10,11 The length of the edge of a regular tetrahedron \(ABCD\) is \(a\). Point \(E\) is the midpoint of edge \(CD\), and point \(F\) is the midpoint of the height \(BL\) of face \(ABD\). Segment \(MN\) has endpoints on lines \(AD\) and \(BC\) and intersects line \(EF\) perpendicularly. Find the length of this segment.
Consider intersecting lines $A D$ and $B C$. Let point $X$ lie on line $A D$, and point $Y$ lie on line $B C$. It is known that the geometric locus of the midpoints of segments $X Y$ is a plane parallel to lines $A D$ and $B C$ and passing through the midpoint of some such segment. In our case, this is the plane $y$, passing through the midpoint $E$ of segment $C D$ parallel to the edges $A D$ and $B C$ of the tetrahedron. In this plane lies point $F$. By the problem's condition, the point $P$ of intersection of line $M N$ with plane $\gamma$ is the midpoint of segment $M N$. Consider the orthogonal projection of the tetrahedron onto plane $\gamma$. We obtain a square $A^{\prime} B^{\prime} D^{\prime} C^{\prime}$, where $A^{\prime}, B^{\prime}, D^{\prime}$, and $C^{\prime}$ are the projections of points $A, B, D$, and $C$ respectively. Let $M^{\prime}$ and $N^{\prime}$ be the projections of points $M$ and $N$. By the theorem of three perpendiculars, $M^{\prime} N^{\prime} \perp E F$, and since $M P = N P$, then $M^{\prime} P = N^{\prime} P$. Let $Q$ be the foot of the perpendicular dropped from point $E$ to $B^{\prime} C^{\prime}$. Then

$$
E Q = \frac{1}{2} D L = \frac{1}{2} \cdot \frac{\frac{a}{2}}{2} = \frac{a}{4}, \quad F Q = F L + L Q = \frac{1}{2} B^{\prime} L + \frac{1}{2} C^{\prime} L = \frac{\frac{a}{4}}{4} + \frac{a}{4} = \frac{a}{2}.
$$

Let $M^{\prime} P = N^{\prime} P = x$, and $\angle L M^{\prime} N^{\prime} = \angle E F Q = \alpha$. From the right triangles $E F Q$ and $L N^{\prime} M^{\prime}$, we find that

$$
\begin{gathered}
\operatorname{tg} \alpha = \frac{E Q}{F Q} = \frac{\frac{a}{4}}{\frac{a}{2}} = \frac{1}{2}, \quad \cos \alpha = \frac{1}{\sqrt{1 + \operatorname{tg}^2 \alpha}} = \frac{2}{\sqrt{5}} \\
L M^{\prime} = M^{\prime} N^{\prime} \cos \alpha = 2 x \cdot \frac{2}{\sqrt{5}} = \frac{4 x}{\sqrt{5}}
\end{gathered}
$$

Let segments $E F$ and $D^{\prime} L$ intersect at point $T$. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_a98533106cc029c284d7g-01.jpg?height=119&width=802&top_left_y=1547&top_left_x=628)

and since $L M^{\prime} = L T + M^{\prime} T$, then

$$
\frac{4 x}{\sqrt{5}} = \frac{a}{8} + \frac{x \sqrt{5}}{2}
$$

from which $M^{\prime} P = x = \frac{\alpha \sqrt{5}}{12}$. The distance from point $M$ to plane $\gamma$ is half the distance between opposite edges of the regular tetrahedron $A B C$, i.e.,

$$
M M^{\prime} = \frac{1}{2} \cdot \frac{a}{\sqrt{2}} = \frac{a}{2 \sqrt{2}}
$$

From the right triangle $M M^{\prime} P$, we find that

| Problem 111195 Topics: | [ Projection helps solve the problem ] |
| :---: | :---: |
|  | $\left[\begin{array}{ll}\text { Ky } \underline{\underline{ }}\end{array}\right]$ |

$$
M P = \sqrt{M M^{\prime 2} + M^{\prime} P^2} = \sqrt{\left(\frac{a \sqrt{2}}{4}\right)^2 + \left(\frac{a \sqrt{5}}{12}\right)^2} = \frac{a \sqrt{23}}{12}
$$

Therefore, $M N = \frac{a \sqrt{23}}{6}$.

## Answer

$\frac{a \sqrt{23}}{a}$

6. 

The edge length of the cube $A B C D A 1 B 1 C 1 D 1$ is $a$. Points $P, K, L$ are the midpoints of edges $A A 1, A 1 D 1, B 1 C 1$ respectively, and point $Q$ is the center of face $C C 1 D 1 D$. Segment $M N$ with endpoints on lines $A D$ and $K L$ intersects line $P Q$ and is perpendicular to it. Find the length of this segment.

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_a98533106cc029c284d7g-02.jpg?height=78&width=112&top_left_y=144&top_left_x=1)

[ Projection helps solve the problem_ ]
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