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
3 years, 10 months ago

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Q1) If

n=1m=1n1m(n+100100)=pq\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 pp and qq. Find p+qp+q.

Bonus:- Remove 100 and generalise it for some positive integer kk.

Q2) Prove that the fraction 21n+414n+3 \dfrac{21n+4}{14n+3} is irreducible for any natural number nn.

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>1n>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 ABCDABCD intersect at point PP. Point QQ lies between the parallel lines BCBC and ADAD such that AQD=CQB\angle AQD = \angle CQB, and line CDCD separates points PP and QQ. Prove that BQP=DAQ\angle BQP = \angle DAQ.

Q6) Let ABCDEABCDE be a convex pentagon such that BCAEBC \parallel AE, AB=BC+AE AB=BC+AE, and ABC=CDE\angle ABC = \angle CDE. Let MM be the midpoint of CECE, and let OO be the circumcenter of triangle BCDBCD. Given that DMO=90oDMO=90^{o}, prove that 2BDA=CDE.2\angle BDA=\angle CDE.

Department 8 - 3 years, 9 months ago

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2) 3(14n+3)2(21n+4)=13(14n + 3) - 2(21n + 4) = 1 this is bezout lemma so proved.

Dev Sharma - 3 years, 9 months ago

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

Dev Sharma - 3 years, 9 months ago

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

21n+414n+3=1+7n+114n+3=1+12[1114n+3]\frac{21n+4}{14n+3}=1+\frac{7n+1}{14n+3}=1+\frac{1}{2}[1-\frac{1}{14n+3}]

Rest depends upon you.

Department 8 - 3 years, 9 months ago

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@Department 8 gcd(21n+4,14n+3)=gcd(14n+3,7n+1)=gcd(7n+1,7n+2)=1\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 nn.

A Former Brilliant Member - 3 years, 9 months ago

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

Department 8 - 3 years, 9 months ago

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Q5) Consider a homothety HH with centre PP that sends AA to CC ( and hence BB to DD). Let HH send QQ to RR. Observe that AQD=CQB=ARD\angle AQD = \angle CQB =\angle ARD , so A,D,Q,RA,D,Q ,R are concyclic points. Therefore BQP=DRP=DRQ=DAQ\angle BQP = \angle DRP =\angle DRQ = \angle DAQ .

Shourya Pandey - 3 years, 9 months ago

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

Department 8 - 3 years, 9 months ago

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Prove that x(x+1)(x+2)(x+3)=y2x(x+1)(x+2)(x+3)=y^{2} has no solution for x,yNx,y \in N.

A Former Brilliant Member - 3 years, 9 months ago

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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 - 3 years, 9 months ago

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Your solution is perfectly correct! Nicely done!

A Former Brilliant Member - 3 years, 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 - 3 years, 10 months ago

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Ok so does anyone want a solution ? @Shourya Pandey are you solving ?

Shrihari B - 3 years, 9 months ago

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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 AD2=AFABAD^2 = AF*AB .

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

Shourya Pandey - 3 years, 9 months ago

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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 - 3 years, 9 months ago

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NiceNice oneone I have a bit simpler proof using pure geometry.

The almighty knows it all. - 2 years, 6 months ago

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@The almighty Knows It All. @Shourya Pandey Let B\angle B = θ\theta

The constructionsconstructions are clearly shown in the above diagram.diagram. In AEF\triangle AEF, Draw EIEI and FIFI the angle bisectors of AEF\angle AEF and AFE\angle AFE respectively.

BED\angle BED = DEF\angle DEF ---- GivenGiven => BED\angle BED = DEF\angle DEF = 9090^{\circ} - θ\theta => AEF\angle AEF = 2θ2\theta => AEI\angle AEI = θ\theta

=>EID=A/2+θ\angle EID = \angle A/2 +\theta But also, EIF\angle EIF = 90+A/290+\angle A/2 Therefore: FID=EIFEID\angle FID = \angle EIF - \angle EID =9090^{\circ} - θ.\theta.

So, DEF\angle DEF = FID\angle FID [= 90θ90 -\theta ] => Quadrilateral EIFDEIFD is a cyclic quad. => FDI\angle FDI = FEI\angle FEI = θ\theta = B\angle B. => ADF=FDI=θ\angle ADF = \angle FDI = \theta

Clearly,, ADC=ABD+BAD\angle ADC = \angle ABD + \angle BAD
=> ADF+FDC=ABD+BAD\angle ADF + \angle FDC = \angle ABD + \angle BAD
=> θ+FDC=θ+BAD\theta + \angle FDC = \theta + \angle BAD => FDC=BAD\angle FDC = \angle BAD

K.I.P.K.I.G.K.I.P.K.I.G.

The almighty knows it all. - 2 years, 6 months 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 - 3 years, 9 months ago

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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 - 3 years, 9 months ago

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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 - 3 years, 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 - 3 years, 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 - 3 years, 9 months ago

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If a1,a2,,anZa_1,a_2,\ldots,a_n \in Z are distinct, prove that (xa1)(xa2)(xan)1(x-a_1)(x-a_2)\ldots(x-a_n)-1 is irreducible.

A Former Brilliant Member - 3 years, 9 months ago

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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 - 3 years, 9 months ago

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Yes

A Former Brilliant Member - 3 years, 9 months ago

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@A Former Brilliant Member Sorry I didn't get u. That "yes" was for what ? c in Z or c in R ?

Shrihari B - 3 years, 9 months ago

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@Shrihari B c in Z

A Former Brilliant Member - 3 years, 9 months ago

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@A Former Brilliant Member Ok thanks. U have got a really nice set of problems....

