There is a distance $AB = a$. Let's construct the point $P$ on this distance so that the sum of the surface areas of the regular tetrahedron formed with the distances $AP$ and the regular hexahedron (cube) formed with the distance $PB$ as edges is the smallest possible.
I. solution: Let $A P=x$ be the edge of the tetrahedron, then the surface area of the tetrahedron is $4 \cdot \frac{x^{2}}{4} \sqrt{3}=x^{2} \sqrt{3}$, and the surface area of the hexahedron is $6(a-x)^{2}$.

Thus, we are looking for the minimum of the quadratic function

$$
y=x^{2} \sqrt{3}+6(a-x)^{2}=(6+\sqrt{3}) x^{2}-12 a x+6 a^{2}
$$

It is known that the extremum of the quadratic function $y=a x^{2}+b x+c$ occurs at $x_{0}=-\frac{b}{2 a}$. Applying this to our case,

$$
x_{0}=\frac{12 a}{12+2 \sqrt{3}}=\frac{6 a}{6+\sqrt{3}}.
$$

Therefore,

$$
\begin{gathered}
A P=\frac{6 a}{6+\sqrt{3}}, \\
B P=a-x=a-\frac{6 a}{6+\sqrt{3}}=\frac{a \sqrt{3}}{6+\sqrt{3}}
\end{gathered}
$$

and thus

$$
\frac{A P}{P B}=\frac{6}{\sqrt{3}}=\frac{2 \sqrt{3}}{1}=\frac{\sqrt{12}}{1}
$$

The division of the distance $A B$ in the ratio $\sqrt{3 \cdot 4}: 1$ is shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_a046f9fb33c3e53a781cg-1.jpg?height=377&width=615&top_left_y=1118&top_left_x=736)

Figure 1

$\sqrt{12}$ was constructed as the geometric mean between 3 and 4 based on the well-known proportional theorem of the right triangle.

Frivaldszky Sándor (Bp. II., Rákóczi g. II. o. t.)

II. solution: The minimal surface area and thus the point $P$ can be constructed without any calculations.

A quarter of the surface area is exactly the sum of one tetrahedron face and 1.5 cube faces. Let's examine how this sum of areas changes as the point $P$ moves along the segment $A B$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a046f9fb33c3e53a781cg-1.jpg?height=713&width=624&top_left_y=1907&top_left_x=729)

We denote an arbitrary position of the point $P$ as $P^{\prime}$ (Figure 2); the corresponding tetrahedron face is the equilateral triangle $A C P^{\prime}$, and the 1.5 cube faces can be visualized as a rectangle with one side $B P^{\prime}$ and the other $B E=P^{\prime} D=1.5 P^{\prime} B$. Transform both shapes into right triangles such that their common leg is the perpendicular from $P^{\prime}$ to $A B$, and the other legs are $P^{\prime} A$ and $P^{\prime} B$. The tetrahedron face area can be replaced by the triangle $A P^{\prime} C^{\prime}$, and the cube face area by the triangle $B P^{\prime} H^{\prime}$ (where $P^{\prime} H^{\prime}=3 P^{\prime} B$). The hypotenuses of these triangles lie on fixed lines independent of the position of $P^{\prime}$, because the angle $\alpha$ formed by the hypotenuse from $A$ and $A B$ has $\tan \alpha=\frac{\sqrt{3}}{2}$, and the angle $\beta$ formed by the hypotenuse from $B$ and $A B$ has $\tan \beta=3$. The intersection of these two fixed lines is denoted by $M$. As $P^{\prime}$ moves, $C^{\prime}$ moves along the hypotenuse $A M$, and $H^{\prime}$ moves along the hypotenuse $B M$.

The quarter of the surface area of the two polyhedra is visualized by the quadrilateral $A C^{\prime} H^{\prime} B$. This quadrilateral is composed of the constant triangle $A M B$ and the variable triangle $M C^{\prime} H^{\prime}$. As $P^{\prime}$ changes, only the area of this latter triangle changes; it is smallest when it is 0, i.e., when $C^{\prime}$ and $H^{\prime}$ coincide with $M$.

The point $M$ and the corresponding point $P$ can be constructed from the diagram.

Remark: Now we can calculate the length of the constructed $A P$ distance based on the construction and the angles $\alpha$ and $\beta$. Indeed, $M P=x \tan \alpha=(a-x) \tan \beta$, so $x \frac{\sqrt{3}}{2}=(a-x) 3$, from which $x=\frac{6 a}{6+\sqrt{3}}$.

Katz Tibor (Hajdúnánás, Körösi Csoma g. III. o. t.)