As some of you may know,a Brilliant member,Srihari B,has qualified RMO 2015 and is now preparing himself for INMO 2016,which is to be held on the 17th of January 2016.Although RMO results have only been posted for Mumbai region,others are on their way.\[\]This note has been created solely for helping INMO 2016 participants.Let us start posting INMO level problems and help each other prepare,in case we qualify :P.\[\]Previous year's INMO participants are kindly requested to help us novices by suggesting books and posting problems.\[\]Brilliant members are requested to post relevant and useful comments or doubts.

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TopNewestQ1) If

\[\Large{\sum _{ n=1 }^{ \infty }{ \frac { \sum _{ m=1 }^{ n }{ \frac { 1 }{ m } } }{ \left( \begin{matrix} n+100 \\ 100 \end{matrix} \right) } } =\frac { p }{ q } }\]

for some relatively prime positive integers \(p\) and \(q\). Find \(p+q\).

Bonus:- Remove 100 and generalise it for some positive integer \(k\).Q2) Prove that the fraction \( \dfrac{21n+4}{14n+3} \) is irreducible for any natural number \(n\).

Q3)In a concert, 20 singers will perform. For each singer, there is a (possibly empty) set of other singers such that he wishes to perform later than all the singers from that set. Can it happen that there are exactly 2010 orders of the singers such that all their wishes are satisﬁed?

Q4) A unit square is dissected into \(n>1\) rectangles such that their sides are parallel to the sides of the square. Any line, parallel to a side of the square and intersecting its interior, also intersects the interior of some rectangle. Prove that in this dissection, there exists a rectangle having no point on the boundary of the square.

Q5) The diagonals of a trapezoid \(ABCD\) intersect at point \(P\). Point \(Q\) lies between the parallel lines \(BC\) and \(AD\) such that \(\angle AQD = \angle CQB\), and line \(CD\) separates points \(P\) and \(Q\). Prove that \(\angle BQP = \angle DAQ\).

Q6) Let \(ABCDE\) be a convex pentagon such that \(BC \parallel AE\), \( AB=BC+AE\), and \(\angle ABC = \angle CDE\). Let \(M\) be the midpoint of \(CE\), and let \(O\) be the circumcenter of triangle \(BCD\). Given that \(DMO=90^{o}\), prove that \(2\angle BDA=\angle CDE.\) – Lakshya Sinha · 9 months, 1 week ago

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– Shourya Pandey · 9 months, 1 week ago

Q5) Consider a homothety \(H\) with centre \(P\) that sends \(A \) to \(C\) ( and hence \(B\) to \(D\)). Let \(H\) send \(Q\) to \(R\). Observe that \(\angle AQD = \angle CQB =\angle ARD \), so \(A,D,Q ,R \) are concyclic points. Therefore \(\angle BQP = \angle DRP =\angle DRQ = \angle DAQ \).Log in to reply

– Lakshya Sinha · 9 months, 1 week ago

Oh nice, just saw it now. I would like to talk to you here's my mail address: batmanbrucew82@gmail.comLog in to reply

– Dev Sharma · 9 months, 1 week ago

2) \(3(14n + 3) - 2(21n + 4) = 1\) this is bezout lemma so proved.Log in to reply

– Dev Sharma · 9 months, 1 week ago

According to Bezout Lemma, if \(ax + by = 1\) then gcd(a,b) = 1. Now if gcd(a,b)=1 then its obvious that a/b is irreducible.Log in to reply

\[\frac{21n+4}{14n+3}=1+\frac{7n+1}{14n+3}=1+\frac{1}{2}[1-\frac{1}{14n+3}]\]

Rest depends upon you. – Lakshya Sinha · 9 months, 1 week ago

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– Svatejas Shivakumar · 9 months, 1 week ago

\(\gcd(21n+4,14n+3)=\gcd(14n+3,7n+1)=\gcd(7n+1,7n+2)=1\). Hence the fraction is irreducible for all natural number \(n\).Log in to reply

– Lakshya Sinha · 9 months, 1 week ago

I know that's easy but if you edit your post to make it easy for the readers to read.Log in to reply

Prove that \(x(x+1)(x+2)(x+3)=y^{2}\) has no solution for \(x,y \in N\). – Svatejas Shivakumar · 8 months, 4 weeks ago

