This is a given.

>A constant size loop don't slip on the cone

Or if the cone is sufficiently flat, the loop will slide up when pulled from the other side.

Let me say this concerning the problem.

To make the problem mathematically sound one has to make some simplifying assumptions. For example,

1. the cone got to be perfect so there is no friction,
   
2. the cone is suffiently tall for the loop not to slip to the ground,
   
3. the height of the cowboy is inessential. The fellow is considered a material point,
   
4. except for gravity, no other forces are present.
   

Feel free to simplify (abstract) the problem further if that is what it takes for the existence of the critical angle to make sense.

I go away for a while, and when I come back I find this delightful puzzle. Thank you.

Here is my take on it:

First, assume that Joe is very small compared to the mountain so that the leader (section of the rope between him and the loop) is not held up off the surface of the mountain. In that case, the entire length of the rope, both the loop and the leader, hug the surface of the mountain, and can be treated like a curve lying on the surface of a cone. Second, assume that there is no friction at all between the surface and the rope, and that Joe's weight keeps the rope tight at every point.

A tight string lying along a surface will form a geodesic, a curve of (locally) minimal length, for the physically obvious reason that if the string could get any shorter, the tension ensures that it would. Now a cone has no intrinsic curvature and can be slit up the side and unrolled into a sector of a circle. Suppose we slit the cone from base to vertex along a straight line that passes through the location of the knot in the rope (cut-at-the-knot, rather than cut-the-knot).

After unrolling the cone into a circle sector, you will see that the point at the location of the knot will appear as two points, one on each of the radii that bound the sector. Of course they are equally distant from the vertex of the sector because they are images of the same original point. The path of the rope will be a line joining these points, and (this is the crucial part) it will be a straight line, because we already know that it is the shortest path joining these points.

If the sector is less than half a circle, then this straight line will fall on the sector, but if the sector is more than half a circle, the straight line will cut across the complementary sector instead. In that case, there is NO geodesic on the surface of the cone for the rope to follow. Any path that follows the cone surface must be longer than the straight path, and so the tension in the rope will cause it to move from any initial path towards the straight line path. But the cone vertex lies inside the closed path formed by the initial path and the straight path, and so when the rope moves to the straight path, it will inevitably cross the vertex of the cone.

Therefore, the rope slips off if and only if the cone is formed by rolling up a sector that exceeds half a circle. The boundary case is of course when the sector is exactly half a circle. In that case, the circumference of the cone base is exactly half the circumference of the circle that would be traced by the slant height, Therefore, the radius of the base is half the slant height, so the angle between the axis of the cone and any line on the cone surface must be 30 degrees. The cone would therefore cast an equilateral triangular shadow at sunset. In the case where the rope does not slip off, the climber has a >60 degree slope to contend with; no wonder he needs a rope!

Here are some further thoughts on the problem:

It is not necessary for the cone to be a right circular cone. The mountain could have any convex simple curve as its base, and as long as the length of a curve traced out along the mountain at a constant distance s from the vertex is less than pi*s, the rope will not slip off. This is because even noncircular cones will unwrap into circular sectors, provided you trim the base so that every point is equidistant from the vertex. This generalization loses some of the charm of the original problem however.

It occurs to me that you can use this problem to survey the mountain. Suppose you want to know the slope of the mountain (which must be a right circular cone of course, for there to be a unique slope), but you have only the rope and a protractor, but no plumb bob or sighting equipment. How do you do it?

Here is another one, inverse to the puzzle in the previous paragraph, and therefore a clue to its solution: Suppose the climber measures the angle formed at the knot between the two ends of the loop, and suppose he knows the length l of the loop. Now he ties himself to his rope a distance d below the knot and walks around the mountain, leaning on the rope the whole way until he returns to his starting point. The mountain is again a frictionless cone, but with a convex but not necessarily circular base. How far did he walk?

Thanks for an interesting puzzle, and I hope you find my extensions interesting as well.

--Stuart Anderson

Stuart, welcome back. The problem is by a Berkeley physics professor passed to me by a Berkeley mathematics professor.

>Here is my take on it:

Very good. I'll wait a little and then put the whole thing at the site.

>Thanks for an interesting puzzle, and I hope you find my
>extensions interesting as well.

Absolutely. It's a very interesting example of Polya's "Looking back" stage in problem solving. All the information was contained in the solution (in the diagram rather) to the original puzzle. But you had to notice that. I'll have it all at the site so as to be able to interlink with pages where Polya is mentioned.

Alex

--Stuart

I see you have posted a nice-looking page on the lasso problem. Thanks for making diagrams for the solution -- much clearer than the verbal explanation alone!

--Stuart

Truth be told I do not know. People are all very different. Perhaps for some the diagram will be helpful.

Some time ago, I gave the problem to a very clever cousin of mine. (A serious retired engineer with a good math background. When me meet we often exchange problems.) We then discussed the solution. He would not accept as a fact that a stretched rope arranges itself along a geodesic.

As a P.S., the problem he gave me was this:

There are six balls of the same shape, but of three colors - two per color. For every pair of monochromatic balls, one is lighter. All light balls are of the same weight, and so are all the heavy balls. Use 2 weighings to determine which balls are light and which are heavy.

Just a balance with a cup on each side.

If they are equal, trade one from the same-colored side with one of the other colors as yet unweighed.

If they are unequal, weigh one from the mixed color pile vs. one from the same colored pile.

Yeah? Yeah!

Weigh A1 + B1 against B2 + C1.

If A1 + B1 < B2 + C1 then

• B1 < B2 because even if A1 < C1, B1 > B2 would lead to A1 + B1 = B2 + C1, at best. Thus B1 = BL, B2 = BH.
  
• If A1 = AH then C1 = CH and if C1 = CL then A1 = AL.
  

Weigh A1 + C1 against B1 + B2.

• A1 + C1 = B1 + B2 implies A1 = AL and C1 = CH.
  
• A1 + C1 < B1 + B2 implies A1 = AL and C1 = CL.
  
• A1 + C1 > B1 + B2 implies A1 = AH and C1 = CH.
  

If A1 + B1 = B2 + C1 then weigh B1 against B2. B1 = BL implies A1 = AH and C1 = CL. B1 = BH implies A1 = AL and C1 = CH.

The case where A1 + B1 > B2 + C1 is similar to the first one.