# This note has been used to help create the INMO Math Contest Preparation wiki

30th Indian National Mathematical Olympiad 2015

1. Let ABC be a right angled triangle with angle B = 90. Let BD be the altitude from B on to AC. Let P, Q and I be incenters of triangles ABD, CBD and ABC respectively. Show that the circumcenter of triangle PIQ lies on hypotenuse AC.

2. For any natural number n > 1, write the infinite decimal expansion of 1/n (for example, we write 1/2 = 0.4999... as its infinite decimal expansion, not 0.5). Determine the length of non-periodic part of the infinite decimal expansion of 1/n.

3. Find all real functions f from R to R satisfying the relation f(x^2 + yf(x)) = xf(x + y).

4. There are four basket-ball players A, B, C, D. Initially the ball is with A. The ball is always passed from one person to a different person. In how many ways can the ball come back to A after seven passes? (For example A -> C -> B -> D -> A -> B -> C -> A and A -> D -> A -> D -> C -> A -> B -> A are two ways in which the ball can come before to A after seven passes.)

5. Let ABCD be a convex quadrilateral. Let the diagonals AC and BD intersect in P. Let PE, PF, PG, PH be the altitudes from P on to sides AB, BC, CD and AD respectively. Show that ABCD has incircle iff 1/PE + 1/PF = 1/PF + 1/PH.

6. Show that from a set of all 11 integers one can select six numbers a^2, b^2, c^2, d^2, e^2, f^2 such that a^2 + b^2 + c^2 == d^2 + e^2 + f^2 (mod 12).

Note by Surya Prakash
5 years, 1 month ago

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## Comments

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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.

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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Did anyone get Question 6?? Can you explain

- 5 years, 1 month ago

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Squares can be only 0,1,4,9 mod 12. Thus by php the modulo with maximum nos will have atleat 3 no.s if it has greater or equal to 6, then we are done. If it has 5 or 4, by php another modulo will have at least 2 no.s thus we are done. The only case with maximal 3 is 3332 hence proved.

- 5 years, 1 month ago

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We just have to take different cases for say $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}$ and $f^{2}$ ... we know that any square is of the form of $3n + 1$ or $3n$, i.e. $3n + 2$ form squares don't exist, so divisibility by 3 can be easily shown, taking different permutations of the $3n + 1$ and $3n$ forms for $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}$ and $f^{2}$. Further,since we have to prove $a^{2} + b^{2} + c^{2} \equiv d^{2} + e^{2} + f^{2} \pmod{12}$ (=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 $k\equiv 0\pmod{12}$

- 5 years, 1 month ago

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Nice. What do you think the cutoff will be?

- 5 years, 1 month ago

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I guess not more than 50.

- 5 years, 1 month ago

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Hmm. I think it might be 50+. Everyone on AOPS got 4 to 3 questions correct.

- 5 years, 1 month ago

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There are only about 10 people there (including me). Most of the students from West Bengal didn't get more than 2 correct. The cutoff last year was 49, and that paper was a tad easier than this year (sigh I feel like kicking myself...).

- 5 years, 1 month ago

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Might be true, though I do think this year's paper had easier questions than last year. Last year's Q6 and Q3 were comparitively harder than any of the questions this year. Plus, everyone has a habit of overestimating their marks. No one in my center solved less than 6 questions in the RMO, even though the highest was 81. xD

- 5 years, 1 month ago

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How come you r 27???Or r u not talking about this year's RMO????????

- 5 years, 1 month ago

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Not my real age.

- 5 years, 1 month ago

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I wouldn't say so. Problems 1-3 last year were completely trivial (daaaaarn why didn't I even look at P1 and P3!!!!!!!!). This year, the trivial problems are P4 and P6. There's a chance the cutoff will be lower than last year by a pretty big margin-- many people probably will spend too much time on P1 thinking it's trivial, and give up when they think it's getting too complicated. Then, many people will lose partials on P3-- I've seen a few flawed solutions in the AoPS thread; for example, one of them just assumed that $f(x)=0$ implies $x=0$ without actually showing it. And I honestly don't think any of the people who claim to have solved P4 by direct counting have actually solved it.