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x(x+1)(x+2)(x+3)={x(x+3)}{(x+1)(x+2)}=(x^2+3.x)(x^2+3.x+2)=y^2. Now note that the gcd(x^2+3x , x^2+3x+2)=2 for all x in N. So x^2+3.x =2.a and x^2+3.x+2=2.b where (a,b)=1. Thus (2.a)(2.b)=y^2. And thus a.b is also a perfect square and (a,b)=1 so we obtain that a=c^2 and b=d^2. where (c,d) is also equal to 1. Thus we have 1+a^2=b^2. So a=0 and b=1 is the only solution but then x will have to be 0 and since x is a natural number this possibility is ruled out. Hence no solution. – Shrihari B · 8 months, 4 weeks ago

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– Svatejas Shivakumar · 8 months, 4 weeks ago

Your solution is perfectly correct! Nicely done!Log in to reply

Hey guys just one day to go for INMO. So i guess its the right time to discuss some basic geometrical lemmas which are strongly used in the tougher problems. So we start posting some lemmas that we know. Its better to go through them before the INMO. So The first lemma: The reflection of the orthocenter on any side of a triangle lies on the circumcircle of the triangle. Please post some useful lemmas like this everyone !!! – Shrihari B · 8 months, 1 week ago

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Hey Svatejas ... can u post solutions of previous questions and post some new questions too ? – Shrihari B · 8 months, 3 weeks ago

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– Svatejas Shivakumar · 8 months, 3 weeks ago

I will be posting the solutions soon. It will be better if you try to solve them yourself. I will also post some more problems soon. Any topic preference you have for problems?Log in to reply

– Shrihari B · 8 months, 3 weeks ago

Yea I will be trying till u post the solutions No priority as such .. but u could give a mixture of NT,Combi,Algebra and Geometry( if u want to and have some qs)Log in to reply

https://brilliant.org/discussions/thread/rmo-2015/,

https://brilliant.org/discussions/thread/rmo-board/,

https://brilliant.org/discussions/thread/rmo-board-part-2/. – Svatejas Shivakumar · 8 months, 3 weeks ago

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Prove that there are infinitely many powers of \(2\) in the sequence \(\lfloor n \sqrt{2} \rfloor\) – Svatejas Shivakumar · 8 months, 4 weeks ago

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– Harsh Shrivastava · 8 months, 3 weeks ago

Set \( n = 2^{x} \sqrt{2} \) for some positive integer x.Log in to reply

– Shrihari B · 8 months, 3 weeks ago

I thought about that too but I guess then the question is trivial. I guess he meant that n is natural no.Log in to reply

– Harsh Shrivastava · 8 months, 3 weeks ago

Yes maybe.Log in to reply

If \(a_1,a_2,\ldots,a_n \in Z\) are distinct, prove that \((x-a_1)(x-a_2)\ldots(x-a_n)-1\) is irreducible. – Svatejas Shivakumar · 8 months, 4 weeks ago

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Let us assume that the given expression is reducible. So there must be at least one factor say (x-c). Thus we have (c-a1)(c-a2)......(c-an)=1.

CASE A: Let n>2 :- Since the product of n factors is 1 and all the factors are integers, we must have each factor either 1 or -1. Now n>2 and hence by PHP there must be two factors which must be both 1 or both -1. Let those two factors be (c-ai) and (c-aj). Since both are equal hence ai=aj ... a contradiction. Hence there are no solutions for n>2.

CASE B:- n=2. So the two factors are (x-a1) and (x-a2). And their product is 1. So both must be 1 or both must be -1 which yields a1=a2 which again is a contradiction. Hence no solutions for n=2.

Thus we conclude that for all n>1 the expression is irreducible. – Shrihari B · 8 months, 4 weeks ago

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– Shrihari B · 8 months, 4 weeks ago

Hi ... can u please explain what irreducible means here ? Does it mean that it cannot have even one factor (x-c) ... where c belongs to Z .... or c could be in the R ?Log in to reply

– Svatejas Shivakumar · 8 months, 4 weeks ago

YesLog in to reply

– Shrihari B · 8 months, 4 weeks ago

Sorry I didn't get u. That "yes" was for what ? c in Z or c in R ?Log in to reply

– Svatejas Shivakumar · 8 months, 4 weeks ago

c in ZLog in to reply

– Shrihari B · 8 months, 4 weeks ago

Ok thanks. U have got a really nice set of problems....Log in to reply

@Shourya Pandey since you are an IMO bronze medalist can you please suggest how to prepare for INMO ? Basically the problem is that I don't know homothety and complicated stuff like that. I just know basics required at the RMO level. So does that suffice for INMO ? What else other than RMO basics must be known ? – Shrihari B · 9 months, 1 week ago

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– Shourya Pandey · 9 months, 1 week ago

Just a lot of problem solving will suffice for INMO. But you may keep studying stuff like homothety,etc. , side-by-side . It is really simple, and is only a formal way of writing down your intuition. You can check out the Mathematical Excalibur, which has very good solved and unsolved problems (but don't give in to the solutions before trying; this teaches you a 1000 ways NOT to approach a certain genre of problem).Log in to reply

