A tetrahedron has one face that is a regular triangle with 6 unit edges, and its other edges are 3, 4, and 5 units long. What is the distance between the straight line of the 3-unit edge and the straight line of the opposite edge?
Place the tetrahedron in a three-dimensional Cartesian coordinate system as shown in the figure. Use the notation from the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_fcb1ce730580af0b2b60g-1.jpg?height=683&width=981&top_left_y=260&top_left_x=561)

Given the edge lengths:

$$
\begin{gathered}
u^{2}+v^{2}+w^{2}=4^{2} \\
(3-u)^{2}+v^{2}+(3 \sqrt{3}-w)^{2}=3^{2} \\
(6-u)^{2}+v^{2}+w^{2}=5^{2}
\end{gathered}
$$

The solution to this system of equations is:

$$
u=\frac{9}{4} ; \quad v=\sqrt{\frac{311}{108}} ; \quad w=\frac{59 \sqrt{3}}{36}
$$

Thus, we also know the coordinates of point \( B \), and we can write the vector \( \overrightarrow{D B} \):

$$
\overrightarrow{D B}=\left(-\frac{3}{4} ; \sqrt{\frac{311}{108}} ;-\frac{49 \sqrt{3}}{36}\right)
$$

Using the direction vector of line \( B D \) and point \( D \), the parametric equations of the line are:

$$
\begin{aligned}
& x=3-\frac{3}{4} t \\
& y=\sqrt{\frac{311}{108}} t \\
& z=3 \sqrt{3}-\frac{49 \sqrt{3}}{36} t
\end{aligned}
$$

Our task is to determine the distance between the lines \( B D \) and \( A C \). This is the length of segment \( E F \) in the figure. Since \( E F \) is perpendicular to the \( x \)-axis and parallel to the \( (y ; z) \)-plane, its projection onto this plane, \( E^{\prime} A \), has the same length as \( E F \). The projection of the right angle formed by \( B D \) and \( E F \) is also a right angle, as one of its sides is parallel to the projection plane. Therefore, the desired distance can also be determined by writing the equation of line \( B^{\prime} D^{\prime} \) in the \( (y ; z) \)-plane and calculating the distance from this line to the origin.

From the system of equations (3), we have \( t=\sqrt{\frac{108}{311}} y \), so the equation of \( B^{\prime} D^{\prime} \) is:

$$
\frac{49 \sqrt{3}}{36} \sqrt{\frac{108}{311}} y+z=3 \sqrt{3}
$$

Now we use the well-known formula for the distance from a point to a line. In the usual notation, the distance from the line \( A x+B y+C=0 \) to the point \( P\left(x_{1} ; y_{1}\right) \) is:

$$
d=\left|\frac{A x_{1}+B y_{1}+C}{\sqrt{A^{2}+B^{2}}}\right|
$$

In our case, the equation of the line is (4), and the point is \( (0 ; 0) \); thus, the desired distance is:

$$
d=\frac{3 \sqrt{3}}{\sqrt{\frac{49^{2} \cdot 3}{36^{2}} \cdot \frac{108}{311}+1}}=\sqrt{\frac{3732}{405}} \approx 3.0356
$$

Zsolt Lázár (Kaposvár, M. Táncsics Gymnasium, 3rd grade)