We attach a cube with an edge of $2 \mathrm{~m}$ to the intersection line $h$ of the floor and one wall of a room, such that one of its faces lies in the plane of the floor and another in the plane of the wall. One end of a $6 \mathrm{~m}$ long ladder rests on the floor, while the other end leans against the wall and touches the edge of the cube opposite to $h$. How high above the floor is the upper end of the ladder?
Our $h$-perpendicular plane sections. Let's use their notations. From the similarity of the two triangles in Figure 1, we have $x: 2=2: y$, from which $y=\frac{4}{x}$. According to the Pythagorean theorem, $(x+2)^{2}+(y+2)^{2}=36$, and thus, using the first relationship as well:

$$
x^{2}+4 x+4+\frac{16}{x^{2}}+\frac{16}{x}+4=36
$$

from which

$$
\left(x+\frac{4}{x}\right)^{2}+4\left(x+\frac{4}{x}\right)=36
$$

By introducing the new unknown $a=x+\frac{4}{x}$, we need to solve the equation $a^{2}+4 a-36=0$, from which $a=-2 \pm 2 \sqrt{10}$. Considering that $x>0$, only the solution $a=-2+2 \sqrt{10}$ is possible. From the equation $a=x+\frac{4}{x}$, we get $x^{2}-a x+4=0$, thus $x^{2}-(-2+2 \sqrt{10}) x+4=0$. From this,

$$
x=\frac{-2+2 \sqrt{10} \pm \sqrt{(-2+2 \sqrt{10})^{2}-16}}{2}=\sqrt{10}-1 \pm \sqrt{7-2 \sqrt{10}}=\sqrt{10}-1 \pm(\sqrt{5}-\sqrt{2})
$$

Thus, the two possible values of $x$ are:

$$
x_{1}=\sqrt{10}-1+\sqrt{5}-\sqrt{2}=2.98 \quad x_{2}=\sqrt{10}-1-\sqrt{5}+\sqrt{2}=1.34
$$

The top end of the ladder is $x_{1}+2=4.98$ or $x_{2}+2=3.34$ meters above the floor.

Julia Pap (Debrecen, Fazekas M. Gymnasium, 8th grade)

Remarks. 1. The equation for $x$ after eliminating the fractions will be a fourth-degree equation. Its 4 roots - all irrational - can be determined with arbitrary precision using approximate methods.

2. The two equations for $x$ and $y$ are a hyperbola and a circle, respectively. The two second-degree curves can be easily plotted, so the solutions can also be given graphically with a certain precision.
3. The negative $x$ values corresponding to the negative $a$ value can be given a geometric meaning if we extend the rays of the right angle formed by the wall and the floor into number lines (Figure 2).
![](https://cdn.mathpix.com/cropped/2024_05_02_94a406687fda4a7e00ecg-1.jpg?height=900&width=518&top_left_y=1432&top_left_x=790)