INMO 2015 question paper

Problem 1: Let the circumcircle of $\triangle PQD$ intersect $AC$ again at $T$ and let $O$ be the circumcenter of $\triangle PIQ$. I claim that $T= O$. Note that $A, P, I$ are collinear since they all lie on the internal angle bisector of $\angle BAC$. Similarly, $C, Q, I$ are also collinear, so $\angle PIQ = 90^{\circ} + \dfrac{\angle ABC}{2} = 135^{\circ}$. Also, $\angle PDQ= 45^{\circ} + 45^{\circ}= 90^{\circ}$. Also, $\angle POQ= 2 \angle PIQ - 180^{\circ}= 90^{\circ}= \angle PDQ$. It hence suffices to show that $PT= QT$, which is true since both sides are equal to $PQ \sin 45^{\circ}$.

Problem 2: The answer is $\max \{v_2 (n), v_5 (n)\}$. I'll post my full proof later.

Problem 3 (the RHS should be $xf(x+y)$) : Set $x=y=0$ to get $f(0)=0$. Now set $y=0$ to get $f(x^2)= xf(x)$. Now setting $x=y$, $f(x^2 + f(x^2))= xf(2x)$. It follows that $-x \cdot f(-2x)= x f(2x) \implies f(2x)= -f(2x)$, or $f(x)=-f(x))$ for all $x$. Now setting $y= -x$, $f(x^2 - f(x^2))= 0$, so $f(x-f(x))=0$ for all positive $x$. Since $f(x-f(x))= -f(-x-f(-x))$, $f(x-f(x)) =0$ for all negative $x$ too. Now setting $x: x-f(x)$ and $y=f(x)$, we see that $f((x-f(x))^2)= (x-f(x)) f(x) \implies (x-f(x))(f(x)=0$. Thus, either $f(x)=0$ or $f(x)= x$. If $f(x)= x$ for some $x \neq 0$, we get that for all $y$, $x f(x+y)= f(x^2)= xf(x)$, or equivalently, $f(x+y)=f(x)$, i.e. $f(x)$ is constant. But it's easy to see that if $f(x)$ is constant, it must be zero. So either $f(x)=0$ for all $x$ or $f(x)=x$ for all $x$.

Problem 4: Let $t_n$ denote the answer for $n$ passes. It's easy to see that $t_{n+1}= 3^n-t_n$, and from $t_1= 0$, we can easily compute $t_7$.

Problem 5: This is probably some tedious trig bash. Haven't tried it yet.

Problem 6: The quadratic residues mod 12 are $0, 1, 4, 9$. If one of them appears six times, we can just take six copies of it. If one of them appears at least four times, say $x$, then another one must appear at least twice, say $y$, then we have $x+x+y= x+x+y$. If all of them appear no more than three times, equality must hold, and modulo $12$, the set must be four copies of $0, 1, 4, 9$ each. Then we can just consider $4+4+1= 9+0+0$. Note that the lower bound is tighter than $11$-- replacing $11$ by $9$ keeps the proof intact.

I scored 84 :)

I solved no. 6 like this i wanna ask that is this correct. x ^{2} is congruent to 0,1,4,9 mod12 =》from set of 11 we can choose a^2 & d^2 such that a ^2 is congruent to d^2 mod12 by PH Thus the problem reduces to choosing 4 no.'s from 9. Again by PHP one can choose b^2 & e^2 such that b ^2 is congruent to e^2 mod12.And again theproblem reduces and againapplying PHP one can see that there exists a^{2}+b^{2}+c ^{2 }is congruent to d^{2}+e^{2}+f^{2 }mod.12 .

I got my performance card and I secured 39 marks in it. Please post ur marks too. @Sreejato Bhattacharya @Siddhartha Srivastava @Shourya Pandey @Rajat Gupta @Rajat Gupta @rangeela ras

Guys do you know any book for Geometry .

It should be good with difficult problems.Same as those which come in RMO,INMO,IMO

Hi ,i got 73 in inmo.

has everyone received his marksheets if yes post your marks

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=362&t=623487&start=260

You can check this thread for my post summarizing all marks.

