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Molson_Hart's profile
Molson Hart
Molson Hart
Molson Hart
@Molson_Hart

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Molson Hart

@Molson_Hart

CEO at http://amazon.com/viahart . CEO at http://edisonlf.com . I tweet about business, e-commerce, supply chain, health, law, & infrastructure

Austin, TX
Joined July 2015

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    1. Alexander Bogomolny‏ @CutTheKnotMath 7 Mar 2018
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      A problem from an old @CruxMathematicorum https://www.cut-the-knot.org/m/Geometry/DiagonalsInPolygon.shtml … #FigureThat #math #geometrypic.twitter.com/HkQ3Mhl2Jc

      2 replies 4 retweets 10 likes
    2. Molson Hart‏ @Molson_Hart 8 Mar 2018
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      Replying to @CutTheKnotMath

      Assume every diagonal is parallel to a side. Number of unique diagonals for a 2n gon is (2n-3)*n. Sides will have an average of (2n-3)*n/(2n) parallel to it. You can simplify and then round this up to say that some side has n-1 diagonals parallel to it. Including the side, ...

      1 reply 0 retweets 0 likes
    3. Molson Hart‏ @Molson_Hart 8 Mar 2018
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      Replying to @Molson_Hart @CutTheKnotMath

      ...we have n-1+1 parallel line segments, each with a pair of unique vertices (2n total) that are shared with the 2n gon. Let's now attempt to draw the shape, starting with the parallel side, connecting it to the diagonals. 1 side, connect to the first diagonal (+2 sides), second

      2 replies 0 retweets 0 likes
    4. Alexander Bogomolny‏ @CutTheKnotMath 8 Mar 2018
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      Replying to @Molson_Hart

      Just for once, trying to read across several tweets, "...we have n-1+1 parallel line segments, each with a pair of unique vertices (2n total)" is suspect: you try to include the second side which you know nothing about

      1 reply 0 retweets 0 likes
    5. Molson Hart‏ @Molson_Hart 8 Mar 2018
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      Replying to @CutTheKnotMath

      I was on phone my phone. I had no choice at the time of writing. If it helps I'm happy make an image and tweet it. I don't understand what you're saying. Let n = 3, there's an average of 9/6 = 1.5 diagonals parallel to each side. Round it up to 2, for one side.

      2 replies 0 retweets 0 likes
    6. Alexander Bogomolny‏ @CutTheKnotMath 8 Mar 2018
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      Replying to @Molson_Hart

      Just read what you wrote (the sentence I quoted). As I see it: 2n vertices is all you have; n parallel segments joining 2n vertices necessarily include 2 sides. Just try to consolidate your tweets when you have time

      1 reply 0 retweets 0 likes
      Molson Hart‏ @Molson_Hart 8 Mar 2018
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      Replying to @CutTheKnotMath

      Does this make it clearer and would you now consider the problem solved by this solution? Thank you.pic.twitter.com/gNV3qO7QuR

      7:50 AM - 8 Mar 2018
      2 replies 0 retweets 0 likes
        1. New conversation
        1. Alexander Bogomolny‏ @CutTheKnotMath 8 Mar 2018
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          Replying to @Molson_Hart

          I am not sure if you can have n-1 diagonals parallel to a side

          1 reply 0 retweets 0 likes
        2. Molson Hart‏ @Molson_Hart 8 Mar 2018
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          Replying to @CutTheKnotMath

          I can explain that with greater clarity. (2n-3)/2 is the average of unique diagonals parallel to each side. It is not a whole number. There must be some side that has more than (2n-3)/2 unique diagonals parallel to it for the average to be lower. Round (2n-3)/2 to n-1. Clear now?

          1 reply 0 retweets 0 likes
        3. 6 more replies
        1. Molson Hart‏ @Molson_Hart 8 Mar 2018
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          Replying to @Molson_Hart @CutTheKnotMath

          Sorry the image didn't work. Here is an edited one.pic.twitter.com/J1eQaezqaz

          0 replies 0 retweets 0 likes
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