RMO 2009

yes govind is correct....... harsh, your method is wrong....

i can say that because i qualified rmo 2009......

any one tell me wha is the cut off in bihar region

i gez it shudnt exceed 40......... bt as the paper ws easy...... now i m feeling it..... it might go to 50s even.......

why the answer of last question can"t b 13..........

it can be 13 if the book is printed only one side..............................think over it

I solved the question using method of contradiction.

Suppose a² - 3a - 19 is divisible by 289.
Then a² - 3a - 19=289k for some a, k ε Z
then a² - 3a - 19 - 289k=0 is a quadratic in a where a ε Z.

for a to be an integer, discriminant must be a perfect square.

i.e. 1156k + 85 must be a perfect square.
i.e. 17(68k + 5) must be a perfect square.
i.e. 68k + 5 must be divisible by 17.

but that is not possible.

Thus our assumption that a² - 3a - 19 is divisible by 289 is false.

There is a very simple and pure geometrical solution to the 1st sum

BM is taken such that BM=AB
AI is produced to make perpendicular on BC
Then the points are joined as shown by the dotted lines.

AB=BM, BI bisects <ABM, Therefore BY perpendicular on AM.

Let <ABY=<XBY=<ACI=<BCI=x^o
ABXY is a cyclic quad.
Therefore <XAY=<YBX

Now <MAI=<ICM=x^o
And they are on same base IM. Therefore AIMC is a cyclic quad.
Therefore <AMI=<ACI=x^o (same arc AI)
Given BC = AB+AI = BM+MC
Therefore AI=MC
Equal arcs subtend equal angles at the centre
Therefore <AMI=<MAC=x^o

Therefore <BAX=<CAX=2x^o
And <BAC=4x^o

But in Triangle ABX <BAX= (90-2x)^o
By the problem,
(90-2x) = 2x,
4x = 90^o

Therefore <BAC = 90^o

No Harsh ur method is not correct..i remeber many ppl whom i know did the same mistake in this question....actually wen u open up a book u will find the find the odd number pages on the right and wen u turn the page u get to see a page with an even number on the left...so if u tear pages with no. 50 and 51 ..then the pages with no 49 and 52 will also be torn off....
btw..where in the question it's given the second page is blank....the second page isnt blank....

But what if a person reads ebooks..

@Sayandas...u can attempt maths olympiad in ur class X itself..the question asked in RMO require the basic concepts that one reads in class IX and X ..but a bit more thinking process is required...there are books available for maths olympiad..u may do those books..and for some good problems on maths olympiad u may visit the maths olympiad section on this site or u may try www.mathlinks.ro : )

isit possible that the book has page printed only one side
in this case i found the possible answers may reach 13 pages : 1,2,3,4,5,6,7,8,9,10,11,12 & 23
i have written about both cases -when book has pageprinted only one side
and when the book has page printed both side
anone plz tell iam right or wrong

Respected seniors,

The RMO 2010 is gonna be conducted on 5th of December 2010. This is me Rahul Mishra a student of Std X. I want to qualify RMO 2010. Please suggest me how should i prepare. I can not get anything of Q1, Q2 & Q3 but i don't want to give up. Please suggest me.... plz plz plz plz.....

Thankyou!
yours obediently
Rahul Mishra.....

Thanks a trillion sir... i've done that....

@ 6,,,,,, 3 pages are torn......

Loads and loads of thanks sir!
got that.....
will take a print out of it...
Hope it will be helpful...

nishant sir ,

may i know the name of the book?

so that it would help others too :)

see people sum of total pages is n(n+1)/2 and sum of remaining pages is 15000 . therefore if rth page is torn then n(n+1)/2 - r(r+1)=15000 solving the above diophantine equation yields no integral solution .. how is this possible....?????

5)The fifth one was unsolved here...I was trying this and so posting what i got.. its simple..

1)When we measure the distance of a line from a point the minimum distance is the perpendicular distance and maximum occurs at either of the end points of a line. So distance between two lines is maximum when the two point we choose is either of the four possible pair of points, one of which is end point of one line and the rest is end point of the other line..But none of those distances are greater than 1, according to the condition of the problem. So distance between any two points on the boundary must be smaller than 1.

2)Join X,Y and extend on either sides. Let it intersect the polygon at Z₁ and Z₂.The points Z₁, Z₂ are in opposite directions of both X and Y.. Since Z₁,Z₂ are in the boundary of the polygon 2|Z₁Z₂|≤ 2, Which means (|XZ₁| + |YZ₁|) + (|XZ₂| + |YZ₂|)≤2. So at least one of the summands in the bracket must be ≤ 1. Choose Z = Z_i, if (|XZ_i| + |YZ_i|)≤ 1.

i m an 8th grader and wen i saw these problems i had no idea abt them can anyone help me qualify for inmo next year

6 was probably the easiest ques of the paper.....
it was an simple application of ap.......
u got sum of torn page nos. = 101 after that taking least possible values, n/2(n+1)<101
which gave n=13
it meant at maximum, 6 pages cud be torn....
even no. of pages wer not possible.....
as odd x even = even
for 1,3,5
for 1 page, values wer 50,51...... (even,odd) pair..... which is not possible.... as page nos. of books are (odd,even) pairs......
then same u needed 2 prove fr 5 pages.......
hence only 3 pages satisfy every value........
hence, 3 pages wer torn........

i was getting 370775 to the fourth one . anyone else getting that? i'll post my solution to 2 and 6 later :)