A...Question :P (1 Viewer)

[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

 

[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