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Topic: Semiregular polygons (Read 1484 times) |
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antkor
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Semiregular polygons
« on: Feb 11th, 2014, 2:33pm » |
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Call semiregular polygon a polygon which has all of its' edges of the same length. Also, all of its' interior or exterior angles must be equal (meaning that any interior angle must be x or 360-x). It must be concave and simple (it should not self-intersect) and only two of its' edges are allowed to meet in each corner.
Find the semiregular polygon that has the minimum number of edges.
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towr
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Re: Semiregular polygons
« Reply #1 on: Feb 11th, 2014, 10:48pm » |
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It has twelve or less edges, so that should limit the options a bit.
For twelve, we have ___
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___| |___
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|___ ___|
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|___|
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| « Last Edit: Feb 11th, 2014, 10:50pm by towr » |
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antkor
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Re: Semiregular polygons
« Reply #2 on: Feb 12th, 2014, 1:15am » |
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your answer for 12 edges is correct but it can be done with less edges, meaning this is not the semiregular polygon with the minimum number of edges we are looking for. try again if you like. i will also give a hint.
It is apparent that it cannot be done with 5 edges or less, because in that case the polygon would be convex. So, the numer of egdes should be 6-11.
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rmsgrey
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Re: Semiregular polygons
« Reply #3 on: Feb 12th, 2014, 3:31am » |
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There are degenerate cases with 8 or 9 edges and an angle of 90 or 60 degrees in 3:1 and 2:1 ratios of interior:exterior angles respectively.
There's also a non-degenerate 10-sided shape with angles of 120 degrees in a 4:1 ratio, which is probably the one intended.
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antkor
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Re: Semiregular polygons
« Reply #4 on: Feb 12th, 2014, 8:58am » |
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I can't understand and draw the degenerate cases you are talking about. Can you provide a sketch? I'm not sure I'm following what you mean.
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towr
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Re: Semiregular polygons
« Reply #5 on: Feb 12th, 2014, 9:44am » |
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Once you think of putting two hexagon together, the ten-edge solution seems rather obvious in hind-sight.
I think those degenerate cases are like ___ __
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|___|___ \/\/
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|___|
So it has touching vertices. Which I assume what you mean to disallow by not having more than two edges meet at a point.
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| « Last Edit: Feb 12th, 2014, 9:47am by towr » |
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antkor
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Re: Semiregular polygons
« Reply #6 on: Feb 13th, 2014, 1:46am » |
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Indeed I wanted to disallow degenerate solutions like that, so I stated that the polygon should be simple and should not self-intesect. What i meant was that the edges should not cross each other. The correct solution is the one that you and rmsgrey mentioned, with a regular hexagon with its' mirror image (10 edges). However, I didn't think that the answer is that obvious, meaning that one has to find how many edges there are, but also the angles should be calculated, otherwise the solution would not be correct. For example, a Pentalfa also has 10 edges but does not satisfly all the restrictions given.
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towr
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Re: Semiregular polygons
« Reply #7 on: Feb 13th, 2014, 9:12am » |
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on Feb 13th, 2014, 1:46am, antkor wrote:| However, I didn't think that the answer is that obvious |
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It's not obvious it's the one with the minimum edges, but it is obvious it satisfies all the other conditions (once you think of it in the first place).
Quote:| but also the angles should be calculated, otherwise the solution would not be correct. For example, a Pentalfa also has 10 edges but does not satisfly all the restrictions given. |
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It's easily seen that that doesn't satisfy the solution, because for three consecutive edges with angles that alternate direction (like /\/ ), the first and last edge must be parallel; which clearly isn't the case for a pentagram.
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| « Last Edit: Feb 13th, 2014, 10:57am by towr » |
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