1wasnt the second one too easy to appear in an RMO :D
53 Answers
62Seeing this paper i will be surprisd if the cut off will go below 55-60 marks in a lot of states!
624th one will be
sum of all numbers from 100 to 999 = (100+999)/2 x 900
sum of numbers will all odd digits = (1+3+5+7+9)x25x111
sum of numbers will all even digits = (2+4+6+8)x20x11 + (2+4+6+8)x25x100
Now find the answer [1]
21ohk 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°
1I 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)
62sahil i dont think the cuttoff will be different
It will be the same ...
but if you have got so many corrects, you should scrape through.
1i 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????
112nd 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?
1what is the solution to 6.............in maharashtra the rmo is always tougher so can there be a retest
62no there would not be a retest ever!
There are schedules for selection to INMO and all.. so i will be very surprised..
and RMO's are supposed to be not very easy :P
1my 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
1my 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
1I got my answer to first as 90 degrees
fourth - 370775
sixth - 1 or 3 or 5
1god...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
11Join 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}...
1shorter 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
114) Page nos are of the form (2a_i-1,2a_i)
where each ai 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}...
11In 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?
39detailed 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
1yes 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
11ok.....
Any other methods for that 2nd sum?
Heard 1 of my friends applied quad. reciprocity.
1guys 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