A Bending Buzz Wire Game

Question

There is a wire connecting an exit and an entry point. At the entry, the wire has height $0$, at the exit, it has height $1$. Since the wire is connected, the wire has height $1/2$ somewhere, whatever its form.

The form of the wire is changed by moving a lever, with the lever taking positions $r∈[0,1]$: The wire is bend and changes its form continuously as $r$ changes from $0$ to $1$ in an arbitrary way, except that it is never cut and kept fixed at the entry and exit points. (For example, at $r=0$ and $r=1$, the wire may be a straight line connecting the entry and exit points, but at $r=1/2$ the wire has multiple hills and valleys.)

You may move the lever back and forth. Your task is to hold a buzz ring around the wire at height $1/2$ and move the lever from $0$ to $1$. You can choose any initial point of the wire to put the ring at height $1/2$. Importantly, you can move the lever back and forth in between.

Can you move the lever from $0$ to $1$ and choose an initial point, such that you can keep the ring at $1/2$ the whole time until the lever reaches $1$?

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Put differently, I am considering the following problem. Take any function $f:[0,1]×[0,1]→[0,1]$ such that

• $f(0,r)=0$ for all $r$ (the entry height is $0$, no matter the lever position)
• $f(1,r)=1$ for all $r$ (the exit height is $1$, no matter the lever position)
• $f(x,r)$ is continuous in $(x,r)$ (for any lever position $r$, the wire connects $0$ and $1$ without interruptions and as we change the lever, the wire changes its form continuously).

I want to show that there exist continuous functions $\hat{x}:[0,1]→[0,1]$ and $\hat{r}:[0,1]→[0,1]$ such that

• $\hat{r}(0)=0$ (the lever is at position $0$ at time $t=0$)
• $\hat{r}(1)=1$ (the lever is at position $1$ at time $t=1$)
• $f(\hat{x}(t),\hat{r}(t))=1/2$ for all $t$ (the ring is held at height $1/2$ at all times)

I think this follows from the fact that the graph of the fixed points is connected.

Answer 1

A sketch, not a formal proof. (Also: my English may lack of strict terminology.)

We can visualize $h = f(x,r)$ as a curved plane (3D plot of the function).

• The edge $x=0$ is at $h=0$.
• The edge $x=1$ is at $h=1$.
• On the edge $r=0$ there is at least one point with $h=\frac 1 2$.
• The same for $r=1$.

Now your question: on our curved plane is there a path connecting the latter two edges, that every point of the path has $h=\frac 1 2$ ?

It seems to me that if there weren't such a path, we could find another path connecting the $x=0$ and $x=1$ edges, that has $h \ne \frac 1 2$ at its every point. In other words: we could circumvent all $h=\frac 1 2$ points, curves and areas, and get from $x=0$ to $x=1$.

But this is equivalent of getting from $h=0$ to $h=1$ in a continuous way without crossing $h=\frac 1 2$. Impossible. So the first path must exist.

I cannot give a more formal proof though.