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.

- 5 years, 1 month ago

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My answer to Q4 is as follows:

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

- 5 years, 1 month ago

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I used PIE on sets in which ONLY P2,P3,P4,P5 is A an the said job is done.Lastly aded to it the no. of ways in which the job can be done without A ocurring in the middle even once. PIE on 4 sets may seem tedious at first but recall that consecutive Pi cant be A so many sets in the expansion work out to 0 and wonderful part is that you have to calculate only 2 times(work it out and see),-------But the recursion way is very good.

- 5 years, 1 month ago

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I dont know how to mention someone on brilliant. You please do it for me so that all INMO experts on brilliant can see my sad performance saga which i have posted and reply to it. Had i got the FE correct INMO wouldhave been really a wonderful journey for me even if i did not qualify.Anyways . . .

- 5 years, 1 month ago

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what do u mean by "PIE" on sets??

- 5 years, 1 month ago

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PIE = Principle of Inclusion and Exclusion.And please dont tell it is not understandable , for i wrote the same acronym in INMO!!!!!!!!!!!!!!!!!!!!!!!!!

- 5 years, 1 month ago

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for example, one of them just assumed that $f(x) = 0$ implies $x= 0$ without actually showing it

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.

Regarding people overestimating their marks, maybe they did solve 6 problems?

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

I solved 5 in RMO (messed up the silly a.p problem) but ended up getting 63...

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

- 5 years, 1 month ago

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That's really strange. What was the cut - off in your region for RMO?

- 5 years, 1 month ago

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@Siddartha: Are you alpha841 on AoPS? Your solution is completely fine, except for that typo at the end. I was talking about the second solution here- http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=623442.

You can file an RTI to see your paper.

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

- 5 years, 1 month ago

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Are you alpha841 on AoPS?

Yes.

- 5 years, 1 month ago

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Sixty five. And I scored 63. :|

- 5 years, 1 month ago

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I think the cut-off is going to be something between 65-70. I managed 4 questions but I don't think I will qualify this year.

- 5 years, 1 month ago

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Oh please don't give such comments after having solved 4. For those like me who have got 2 it seems THE GREAT INDIAN MELODRAMA.Honestly.

- 5 years, 1 month ago

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I wasn't being melodramatic. I was just saying I think I won't qualify. I can of course qualify if things work out but I am not crying about solving 4.

- 5 years, 1 month ago

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Guys Please help here

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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But you can write INMO because u qualified RMO 2013........ @Sreejato Bhattacharya

- 5 years, 1 month ago

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Nothing as such. Even I qualified RMO 13, but due to poor marks in INMO (23 marks), I had to start from scratch.

- 5 years, 1 month ago

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Guys Please help here

- 5 years, 1 month ago

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I could have if I didn't mess up that heavily last INMO (29).

- 5 years, 1 month ago

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I think you have a typo in the reccurence

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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t_2 = 3

A -> B-> A

A -> C -> A

A -> D -> A

- 5 years, 1 month ago

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Oops, fixed.

- 5 years, 1 month ago

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Can you explain the last answer?

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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hi Sreejato i said when u plug x - f(x) in place of x then new x is no more an independent variable in general as in this case x and f(x) are equal so x on lhs merely represent 0 and no more value can be assumed by it.

- 5 years, 1 month ago

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Basically, she means that x-f (x) may not be an onto function, as there may not exist x such that x- f (x) equals , say, 5.

- 5 years, 1 month ago

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I don't see how that's a problem. Whatever the value of $x-f(x)$ is, I'm plugging it in place of $x$.

- 5 years, 1 month ago

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I am not saying that there is a problem with your solution. I was just expressing his argument.

- 5 years, 1 month ago

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yeah again my point is that the value of x -f(x) is zero (as final answer is f(x)=x) so effectively u r substituting x=0.now x can not take any other value. x can be substituted with x-f(x) only when u r sure that range of x-f(x) is R which clearly is not the case here. I hope i m making sense to u.

