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In any convex 2n-gon there is a diagonal not parallel to any side

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Prove that in any convex 2n-gon there is a diagonal not parallel to any side.
 

Sudharaka

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MHB Math Helper
Feb 5, 2012
1,621
Prove that in any convex 2n-gon there is a diagonal not parallel to any side.
Hi caffeinemachine, :)

I don't think that this is a true statement. For example, if you take a regular hexagon which is a convex 2n-gon, it is clear that every diagonal is parallel to a side.

Kind Regards,
Sudharaka.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hi caffeinemachine, :)

I don't think that this is a true statement. For example, if you take a regular hexagon which is a convex 2n-gon, it is clear that every diagonal is parallel to a side.

Kind Regards,
Sudharaka.
Hey Sudharaka.

Let us number the vertices as A,B,C,D,E,F. Now AC ain't parallel to any side. Isn't it?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hey Sudharaka.

Let us number the vertices as A,B,C,D,E,F. Now AC ain't parallel to any side. Isn't it?
Yep. Sorry for the mistake, I somehow had overlooked such a obvious thing. :)

I thought about this problem for a while and here is a solution that I came up with,

Suppose there exists a convex 2n-gon where every diagonal is parallel to at least one side. From one vertex(say A) we can draw 2n-2 diagonals(excluding the sides that are adjacent to that vertex). Note that each of these diagonals cannot be parallel to any other diagonal (If any two of them are parallel then the 2n-gon will not be convex, because the line joining the two end points of these diagonals is not inside the 2n-gon). Let B and C be the adjacent vertices of A (AB and AC are sides of the 2n-gon). Then BC should be should be parallel to at least one of the diagonals described above(as it cannot be parallel to AB or AC). Let this diagonal be AE. Then either BC or AE will not lie inside the 2n-gon. The 2n-gon will not be convex and hence our assumption is not correct.

If you have any comments on this reasoning please don't hesitate to tell me. :)

Kind Regards,
Sudharaka.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Yep. Sorry for the mistake, I somehow had overlooked such a obvious thing. :)

I thought about this problem for a while and here is a solution that I came up with,

Suppose there exists a convex 2n-gon where every diagonal is parallel to at least one side. From one vertex(say A) we can draw 2n-2 diagonals(excluding the sides that are adjacent to that vertex). Note that each of these diagonals cannot be parallel to any other diagonal (If any two of them are parallel then the 2n-gon will not be convex, because the line joining the two end points of these diagonals is not inside the 2n-gon). Let B and C be the adjacent vertices of A (AB and AC are sides of the 2n-gon). Then BC should be should be parallel to at least one of the diagonals described above(as it cannot be parallel to AB or AC). Let this diagonal be AE. Then either BC or AE will not lie inside the 2n-gon. The 2n-gon will not be convex and hence our assumption is not correct.

If you have any comments on this reasoning please don't hesitate to tell me. :)

Kind Regards,
Sudharaka.
I didn't understand when you said "Note that each of these diagonals cannot be parallel to any other diagonal" since if we take a regular hexagon and label its vertices as A,B,C,D,E,F. Then AE||BD.

Also "Let B and C be the adjacent vertices of A (AB and AC are sides of the 2n-gon). Then BC should be should be parallel to at least one of the diagonals described above". Why is this true (taking 'the diagonals described above' as 'the diagonals passing through A' )?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I didn't understand when you said "Note that each of these diagonals cannot be parallel to any other diagonal" since if we take a regular hexagon and label its vertices as A,B,C,D,E,F. Then AE||BD.
The diagonals that I am talking about have A as an endpoint.

Also "Let B and C be the adjacent vertices of A (AB and AC are sides of the 2n-gon). Then BC should be should be parallel to at least one of the diagonals described above". Why is this true (taking 'the diagonals described above' as 'the diagonals passing through A' )?
There are 2n-2 diagonals starting from A. According to the assumption each diagonal is parallel to a side of the 2n-gon. But no two diagonals(starting from A) are parallel to each other. Therefore for each side of the 2n-gon(excluding AB and AC) there is a diagonal(starting from A) which is parallel to it. BC should be parallel to at least one side of the 2n-gon. But BC is not parallel to AC or AB. Therefore BC should be parallel to a diagonal starting from A. Does that make sense? :)
 

caffeinemachine

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MHB Math Scholar
Mar 10, 2012
834
The diagonals that I am talking about have A as an endpoint.