1) Number Theory-structures, problems , examples by Titu Andreescu 2)Challenges and thrills from the pre-college mathematics-HBCE,TIFR 3)Problem Solving strategies -Arthur engel 4)Concise geometry-MAA 5)101 problems in algebra from MOSP-titu 6)IMO COMPENIUM 7)LAST 15 YEARS INMO PAPERS 8)MATHEMATICAL OLYMPIAD TREASURES-TITU 9)Problem primer for the olympiads-BJ VENKATCHALA – Priyanshu Mishra · 9 months ago

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– Shourya Pandey · 9 months ago

Wow. You may even make it to the IMO, if you have done ALL of that.Log in to reply

You can also solve number theory problems from the set "INMO 2016 PRACTICE SET-1" posted by me, if you want. It contains only 6 number theory problems. – Priyanshu Mishra · 9 months ago

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So here is a geometry problem. A triangle ABC has angle ACB > angle ABC. The internal bisector of angle BAC meets BC at D. The point E on AB is such that angle EDB=90 degrees. The point F on AC such that angle BED=angle DEF. Show that angle BAD=angle FDC – Shrihari B · 9 months, 1 week ago

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Show that the problem is equivalent to proving that \(AD^2 = AF*AB \).

\( \frac {AF}{AD} = \frac {\frac {AF}{AE}}{\frac {AD}{AE}} \), and simplify the expression, using sine rule and half-angle formulae, to \( \frac {AD}{AB} \).

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– Shrihari B · 9 months, 1 week ago

Ya that is a nice solution. This is a BMO Level 2 question. I don't know what the official solution is, but i solved this question using pure geometry as follows. Observe that D is the excentre opposite to A for triangle AFE. Hence we obtain that angles DFC and DFE are equal. Extend EF to meet BC at X. Note that triangles BED and XED are congruent. Now do some angle chasing and you are done. :)Log in to reply

@Shourya Pandey are you solving ? – Shrihari B · 9 months, 1 week ago

Ok so does anyone want a solution ?Log in to reply

Prove that the equation \(x^2+y^2+z^2=2xyz\) has no integral solution except \(x=y=z=0\).

ExtensionProve that the equation \(x^2+y^2+z^2=kxyz\) has infinitely many solutions only for \(k=1\) and \(k=3\). – Svatejas Shivakumar · 8 months, 2 weeks agoLog in to reply

– Priyanshu Mishra · 8 months, 2 weeks ago

Same problem is in MATHEMATICAL OLYMPIAD CHALLENGES by TITU ANDREESCU.Log in to reply

– Himanshu Singh · 8 months, 2 weeks ago

A simple use of fermat s infinite descentLog in to reply

If \(q=\dfrac{a^2+b^2}{ab+1}\) where \(a,b,q \in Z\), prove that \(q\) is a perfect square. (very hard) – Svatejas Shivakumar · 8 months, 2 weeks ago

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Counterexample: \( a = 4, b = 3 \). – Karthik Venkata · 8 months, 2 weeks agoLog in to reply

– Svatejas Shivakumar · 8 months, 2 weeks ago

q must be an integerLog in to reply

– Karthik Venkata · 8 months, 2 weeks ago

Then please edit the question.Log in to reply

– Svatejas Shivakumar · 8 months, 2 weeks ago

Sorry I put Q for ZLog in to reply

If \(n \ge 3\), then prove that \(2^n\) can be represented in the form \(7x^2+y^2\) where x and y are odd. (very hard) – Svatejas Shivakumar · 8 months, 2 weeks ago

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– Himanshu Singh · 8 months, 2 weeks ago

Both these ques are ditto from problem solving strategies by arthur EngelLog in to reply

Prove that if a positive integer can be expressed as a sum of three squares, then its square can also be expressed as a sum of three squares. In other words, prove that if \(n=a^2+b^2+c^2\), then \(n^2=x^2+y^2+z^2\) where \(n,a,b,c,x,y,z \in N\). – Svatejas Shivakumar · 8 months, 2 weeks ago

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Prove that \(y^2=x^3+7\) has no integral solutions. – Svatejas Shivakumar · 8 months, 2 weeks ago

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@Shourya Pandey I want a demonstration of how to solve this question using homothety..can u pls help me with a solution ? Question is as follows :

X and Y are two circles touching internally at T(X is inside Y). A chord AB of Y is tangent to X at P. Line TP meets Y again at Q. Prove that Q is the midpoint of arc AQB of Y. – Shrihari B · 8 months, 3 weeks ago

