Question

Suppose five of the nine vertices of a regular nine-sided polygon are arbitrarily chosen. Show that one can select four among these five such that they are the vertices of a trapezium.

Answer 1

Well, I think I have a solution, but it might not be the most elegant, and I might be leading you down the "rabbit hole" from Alice in Wonderland, but I'll tell you what I've got.

Answer 2

Sure

Answer 3

My method is a matter of limiting or defining the finite cases and then you have to show that for each case of selection of five vertices you can get a (British definition) trapezium. From:

http://mathworld.wolfram.com/Trapezium.html

Answer 4

*

Answer 5

Feel free to reject this at any time if it is too cumbersome, but maybe it isn't. To start off, I have that regular nonagon where I arbitrarily require (valid by rotation) that we start at vertex #1, which I also call vertex #10 because we are really trying to get an arrangement of:

1 w x y z 10

where w - 1 = either z - y or z - x

To start off, by using symmetry, if we take the case that the closest vertex to 1 is 3-away, making it vertex 4, we have only one case of that and by inspection and assignment of the other vertices and the values or w, x, y, and z we fulfill the requirement.

Answer 6

If we take the case that the closest vertex to 1 is 2-away and that other vertex in the other direction from 1 is 2-away, we have only 4 cases. And, bear in mind, that we can possibly create some redundancies, but if we are fairly finite, that won't matter much.

Answer 7

If we take the case that the closest vertex to 1 is 2-away and that other vertex in the other direction from 1 is 3-away, we have only 3 cases.

Answer 8

btw, we cannot have "closest vertex to #1" being more than 3 away, because we can rotate.

Answer 9

If we take the case that the closest vertex to 1 is 2-away and that other vertex in the other direction from 1 is 4-away, we have only 1 case.

Answer 10

So, that takes care of 3-away (1 case), and 2-away (1 sub-case of 4, a second sub-case of 3, and a third sub-case of 1) for 9 cases so far. We are left with what happens if the closest vertex to #1 is 1-away. That might look like the hardest, but because of our work, it's the easiest.

Answer 11

I'm thinking it through as I go, but I think there might be only one case there. Where there are 5 vertices in a row. Because once you introduce a gap where there are 2-away, you can rotate and get back to the case of considering a 2-away vertex from vertex #1.

Answer 12

So, I've come up with a finite 10 cases, with some possible redundancies that might make that number even smaller. But 10 is manageable. I'll sit here and try to see if I can come up with more than my 10 cases using my rotation and symmetry method.

Answer 13

I can't find any other cases. And again, there are quite possibly some redundancies in my listing, but even if you take each one and now easily find the trapezium, and "duplicate" your effort somewhere along the way, it won't be much of wasted effort.

Answer 14

This is sort of a brute force method, but it will work.

Answer 15

Do you see any more than the 10 cases? And if not, are you able to find the trapezium by eliminating one vertex from within each case? Once you posit the 10 cases, it's pretty straight-forward.

Answer 16

Well, no one's answering, so I'll flesh it out a little more. Take the case of where the closest vertex to vertex #1 is 3-away. There is only one arrangement: 1, 4, 5, 6, and 7 That's perhaps the easiest one to see. 4 - 1 = 3 and 10 - 7 = 3 That forces us to select 5 and 6 as the other vertices, in this particular case.

Answer 17

Looking at the case of where the closest vertex to vertex #1 on one side is 2-away and on the other side is 4-away also forces us into one arrangement only, which of course is constrained by use of rotation and symmetry:

1, 3, 4, 5, and 6

3 - 1 = 2 and 10 - 6 = 4 and vertices 4 and 5 are forced as far as selection. So, this case is as fundamental as the previous one expanded upon.

Answer 18

Do you want me to expand on any of the other cases?

Answer 19

I think that the only thing I would alter or retract or modify is the statement in my 3rd post where I said:

where w - 1 = either z - y or z - x

That is too constraining and I would instead replace that by merely listing the 10 case numbers and finding for each which number to eliminate. Since no one's writing, I'm going to proceed with listing the cases and doing the elimination. Probably take a little while.

Answer 20

case # arrangement eliminate
------ ----------- --------
1 1 4 5 6 7 1
2 1 3 4 5 8 3
3 1 3 4 6 8 4
4 1 3 4 7 8 1
5 1 3 5 6 8 1
6 1 3 4 5 7 4
7 1 3 4 6 7 4
8 1 3 5 6 7 6
9 1 3 4 5 6 1
10 1 2 3 4 5 1 (or 3 or 5)

Answer 21

An example of a non-listed case is: 1 3 5 7 8 which is the same as
1 3 4 6 8 by symmetry.

Answer 22

Well, I can't think of anything else to write at this time, because I believe these are the only cases. At this point, I can only think of answering any questions.