There are 2n-2 diagonals starting from A. According to the assumption each diagonal is parallel to a side of the 2n-gon. But no two diagonals(starting from A) are parallel to each other. Therefore for each side of the 2n-gon(excluding AB and AC) there is a diagonal(starting from A) which is parallel to it. BC should be parallel to at least one side of the 2n-gon. But BC is not parallel to AC or AB. Therefore BC should be parallel to a diagonal starting from A. Does that make sense? :)
Looks correct. Brilliant! I didn't think of this solution.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
There are 2n-2 diagonals starting from A. According to the assumption each diagonal is parallel to a side of the 2n-gon. But no two diagonals(starting from A) are parallel to each other. Therefore for each side of the 2n-gon(excluding AB and AC) there is a diagonal(starting from A) which is parallel to it. BC should be parallel to at least one side of the 2n-gon. But BC is not parallel to AC or AB. Therefore BC should be parallel to a diagonal starting from A. Does that make sense? :)
Sorry about this, but I am not convinced by that ingenious argument, for two reasons.

First, the number of diagonals starting from A is 2n–3, not 2n–2 (there is no diagonal from A to A itself, or to B or C).

Second, there does not seem to be anything about Sudharaka's argument that would not apply equally well to a polygon with an odd number of vertices. The result is false in that case, as you can see from the example of a regular pentagon. So there has to be some feature in the proof that depends on the number of vertices being even.
 

Sudharaka

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MHB Math Helper
Feb 5, 2012
1,621
Sorry about this, but I am not convinced by that ingenious argument, for two reasons.

First, the number of diagonals starting from A is 2n–3, not 2n–2 (there is no diagonal from A to A itself, or to B or C).

Second, there does not seem to be anything about Sudharaka's argument that would not apply equally well to a polygon with an odd number of vertices. The result is false in that case, as you can see from the example of a regular pentagon. So there has to be some feature in the proof that depends on the number of vertices being even.
Indeed. Thanks for pointing that out. Back to square one. :p
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Prove that in any convex 2n-gon there is a diagonal not parallel to any side.
In a convex $k$-gon there are $k-3$ diagonals from each vertex. There are $k$ vertices, and each diagonal connects two vertices. So there are $\frac12k(k-3)$ diagonals in all.

Now suppose that each diagonal is parallel to a side. There are $k$ sides, so the average number of diagonals parallel to each side will be $\frac12(k-3)$. The situation will be more complicated if some sides are parallel to other sides, but in any case there must exist at least one side that has at least $\frac12(k-3)$ distinct diagonals parallel to it.

If $k$ is odd then that causes no problems. For example, if $k=7$ there is the example of of the regular heptagon, as pictured below. In this case, $\frac12(k-3)=2$, and you can see that each coloured side is parallel to two diagonals of that same colour.

heptagon.gif

But if $k$ is even then $\frac12(k-3)$ is not an integer. Since each side must be parallel to an integer number of diagonals, it follows that there must be some edge that has at least $\frac12(k-2)$ diagonals parallel to it. Call that edge $A_1A_k$, where the vertices have been labelled (consecutively) $A_1,A_2,\ldots,A_k$. Each of those parallel diagonals connects two vertices, thus accounting for a total of $k-2$ vertices. The edge $A_1A_k$ also connects two vertices, so altogether that uses up the entire number of $k$ vertices. The only way for that to happen without any of those diagonals crossing each other is if the diagonals are $A_2A_{k-1}$, $A_3A_{k-2}$ and so on up to $A_{k/2}A_{(k/2)+1}$. But that last one is not a diagonal at all, since it connects two consecutive vertices and is therefore an edge.

That contradiction shows that the construction is not possible in the case when $k$ is even.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
In a convex $k$-gon there are $k-3$ diagonals from each vertex. There are $k$ vertices, and each diagonal connects two vertices. So there are $\frac12k(k-3)$ diagonals in all.

Now suppose that each diagonal is parallel to a side. There are $k$ sides, so the average number of diagonals parallel to each side will be $\frac12(k-3)$. The situation will be more complicated if some sides are parallel to other sides, but in any case there must exist at least one side that has at least $\frac12(k-3)$ distinct diagonals parallel to it.

If $k$ is odd then that causes no problems. For example, if $k=7$ there is the example of of the regular heptagon, as pictured below. In this case, $\frac12(k-3)=2$, and you can see that each coloured side is parallel to two diagonals of that same colour.


But if $k$ is even then $\frac12(k-3)$ is not an integer. Since each side must be parallel to an integer number of diagonals, it follows that there must be some edge that has at least $\frac12(k-2)$ diagonals parallel to it. Call that edge $A_1A_k$, where the vertices have been labelled (consecutively) $A_1,A_2,\ldots,A_k$. Each of those parallel diagonals connects two vertices, thus accounting for a total of $k-2$ vertices. The edge $A_1A_k$ also connects two vertices, so altogether that uses up the entire number of $k$ vertices. The only way for that to happen without any of those diagonals crossing each other is if the diagonals are $A_2A_{k-1}$, $A_3A_{k-2}$ and so on up to $A_{k/2}A_{(k/2)+1}$. But that last one is not a diagonal at all, since it connects two consecutive vertices and is therefore an edge.

That contradiction shows that the construction is not possible in the case when $k$ is even.
Nice. This is the solution I had in mind. Thanks for pointing out the flaw in Sudharaka's argument.