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INMO Practice Board 2015-16

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.

Note by Adarsh Kumar
9 months, 2 weeks ago

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Q1) 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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@Lakshya Sinha 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 \(\angle BQP = \angle DRP =\angle DRQ = \angle DAQ \). Shourya Pandey · 9 months, 1 week ago

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@Shourya Pandey Oh nice, just saw it now. I would like to talk to you here's my mail address: batmanbrucew82@gmail.com Lakshya Sinha · 9 months, 1 week ago

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@Lakshya Sinha 2) \(3(14n + 3) - 2(21n + 4) = 1\) this is bezout lemma so proved. Dev Sharma · 9 months, 1 week ago

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@Dev Sharma 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. Dev Sharma · 9 months, 1 week ago

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@Dev Sharma Let me show it to you using the basics

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

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

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@Lakshya Sinha \(\gcd(21n+4,14n+3)=\gcd(14n+3,7n+1)=\gcd(7n+1,7n+2)=1\). Hence the fraction is irreducible for all natural number \(n\). Svatejas Shivakumar · 9 months, 1 week ago

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@Dev Sharma I know that's easy but if you edit your post to make it easy for the readers to read. Lakshya Sinha · 9 months, 1 week ago

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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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@Svatejas Shivakumar Maybe I got the solution for this question .... please check though as I am quite bad at proof writing. My solution is as follows:

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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@Shrihari B Your solution is perfectly correct! Nicely done! Svatejas Shivakumar · 8 months, 4 weeks ago

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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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@Shrihari B 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? Svatejas Shivakumar · 8 months, 3 weeks ago

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@Svatejas Shivakumar 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) Shrihari B · 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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@Svatejas Shivakumar Set \( n = 2^{x} \sqrt{2} \) for some positive integer x. Harsh Shrivastava · 8 months, 3 weeks ago

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@Harsh Shrivastava I thought about that too but I guess then the question is trivial. I guess he meant that n is natural no. Shrihari B · 8 months, 3 weeks ago

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@Shrihari B Yes maybe. Harsh Shrivastava · 8 months, 3 weeks ago

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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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@Svatejas Shivakumar Hey I have also attempted this question. Please check my solution.....

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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@Svatejas Shivakumar 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 ? Shrihari B · 8 months, 4 weeks ago

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@Shrihari B Yes Svatejas Shivakumar · 8 months, 4 weeks ago

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@Svatejas Shivakumar Sorry I didn't get u. That "yes" was for what ? c in Z or c in R ? Shrihari B · 8 months, 4 weeks ago

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@Shrihari B c in Z Svatejas Shivakumar · 8 months, 4 weeks ago

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@Svatejas Shivakumar Ok thanks. U have got a really nice set of problems.... Shrihari B · 8 months, 4 weeks ago

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@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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@Shrihari B 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). Shourya Pandey · 9 months, 1 week ago

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@Shourya Pandey Shourya,is it enough if to qualify INMO if have completed these books sufficiently ?:

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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@Priyanshu Mishra Wow. You may even make it to the IMO, if you have done ALL of that. Shourya Pandey · 9 months ago

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@Shourya Pandey Ok, then it will be my target and by next year (in class 11th) i will try to complete all these books and then go for RMO.

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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@Shrihari B I have a trigonometry solution that is simple. I'll just post a brief outline.

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

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

Shourya Pandey · 9 months, 1 week ago

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@Shourya Pandey 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. :) Shrihari B · 9 months, 1 week ago

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@Shrihari B Ok so does anyone want a solution ? @Shourya Pandey are you solving ? Shrihari B · 9 months, 1 week ago

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

Extension Prove 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 ago

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@Svatejas Shivakumar Same problem is in MATHEMATICAL OLYMPIAD CHALLENGES by TITU ANDREESCU. Priyanshu Mishra · 8 months, 2 weeks ago

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@Svatejas Shivakumar A simple use of fermat s infinite descent Himanshu Singh · 8 months, 2 weeks ago

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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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@Svatejas Shivakumar Counterexample : \( a = 4, b = 3 \). Karthik Venkata · 8 months, 2 weeks ago

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@Karthik Venkata q must be an integer Svatejas Shivakumar · 8 months, 2 weeks ago

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@Svatejas Shivakumar Then please edit the question. Karthik Venkata · 8 months, 2 weeks ago

