What is the minimum length a piece of wire must have so that a cube frame with an edge of 10 cm can be bent from it?

(The wire can pass along the same edge twice, can be bent at $90^{\circ}$ and $180^{\circ}$, but it cannot be broken.)
How to bend a frame from a wire of length 150 cm, as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_8b1cc902b566a3f27ee1g-60.jpg?height=592&width=663&top_left_y=433&top_left_x=709)

Let's assume for a moment that we are only allowed to bend the wire at the vertices of the cube, and that some frame starts and ends at the vertices of the cube. Let's count the number of edges in this frame. In the frame, there is exactly one starting vertex and one where the wire ends (these vertices can coincide). We will call these vertices passing. The wire, entering a passing vertex, must necessarily exit from it. Therefore, such a vertex has an even number of edges. The terminal vertex has at least three edges. Therefore, the total number of edges is at least ( $6 \cdot 4 + 2 \cdot 3$ ) : 2 : at least 15 edges, and since the length of one edge is 10 cm, at least 150 cm of wire is required.

Now we need to prove that if the wire is bent at internal points of the edges, it will still require at least 150 cm. For this, we will show that without increasing the length of the wire, any frame can be reconstructed so that all bends occur at the vertices. We will not describe this reconstruction in detail, but will provide several explanatory pictures (see fig.).

![](https://cdn.mathpix.com/cropped/2024_05_06_8b1cc902b566a3f27ee1g-60.jpg?height=186&width=1584&top_left_y=1462&top_left_x=243)

## Answer

150 cm.