A Satisfying Property of Diagonals
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- Опубліковано 12 чер 2024
- We prove that every convex polygon with an even number of sides must have at least 1 diagonal which is not parallel to any of its sides.
00:00 Intro
00:44 Idea for the proof
01:12 Counting diagonals
04:38 Improving our upper bound
Another way to count the total number of diagonals is that, with a total of 2n vertices, we can choose any two to form a diagonal. However, there would be a total of 2n cases where such a line is one of its sides. Therefore, the total number is C(2n , 2) - 2n = 2n^2 - 3n as desired.
Typo? I guess an "n" is accidentally missed in the last equation.
yup! @@ranshen1486
I love your videos, they are very informative and you have a talent for explaining things in a very coherent way. Keep up the good work!
Thank you!
A diagonal is good if it is parallel to a side. For a regular polygon with an even number of sides, a diagonal is good if and only if the distance on the polygon between its endpoints is odd. So about half of the diagonals are good. The video does indeed overlook a second double counting. Let us number the vertices from 0 to 2n-1. Consider the pair of opposite sides (0, 2n-1) and (n-1, n). Every diagonal that is parallel to these sides is determined by a vertex between 0 and n-1. There are n-2 of them and there are n pairs of opposite sides, so there are n*(n-2) good diagonals.
Pretty interesting result!
3:00 I was expecting you to cite the Parallel Postulate here, which maybe more precisely guarantees why there can't be another diagonal parallel to the isolated side.
unless you use Hilbert's version of it, then nothing is guaranteed, but it's still alleged to be identical... go Logicism!
The difference is just n, then.
i am not sure that u have used the convexity argument properly at 3:16. The figure of three vertices being coliner does not violate convexity. It is still convex. I think u might be looking for all the angles being less than 180.
He didn't draw a very informative sketch at that step, but if those 3 vertices are collinear along diagonals (not edges), then the polygon would have to be concave to push the middle vertex into line with the other two.
Yup you r right, he does mention diagonal. I thought he said side. Yes convexity does seem to take care of that. Thanks.
Not sure i follow the reasoning for 5:55. Why do i need it to be an edge. It seems like u should be using some relations between the no. of edges and diagonals to conclude that something is an edge and not a diagonal. However, you haven't proved anything.
Being the furthest away => it is an edge. This needs to be proved.
The argument was that ALL the vertices have been paired off here, so there is no other vertex beyond (bottom right on the diagram) these two. That means they must be adjacent vertices on the polygon, so the line segment between them must be an edge, not a diagonal.
@@RGP_Maths
Edit:- Ignore the "counterexample" below. It is clearly not a counter example.
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To be an edge, it must be the furthest away. But there is no argument proving this. It says there must be a smallest diagonal and that must be the smallest edge. HEre is my counter example. Consider a truncated triangle (two ends parallel making it a trapezoid). Put vertices on the ends of the two flat edges and many along the sides of the truncated triangle (trapezoid); make sure its 2n, wont hurt. Pick as the edge, the smaller of the two ends as your side, then all the other diagonals parallel to it increase in size.