10.28.2004
manhole covers
amended at 1:00 am on November 2nd. Click on the links next to the diagrams if you can't see the diagrams.
My friend P.L. was discussing a problem with our mutual friend Diana, who relayed this onward to me:
They were hashing out answers to the rather simple question of "why are manhole covers round?" The answers given were surprisingly diverse, ranging from ease of transport to infinite degrees of rotational symmetry (thus allowing the manhole cover replace-ers to replace said manhole covers without regard to initial orientation). Diana's answer was that round manhole covers and only round manhole covers are physically incapable of falling through their holes. A proof, of sorts, was derived, but only by way of a few examples. I submit mine for your, (and their) approval: This requires two parts: one in which the number of sides is even, and one in which the number of sides is odd.
Diagram 1: The hexagon (number of sides is even)
Consider regular hexagon ABCDEF. If we take S to be the midpoint of side AB, and T to be the midpoint of side DE, and G to be the center, then we can draw the following lines: AD, BE, and ST, all going through point G. If we inscribe a circle with radius GS, it is tangent only at the midpoints of each side: thus making ST the shortest distance across the polygon while still traversing center point G. We can then circumscribe circle with radius AG, which touches the polygon only at the corners, thus making AD (and BE and CF) the longest distance across the polygon while still traversing center point G. Since AG > GS, and by extension AD > ST, then we can always maneuver the polygon into the manhole. As the number of sides (n) approaches infinity, SG will approach the length of AG. But only when n equals infinity is the difference between SG and AG 0, and the manhole cover (now a circle) will not be able to fall through the hole.
Diagram 2: The pentagon (number of sides is odd)
Consider now regular pentagon ABCDE. If we take S to be the midpoint of side AB, then we may draw the following lines: AD and BD (which are equal) and SD. Circumscribe a circle with center D and radius AD: it touches the pentagon only at point B. This is the longest distance across the pentagon. Inscribe circle with center D and radius SD: it is tangent to the pentagon only at point S, and is thus the shortest distance across the pentagon while still traversing the center point (not shown). Since AD (and BD) > SD, then we can always maneuver this polygon into the manhole. As the number of sides (n) approaches infinity, SD will approach the length of AD. But only when n equals infinity is the difference between AD and SD 0, and the manhole cover (now a circle) will not be able to fall through the hole.
Diagram 3: The nonagon (number of sides is odd - superfluous diagram, but I was on a roll)
Same deal as above. enjoy. Diagrams created using Macromedia FreeHand.
So there it is. I'm going back to work.
My friend P.L. was discussing a problem with our mutual friend Diana, who relayed this onward to me:
They were hashing out answers to the rather simple question of "why are manhole covers round?" The answers given were surprisingly diverse, ranging from ease of transport to infinite degrees of rotational symmetry (thus allowing the manhole cover replace-ers to replace said manhole covers without regard to initial orientation). Diana's answer was that round manhole covers and only round manhole covers are physically incapable of falling through their holes. A proof, of sorts, was derived, but only by way of a few examples. I submit mine for your, (and their) approval: This requires two parts: one in which the number of sides is even, and one in which the number of sides is odd.
Diagram 1: The hexagon (number of sides is even)
http://users.ox.ac.uk/~kebl2515/Hexagon.tiff
Consider regular hexagon ABCDEF. If we take S to be the midpoint of side AB, and T to be the midpoint of side DE, and G to be the center, then we can draw the following lines: AD, BE, and ST, all going through point G. If we inscribe a circle with radius GS, it is tangent only at the midpoints of each side: thus making ST the shortest distance across the polygon while still traversing center point G. We can then circumscribe circle with radius AG, which touches the polygon only at the corners, thus making AD (and BE and CF) the longest distance across the polygon while still traversing center point G. Since AG > GS, and by extension AD > ST, then we can always maneuver the polygon into the manhole. As the number of sides (n) approaches infinity, SG will approach the length of AG. But only when n equals infinity is the difference between SG and AG 0, and the manhole cover (now a circle) will not be able to fall through the hole.
Diagram 2: The pentagon (number of sides is odd)
http://users.ox.ac.uk/~kebl2515/Pentagon.tiff
Consider now regular pentagon ABCDE. If we take S to be the midpoint of side AB, then we may draw the following lines: AD and BD (which are equal) and SD. Circumscribe a circle with center D and radius AD: it touches the pentagon only at point B. This is the longest distance across the pentagon. Inscribe circle with center D and radius SD: it is tangent to the pentagon only at point S, and is thus the shortest distance across the pentagon while still traversing the center point (not shown). Since AD (and BD) > SD, then we can always maneuver this polygon into the manhole. As the number of sides (n) approaches infinity, SD will approach the length of AD. But only when n equals infinity is the difference between AD and SD 0, and the manhole cover (now a circle) will not be able to fall through the hole.
Diagram 3: The nonagon (number of sides is odd - superfluous diagram, but I was on a roll)
http://users.ox.ac.uk/~kebl2515/Nonagon.tiff
Same deal as above. enjoy. Diagrams created using Macromedia FreeHand.
So there it is. I'm going back to work.
