A...Question :P (1 Viewer)

[airie]

OK actually don't really know where this question belongs...but here goes...

Prove that for any integer n>=5 and for any convex n-gon, the arithmetic mean of its sidelengths is no greater than the arithmetic mean of its diagonals.

How would you attack such problems? I only got up to...uhh...there are n sides and n(n-3)/2 diagonals in an n-gon...pretty pathetic Can anyone please help?

EDIT: changed 'n>4' to 'n>=5' so it's more clear Now you can't have started with a quadrilateral!

 

[KeypadSDM]

OK actually don't really know where this question belongs...but here goes...

Prove that for any integer n>4 and for any convex n-gon, the arithmetic mean of its sidelengths is no greater than the arithmetic mean of its diagonals.

How would you attack such problems? I only got up to...uhh...there are n sides and n(n-3)/2 diagonals in an n-gon...pretty pathetic Can anyone please help?

Some thoughts of mine:

For any diagonal which has its ends connected by 3 or more sides around the polygon (going both ways), it must needs be longer than at least 1 side of that polygon. [Due to the convexity which makes at least one set of sides the diagonal cuts across makes the sum of the interior angles with the diagonal & 2 connecting sides strictly LESS than 180 degrees, which means there MUST be a shorter side on that side, as if there wasn't, the angle would be GREATER (or equal) than 180 degrees]

For any diagonal making a triangle with 2 sides, as long as the vertex of the triangle is more than 60 degrees, that particular diagonal is longer than both sides.

The longest diagonal always intersects the shortest diagonal? (dunno if that's right). This means that if you find a diagonal which isn't greater than the sides it's next to, then the longest diagonal intersects it, and taking this into account will kindof weight the mean of the diagonals in your favour.

You can have at most 2 angles less than 60 degrees in the polygon. Leading to at most 2 "short" diagonals as described above.

For every side, there is always a diagonal longer than it, except for at most 1 side. [That's a guess].

Don't know if I'm on the right track. You might want to put this in the extra-curricular forum, and get some IMO/AMO masters onto it.

 

[vizman]

rofl. is this type of stuff in the maths syllabus?

 

[insert-username]

Well, it's not 2-unit or 3-unit. It's more a "I love maths" type of question... sorry I can't be of more help than that...


I_F

 

[Slidey]

This type of question does occur in the 4unit maths syllabus, however probably not to this level of difficulty. The ones in the 4u syllabus usually require induction.

I'm going to move this to the extracurricular forum. It's good to see you trying to solve questions like this, but don't be disillusioned if you can't answer them.

 

[KeypadSDM]

Omg, I look like a jackass. How on earth do you even approach this? If I was thrown this in 4 unit ... I canna bear ta watch!

 

[Templar]

Omg, I look like a jackass. How on earth do you even approach this? If I was thrown this in 4 unit ... I canna bear ta watch!

That's why you get either an average of 48min (AMOC Senior/APMO), 60min (AMO) or 90min (IMO) to solve it.

 

[KeypadSDM]

But dude, I can't even see where to start. Seriously. I mean those were some arbitrary thoughts in my head, nothing concrete. Let alone on the way to proving it.

 

[Templar]

Where did this question come from?

I keep on thinking it could be solved using graph theory for some reason, even though since it requires distance it can't be solved topologically.

 

[SeDaTeD]

Considering Airie's hsc is in 2007, it might be a year 10/11 level olympiad style q. Half an hour and all I've got is squiggles andcrossing out.

 

[KeypadSDM]

Ok, well I'll begin.

For all regular polygons - trivial.

Now you do the other ones.

 

[icycloud]

I think the triangle inequality is paramount to solving this one.
Saw this question this morning, best I can do is prove:

AM_l <= 2AM_d

Where AM_l and AM_d are the arithmetic means of the side-lengths and diagonals, respectively.

However, we are required to prove AM_l <= AM_d... I'll try again when I come back tonight -- gotta run now.

 

[Riviet]

Hmm, a very interesting question, but way beyond the scope of high school maths I must say. Maybe start with a pentagon and try and prove it for that?

 

[Stan..]

Interesting, is it worth considering limits?

 

[Mountain.Dew]

this might work for a REGULAR n-sided convex polygon. this would prolly make the question easier, since u have equal length sides and equal interior angle sizes.

 

[Sober]

Hmm, a very interesting question, but way beyond the scope of high school maths I must say. Maybe start with a pentagon and try and prove it for that?

Agh! I read it as n >= 4 so I was trying to solve it for a quadrilateral, I suppose it isnt true for a quadrilateral otherwise surely the question would have been restated.

 

[KeypadSDM]

this might work for a REGULAR n-sided convex polygon. this would prolly make the question easier, since u have equal length sides and equal interior angle sizes.

I already did the proof. It's trivial. The shortest diagonal spans 2 sides, and is hence longer than them both, as the angle subtended by the 2 opposite sides needs be greater than 90 degrees.

 

[Templar]

No solution, but here are two points on how to possibly attack the question:

1. Prove it holds for n=5 (pentagon).
2. Prove that the sum of diagonals that span two sides is greater than the sum of the sides for n>4 (not sure if it's true).

If those two can be proven, then there might be a solution.

 

[airie]

No solution, but here are two points on how to possibly attack the question:

1. Prove it holds for n=5 (pentagon).
2. Prove that the sum of diagonals that span two sides is greater than the sum of the sides for n>4 (not sure if it's true).

If those two can be proven, then there might be a solution.

...So you mean maybe induction is the way to go? Hmmm...but I still have no idea how

Yep I was trying to do this question to train myself for the AMO coming up...and now I'm totally freaking out cos I seem to stumble every few steps Just hoping that I won't just go in there and sit there for the whole 4 hours twiddling my thumb cos I don't know any of the probs...>.<

 

[Templar]

My current take on the question would be to prove it through induction, but it might not yield any result.

It would be best to train yourself for the AMO not by doing past questions straight away if you are not familiar with it. AMT does offer several books on how to tackle olympiad style questions, whether they are effective I am not sure.

Have you done the AMOC Correspondence Program 4 last year? It is used to give people some practice at solving olympiad style questions, by providing hints to guide you. Certainly easier than the AMO, but it is something to start with.

As for your PM, I'll answer it here just in case there is anyone else that is after the same information.

The 2006 AMO will be held on the 7th and 8th Feb. Each day consists of 4 questions, with 4 hours to do them. Like standard international olympiads each question is marked out of 7, and the marks are gained exponentially (ie relatively easy to get 1 mark, harder to get the next mark etc).

There isn't much specific advice I can give you beyond what AMT recommends, since the questions can cover quite a wide range of topics. The most important thing is to realise that you do have all those time, and it's not uncommon to be stuck on a question and not realise any possible ways to solve it until later. Keep on thinking about them.

A useful tip is to draw large sized, proper scaled diagram for geometric questions. It might not give you the answer (or the answer but no working), but can provide you insights into how to solve it. Or in the case of giving you the answer, you can work from both ends to meet in the middle, or know what you're looking out for when writing out your working.

Write every idea you have on the question down and try to justify everything. You never know what you might be given a mark for.

If this question seems very different, don't stress out, because it is certainly one of the harder ones. However, even if you do find the questions in the AMO intimidating, don't worry either, you're already within an elite group of people in Australia to participate in the AMO (roughly 100 or so a year).

That's pretty much all I can think of at the moment. SeDaTeD did the AMO in 2003 and got a bronze, so he might have some inputs of his own.