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A cone has base diameter $1$ and slant height $3$ units. From a point $A$ halfway up the side of the cone, a string is passed twice around it to come to a point $B$ on the circumference of the base, directly below $A$. The string is then pulled until taut. How far is it from A to B along this taut string?

diagram

Unsure of how to approach this strange looking question, any help would be appreciated.

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Each circle sector in diagram below is a representation of the lateral surface of your cone, flattened-out on a plane. Notice that center angle is $\pi/3$.

Two copies of the lateral surface are needed because the string must make two turns: a tight string is a minimum-length path, that is a line joining $A$ with $B$ (red line in the diagram). Using cosine law it is easy to find its length.

Notice that this won't work for three turns: the result in that case would be an impossible path passing through the cone vertex. It happens that in that case it is not possible to place a taut string from $A$ to $B$.

enter image description here

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  • $\begingroup$ What's missing here is the essential fact that this unrolling locally preserves geodesics...without that, such a trick would let you find geodesics on a standard torus in 3-space by drawing lines on a square (or a few squares) in the plane. $\endgroup$ – John Hughes Jul 22 '17 at 11:29
  • $\begingroup$ @JohnHughes Of course a cone has zero curvature and its lateral surface is developable. That's why the trick works. $\endgroup$ – Aretino Jul 22 '17 at 12:45
  • $\begingroup$ I know that. Do you think that OP did/does? $\endgroup$ – John Hughes Jul 22 '17 at 12:53
  • $\begingroup$ I added for him/her the Wikipedia link. $\endgroup$ – Aretino Jul 22 '17 at 12:55
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The string is taut, so it's a geodesic. Try cutting the cone along the line AB and laying it out as a disk-with-wedge-removed in the plane (sort of a pac-man shape). The line will then appear as a pair of straight lines in the plane (because it got cut in half).

If you can solve the problem for the "once around" case, the "twice around probably won't be too tough.

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  • $\begingroup$ Could you elaborate a bit further please? @John Hughes $\endgroup$ – oscquito Jul 22 '17 at 7:50
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    $\begingroup$ Apologies...I thought you wanted a hint rather than a complete answer. $\endgroup$ – John Hughes Jul 22 '17 at 11:26

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