RMO 2009

wasnt the second one too easy to appear in an RMO :D

ohk here goes the 1st one............tats only i tried :P

it was a sitter actually :P

let teh lengths of side AB and AC be t

angle ABC = angle ACB=θ

angle ABI =θ/2

angle BAI=90-θ

angle AIB=90+ (θ/2)

now
AIsin(θ/2)=ABsin(90+(θ/2))

so AI=t.tan (θ/2)

further BC=2tcos(θ)

since BC=AB+AI

2tcos(θ)=t+t.tan (θ/2)

2cos(θ)=1+tan(θ/2)

2(2cos²(θ/2)-1)=1+tan(θ/2)

2(2sec²(θ/2) -1)=1+tan (θ/2)

tan (θ/2)=n

2(2(1+n²) -1)=1+n

solving u will get n=√2-1

so tan(θ/2)=√2-1=tan(45/2)
θ=45

so angle BAC=90°

i think the answer of ques no. 6 is three pages...... and in ques no. 4 i 2 got 370775.......

anyone solved ques no. 5??? i proved the first part.......

i am expecting around 40-42...... is that gonna be sufficient????

2nd part of q-5 was quite easy....u trivially need a slight application of PHP like this :-
If a+b≤1, then at least 1 of a and b must be less than half....rest can be done by arguments of simple symmetry.....
I managed 4 sums and a small amt of the 2nd one as well.....(It was simple application of quadratic reciprocity)....Shall this be enough for a 12th grader?

my solution to 3

3^{2008}+4^{2009}=(3^{1004})^2+(2\times 4^{1004})^2+4(3^{1004})(4^{1004})-4(3^{1004})(4^{1004})
=(3^{1004}+2 \times 4^{1004})^2-4(3^{1004})(4^{1004})
=(3^{1004}+2 \times4^{1004}-2\times 3^{502} \times 4^{502})(3^{1004}+2 \times4^{1004}+2\times 3^{502} \times 4^{502})
it is sufficient to show
3^{1004}+2 \times4^{1004}-2\times 3^{502} \times 4^{502} \ge 2009^{182}
or 3^{1004}+2 \times4^{1004}\ge 2009^{182}+2\times 3^{502} \times 4^{502}
now i will show 4^{1004}\ge 2\times 3^{502} \times 4^{502}
or 4^{502}\ge 2\times 3^{502} or \left(\frac{4}{3} \right)^{502}\ge 2
\left(\frac{4}{3} \right) \approx 1.3 so \left(\frac{4}{3} \right)^4 \approx (1.6)^2>2
so it is sufficient to show 3^{1004}+4^{1004} > 2009^{182}
3^{1004}+4^{1004} > (2187)^{143}+ (4096)^{167}
applying AM-GM
(2187)^{143}+ (4096)^{167} > (2187 \times 4096)^{165}>(2009)^{310}>(2009)^{82}
as the smaller of the two factors is greater than 2009^{82}, both the factors are greater than 2009^{82}. hence proved

god...expecting RMO selection is not a good thing...i mean if its in ur kismat..it'll come...if not..tough luck...althou i think i got 3.5-4 Q's correct..i dunno what the examinors think..BTW i'm in Xth.....so if the examinors are liinent....i wud deff get thru..rest as they say....wud be history :) :p

4) Page nos are of the form (2a_i-1,2a_i)
where each a_i is natural.
If the number of pages torn be n then we have 4\sum_{i=1}^{n}a_i=101+n
If n> 3 then (suppose n=7) \sum_{i=1}^{7}a_i=27
But \sum_{i=1}^{7}a_i\ge 28 hence a contradiction for all n>3.
Thus \boxed{n=3}...

In the 2nd sum.....what propmpted that split fibonacci?
I mean it's difficult to hit at that directly in an exam........how did the sum hint at that?

BTW are u sure abt ur soln - i mean u are confident with ur logic?

yes i'm confident with my solution. i thought if 1 could express the polynomial as a multiple of 17 + a multiple of 289(or a multiple of not even 17)
you can say it was more of a guesswork :) first i added 17 then 34 the 51. i was lucky with 51

ok.....
Any other methods for that 2nd sum?
Heard 1 of my friends applied quad. reciprocity.

guys would i get through in maharashtra if i got 2 complete 1 half..................would i get any marks for the first one if i took that each page is numbered just once ike both sides of a page have the same number...........reply soon im nervous