RMO 2009

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)

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????

what is the solution to 6.............in maharashtra the rmo is always tougher so can there be a retest

my solution to 3

3​2008​​+4​2009​​=(3​1004​​)​2​​+(2×4​1004​​)​2​​+4(3​1004​​)(4​1004​​)−4(3​1004​​)(4​1004​​)
=(3​1004​​+2×4​1004​​)​2​​−4(3​1004​​)(4​1004​​)
=(3​1004​​+2×4​1004​​−2×3​502​​×4​502​​)(3​1004​​+2×4​1004​​+2×3​502​​×4​502​​)
it is sufficient to show
3​1004​​+2×4​1004​​−2×3​502​​×4​502​​≥2009​182​​
or 3​1004​​+2×4​1004​​≥2009​182​​+2×3​502​​×4​502​​
now i will show 4​1004​​≥2×3​502​​×4​502​​
or 4​502​​≥2×3​502​​ or (​3​​4​​)​502​​≥2
(​3​​4​​)≈1.3 so (​3​​4​​)​4​​≈(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​​>(2187×4096)​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

my solution to 2

a​2​​−3a−19=a​2​​−10a+7a+51−70=(a−7)(a+10)+51
we also observe
so from here we see (a−7)(a+10) is either a multiple of 289 or a non multiple of 17
if it is a multiple of 289, adding 51 will make it a non multiple of 289
if it is a non multiple of 17, the status wouldnt change

i also got ans to 1st one 90 degrees

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

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

Join BI and CI.
​sinx​​AI​​=​sin2x​​CI​​ where angle ICA=x.

Also ​sin2x​​AB​​=​sin4x​​BC​​

AI=​2cosx​​CI​​

CI=​2cosx​​BC​​

We thus have

AB=​2cos2x​​BC​​ & AI=​4cos​2​​x​​BC​​

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

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∑​i=1​n​​a​i​​=101+n
If n>3 then (suppose n=7) ∑​i=1​7​​a​i​​=27
But ∑​i=1​7​​a​i​​≥28 hence a contradiction for all n>3.
Thus ...

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