Please post solutions. @Sreejato Bhattacharya @brian charlesworth @Sravan Chinta @Sanjeet Raria @megh choksi

Typo in the 3rd question.

Typo in Q3. It should be $xf(x+y)$, not $x(x+y)$.

Also, how did your paper go? @Surya Prakash

EDIT:- Enclose only the numerical terms in LaTeX. Try not to enclose text.

I just messed the question 4.

how many marks will be removed if for example the whole answer is correct but at the conclusion, 54+54+72+72+72+96+54+72=492 is written but its 546. @Sreejato Bhattacharya @Siddhartha Srivastava

My performance summary: <<<1>>>Proceeded a little going by Sreejato's soln. Found all the values you got by angle chasing(PIQ,PDQ,POQ). <<<2>>>Did only trivial case n=2^p5^q which will be max(p,q).For primes for which i exists such that 5*10^i<p<10^i+1 ans is 0 by fermat's little theorem but (GOD knows how)wrote the length of the periodic part instead(order of 10 mod p).Anyways so got only the composite powers of 2 and 5.(Please post your solution fast, your answer shows it must be a beautiful solution) <<<3>>>proved only f(0)=0,f(x^2)=xf(x) and that f is odd.Could not proceed furthur. <<< 4>>>Got it full correct.Used PIE. <5>Just left it. <<<6>>>Got full correct.Used PHP 3 times.--------------------- So that was all about my performance in the first and last INMO of my life.At present i am pretty much disappointed about it. Shudve got 3 correct.Whatever i just want you to tell me how much partial marks i can get for 1,2,3.Please post your opinions fast.I just want to clear my dilema and let myself move on from this sad stage.

Hey, can u all simply post your expected marks and expected cut-off please...

Hi for the fourth question can the following be the solution - Starting from A every person has three options to pass to (since he cannot pass to himself ) except the last person who has only option A to pass to and hence there are 3^6 ways of doing so ! However if A ends up to be at the second last position this is not possible and hence the no. of ways A ends to be second last is 3^5 . Also a s palindromes are not possible the total no of ways is 2*(3^5)

Also for the fifth question _ Since AB + CD = BC + AD Area of quadrilateral =$\sqrt{abcd - abcd cos^{2}((A+C)/2)}$ = sum of areas of triangles . I think equating we will get it

You did How many Questions Drumil and in which class you are??

I decided to take a short break and write my analysis. I'm expecting 40 - 45; please comment on what you think i should get.

Hidden Text Q1 --> I could not proceed at all in this question. I drew the diagram, wrote formula for inradius and circumradius, tried to simplify it and left it at that. Expecting 0-1 marks.

Q2 --> I took that 1/n has periodic length 0 if it is coprime to 2 and 5 as a lemma ( I Think it was in some NCERT that it starts repeating immediately, so i thought it was obvious ) and then proved for the other case, when n has powers of 2 and/or 5 as a factor. I think they should give me 9 - 10 for one case, but to be safe, i consider that i get 4 marks.

Q3 --> I can't believe i messed this up so bad. I got f(xsq) = xf(x) and f(0) = 0, then f(xsq. - f(xsq.)) = 0, and then used an argument that the function is increasing, by putting n and 2n or something (i can't exactly remember) and stated that if it was strictly increasing, it must be one to one, thus xsq. = f(xsq.) = xf(x) and thus, f(x) = x. i forgot that f(x) could be 0, since the argument for strictly increasing had a product relation, and thus could be 0. Even assuming my argument was flawed, i feel that i deserve 4 marks atleast, since i had essentially derived everything to complete the proof. (As someone pointed out, using f(xsq) = xf(x), if f(xsq) = 0, either x = 0 or f(x) = 0. if x = 0, xsq = 0 therefore only f(0) = 0 if f(x) unequal to 0.)

Q4 --> Super silly, since i pride myself on being able to apply basic combinatorics. I guess not being able to do maths the last 15 days took its toll. anyways, i did 7 cases and forgot one with 54 possibilities, thus getting 492. i think i deserve 13, since the approach was correct and for a case they usually only cut 3-4 per case.

