Right cone, you are at A and need to complete a revolution before reaching the bottom B. What is shortest distance AB?

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You are on a mountain that is a right cone shape. You are trying to get to B and you are somewhere up the mountain A such that you lie on the line OB.

The line AB must do one full revolution of the mountain and is the shortest distance.

What is the general formula for finding any AB distance? How much of that distance is uphill/downhill?

enter image description here


When uncurled the cone looks like the shape below with A and B demonstrated. With A being a distance $x$ along the line OB. I appreciate that the straight line is the shortest distance but how do I work that distance out?

I am not sure where to begin with uphill/downhill part.

enter image description here

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There are 4 best solutions below

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Henry has answered most of your questions, except how to calculate the distance of the path. To find the distance you need to find the angle $\theta$ of the circular sector.

enter image description here

We know the circular arc of the sector has the same length as the circumference of the cone's base circle, so $\theta s = 2\pi r$ giving: $$\theta = \frac{2 \pi r}{s}$$ If the point $A$ is located a distance $d$ from the cone vertex, we can now use the Law of Cosines to find the path distance x: $$x^2 = s^2 + d^2 - 2sd \cos(\frac{2 \pi r}{s})$$

EDIT

To find the uphill distance $x_u$ we use the method given by Henry, i.e. drop a perpendicular from $O$ to $AB$, meeting at $C$. The height $h$ of $OC$ can be found by the area formulas for a triangle: $$\text{Area} = \frac{1}{2}\cdot s \cdot d \cdot \sin(\theta) = \frac{1}{2}\cdot x \cdot h$$ giving $$h = \frac{s \cdot d \cdot \sin(\theta)}{x}$$ We then see that $x_u^2+h^2=d^2$ or $$x_u = \sqrt{d^2-h^2}$$

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Hint:

Cut along the dotted line $OB$, allowing the cone to be flattened. The quickest winding route $AB$ is then a straight line. (Added) to find the uphill part, drop a perpendicular from $O$ to $AB$, meeting at $C$. On $AC$ the path is approaching $O$ and so uphill,while on $CB$ it is moving away from $O$ and so downhill

enter image description here enter image description here But if the angle is too big, that straight line would be outside the cone, so the alternative would be to go from $A$ to $O$, round the cone at $O$ and then down to $B$

enter image description here

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One might solve this numerically using a path of n steps. Let a be the direct distance from the cone peak of point A. Then the minimal distance between A and B is given by the solution of enter image description here

Components x[i] give the path from B to A which might get up and down.

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Is there not still an issue with the case where the sector angle is greater than 180 deg, as the solution given does not strictly go around the cone.

I came to this problem via the 'Korean' exam question, where the problem said that it was a sight seeing tour around a mountain and you had to 'go around the mountain'.

Can anyone shed any further light on this?