There is an ice mountain which is a perfect circular cone. A climber intends to go all the way to the top. He has a rope and his strategy is to create a lasso, throw it over the peak, and pull until the rope tightens up, and then pull himself up. He repeats the process until he reaches the top (of course, mathematically speaking he will reach the top only in the limit as the number of steps tends to infinity). What is the limit angle at the peak of the mountain for which this strategy works? (equivalently, what is the limit slope of the mountain, or the ratio of the height and the radius of the base)In this problem friction is negligible. (Of course, this is absurd, since the climber would not be able to pull himself up the rope, or his shoe-laces would not form a knot, etc... Make whatever assumption you want that doesn't render the problem trivial)

this might be a little unexpected, but the question seems to be similar to one of these bug-crawling-on-the-wall problem. the answer would be 30 degree by flattening the cone into a circular sector

Rumour has it that an esteemed member of TCM at Cambridge used to ask this question to potential PhD students, then leave the room for half an hour. If they had an answer by the time he returned, he would accept them.It's too long since I did any general relativity to recall all the details, but you can solve it as follows:1. Parameterise the surface of the cone using your favourite co-ordinate system.2. Write down the Euclidean metric tensor for your chosen system.3. Search for closed geodesics - this is the bit I can't remember how to do... Even if I could, all those Christoffel symbols are a bugger to LaTeX.4. If any closed geodesics exist, you can climb the mountain by throwing the rope around the geodesic line and knotting it (let's assume I have some rope on rope friction). The rope can't move - by definition of a geodesic, any small variation to the position of the rope will give a longer loop (OK - I'll have an inextensible rope please). Since the rope is knotted and inextensible, it can't slip on to a loop of longer length.5. Therefore my rope will stay in place and I can pull myself up on it. Better have an infinitely strong rope just to be on the safe side.

Hi joet, wileysw,I heard the problem from a friend who came with it after he visited Cambridge University, so the rumor might be true. I personally didn't try to solve it in half an hour, I put it on the back of my mind, but I gave my friend the solution the next day. He said the solution is wrong. I thought about it some more, and I came with a different answer. We both agreed that the second answer is the "correct" one. However, it turns out you can ask the problem in two ways: either you can form a tight knot in the rope, or you can't. If it's a lasso where the knot can glide, then the problem is slightly more amusing; that's the reason I proposed that there is no friction. So to summarize, here are the 2 problems: 1. what's the critical angle assuming you can make a tight knot2. what's the critical angle assuming the knot glides without frictionFor the first problem, Joet's plan of attack is perfect, but, as wilesyw mentioned you don't need to do any calculations since the surface can be flattened. For some reason I am not getting 30 degrees for either problem, but I'm hesitant to say that wileysw is wrong. I'll get back and think about the problem some more myself. Best,V.

Hmmm, maybe I'm actually trying to answer the lasso version? If you can tie a tight knot, I could just tie the rope around a circular cross section of the cone (even though this isn't a geodesic). But so long as the rope is above me, I will be able to pull myself up.If the rope is around a geodesic, and it's a lasso, as long as I keep the lasso tight I'll be able to grip the cone, no matter what direction I pull in.

- CrashedMint
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Hmmm, could somebody walk me through the solution as I don't get it. It seems like a underdefined problem to me: How would I even know how I could throw the lasso? How/Where it would get a hold of the mountain? Am I supposed to do this with one throw? If not would the climber throw again?

At least for the case where the lasso has fixed size, this problem seems to have crossed the Atlantic.

Nice link. And it looks like the answer for the fixed size is indeed 30 degrees.For the gliding knot case, the problem is somewhat a physics problem.

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