\(30th\quad Indian\quad National\quad Mathematical\quad Olympiad-2015\\ \\ 1.\quad Let\quad ABC\quad be\quad a\quad right\quad angled\quad triangle\quad with\quad \angle B={ 90 }^{ 0 }.\\ Let\quad BD\quad be\quad the\quad altitude\quad from\quad B\quad on\quad to\quad AC.\quad Let\quad P,\quad Q\\ and\quad I\quad be\quad incenters\quad of\quad triangles\quad ABD,\quad CBD\quad and\quad ABC\\ respectively.\quad Show\quad that\quad the\quad circumcenter\quad of\quad triangle\quad \\ PIQ\quad lies\quad on\quad hypotenuse\quad AC.\\ \\ 2.\quad For\quad any\quad natural\quad number\quad n>1,\quad write\quad the\quad infinite\quad \\ decimal\quad expansion\quad of\quad 1/n\quad (for\quad example,\quad we\quad write\\ 1/2\quad =\quad 0.4\overset { \_ }{ 9 } \quad as\quad its\quad infinite\quad decimal\quad expansion,\quad not\quad \\ 0.5).\quad Determine\quad the\quad length\quad of\quad non-periodic\quad part\quad of\\ \quad the\quad infinite\quad decimal\quad expansion\quad of\quad 1/n.\\ \\ 3.\quad Find\quad all\quad real\quad functions\quad f\quad from\quad R\rightarrow R\quad satisfying\\ the\quad relation\\ \qquad \qquad \qquad \qquad f({ x }^{ 2 }+yf(x))\quad =\quad xf(x+y).\\ \\ 4.\quad There\quad are\quad four\quad basket-ball\quad players\quad A,B,C,D.\\ Initially\quad the\quad ball\quad is\quad with\quad A.\quad The\quad ball\quad is\quad always\\ passed\quad from\quad one\quad person\quad to\quad a\quad different\quad person.\\ In\quad how\quad many\quad ways\quad can\quad the\quad ball\quad come\quad back\quad to\quad A\\ after\quad seven\quad passes?\quad (For\quad example\quad A\rightarrow C\rightarrow B\rightarrow \\ D\rightarrow A\rightarrow B\rightarrow C\rightarrow A\quad and\quad A\rightarrow D\rightarrow A\rightarrow D\rightarrow C\rightarrow A\rightarrow B\rightarrow A\quad \\ are\quad two\quad ways\quad in\quad which\quad the\quad ball\quad can\quad come\quad \\ before\quad to\quad A\quad after\quad seven\quad passes.)\\ \\ 5.\quad Let\quad ABCD\quad be\quad a\quad convex\quad quadilateral.\quad Let\quad the\quad \\ diagonals\quad AC\quad and\quad BD\quad intersect\quad in\quad P.\quad Let\quad PE,\quad \\ PF,\quad PG,\quad PH\quad be\quad the\quad altitudes\quad from\quad P\quad on\quad to\quad \\ sides\quad AB,\quad BC,\quad CD\quad and\quad AD\quad respectively.\quad Show\quad \\ that\quad ABCD\quad has\quad incircle\quad iff\quad \\ \qquad \qquad \qquad \frac { 1 }{ PE } +\frac { 1 }{ PG } =\frac { 1 }{ PF } +\frac { 1 }{ PH } .\\ \\ 6.\quad Show\quad that\quad from\quad a\quad set\quad of\quad all\quad 11\quad integers\quad \\ one\quad can\quad select\quad six\quad numbers\quad { a }^{ 2 },\quad { b }^{ 2 },\quad { c }^{ 2 },\quad { d }^{ 2 },\quad \\ { e }^{ 2 },\quad { f }^{ 2 }\quad such\quad that\quad \\ \qquad \qquad { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }\equiv { d }^{ 2 }+{ e }^{ 2 }+{ f }^{ 2 }\quad (mod\quad 12).\)

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## Comments

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TopNewestProblem 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.

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A lengthier computation shows that \( n=8\) works too, but \(n=7\) doesn't ; e.g \( 4,1,0,0,0,0,0\).

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Did anyone get Question 6?? Can you explain

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k, say), as divisibility by 3 is already shown, we can suppose \((a^{2} - d^{2}) + (b^{2} - e^{2}) + (c^{2} - f^{2})\) is divisible by 4 by a little PHP and different cases for \(a, b, c, d, e, f\) being odd and/or even.... So we have proved \(k\equiv 0\pmod{12} \)Log in to reply

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Hi Sreejato in question 3 when you set x : x-f(x) then as f(x) = x , new variable x is identically 0 so can not be taken as variable capable of taking all values.Is this step of your correct ? Thanks for providing the solutions

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I don't understand your objection; I just plugged \(x-f(x)\) in place of \(x\).

