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        riddles >> medium >> Semiregular polygons
        
(Message started by: antkor on Feb 11^th, 2014, 2:33pm)

Title: Semiregular polygons
Post by antkor on Feb 11^th, 2014, 2:33pm

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.

Title: Re: Semiregular polygons
Post by towr on Feb 11^th, 2014, 10:48pm

It has twelve or less edges, so that should limit the options a bit.

For twelve, we have

      ___
    |   |
 ___|   |___
|           |
|___     ___|
    |   |
    |___|

Title: Re: Semiregular polygons
Post by antkor on Feb 12^th, 2014, 1:15am

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 [hide]5 edges or less, because in that case the polygon would be convex. So, the numer of egdes should be 6-11.[/hide]

Title: Re: Semiregular polygons
Post by rmsgrey on Feb 12^th, 2014, 3:31am

There are degenerate cases with [hide] 8 or 9 [/hide] edges and an angle of [hide] 90 or 60 [/hide] degrees in [hide] 3:1 and 2:1 [/hide] ratios of interior:exterior angles respectively.

There's also a non-degenerate [hide] 10[/hide]-sided shape with angles of [hide] 120[/hide] degrees in a [hide] 4:1[/hide] ratio, which is probably the one intended.

Title: Re: Semiregular polygons
Post by antkor on Feb 12^th, 2014, 8:58am

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.

Title: Re: Semiregular polygons
Post by towr on Feb 12^th, 2014, 9:44am

Once you think of [hide]putting two hexagon together[/hide], the [hide]ten-edge[/hide] solution seems rather obvious in hind-sight.

I think those degenerate cases are like

  ___           __
|   |         _\/_
|___|___ \/\/
    |   |
    |___|

So it has touching vertices. Which I assume what you mean to disallow by not having more than two edges meet at a point.

Title: Re: Semiregular polygons
Post by antkor on Feb 13^th, 2014, 1:46am

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.

Title: Re: Semiregular polygons
Post by towr on Feb 13^th, 2014, 9:12am

on 02/13/14 at 01:46:41, antkor wrote:

However, I didn't think that the answer is that obvious

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.

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.