Given a cube with edge length $x$. Construct segments of lengths $y$ and $z$ such that the sum of the volumes of the cubes with edge lengths $y$ and $z$, which have no common points, is equal to the volume of the original cube.
I. solution. If $m$ and $n$ are given positive integers, and $x$ is a given segment, then $\frac{m}{n} \cdot x$ can easily be constructed using similar triangles. For example, on one of two rays starting from a common point $A$, we measure the segment $A B = m x$, and on the other, we measure the $n$-fold of an arbitrary segment $A C_{0}$, which we denote as $A C$. The line through $C_{0}$ parallel to $C B$ intersects the segment $A B$ at a point that is exactly $A B_{0} = \frac{m}{n} \cdot x$ long (Figure 1).

# 1984-11-366-1.eps 

## Figure 1

1984-11-366-2.eps

Figure 2

Let $r$ be any positive rational number. $\sqrt{r} \cdot x$ is the altitude to the hypotenuse of a right triangle that divides the hypotenuse into segments of lengths $r x$ and $x$ (Figure 2). Based on this, $\sqrt{r} \cdot x$ can be constructed as follows: draw a Thales circle over the segment of length $(r+1) x$, and draw a perpendicular line from one endpoint of the diameter at a distance of $x$. This line intersects the circle at a point that is exactly $\sqrt{r} x$ away. The problem requires constructing segments of lengths $y$ and $z$ such that $y^{3} + z^{3} = x^{3}$. Let

$$
y + z = t \cdot x
$$

With this notation, the equation $x^{3} = y^{3} + z^{3} = (y + z)(y^{2} - yz + z^{2})$ can be written as:

$$
y^{2} - yz + z^{2} = \frac{x^{2}}{t}
$$

Square (1), subtract (2) from it, divide by 3, and then take the square root:

$$
\sqrt{yz} = x \sqrt{\frac{t^{2}}{3} - \frac{1}{3t}}
$$

It is clear that (3) and (1) imply (2), and (1) and (2) imply $x^{3} = y^{3} + z^{3}$. Therefore, it is sufficient to construct segments of lengths $y$ and $z$ such that (1) and (3) hold. If $t$ is a positive rational number and

$$
\frac{t^{2}}{3} - \frac{1}{3t} > 0, \quad \text{or} \quad t > 1
$$

then, according to the initial remarks, we can construct segments of lengths $t \cdot x$ and $x \cdot \sqrt{t^{2} / 3 - 1 / (3t)}$. The right triangle $ABC$ with hypotenuse $y + z = t x$ and altitude $\sqrt{yz}$ divides the hypotenuse into segments of lengths $y$ and $z$.

If $t > 1$, then the following construction provides the required segments $y$ and $z$: construct the segment $AB = t x$ and draw the Thales circle over it. The line $e$ at a distance of $x \sqrt{t^{2} / 3 - 1 / (3t)}$ from the line $AB$ can be constructed according to the initial remarks. If $e$ intersects the Thales circle, then the intersection points correspond to $C$. The perpendicular projections of these points onto $AB$ divide the hypotenuse into segments $AD$ and $DB$ of lengths $y$ and $z$, respectively, which we were looking for (Figure 3).

$1984-11-366-3 . \mathrm{eps}$

Figure 3

We need to determine whether $t$ can be chosen such that $e$ intersects the Thales circle. This is precisely the condition that the distance $x \sqrt{t^{2} / 3 - 1 / (3t)}$ does not exceed half the diameter $t x$, i.e., the following inequality must hold:

$$
\sqrt{\frac{t^{2}}{3} - \frac{1}{3t}} \leq \frac{t}{2}, \quad \text{or} \quad t^{3} \leq 4
$$

Thus, our construction is valid for any $1 < t \leq \sqrt[3]{4}$.

II. solution. We seek $y$ and $z$ in the form $y = (a + \sqrt{b}) x$ and $z = (a - \sqrt{b}) x$, where $a$ and $b$ are positive rational numbers. (Note that the previous solution also leads to such forms for $y$ and $z$.) Then,

$$
y^{3} + z^{3} = \left((a + \sqrt{b})^{3} + (a - \sqrt{b})^{3}\right) x^{3} = \left(2a^{3} + 6ab\right) x^{3}
$$

Thus, $a$ and $b$ satisfy the equation if and only if $2a^{3} + 6ab = 1$, or

$$
b = \frac{1 - 2a^{3}}{6a}
$$

From (4) and (6), we have $a^{2} > \frac{1 - 2a^{3}}{6a} > 0$, which implies $\frac{1}{2} < a < 1$. We need to find $a$ and $b$ such that $a > 0$, $b$ is rational, and $a - \sqrt{b} > 0$. Thus, the segments $y = (a + \sqrt{b}) x$ and $z = (a - \sqrt{b}) x$ can be constructed, and for $y$ and $z$, we have $y^{3} + z^{3} = x^{3}$, solving the problem. For example, $a$ can be chosen as $\frac{2}{3}$.

III. solution. This time, we seek $y$ and $z$ in the form $y = x \sin \alpha$ and $z = x \sin \beta$, where $\alpha$ and $\beta$ are constructible angles. The condition $x^{3} = y^{3} + z^{3}$ can be written as:

$$
1 = \sin^{3} \alpha + \sin^{3} \beta = \frac{3 \sin \alpha - \sin 3\alpha + 3 \sin \beta - \sin 3\beta}{4} = \frac{3}{2} \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} - \frac{1}{2} \sin \frac{3\alpha + 3\beta}{2} \cos \frac{3\alpha - 3\beta}{2}
$$

Notice that if $\alpha + \beta = 120^\circ$, then $\sin \frac{3\alpha + 3\beta}{2} = 0$ and $\sin \frac{\alpha + \beta}{2} = \frac{\sqrt{3}}{2}$. The above equation simplifies to:

$$
\frac{3\sqrt{3}}{4} \cos \frac{\alpha - \beta}{2} = 1
$$

or

$$
\cos \frac{\alpha - \beta}{2} = \frac{4}{3\sqrt{3}}
$$

The angle opposite the side of length 4 in a right triangle with hypotenuse $3\sqrt{3} > 4$ is exactly $\frac{\alpha - \beta}{2}$, which is less than $60^\circ$. Additionally, $\frac{\alpha + \beta}{2} = 60^\circ$ by assumption, so $\alpha$ and $\beta$ can be constructed, and from this, the positive $y = x \sin \alpha$ and $z = x \sin \beta$ can also be constructed. The above calculations can be reversed from (7), leading to the conclusion that for the constructed angles, $\sin^{3} \alpha + \sin^{3} \beta = 1$, or $y^{3} + z^{3} = x^{3}$, as required by the problem.

Remark. Several people noticed that the problem is related to the Delian problem, which challenged the Greeks to construct a cube with twice the volume of a given cube. After many ingenious attempts, it was only proven in the late 19th century that such a cube's edge cannot be constructed with a compass and straightedge. In the language of our problem, this means that the problem cannot be solved if $y = z$. The