A problem from an old @CruxMathematicorum https://www.cut-the-knot.org/m/Geometry/DiagonalsInPolygon.shtml … #FigureThat #math #geometrypic.twitter.com/HkQ3Mhl2Jc
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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, ...
...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
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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
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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.
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Diagonal (+2 sides)... And so on. When you reach the final diagonal you have accumulated 3+(n-1)*2 sides = 2n -1 sides. You have one side left to connect the two vertices but there is no way to do that without overlapping the diagonal completely making it not a diagonal...
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By contradiction, not all diagonal lines in a 2n gon have a side to which they are parallel.
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