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 dont think this one was very tough...some over-excitedness ......mindlessness....folishness....ensured i didnt get all the 6 Q's....BTW can any1 tell what is (was) the cutoff in delhi region for class Xth students??(if its any different from XIIthies)

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

There's a very easy and short soln to number 1.....I'll post it soon.

Join BI and CI.
\frac{AI}{sinx}=\frac{CI}{sin2x} where angle ICA=x.

Also \frac{AB}{sin2x}=\frac{BC}{sin4x}

AI=\frac{CI}{2cosx}

CI=\frac{BC}{2cosx}

We thus have

AB=\frac{BC}{2cos2x} & AI=\frac{BC}{4cos^2x}

Applying AB+AI=BC, we have a quad in cos²x, so x turns out to be 2212deg.
Thus angle BAC=90^{0}...

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