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Ah, again, the difference between “it is obvious to me” and “I have a proof”….
It seems to me if I were to take that cylinder and treat it as if it were a hollow tube, then make a vertical cut along it, and then “unroll” it so it lays flat, then:
Line AB would be the hypotenuse of a right triangle CAB where the right-angle corner C is the same as the lower right of the above picture, where the dotted line meets the bottom edge of the cylinder. The two sides of triangle CAB (adjacent to the right angle corner C) would have lengths 3, and PI. The length of the hypotenuse of right triangle CAB (i.e., line AB), easily calculated using Pythagoras from the known geometry and dimensions of the triangle, is the answer to the question posed….
But all that is assuming that the line AB, once the cylinder is unrolled flat, is indeed a straight line. I’m certain it is, but I sure have been wrong about my assumptions here before sometimes. And I don’t have a proof of my belief. Ah, but this is sure a fun blog.
The line AB on the flat transform of the cylinder would indeed be a strait line. The question I’m unsure of would then be best posed in terms of the inverse transform, specifically, rolling that right triangle back onto the cylinder. AB is indeed the “distance” (shortest path) from A to B when the triangle is on the plane, but is it true that once the triangle is rolled around the cylinder, is that resulting path from A to B still the shortest possible on the cylinder’s surface? (If so, it would be the distance from A to B on the cylinder.) I think it would be so, but still I have no argument that proves it so.
December 14th, 2011 at 8:53 am
Ah, again, the difference between “it is obvious to me” and “I have a proof”….
It seems to me if I were to take that cylinder and treat it as if it were a hollow tube, then make a vertical cut along it, and then “unroll” it so it lays flat, then:
Line AB would be the hypotenuse of a right triangle CAB where the right-angle corner C is the same as the lower right of the above picture, where the dotted line meets the bottom edge of the cylinder. The two sides of triangle CAB (adjacent to the right angle corner C) would have lengths 3, and PI. The length of the hypotenuse of right triangle CAB (i.e., line AB), easily calculated using Pythagoras from the known geometry and dimensions of the triangle, is the answer to the question posed….
But all that is assuming that the line AB, once the cylinder is unrolled flat, is indeed a straight line. I’m certain it is, but I sure have been wrong about my assumptions here before sometimes. And I don’t have a proof of my belief. Ah, but this is sure a fun blog.
December 14th, 2011 at 9:06 am
My last paragraph may need rewording:
The line AB on the flat transform of the cylinder would indeed be a strait line. The question I’m unsure of would then be best posed in terms of the inverse transform, specifically, rolling that right triangle back onto the cylinder. AB is indeed the “distance” (shortest path) from A to B when the triangle is on the plane, but is it true that once the triangle is rolled around the cylinder, is that resulting path from A to B still the shortest possible on the cylinder’s surface? (If so, it would be the distance from A to B on the cylinder.) I think it would be so, but still I have no argument that proves it so.
December 14th, 2011 at 7:32 pm
…5?
December 24th, 2011 at 2:36 pm
$\sqrt(9 + \pi^2)$