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You first choose any two positive integers \(a,b\) and replace them with \(\text{gcd(a,b)}\) and \( \text{lcm(a,b)}\). Prove that eventually, the numbers will stop changing. – Svatejas Shivakumar · 8 months, 3 weeks ago

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– Shrihari B · 8 months, 3 weeks ago

Hey in this question once we replace a,b by lcm(a,b) and gcd(a,b) then won't they be constant after that forever ? Because gcd(gcd(a,b),lcm(a,b))=gcd(a,b) and lcm(gcd(a,b),lcm(a,b))=lcm(a,b). Or am i misinterpreting the question ?Log in to reply

– Svatejas Shivakumar · 8 months, 3 weeks ago

Yes you have. I guess the problem hasn't been framed properly. You replace one with gcd and the other with lcm.Log in to reply

– Shrihari B · 8 months, 3 weeks ago

Sorry for disturbing but could u add an example for the above question ? Taking some random but nice example ? I haven't understood it properly....Log in to reply

– Svatejas Shivakumar · 8 months, 3 weeks ago

Take a as 20 and b as 16. gcd(20,16)=4 and lcm(20,16)=80. gcd(4,80)=4 and lcm(4,80)=80. This continues.Log in to reply

Find all polynomials \(p\) such that \(p(x+1)=p(x)+2x+1\). – Svatejas Shivakumar · 8 months, 3 weeks ago

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– Shrihari B · 8 months, 3 weeks ago

Is x any real number in this question ?Log in to reply

Does there exist a solution for the equation \(x^{2}+y^{3}=z^{4}\) for prime \(x,y,z\)? – Svatejas Shivakumar · 8 months, 3 weeks ago

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Let x,y,z all be odd primes. Hence LHS would be even while the RHS would be odd which is absurd. Also note that if two primes are equal then it again leads to a contradiction...check the parity and if all three are 2's then it does not yield a solution..... Hence we conclude that only one of x,y,z is 2.

CASE A: x=2..... Then we have y^3+4=z^4. We know that any prime is either 1 or -1 modulo 6. Thus y^3 is congruent to either 1 or -1 modulo 6. Thus y^3+4 is congruent to either 3 or 5 modulo 6. But note that z^4 is always congruent to 1 modulo 6... which is a contradiction. Hence x=2 is impossible.

CASE B: y=2.... In that case we can factorize the above expression to (z^2+x)(z^2-x)=8. Checking various possibilities we can show that there is no solution for this case too....

CASE C: z=2. Thus we have x^2+y^3=16. x cannot be 2 we have shown that earlier. Thus if x=3 or x=4 then there is no solution and for x>=5 the LHS exceeds the RHS.

Hence there is no solution for any primes x,y,z. – Shrihari B · 8 months, 3 weeks ago

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– Svatejas Shivakumar · 8 months, 3 weeks ago

perfectly correct!Log in to reply

– Shrihari B · 8 months, 3 weeks ago

Thanks ... I am trying the other now ..... U have got a really nice problem collectionLog in to reply

– Svatejas Shivakumar · 8 months, 3 weeks ago

and BTW, Happy New Year!Log in to reply

– Shrihari B · 8 months, 3 weeks ago

Oh yea .... I completely forgot about that .. HAPPY NEW YEAR !!!Log in to reply

– Svatejas Shivakumar · 8 months, 3 weeks ago

I only have some of the books recommended for the olympiads. How many books do you have?Log in to reply

I mostly never open these books though ... as U may have seen I am almost always online .... that's the best place to study for olympiads – Shrihari B · 8 months, 3 weeks ago

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\(a,b \in N\) are such that \(\dfrac{a+1}{b}+\dfrac{b+1}{a} \in N\). Let \(d=\gcd(a,b)\). Prove that \(d^{2} \le a+b\) – Svatejas Shivakumar · 8 months, 3 weeks ago

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This problem is quite similar to one problem of this year's RMO. Prove that there are infinitely many integer solutions to the equation \(x^{2}+y^{2}+z^{2}=x^{3}+y^{3}+z^{3}\). – Svatejas Shivakumar · 8 months, 3 weeks ago

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– Shrihari B · 8 months, 3 weeks ago

Thanks for your contributions to the INMO practise board. I hope u will keep posting such wonderful problems and solutions ...Log in to reply

– Svatejas Shivakumar · 8 months, 3 weeks ago

Check out the Brilliant Inequality Contest and the Brilliant Polynomial Roots Contest. There have been some inequality and polynomial related questions in INMO so you can expect them too!Log in to reply