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@Karthik Venkata Sorry I put Q for Z Svatejas Shivakumar · 8 months, 2 weeks ago

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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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@Svatejas Shivakumar Both these ques are ditto from problem solving strategies by arthur Engel Himanshu Singh · 8 months, 2 weeks ago

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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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@Svatejas Shivakumar 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 ? Shrihari B · 8 months, 3 weeks ago

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@Shrihari B Yes you have. I guess the problem hasn't been framed properly. You replace one with gcd and the other with lcm. Svatejas Shivakumar · 8 months, 3 weeks ago

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@Svatejas Shivakumar 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.... Shrihari B · 8 months, 3 weeks ago

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@Shrihari B 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. Svatejas Shivakumar · 8 months, 3 weeks ago

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

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@Svatejas Shivakumar Is x any real number in this question ? Shrihari B · 8 months, 3 weeks ago

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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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@Svatejas Shivakumar Ok I tried this one too.... but please check my solution.....

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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@Shrihari B perfectly correct! Svatejas Shivakumar · 8 months, 3 weeks ago

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@Svatejas Shivakumar Thanks ... I am trying the other now ..... U have got a really nice problem collection Shrihari B · 8 months, 3 weeks ago

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@Shrihari B and BTW, Happy New Year! Svatejas Shivakumar · 8 months, 3 weeks ago

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@Svatejas Shivakumar Oh yea .... I completely forgot about that .. HAPPY NEW YEAR !!! Shrihari B · 8 months, 3 weeks ago

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@Shrihari B I only have some of the books recommended for the olympiads. How many books do you have? Svatejas Shivakumar · 8 months, 3 weeks ago

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@Svatejas Shivakumar I have the following books only ... 1) Challenges and Thrills of Pre-college mathematics 2) An excursion in mathematics 3) I am ordering a book called Inequalities - An approach through problems by B.J.Venkatachala

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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@Svatejas Shivakumar Thanks for your contributions to the INMO practise board. I hope u will keep posting such wonderful problems and solutions ... Shrihari B · 8 months, 3 weeks ago

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@Shrihari B 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! Svatejas Shivakumar · 8 months, 3 weeks ago

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@Svatejas Shivakumar Sure thanks ! Shrihari B · 8 months, 3 weeks ago

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@Svatejas Shivakumar Maybe I got this question. But I don't know if such an approach is allowed or no. So please check.My solution is as follows :

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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@Shrihari B 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. Svatejas Shivakumar · 8 months, 3 weeks ago

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@Svatejas Shivakumar Ok thanks for that. Shrihari B · 8 months, 3 weeks ago

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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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@Priyanshu Mishra Loved your problem set and i am dying to hear about geometry and algebra problem sets too :) Shrihari B · 9 months ago

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@Shrihari B Thanks! I will be posting the sets by next 5 days. Priyanshu Mishra · 9 months ago

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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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@Rajdeep Dhingra This is the Brocard's problem, and is still open. Shourya Pandey · 9 months, 1 week ago

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@Shourya Pandey Ok, Thanks Rajdeep Dhingra · 9 months, 1 week ago

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@Rajdeep Dhingra I believe the answer is no such n exist for n>6 Lakshya Sinha · 9 months, 1 week ago

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@Lakshya Sinha \(7! +1 = 71^2\) . It is the only solution known till date (other than \( 25 = 4! +1 \) and \(121= 5! +1\), but here Rajdeep asked for solutions where \(n>6\)). Shourya Pandey · 9 months, 1 week ago

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@Shourya Pandey Thanks for the solution, but I think I went some where wrong in my solution. Lakshya Sinha · 9 months, 1 week ago

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@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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@Shrihari B Check these sets

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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@Aakash Khandelwal You can check it on hbcse site. Adarsh Kumar · 9 months, 1 week ago

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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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@Hemant Kumae It seems that u ve done a typographical error the ans seems trivial Himanshu Singh · 7 months, 4 weeks ago

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@Himanshu Singh there is no error Hemant Kumae · 7 months, 4 weeks ago

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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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@Svatejas Shivakumar try the new question for proof contest (https://brilliant.org/profile/lakshya-a71u6y/sets/proof-contest/) Lakshya Sinha · 8 months, 2 weeks ago

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@Lakshya Sinha Thanks. I will take a look at it. Svatejas Shivakumar · 8 months, 2 weeks ago

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