One-step problems in geometry

Here's a cute question which Frederic Bourgeois asked me on a train journey recently. He was asked it by Givental, if my memory serves correctly, but I've no idea where it came from originally. Anyway, the question:

There is a mountain of frictionless ice in the shape of a perfect cone with a circular base. A cowboy is at the bottom and he wants to climb the mountain. So, he throws up his lasso which slips neatly over the top of the cone, he pulls it tight and starts to climb. If the mountain is very steep, with a narrow angle at the top, there is no problem; the lasso grips tight and up he goes. On the other hand if the mountain is very flat, with a very shallow angle at the top, the lasso slips off as soon as the cowboy pulls on it. The question is: what is the critical angle at which the cowboy can no longer climb the ice-mountain?

To solve it, you should think like a geometer and not an engineer. (And yes, it needs just one trick which is certainly applicable elsewhere.)

P.S. When I was asked the question, I failed miserably!

Here is a problem that I learned from W. Thurston. I do not remember whose problem it was originally. Possibly Conway?

Suppose that you have a finite collection of round circles in round $S^3$, not necessarily all of the same radius, such that each pair is linked exactly once. (In particular, no two intersect.) Prove that there is an isotopy in the space of such collections of circles so that afterwards, they are all great circles.

Suppose that we have two simple closed curves in $R^3$ which are linked. And suppose that the distance between these curves is $1$. Prove that the length of each curve is at least $2\pi$.

This problem has an interesting history. It was published in the book by W. Hayman, Unsolved problems in Function theory, where it was attributed to F. Gehring. I solved it in 1977, jointly with Oleg Vinkovski, prepared a paper and gave a seminar talk. After the talk, I was approached by an undergraduate student, who proposed a ridiculously simple solution. Just two lines, using nothing. So I did not submit my paper. Later I've seen several published solutions, but none of them was so simple.

EDIT. Here is this proof (due to Igor Syutrik). Fix a point $M$ on $A$. Then one can find another point $M'$ on $A$ such that the interval $[M,M']$ intersects $B$. Indeed, otherwise we can deform $A$ to $M$ moving straight along these intervals $[M,M']$ and deformation will not cross $B$. Let $O$ be a point on $[M,M']$ that belongs to $B$. Let $A'$ be the central projection of $A$ from $O$ onto the unit sphere around $O$. Then $A'$ passes through two diametrically opposite points of the sphere and thus its length is at least $2\pi$.

EDIT2. Exactly the same proof is published in the paper Criticality for the Gehring link problem, Geometry & Topology 10 (2006) 2055–2115, where it is credited to Marvin Ortel.

EDIT3. Our original solution with Vinkovski also has been rediscovered since then. It can be seen in this file: http://www.math.purdue.edu/~eremenko/dvi/gehring.pdf Thanks to Anton Petrunin for finding this file on my computer:-)