Let $A B C$ be an equilateral triangle with unit side length. Let the reflection of $C$ over $A$ be $D$. We know that from point $F$ on the extension of line $A B$ beyond $B$, $E F=1$, where $E$ is the intersection of lines $F C$ and $D B$. Determine the lengths of segments $C E=x$ and $B F=y$.
I. solution: In the isosceles triangle $ABD$, the angle at vertex $A$ is $120^{\circ}$ (see the figure), and thus $\angle ABD = \angle EBF = 30^{\circ}$ and $\angle CBE = 90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_704288a5ab5fe5a82a2dg-1.jpg?height=494&width=819&top_left_y=314&top_left_x=642)

In the right triangle $CBE$, denoting $\angle CEB$ by $\varphi$,

$$
x=\frac{1}{\sin \varphi}
$$

In the triangle $BEF$, by the Law of Sines,

$$
1: y=\sin 30^{\circ}: \sin (180-\varphi)=\frac{1}{2}: \sin \varphi
$$

from which

$$
y=2 \sin \varphi
$$

The product of (1) and (2)

$$
x y=2
$$

In the triangle $ABC$, the altitude from $C$ is $CG=\frac{\sqrt{3}}{2}$. Applying the Pythagorean theorem to the right triangle $CGF$,

$$
\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=(x+1)^{2}
$$

Expanding the parentheses, simplifying, and multiplying by $x^{2}$ (since $x^{2} \neq 0$) and reducing to 0,

$$
y^{4}+y^{3}-4 y-4=0
$$

On the left side, $(y+1)$ can be factored out from the first two terms and also from the last two terms,

$$
(y+1)\left(y^{3}-4\right)=0
$$

This equation has only one positive root,

$$
y=\sqrt[3]{4}, \text { and thus from (3) } x=\frac{2}{\sqrt[3]{4}}=\sqrt[3]{\frac{8}{4}}=\sqrt[3]{2}
$$

Danassy Károly (Mosonmagyaróvár, Kossuth g. II. o. t.)

II. solution: In the right triangle $CBD$, $BD=\sqrt{DC^{2}-CB^{2}}=\sqrt{4-1}=\sqrt{3}$, and from the right triangle $CBE$, $BE=\sqrt{x^{2}-1}$.

Applying Menelaus' theorem to triangle $CAF$ and line $DE$,

$$
(CAD)(AFB)(FCE)=(-2) \cdot \frac{1}{y} \cdot \frac{1}{x}=-1
$$

which simplifies to

Similarly, applying Menelaus' theorem to triangle $CDE$ and line $AF$,

$$
(CDA)(DEB)(ECF)=1 \cdot \frac{\sqrt{3}}{\sqrt{x^{2}-1}} \cdot\left(-\frac{1}{x+1}\right)=-1
$$

which simplifies to

$$
(x+1) \sqrt{x^{2}-1}=\sqrt{3}
$$

Squaring and rearranging,

$$
x^{4}+2 x^{3}-2 x-4=0
$$

which factors as

$$
(x+2)\left(x^{3}-2\right)=0
$$

This equation has only one positive root,

$$
x=\sqrt[3]{2} \text { and thus from (5) } y=\frac{2}{\sqrt[3]{2}}=\sqrt[3]{\frac{8}{2}}=\sqrt[3]{4}
$$

Argay Gyula (Balassagyarmat, Balassa g. II. o. t.)

Note: The Delian problem is well-known, which seeks to construct the edge of a cube whose volume is twice that of a given cube. Considering the edge of the given cube as a unit, the equation $x^{3}=2 \cdot 1^{3}=2$ yields $x=\sqrt[3]{2}$, which is known to be non-constructible with a compass and straightedge. Based on the present problem, we find a method to construct the distance $\sqrt[3]{2}$, provided that instead of the common single-edged straightedge, we allow the use of a "distance-carrying straightedge," i.e., a straightedge on which a unit segment is marked.

The construction process (using the figure): On one ray of a $120^{\circ}$ angle at vertex $B$, measure the segment $BC=1$ from $B$. Draw a line through $B$ perpendicular to $CB$, which thus forms a $30^{\circ}$ angle with the other ray of the $120^{\circ}$ angle. Place the distance-carrying straightedge (rotating around $C$ and sliding - everyone can try this on a flat sheet of paper) so that it touches $C$, and at the same time, the two endpoints of a unit segment $E$ and $F$ lie on the rays of the $30^{\circ}$ angle. In this position - as we have proven - $CE=x=\sqrt[3]{2}$. (This is also a construction, but not Euclidean.)