A 1 F$ intersect the lines $D D 1$ and $E D 1$ at points $G$ and $K$ respectively. Denote $\angle D D 1 E=\angle G A 1 D 1=\alpha$. Then

$$
\operatorname{tg} \alpha=\frac{E D}{D D_{1}}=\frac{1}{2}, \cos \alpha=\frac{2}{\sqrt{5}}
$$

$$
G D 1=A 1 D 1 \operatorname{tg} \alpha=\frac{1}{2} a, A 1 K=A 1 D 1 \cos \alpha=\frac{2 a}{\sqrt{5}}
$$

From the equality of the right triangles $D G F$ and $D 1 G A 1$, it follows that $D F=A 1 D 1=a$. Therefore,

$$
\frac{B H}{B C}=\frac{A F^{\prime}}{A D}=2
$$

Moreover,

$$
A 1 F=2 A 1 G=2 \sqrt{A D^{2}+D G^{2}}=2 \sqrt{a^{2}+\frac{a^{2}}{4}}=a \sqrt{5}
$$

Drop the perpendicular $K H$ from point $K$ to the line $A 1 R$. The line $E D 1$ is perpendicular to two intersecting lines $A 1 R$ and $A 1 F$ in the plane $A 1 F R$, so $K H \perp E D 1$. Thus, $K H$ is the common perpendicular of the skew lines $E D 1$ and $A 1 R$. Since the segment connecting the midpoints of opposite edges of a regular tetrahedron is the common perpendicular of the lines containing these edges, the length of $K H$ is the desired distance. Consider the right triangle $A 1 F R$. Denote $\angle F A 1 R=\beta$. Then

$$
\operatorname{tg} \beta=\frac{R F^{\prime}}{A_{1} F^{\prime}}=\frac{a}{a \sqrt{5}}=\frac{1}{\sqrt{5}}, \cos \beta=\sqrt{\frac{5}{6}}, \sin \beta=\frac{1}{\sqrt{6}}
$$

$$
H K=A 1 K \sin \beta=\frac{2 a}{\sqrt{5}} \cdot \frac{1}{\sqrt{6}}=\frac{2 a}{\sqrt{30}}=a \sqrt{\frac{2}{15}} .
$$

## Answer

$2: 1 ; a^{\sqrt{\frac{2}{15}}}$