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I think you have a typo in the reccurence

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Where? I can't find anything wrong...

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A -> B-> A

A -> C -> A

A -> D -> A

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So you too appeared for the INMO? How many did you get?

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Nope, didn't qualify RMO. (Don't know why, I solved 5 problems there; probably because of my indecipherable handwriting.)

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@Sreejato Bhattacharya

But you can write INMO because u qualified RMO 2013........Log in to reply

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here

Guys Please helpLog in to reply

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Nice. What do you think the cutoff will be?

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I guess not more than 50.

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here

Guys Please helpLog in to reply

Comment deleted Feb 21, 2015

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sighI feel like kicking myself...).Log in to reply

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Regarding people overestimating their marks, maybe they did solve 6 problems? I solved 5 in RMO (messed up the silly a.p problem) but ended up getting 63... I really have no idea why. :( But yeah, I assure you, most of the guys who claim to have solved 5-6 problems have solved no more than 2 (if that weren't the case last year, the cutoff would have been around 70). This happens every time.

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Let X denote the set of members {B,C,D}. From given hypothesis we draw out the possible ways of passings.

\(A\rightarrow X\rightarrow A\rightarrow X\rightarrow X\rightarrow A\rightarrow X\rightarrow A\quad -\quad P1\\ A\rightarrow X\rightarrow X\rightarrow A\rightarrow X\rightarrow A\rightarrow X\rightarrow A\quad -\quad P2\\ A\rightarrow X\rightarrow X\rightarrow A\rightarrow X\rightarrow X\rightarrow X\rightarrow A\quad -\quad P3\\ A\rightarrow X\rightarrow X\rightarrow X\rightarrow A\rightarrow X\rightarrow X\rightarrow A\quad -\quad P4\\ A\rightarrow X\rightarrow X\rightarrow X\rightarrow X\rightarrow A\rightarrow X\rightarrow A\quad -\quad P5\\ A\rightarrow X\rightarrow X\rightarrow X\rightarrow X\rightarrow X\rightarrow X\rightarrow A\quad -\quad P6\\ A\rightarrow X\rightarrow A\rightarrow X\rightarrow A\rightarrow X\rightarrow X\rightarrow A\quad -\quad P7\\ A\rightarrow X\rightarrow A\rightarrow X\rightarrow X\rightarrow X\rightarrow X\rightarrow A\quad -\quad P8\)

no. of ways in P1 = 1x3x1x3x2x1x3x1 = 54........... no. of ways in P2 = 1x3x2x1x3x1x3x1 = 54........... no. of ways in P3 = 1x3x2x1x3x2x2x1 = 72........... no. of ways in P4 = 1x3x2x2x1x3x2x1 = 72........... no. of ways in P5 = 1x3x2x2x2x1x3x1 = 72........... no. of ways in P6 = 1x3x2x2x2x2x2x1 = 96........... no. of ways in P7 = 1x3x1x3x1x3x2x1 = 54........... no. of ways in P8 = 1x3x1x3x2x2x2x1 = 72...........

Total no. of ways = 546

How many marks can i get??? @Sreejato Bhattacharya @Siddhartha Srivastava @Chandrachur Banerjee

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I tried; couldn't convince my parents it was a mistake on the grader's part.

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Yes.

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You aren't talking about my solution, are you? I think I'm the only one who posted a solution for the FE on AOPS, at least the only one in the INMO Prep thread.

The topper got 81 marks. Even then, I'm from North Bihar. I honestly doubt more than 2 to 3 people solved all 6.

You can file an RTI to see your paper. A guy at my center did that.

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I scored 84 :)

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Congratulations!

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Awesome! Did you get 5 completely correct?

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I think I was just overimagining things by thinking that I had messed up the second question. My solution to that question was correct, so I solved 5 questions correctly (All except the fifth question).

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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

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I got 45, in 10 In which class are you?

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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 .

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That should be correct. Why didn't I think of that. :/

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

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

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has everyone received his marksheets if yes post your marks

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I got 51. You?

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messed it up 24 but hope to do better in 11th class , any other students of whom you know the mark

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ya two of micheals have got 16 and 1 got 19

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Hi ,i got 73 in inmo.

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Guys do you know any book for Geometry .

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

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You should see Geometry Revisited , it is a fantastic book. You can also get the free pdf here.

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Try Sharygin, it's the best book for IMO Geometry, and also for INMO. Our teacher at INMOTC referred us that book.

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Any other books or resources or tips given to you at INMOTC

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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

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even i did the same and i proved that by php that we can find 6 integers such that the above expression is div. by 3 and 4 but i did a conceptual error that these six integers may not be the same , meanwhile how much did you get in which class are you and from which region

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im in class 11 from jharkhand region ...

