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19-gon and Pigeonhole principle

by Numeriprimi
Tags: 19-gon, pigeonhole principle, tops, trapezoid
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Numeriprimi
#1
Dec16-12, 05:42 PM
P: 138
Hey!
I have got some question for you.

Decide if you can choose seven tops of the regular 19-gon and four of them are tops of trapezoid.
(I think - Pigeonhole principle, but how?)
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tiny-tim
#2
Dec17-12, 01:39 PM
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Hi Numeriprimi!
What do you mean by "tops"?
0xDEADBEEF
#3
Dec17-12, 03:18 PM
P: 825
I think he means the vertices. How I would try to prove it. Feel free to stop reading once you think that you have the answer...

1) A trapezoid is defined by having two parallel sides. So you want to construct a set of points and none of the connections between the points are to be parallel.

2) If we numerate the points we can start forming all the families of parallel lines.

3) If we enumerate in a circle one family is {(2,19),(3,18),(4,17),(5,16),..., (10,11)} You see that even one "length 1" pair is included.

4) There are 19 of these families, and they account for all the possible connections there are.

5) The possible connections between n points are (n^2-n)/2

6) Pidgeonhole

Numeriprimi
#4
Dec17-12, 05:04 PM
P: 138
19-gon and Pigeonhole principle

Yes, I mean vertices... Sorry for my English because is quite hard to choose right word with same meaning in my language when is a lot of words :-)

So, I will read and understand your answer after school because I going to sleep. Then I will write when I won't understand you.

For now... thanks very much :-)
0xDEADBEEF
#5
Jan2-13, 07:23 PM
P: 825
Numeriprimi. You PM me, but maybe others are interested in the answer as well. So I'll discuss the questions here. I hope that is ok.
I understand to 4), it is okay, but no 5) and 6). How you know it and how do you prove it for choose seven of them?
A connection between two points is the same whether it is between say points (1,2) and or (2,1). Connections between a point and itself (1,1) don't count. There is more than one way to count the number of unordered dissimilar pairs of N numbers. What I did was taking a square NxN matrix, which has N^2 entries. Remove the diagonal where the indices are identical, and then take half of what is left. Similar to a distance table like this http://www.kznded.gov.za/portals/0/e...nce-table.html where you can see the distances between any two South African cities in the table, and there are no double entries.

So between 7 points there are (7*7-7)/2=21 connections. If two of them are in parallel you are done. Connections are in parallel if they are in the same family. There are only 19 families.


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