11-4. In the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with side 1, the diagonal $A C_{1}$ is drawn. For each point $N$ lying on this diagonal, a section of the cube is constructed by a plane perpendicular to $A C_{1}$ and passing through $N$. Let $P$ be the perimeter of this section. Construct the graph of the dependence of $P$ on the value $x=A N$.
Solution. Note that the sections $B D A_{l}$ and $B_{l} D_{l} A$ are perpendicular to the diagonal $A C_{l}$ and divide it into three equal parts, each of length $\frac{\sqrt{3}}{3}$. It is sufficient to consider the trace that the section leaves on two faces, for example, $A B B_{l} A_{l}$ and $B C C_{l} B_{l}$, the entire perimeter of the section is three times larger. All sections are parallel to each other. The traces pass at an angle of $45^{\circ}$ to the edges.

As $x$ increases, the section initially has the shape of an equilateral triangle, the perimeter of which grows proportionally to $x$. The largest section is the one that passes through points $B, D$, and $A_{1}$, its perimeter is $3 \sqrt{2}$, the same perimeter is for the triangle $B_{l} D_{l} C$.
![](https://cdn.mathpix.com/cropped/2024_05_06_95e9b67866e8638555fag-2.jpg?height=442&width=1666&top_left_y=1058&top_left_x=206)

Thus, for $0 \leq x \leq \frac{\sqrt{3}}{3}$ we have $P=3 \sqrt{6} x$, for $\frac{\sqrt{3}}{3} \leq x \leq \frac{2 \sqrt{3}}{3}$ we have $P=3 \sqrt{2}$, for $\frac{2 \sqrt{3}}{3} \leq x \leq \sqrt{3}$ we have $P=3 \sqrt{6}(\sqrt{3}-x)$.

Criteria. A complete answer without sufficient justification - 2 points. Full solution - 7 points.