# A Problem on an Icy Cone

An engaging problem has been communicated to me by Alexander Givental, a math professor at Berkeley, and to him by the physics professor at Berkeley, Dmitry Budker.

Cowboy Joe wants to climb an absolutely slippery glacier of the round conical form. He has a lasso (with a loop of a fixed size) which he can throw over the top and pull himself up - if only the loop won't slip off the vertex of the cone. It seems clear that when the cone is almost as flat as the plane, it will slip off, and when the cone is as sharp as a needle, the loop will hang. Thus there should exist some intermediate critical angle. What is it?

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Copyright © 1996-2018 Alexander BogomolnyCowboy Joe wants to climb an absolutely slippery glacier of the round conical form. He has a lasso (with a loop of a fixed size) which he can throw over the top and pull himself up - if only the loop won't slip off the vertex of the cone. It seems clear that when the cone is almost as flat as the plane, it will slip off, and when the cone is as sharp as a needle, the loop will hang. Thus there should exist some intermediate critical angle. What is it?

### Solution

The following solution is by Stuart Anderson.

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 length^{(1)}. Therefore, the radius of the base is half the slant length, so the angle between the axis of the cone and any line on the cone surface must be 30°. 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 steeper than the 60° 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 πs, 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?

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Copyright © 1996-2018 Alexander Bogomolny### Remarks

**(1)**. Let L be the length of the slant and r the radius of the cone's bases. When cut and unrolled the cone becomes a sector:

In the critical configuration, where the sector is a semicircle, 2πr = πL, i.e. r = L/2.

Half the angle at the apex in the cone vertical cross-section is 30°.

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Copyright © 1996-2018 Alexander Bogomolny