We place a rectangular parallelepiped on a horizontal plane, one by one, with three different faces. In the three cases, the free surface areas of the body are in the ratio of the numbers $p, q, r$. Calculate the ratio of the edges of the parallelepiped. - What restriction must hold for the numbers $p, q, r$ for the problem to have a solution?
I. We denote the edges converging at one vertex of the body by $a, b, c$, so its surface area is $F=2(a b+a c+b c)$, and its free surface area in the mentioned 3 positions is:

$$
F-a b=k r, \quad F-b c=k p, \quad F-c a=k q
$$

where $k$ is some suitable, non-zero number. Each of the left sides is greater than $F / 2$, so $p, q, r$ are always non-zero and of the same sign. It suffices to deal with the case where $p, q, r$ are all positive, thus $k>0$ also holds.

By adding the equations (1) and substituting the expression obtained for $F$ into each,

$$
\begin{gathered}
3 F-F / 2=5 F / 2=k(p+q+r), \quad F=0.4 \cdot k(p+q+r) \\
b c=F-k p=0.2 k(-3 p+2 q+2 r) \\
c a=0.2 k(2 p-3 q+2 r), \quad a b=0.2 k(2 p+2 q-3 r)
\end{gathered}
$$

Thus, the ratio of the edges, taking into account the simplification possibilities,

$$
\begin{aligned}
& a: b: c=\frac{a b c}{b c}: \frac{a b c}{c a}: \frac{a b c}{a b}=\frac{1}{b c}: \frac{1}{c a}: \frac{1}{a b}= \\
= & \frac{1}{-3 p+2 q+2 r}: \frac{1}{2 p-3 q+2 r}: \frac{1}{2 p+2 q-3 r}
\end{aligned}
$$

II. Here, since $a, b, c>0$, the three denominators must have the same sign. And since their sum $p+q+r>0$, each is positive:

$$
\begin{aligned}
& 2 p+2 r>3 q \\
& 2 p+2 q>3 r \\
& 2 q+2 r>3 p
\end{aligned}
$$

Gabriella Varga (Szombathely, Savaria g. III. o. t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_e277d2ff7f72295310c8g-1.jpg?height=651&width=659&top_left_y=1390&top_left_x=711)

Remark. We get a visual representation of the conditions (2)-(4) as follows. Without loss of generality, we can choose the value of one of the ratios; let $r=2$. The modified conditions are satisfied by those pairs of numbers $p, q$ which, when considered as the coordinates of points in the $P, Q$ Cartesian coordinate system, lie below the line $q=(2 p+4) / 3$ and above the lines $q=3-p$ and $q=(3 p-4) / 2$ (Figure 1), i.e., the coordinate pairs of the points inside the triangle defined by the vertices $(4, 4)$, $(1,2)$, and $(2,1)$. - It is also easy to read from the figure that none of the $p, q, r$ ratios can reach the double of the smaller (smaller ones).