In an apartment, the length of room I and room II is equal, and the width of room II and room III is equal, finally, the diagonal of room I and room III is equal. (The base of the rooms is, of course, rectangular.) The floor area of the rooms is $a, b, c$ square meters, respectively. Calculate the dimensions of the individual rooms.
Let the dimensions of the rooms (always the width first) be I: $x, y$, II: $x, z$, III: $u, z$, thus the equations applicable to them are:

$$
\begin{aligned}
x y & =a, \\
x z & =b, \\
u z & =c, \\
x^{2}+y^{2} & =z^{2}+u^{2}
\end{aligned}
$$

(The arrangement of the rooms and their mirror images in the figures is arbitrary.)

![](https://cdn.mathpix.com/cropped/2024_05_02_5784c27b31979a0e3601g-1.jpg?height=293&width=512&top_left_y=542&top_left_x=793)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_5784c27b31979a0e3601g-1.jpg?height=298&width=507&top_left_y=903&top_left_x=798)

Figure 2

Express the other unknowns in terms of $x$ from equations (1)-(3) (express $u$ in terms of $z$ first), and substitute these expressions into (4):

$$
x^{2}+\frac{a^{2}}{x^{2}}=\frac{b^{2}}{x^{2}}+\frac{c^{2} x^{2}}{b^{2}}
$$

From this (since the unknowns in our problem are positive),

$$
x^{4}\left(1-\frac{c^{2}}{b^{2}}\right)=b^{2}-a^{2}
$$

and if the coefficient of $x^{4}$ is not 0, that is, $c \neq b$, then

$$
x^{4}=\frac{b^{2}\left(b^{2}-a^{2}\right)}{b^{2}-c^{2}}
$$

This has a real (positive) solution only if the differences in the numerator and denominator have the same sign, that is, if (Figure 1, or Figure 2)

$$
\begin{aligned}
& \text { either } b > a \text { and } b > c
\end{aligned}
$$

and in this case the solution is:

$$
x=\sqrt[4]{\frac{b^{2}\left(b^{2}-a^{2}\right)}{b^{2}-c^{2}}}, \quad y=\frac{a}{x}, \quad z=\frac{b}{x}, \quad u=\frac{c x}{b}
$$

If, however, the coefficient of $x^{4}$ in (5) is 0, that is,

$$
b=c
$$

then (5) can only have a solution if the right-hand side is also 0, that is,

$$
b=a
$$

In this case, (5) and the problem have the solution

$$
x=u=t, \quad y=z=\frac{b}{t}
$$

where $t$ is any positive number. It is easy to see that in this case, not only are the three rooms of equal area, but they are also congruent.

Julia Kiraly (Budapest, Fazekas M. High School, 1st year) Antal Somogyi (Budapest, Moricz Zs. High School, 2nd year)

Note. In practical problems, the length of a rectangle is usually understood to be the longer of its sides. In this sense, (8) is only a solution if in I and II $x>y, z$, and in III $z>x$ and $z>y$ and $u \geq z$ conditions:

$$
a \leq \frac{b}{c} \sqrt{2 c^{2}-b^{2}}
$$

(when it is even more true that $a<c$).