- 5 years, 1 month ago

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No such thing. Allow me to be a bit more formal: let $P(x, y)$ be the proposition $f(x^2+yf(x))= xf(x+y)$. I'm just considering $P(x-f(x), f(x))$. The range of $x-f(x)$ is irrelevant.

- 5 years, 1 month ago

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its not SHE its HE by the way :-)

- 5 years, 1 month ago

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I scored 84 :)

- 5 years, 1 month ago

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Awesome! Did you get 5 completely correct?

- 5 years, 1 month ago

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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).

- 5 years, 1 month ago

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Congratulations!

- 5 years, 1 month ago

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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 .

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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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.

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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For combinatorics, we were provided Combinatorial Problems in Mathematical Competitions, and yeah Problem Solving Strategies is also one of the best for INMO and IMO.

- 4 years, 2 months ago

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How was INMO ?

- 4 years, 2 months ago

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Ya, it was alright!! I solved 2 and a half :3

- 4 years, 2 months ago

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Did u also give INMO this time??

- 4 years, 2 months ago

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Nope.

- 4 years, 2 months ago

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Hi ,i got 73 in inmo.

- 5 years, 1 month ago

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has everyone received his marksheets if yes post your marks

- 5 years, 1 month ago

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I got 51. You?

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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From North Bihar? Ishan Tarunesh got 43.(He posted somewhere in this threa.). Two students from DAV got ~30 and ~10. Don't know about the rest. I'll ask them about other people from DAV.

- 5 years, 1 month ago

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ya two of micheals have got 16 and 1 got 19

- 5 years, 1 month ago

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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.

- 5 years, 1 month ago

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Typo in the 3rd question.

- 5 years, 1 month ago

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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.

- 5 years, 1 month ago

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I just messed the question 4.

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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I got the recursion correct

- 5 years, 1 month ago

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Then it should be at least a 15.

- 5 years, 1 month ago

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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.

- 5 years, 1 month ago

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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.

- 5 years, 1 month ago

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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.

- 5 years, 1 month ago

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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.

- 5 years, 1 month ago

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Wow. That seems long. Though if you did get the answer, I doubt they'll cut marks for length.

- 5 years, 1 month ago

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Ya but though it was long i had to calculate at only 3 places. As far as i have heard they don't go by length ever.The recursion way was definitely the best however.

- 5 years, 1 month ago

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Please Sreejato reply to this thread.

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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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)

- 5 years, 1 month ago

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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.

- 5 years, 1 month ago

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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 .

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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3.5 and i'm in tenth

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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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).

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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Then you should surely get CoM, if not selection... :-)

- 5 years, 1 month ago

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when does the result of inmo come out?

- 5 years, 1 month ago

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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 :/

- 5 years, 1 month ago

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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?

- 5 years, 1 month ago

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Yes I am the same Utkarsh Gupta from AoPS Are u on AoPS ???

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

- 5 years, 1 month ago

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Yes. I'm on AOPS as well, but don't go there often. Also, I thought you meant scorecards. When do you think we will get scorecards?

- 5 years, 1 month ago

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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.

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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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 ?

- 5 years, 1 month ago

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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 ?

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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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 .

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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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?

- 5 years, 1 month ago

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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.

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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CBSE GMO. The guy who was with Anunay. I think I solved 4. But only 1 in which I'm 100% sure I've done nothign wrong. Also, I haven't gotten my score card yet.

Edit:- Just got it. 51/102

- 5 years, 1 month ago

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how much did you get in inao

- 5 years, 1 month ago

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Oh. Do you know the marks of the other FIITJEE people?

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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im in class 11 from jharkhand region ...

- 5 years, 1 month ago

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how much marks did you get

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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Writing them well is the trick.

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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Has anyone solved the 5th problem both ways ?

- 5 years, 1 month ago

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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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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I got 45 too. Expected is around 55.

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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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 $x$.

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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