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

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

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Dude, nice. You'll have to wait for some time though for everyone to receive theirs. Some people were saying that the results/cutoff will come in Mid March.

Also, I met you I think. Weren't you the guy who had also given the INAO? How did that go?

EDIT:- Also, how many questions did you solve? I wasn't able to talk to you after the exam.

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Yes, That did not go very well, How much did you score in INMO?, You were the one in KV GMO right? As far problems, I think 2 completely and one almost completely

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Edit:- Just got it. 51/102

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@Sreejato Bhattacharya @Surya Prakash @Prasun Biswas

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

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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 .

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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

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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 ?

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has anyone proved that a quadrilateral has an incircle if the property given is true in the 5th problem?If so how ?

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when does the result of inmo come out?

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Mid March it should be :) I am more worried as the official solutions haven't be released yet which is weird considering that all other INOs' answer keys were uploaded the very next day :/

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they have been released now... Check past year's question papers and solutions

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Um, I got my scorecard last time before March. It all depends-- it probably takes some time for the scorecard to reach West Bengal. Unless the HBCSE guys mess up heavily (like last time, they sent the scorecard of one of my friends to Aurangabad instead of Durgapur), you should get them within the first week of March. The results, however, will probably be officially declared mid-March, just like last time.

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Mid March? That's very far away... I thought late Feb at worst. And forget INMO, they haven't released RMO answer keys yet. Also, are you utkarshgupta on AOPS?

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And yes, after the reevaluation and everything, the results are declared by Mid March

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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

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We cannot assume whether one shall qualify or not but 40-45 is good in INMO. Certificate of Merit is given up to class 10th only (as of what I know, maybe...) and perhaps you are in 11th (because your age is 17).

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I am in 10th and my real age is 15 :)

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I have not done probability so well, so cant tell.

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

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3.5 and i'm in tenth

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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

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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)

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It's not that simple. Consider the case when A has the ball after the third last pass. In this case, A has three choices for the second last pass. However, if the ball were with someone else after the third last pass, he/she would have two choices for the second last pass (can't give the ball to A). Your argument is flawed there.

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I would say his approach is correct. We can proceed by inclusion exclusion and thus the answer is 3^6 - 3^5 + 3^4 - 3^3 +3^2 - 3. BTW, you are anonymous bunny on aops right? Big fan :P u had bad luck this year .

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

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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.

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I believe you're getting around 10 in the first problem. Not sure about your score on problem 2, but tacking the "trivial case" won't fetch more than 5. In problem 3, you just solved 80% of that problem, so I guess at least 10. I'm not sure how you used PHP in problem 4, care to post your full solution? Yeah, problem 5 was a bit too hard.

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Consider A>P1>P2>P3>P4>P5>P6>A.------------------------------- Now P1 and P6 cant be A.So ------------------------------ Let Ai={No. of ways in which the said job is done when ONLY Pi is A}for i=2,3,4,5---------------------------------- Then i used PIE on A2,A3,A4,A5, i.e calculated |A2 U A3 U A4 U A5|.(See thats easy as consecutive Pi cant be A so many sets contribute 0 to the counting by definition)--------------------Lastly added the no. of ways in which no Pi is A and got 546.

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Not PHP yaar, i used PIE(Principle of Inclusion and Exclusion).I have given a sketch of my solution to P4 in this note (see a bit above).Well i am totally confident in P4 and P6.Just tell me what to expect on the scorecard.

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Please Sreejato reply to this thread.

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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

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If you got the recursion correct, trivial computation errors won't cost you more than 2-3.

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I got the recursion correct

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I just messed the question 4.

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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.

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Typo in the 3rd question.

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

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

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

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I got 45 too. Expected is around 55.

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

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Has anyone solved the 5th problem both ways ?

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It is solved on aops by both synthetic and other methods http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=623459 BTW official solution is also good

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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

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You proved that \( f(0) = 0 \). You haven't proved that \( f(x) = 0 \) implies \( x = 0 \).

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

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Writing them well is the trick.

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Depends on how many marks you get. Though with 5 questions correct, you should get in easily.

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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 \(x\).

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You don't know if the function is continuous. Therefore you can't apply limit.

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Comment deleted Feb 22, 2015

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I don't even know what to say. :3

Anyways, serious answer: INMO.

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Ya it was just a spell-binding question!!!!Hee-Hee

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Are you kidding?

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I think INMO is the toughest exam on maths in India at the school level. Its full-form explains it!!

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INMO , definitely!

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@Ashutosh Kaul; INMO is the answer.

JEE Advanced is tough due to immense competition and the time constraint.

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I think as a whole JEE advanced is more tough......as you have to study all maths ,physics and chemistry properly.........but if you compare only maths then obviously INMO is the hardest..

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