Answer: $\frac{1}{2} \cdot(2+\sqrt[3]{4})^{\frac{3}{2}}$.

## Solution.

Let $A$ and $B$ be the ends of the beam, with $B$ being the lower end, which first enters the lower corridor (we assume the beam is being transferred from the upper corridor to the lower one). For brevity, let $d=\frac{1}{2} \cdot(2+\sqrt[3]{4})^{\frac{3}{2}}$ be the length of the beam specified in the answer, and $P Q R S$ be the square forming the hole.

The solution will consist of two parts: I - we will show how to transfer the beam of the specified length, and II - we will prove that a beam of greater length cannot be transferred in any way.

I. We will need the following additional notations. The wall of the upper corridor passing through points $Q, R$ intersects the ceiling of the upper corridor along the line $a$ (see figure). The wall of the lower corridor containing points $R, S$ intersects the floor of the lower corridor along the line $b$. $Q_{1}, R_{1}$ are the orthogonal projections of points $Q, R$ onto the line $a$, and $R_{2}, S_{2}$ are the orthogonal projections of points $R, S$ onto the line $b$. $A_{0}, B_{0}$ are points on the lines $a, b$ respectively, such that $\overrightarrow{A_{0} Q_{1}}=\overrightarrow{Q_{1} R_{1}}$ and $\overrightarrow{B_{0} S_{2}}=\overrightarrow{S_{2} R_{2}}$. Segments $P A_{0}$ and $P B_{0}$ are parallel to segment $Q_{1} S$, so points $P, A_{0}, B_{0}$ lie on the same line, and $P A_{0}=P B_{0}=\sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e7cec4165a007c52c5a3g-09.jpg?height=571&width=765&top_left_y=1642&top_left_x=246)

$b$

$a$

![](https://cdn.mathpix.com/cropped/2024_05_06_e7cec4165a007c52c5a3g-09.jpg?height=568&width=685&top_left_y=2280&top_left_x=320)

$a$

![](https://cdn.mathpix.com/cropped/2024_05_06_e7cec4165a007c52c5a3g-09.jpg?height=531&width=677&top_left_y=1648&top_left_x=1141)

$b$

$a$

![](https://cdn.mathpix.com/cropped/2024_05_06_e7cec4165a007c52c5a3g-09.jpg?height=566&width=682&top_left_y=2284&top_left_x=1144)

The movement of the beam will consist of the following steps (see figures (1)-(4) above):

(1) Place the beam so that its end $A$ lies on the line $a$, and the end $B$ is at point $P$.

(2) Move the end $A$ along the line $a$ such that the beam always passes through point $P$.

(3) Continue this movement until point $A$ coincides with $A_{0}$. At this moment, the beam will be aligned along the line $A_{0} B_{0}$.

(4) Then move the beam along the line $A_{0} B_{0}$ until point $B$ coincides with $B_{0}$.

(5) Finally, move point $B$ along the line $b$ such that the entire beam still passes through $P$, until the entire beam is in the lower corridor.

The key fact to verify is that such movement is feasible, i.e., the beam will always be entirely within the corridors. It is sufficient to check this only for steps (1) - (3).

Draw the plane $\delta$ through the line $a$ and point $P$. During all steps (1) - (4), the beam is within the plane $\delta$, so we can study the movement of the beam separately in this plane (see figure below). Let $A Q_{1}=x$. Extend the line $A P$ to intersect the wall of the lower corridor at point $X$. Then, from the similarity of triangles $A P Q_{1}$ and $A X R_{1}$, it is not difficult to find $A X=\frac{x+1}{x} \cdot \sqrt{2+x^{2}}$. The derivative of this function is $\frac{x^{3}-2}{x^{2} \sqrt{2+x^{2}}}$. From this, it is clear that the minimum of the function is achieved at $x=\sqrt[3]{2}$. Substituting this value into the expression for $A X$, we get after transformations $\min (A X)=d$ (recall, $d$ denotes the number given in the answer). Thus, the length $A X$ is always not less than the length of the beam (in particular, when $x=1, A X=2 \sqrt{3}>d$), i.e., point $B$ never goes beyond the segment $A X$, and therefore never goes beyond the corridors, as required.

![](https://cdn.mathpix.com/cropped/2024_05_06_e7cec4165a007c52c5a3g-10.jpg?height=911&width=945&top_left_y=1735&top_left_x=561)

II. Now we will prove that a beam longer than $d$ cannot be transferred from one corridor to another in any way. Let $O$ be the point on the beam that divides it in the ratio $A O: O B=\sqrt[3]{2}: 1$. Since the movement of the beam is continuous, at any transfer method, there will be a moment when point $O$ is in the plane $P Q R S$. We will show that at this moment, the length of the beam cannot be greater than $d$.

Let $\alpha$ and $\beta$ be the vertical planes passing through $O$ parallel to the axes of the upper and lower corridors, respectively, $z_{A}$ be the distance from $A$ to the plane $P Q R S$, $x_{A}$ be the distance from $A$ to $\alpha$, $y_{A}$ be the distance from $A$ to $\beta$, and $z_{B}, x_{B}, y_{B}$ be the corresponding distances for point $B$. (In the figure below on the left, a "top view" is shown in projection onto the plane $P Q R S$, the planes $\alpha$ and $\beta$ project into lines, and the segments $z_{A}$ and $z_{B}$ are not visible as they project into points $A$ and $B$ respectively. Point $O$ is in the plane $P Q R S$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e7cec4165a007c52c5a3g-11.jpg?height=631&width=868&top_left_y=838&top_left_x=594)

Clearly, $z_{A}: z_{B}=x_{A}: x_{B}=y_{A}: y_{B}=\sqrt[3]{2}: 1$. Additionally, $x_{A} \leq 1, y_{B} \leq 1, z_{A} \leq 1$. Then the length of the entire beam

$A B=(1+\sqrt[3]{2}) \cdot \sqrt{x_{B}^{2}+y_{B}^{2}+z_{B}^{2}}=(1+\sqrt[3]{2}) \cdot \sqrt{\frac{x_{A}^{2}}{\sqrt[3]{4}}+y_{B}^{2}+\frac{z_{A}^{2}}{\sqrt[3]{4}}} \leq(1+\sqrt[3]{2}) \cdot \sqrt{\frac{1}{\sqrt[3]{4}}+1+\frac{1}{\sqrt[3]{4}}}=d$.

Criteria for evaluating solutions.

(-) A solution that does not contain any progress in the correct direction, as well as any solution in which the answer $2 \sqrt{3}$ is given.

(-/+) There is a correct idea of the solution (algorithm for passing the beam through the hole) and an attempt (not completed) to formulate a function expressing the length of the segment $A X$ in terms of a variable.

(+/-) In the problem, the correct answer is obtained and it is shown that a beam of this length can be transferred. However, it is not proven that a beam of greater length cannot be transferred.

(+) Correct solution (answer + method of passing the beam + proof of maximality).
5. Two corridors, each 1 m high and 1 m wide, run perpendicular to each other on the first and second floors of a building. The separating floor has been removed, creating a 1 m x 1 m hole in the floor of one and the ceiling of the other. What is the maximum length of the beam that can be passed from one corridor to the other through the hole? (Consider the beam as a non-bending segment of zero thickness. The thickness of the floor is also zero, i.e., the floor of the upper corridor and the ceiling of the lower corridor are in the same plane.)