Q5 --> i thought this was easy when i saw the paper, but was soon proved wrong.i wrote that ab + cd = bc + ad and then noted that PEAH etc. were cyclic quadrilaterals, then named 4 angles as w,x,y,z and did some angle chasing, expressed ab,bc,da in PE,PF,PH but could not get the expression for PG from CD. i think i could not complete the angle chasing, since otherwise i'd be done. Thus, used sine rule, left the expression unsimplified. Expecting 2 marks.

Q6 --> 17.

so that would be why i expect 40 marks. Please do tell me what you think ill get :P i think i have a 5% chance at merit certificate and 1% chance at camp. Guess im overestimating :D

when does the result of inmo come out?

I did something in the 5th one... I used the incircles' radii characterization to solve the problem :P

I don't think I will get marks on the 5th one...

What do you think ?

Someone told me that Mumbai region people get their scorecards by end of 2nd or start of 3rd week. Anyway, HBCSE told me that everyone will get the cards by 28 Feb.. Can't wait

I solved no. 6 like this i wanna ask that is this correct. x ^2 is congruent to 0,1,4,9 mod12 =》from set of 11 we can choose a^2 & d^2 such that a ^2 is congruent to d^2 mod12 by PH Thus the problem reduces to choosing 4 no.'s from 9. Again by PHP one can choose b^2 & e^2 such that b ^2 is congruent to e^2 mod12.And again theproblem reduces and againapplying PHP one can see that there exists a^2+b^2+c ^2 is congruent to d^2+e^2+f^2 mod.12 .

@Sreejato Bhattacharya @Surya Prakash @Prasun Biswas

Guys could you tell me how to prepare for RMO.Books and resources

Hey everyone, I just received my performance card, and it says I scored 43/100, How do I know the cutoff?

did anyone got inmo performance card ?? what would be expected cutoff? i scored 45,any chances

in the last ques. cant we apply php ??? we can directly conclude the solution..... for divisiblity of a perfect square we can be use the fact that a perfect square is of form 3k+1 or 3k and such nos are of form 4k-1 or 4k...... and since div by 12 is reqd
we show by cases that the difference of the expression is individually divisible by 3 and 4 plss reply if its orrect... i have also appeared for INMO and want to know if im correct

i've got 5 correct answers . is there any chance of me being selected for the camp

Is the following solution of problem 3 of INMO correct.

Let f be const. function. =>f(x)=0 Let f is not a const. function. Putting x=0 & f(0)=a, =>f(ay)=0 'y 'is a variable and' a'is const. & f is not a constant function. =>ay=const. =>a=0 Thus if f(x)=0 then x=0 &f(0)=0 Now putting y= -x, We get f(x^2 - x.f(x))=0 =>x^2 - x.f(x)=0 =>f(x)=x Therefore f(x)=0,x

Has anyone solved the 5th problem both ways ?

I recieved my Performance Card today. Secured only 41. Any chances?

My score is 45. What will be the expected cutoff ?

i am class 10th scored 35 can i expect merit certificate.( i know i wont qualify for sure.)

Guys, Please tell me if my solution is correct.

Q3. Putting $y = 0$ in the equation, we get $f(x^{2}) = xf(x)$. Therefore, $f(x) = \sqrt{x}f(\sqrt{x}) = \sqrt{x} \cdot \sqrt{\sqrt{x}}f(\sqrt{\sqrt{x}}) =............= x^{\frac {1}{2} + \frac {1}{4} +.............+ \frac {1}{2^{n}}} \cdot f(x^{\frac {1}{2^{n}}})$. Now applying $lim$ $n\rightarrow \infty$ on both sides, we get $f(x) = x \cdot f(1)$. Taking $f(1) = c$, we get $f(x) = cx$. Putting this in the original equation, we get $cx^{2} + c^{2}xy = cx^{2} + cxy$ $\Rightarrow$ $c^{2} = c$ $\Rightarrow$ $c = 0$ or $1$. Thus we get $f(x) = 0$ or $f(x) = x$ for all $x$.