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MathGroup Archive 2001

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Re: triangles in circles

  • To: mathgroup at smc.vnet.net
  • Subject: [mg26829] Re: [mg26813] triangles in circles
  • From: BobHanlon at aol.com
  • Date: Thu, 25 Jan 2001 01:13:11 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Needs["DiscreteMath`Combinatorica`"];

n = 5;

ptlist = Table[{Cos[i 2 \[Pi]/n], Sin[i 2 \[Pi]/n]}, {i, 1, n}];

triList = Append[#, First[#]]& /@ KSubsets[ptlist, 3];

Length[triList]

10

Show[Graphics[{
      Circle[{0, 0}, 1],
      {PointSize[0.02], Line /@ triList}
      }], AspectRatio -> Automatic];

Graphically, you will get the same result by just drawing a line with each 
pair of points

Show[Graphics[{
      Circle[{0, 0}, 1],
      {PointSize[0.02], Line /@ KSubsets[ptlist, 2]}
      }], AspectRatio -> Automatic];

Bob Hanlon

In a message dated 2001/1/24 5:22:16 AM, tdevries at shop.westworld.ca writes:

>I'm teaching a high school math class and we are doing permutations and
>combinations.  One of the "standard" questions is ..."given a certain number
>of points located around a circle, how many triangles can be formed...."
>
>The simple line below creates a circle with 5 points arranged around it.
>  
>Could someone help me with a way to generate the lists of points that would
>create all the triangles.   I know that for more points it would get kind
>of
>messy, but I wanted to actually draw all the triangles as I thought it
>might
>be an interesting graphic...  
>
>Thanks for any help you might have....
>
>
>n = 5;
>
>ptlist = Table[{Cos[i 2 \[Pi]/n], Sin[i 2 \[Pi]/n]}, {i, 1, n}];
>
>Show[Graphics[{
>      Circle[{0, 0}, 1],
>      {PointSize[0.02], Point /@ ptlist}
>      }], AspectRatio -> Automatic]
>


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