– Shrihari B · 8 months, 3 weeks ago

Sure thanks !Log in to reply

We look at (x,y,z) which are in AP. Let x=a-d, y=a ,z=a+d. Now when we substitute these values in the given equation we obtain another equation given by : 3a^2 + 2d^2 = 3a^3+6.a.d^2. Thus d^2 comes out to be 3.a^2.(a-1)/2. Now a^2 is already a perfect square thus 3.(a-1)/2 must be a perfect square. Now note that a=2.3^(2n+1)+1 satisfies the need.From this we can easily find the value of d and hence x,y,z. Already we have a trivial solution (x,y,z)=(1,1,1),(1,0,1).Hence by induction we have infinite solutions for this equation. – Shrihari B · 8 months, 3 weeks ago

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– Svatejas Shivakumar · 8 months, 3 weeks ago

Your solution is correct but there is a much easier may to solve this problem. Such approaches are perfectly allowed. For example, in a RMO question for this years paper, it asked to prove that there are infinitely many integral solutions for the equation \(x^{3}+y^{4}=z^{31}\). One can simply substitute \(y\) as \(0\) and \(x\) as \(n^{31}\) and \(z\) as \(n^{3}\) where \(n\) is an integer. Similar substitutions of setting one variable to \(0\) also works.Log in to reply

– Shrihari B · 8 months, 3 weeks ago

Ok thanks for that.Log in to reply

Students selected for INMO 2016 can solve problems from the set "INMO 2016 PRACTICE SET-1" posted by me. It contains 6 number theory problems only. I will be posting other sets for different topics. – Priyanshu Mishra · 9 months ago

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– Shrihari B · 9 months ago

Loved your problem set and i am dying to hear about geometry and algebra problem sets too :)Log in to reply

– Priyanshu Mishra · 9 months ago

Thanks! I will be posting the sets by next 5 days.Log in to reply

Find the values of n>6 for which n! + 1 is a perfect square. n belongs to Integers – Rajdeep Dhingra · 9 months, 1 week ago

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– Shourya Pandey · 9 months, 1 week ago

This is the Brocard's problem, and is still open.Log in to reply

– Rajdeep Dhingra · 9 months, 1 week ago

Ok, ThanksLog in to reply

– Lakshya Sinha · 9 months, 1 week ago

I believe the answer is no such n exist for n>6Log in to reply

– Shourya Pandey · 9 months, 1 week ago

\(7! +1 = 71^2\) . It is the only solution known till date (other than \( 25 = 4! +1 \) and \(121= 5! +1\), but here Rajdeep asked for solutions where \(n>6\)).Log in to reply

– Lakshya Sinha · 9 months, 1 week ago

Thanks for the solution, but I think I went some where wrong in my solution.Log in to reply

@Nihar Mahajan Your geometry is excellent. Could you post some geometry problem ? Not from previous papers – Shrihari B · 9 months, 1 week ago

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Hint needed ? Or do you guys want to try more ? – Shrihari B · 9 months, 1 week ago

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Someone please post some good geometry problem. Not from the past RMO or INMO papers please ! – Shrihari B · 9 months, 1 week ago

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https://brilliant.org/profile/xuming-66nh7b/sets/my-geometry-challenges/

[https://brilliant.org/profile/xuming-66nh7b/sets/my-geometry-proof-challenges/] (https://brilliant.org/profile/xuming-66nh7b/sets/my-geometry-proof-challenges/) – Svatejas Shivakumar · 8 months, 2 weeks ago

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Any news about result of RMO of rajasthan region – Aakash Khandelwal · 9 months, 1 week ago

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– Adarsh Kumar · 9 months, 1 week ago

You can check it on hbcse site.Log in to reply

Question:Consider two circles S and R.Let the center of circle S lie on R.Let S and R intersect at A and B.Let C be a point on S such that AB=AC.Then prove that the point of intersection of AC and R lies in or on S. – Hemant Kumae · 7 months, 4 weeks ago

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– Himanshu Singh · 7 months, 4 weeks ago

It seems that u ve done a typographical error the ans seems trivialLog in to reply

– Hemant Kumae · 7 months, 4 weeks ago

there is no errorLog in to reply

If \(n \in N\) and \(4^{n}+2^{n}+1\) is prime, prove that \(n\) is a power of \(3\). – Svatejas Shivakumar · 8 months, 4 weeks ago

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– Lakshya Sinha · 8 months, 2 weeks ago

try the new question for proof contest (https://brilliant.org/profile/lakshya-a71u6y/sets/proof-contest/)Log in to reply

– Svatejas Shivakumar · 8 months, 2 weeks ago

Thanks. I will take a look at it.Log in to reply