A...Question :P (1 Viewer)

[airie]

sure, if you mean by 'dumb old me' as a girl who is going to participate in the AMO, and is probably on her way to killing the HSC maths exams when she does them.

lol wow thanks for your confidence in me Yep I'm gonna do the AMO very soon...*sees 2001 AMO paper* Ahh shoot *hides*

 

[Templar]

Try 2004 and 2005. They were the easier years in recent times.

 

[m0ofin]

Congratulations and I wish you all the best in the AMO Airie Btw, who_loves_maths, your handwriting is so uber-cute :uhhuh:

 

[airie]

Try 2004 and 2005. They were the easier years in recent times.

...Really? ... The reason being...? Wow. We're getting dumber. x.x

Congratulations and I wish you all the best in the AMO Airie

Thanks m0ofin

 

[buchanan]

dumb old me stared at it for like 5 whole minutes

You're not dumb!

Here's another way to explain the solution to your problem. Since you've spent some time trying to understand my previous solution, I think this time you will find this one easier to understand.

Consider the convex polygon A₀A₁A₂...A_n-1, with the subscripts reduced (mod n). Let A_iA_j be a diagonal. Then by the triangle inequality,

A_iA_j+A_i+1A_j+1 > A_iA_i+1+A_jA_j+1.

When we sum these inequalities over all n(n-3)/2 diagonals A_iA_j, each diagonal occurs twice on the left, and each side n-3 times on the right; so we get
2d > (n-3)p or n-3 < 2d/p.

This is from the first part of the 1984 IMO Question 5.

You can get all the 1959-2005 IMO's on my website www.fourunitmaths.cjb.net :
http://www.angelfire.com/ab7/fourunit/imo-all.pdf (1959-2003)
http://www4.tpgi.com.au/nanahcub/imo2004.pdf (2004)
http://www4.tpgi.com.au/nanahcub/imo2005.pdf (2005)

Hope this helps.

Congratulations on getting into the AMO, and I hope it goes well for you.

 

[airie]

You're not dumb!

Here's another way to explain the solution to your problem. Since you've spent some time trying to understand my previous solution, I think this time you will find this one easier to understand.

Consider the convex polygon A₀A₁A₂...A_n-1, with the subscripts reduced (mod n). Let A_iA_j be a diagonal. Then by the triangle inequality,

A_iA_j+A_i+1A_j+1 > A_iA_i+1+A_jA_j+1.

When we sum these inequalities over all n(n-3)/2 diagonals A_iA_j, each diagonal occurs twice on the left, and each side n-3 times on the right; so we get
2d > (n-3)p or n-3 < 2d/p.

This is from the first part of the 1984 IMO Question 5.

You can get all the 1959-2005 IMO's on my website www.fourunitmaths.cjb.net :
http://www.angelfire.com/ab7/fourunit/imo-all.pdf (1959-2003)
http://www4.tpgi.com.au/nanahcub/imo2004.pdf (2004)
http://www4.tpgi.com.au/nanahcub/imo2005.pdf (2005)

Hope this helps.

Congratulations on getting into the AMO, and I hope it goes well for you.

Thanks for that Well the part that I had to think about for quite some time was the 'summing over' part See I spent a while trying to picture it in my head, to see how many times each diagonal and side appear on the sides of the inequality And thanks for the links