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

my solution to 2

a^2-3a-19=a^2-10a+7a+51-70=(a-7)(a+10)+51
we also observe a-7 \equiv a+10\ mod(17)
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

I got my answer to first as 90 degrees
fourth - 370775
sixth - 1 or 3 or 5

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.
\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}...

shorter soln to 4

sum of torn pages=101
note sum of 1 leaf(2 consecutive pages)=4n+1 for some n
101 is of the form 4n+3
so we conclude no of torn leaves are of the form 4k+3
so possible outcomes 3,7,11,...
consider 7 leafs ie 14 pages
min sum we have is (14)(15)/2=105>101
so ans is 3

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?

detailed solutions can be seen here :

question 1:

http://www.mathlinks.ro/viewtopic.php?p=1695934#1695934

question 2:

http://www.mathlinks.ro/viewtopic.php?p=1695941#1695941

question 3:

http://www.mathlinks.ro/viewtopic.php?p=1695943#1695943

question 4:

http://www.mathlinks.ro/viewtopic.php?p=1695944#p1695944

question 5:

http://www.mathlinks.ro/viewtopic.php?p=1695949#1695949

question 6:

http://www.mathlinks.ro/viewtopic.php?p=1695946#1695946

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