Let $f:S^1\to S^1$ be continous. Show that if $f$ is a homotopy to the Constant function, there exists a point $x\in S^1$ such that the length between $x$ and $f(x)$ is $\frac{\pi}{6}$.
Hint: You can lift a Path from $S^1$ to $\mathbb{R}$.
I have no Idea how to even start solving this. Any intuition would be very helpful.
Throughout I assume that by "length" you mean the Euclidean distance. This is important because on the unit sphere the maximal Euclidean distance is achieved on a pair of antipodal points. And also because that maximal distance is $2$ which is greater than $\pi/6$.
So first of all we need the following
The proof can be found here.
So let $p_1$ be a fixed point of $f$.
Now let $\Lambda:S^1\to S^1$ be the antipodal map $\Lambda(x)=-x$. It follows that $\Lambda\circ f$ is also null homotopic and thus by our Lemma it has a fixed point as well. I.e. there exists $p_2\in S^1$ such that $f(p_2)=-p_2$.
Now consider the function
$$D:S^1\to\mathbb{R}$$ $$D(x)=d(x,f(x))$$
where $d$ is the Euclidean distance. It follows that
$$D(p_1)=0$$ $$D(p_2)=2$$
and since $D$ is continous and $S^1$ is connected then the image of $D$ is connected as well. In this case it is equal to $[0,2]$. In particular for any $r\in[0,2]$ there is $x\in S^1$ such that $d(x,f(x))=r$.