Shrihari B - 3 years, 9 months ago

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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 - 3 years, 9 months ago

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Prove that there are infinitely many powers of 22 in the sequence n2\lfloor n \sqrt{2} \rfloor

A Former Brilliant Member - 3 years, 9 months ago

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Set n=2x2 n = 2^{x} \sqrt{2} for some positive integer x.

Harsh Shrivastava - 3 years, 9 months ago

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

Shrihari B - 3 years, 9 months ago

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@Shrihari B Yes maybe.

Harsh Shrivastava - 3 years, 9 months ago

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Hey Svatejas ... can u post solutions of previous questions and post some new questions too ?

Shrihari B - 3 years, 9 months ago

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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?

A Former Brilliant Member - 3 years, 9 months ago

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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 - 3 years, 9 months ago

Prove that y2=x3+7y^2=x^3+7 has no integral solutions.

A Former Brilliant Member - 3 years, 9 months 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 - 3 years, 9 months ago

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Any news about result of RMO of rajasthan region

Aakash Khandelwal - 3 years, 10 months ago

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You can check it on hbcse site.

Adarsh Kumar - 3 years, 10 months ago

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Someone please post some good geometry problem. Not from the past RMO or INMO papers please !

Shrihari B - 3 years, 10 months ago

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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/)

A Former Brilliant Member - 3 years, 9 months ago

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Hint needed ? Or do you guys want to try more ?

Shrihari B - 3 years, 10 months ago

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@Nihar Mahajan Your geometry is excellent. Could you post some geometry problem ? Not from previous papers

Shrihari B - 3 years, 10 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 - 3 years, 10 months ago

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I believe the answer is no such n exist for n>6

Department 8 - 3 years, 9 months ago

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7!+1=7127! +1 = 71^2 . It is the only solution known till date (other than 25=4!+1 25 = 4! +1 and 121=5!+1121= 5! +1, but here Rajdeep asked for solutions where n>6n>6).

Shourya Pandey - 3 years, 9 months ago

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@Shourya Pandey Thanks for the solution, but I think I went some where wrong in my solution.

Department 8 - 3 years, 9 months ago

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This is the Brocard's problem, and is still open.

Shourya Pandey - 3 years, 9 months ago

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Ok, Thanks

Rajdeep Dhingra - 3 years, 9 months 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 - 3 years, 9 months ago

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

Shrihari B - 3 years, 9 months ago

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Thanks! I will be posting the sets by next 5 days.

Priyanshu Mishra - 3 years, 9 months 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 x2+y2+z2=x3+y3+z3x^{2}+y^{2}+z^{2}=x^{3}+y^{3}+z^{3}.

A Former Brilliant Member - 3 years, 9 months ago

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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 - 3 years, 9 months ago

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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 x3+y4=z31x^{3}+y^{4}=z^{31}. One can simply substitute yy as 00 and xx as n31n^{31} and zz as n3n^{3} where nn is an integer. Similar substitutions of setting one variable to 00 also works.

A Former Brilliant Member - 3 years, 9 months ago

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@A Former Brilliant Member Ok thanks for that.

Shrihari B - 3 years, 9 months ago

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

Shrihari B - 3 years, 9 months ago

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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!

A Former Brilliant Member - 3 years, 9 months ago

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@A Former Brilliant Member Sure thanks !

Shrihari B - 3 years, 9 months ago

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a,bNa,b \in N are such that a+1b+b+1aN\dfrac{a+1}{b}+\dfrac{b+1}{a} \in N. Let d=gcd(a,b)d=\gcd(a,b). Prove that d2a+bd^{2} \le a+b

A Former Brilliant Member - 3 years, 9 months ago

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Does there exist a solution for the equation x2+y3=z4x^{2}+y^{3}=z^{4} for prime x,y,zx,y,z?

A Former Brilliant Member - 3 years, 9 months ago

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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 - 3 years, 9 months ago

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perfectly correct!

A Former Brilliant Member - 3 years, 9 months ago

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@A Former Brilliant Member Thanks ... I am trying the other now ..... U have got a really nice problem collection

Shrihari B - 3 years, 9 months ago

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@Shrihari B I only have some of the books recommended for the olympiads. How many books do you have?

A Former Brilliant Member - 3 years, 9 months ago

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@A Former Brilliant Member 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 - 3 years, 9 months ago

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@Shrihari B and BTW, Happy New Year!

A Former Brilliant Member - 3 years, 9 months ago

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@A Former Brilliant Member Oh yea .... I completely forgot about that .. HAPPY NEW YEAR !!!

Shrihari B - 3 years, 9 months ago

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Find all polynomials pp such that p(x+1)=p(x)+2x+1p(x+1)=p(x)+2x+1.

A Former Brilliant Member - 3 years, 9 months ago

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Is x any real number in this question ?

Shrihari B - 3 years, 9 months ago

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You first choose any two positive integers a,ba,b and replace them with gcd(a,b)\text{gcd(a,b)} and lcm(a,b) \text{lcm(a,b)}. Prove that eventually, the numbers will stop changing.

A Former Brilliant Member - 3 years, 9 months ago

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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 - 3 years, 9 months ago

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

A Former Brilliant Member - 3 years, 9 months ago

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@A Former Brilliant Member 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 - 3 years, 9 months 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.

A Former Brilliant Member - 3 years, 9 months 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 - 3 years, 9 months 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=a2+b2+c2n=a^2+b^2+c^2, then n2=x2+y2+z2n^2=x^2+y^2+z^2 where n,a,b,c,x,y,zNn,a,b,c,x,y,z \in N.

A Former Brilliant Member - 3 years, 9 months ago

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If n3n \ge 3, then prove that 2n2^n can be represented in the form 7x2+y27x^2+y^2 where x and y are odd. (very hard)

A Former Brilliant Member - 3 years, 9 months ago

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Both these ques are ditto from problem solving strategies by arthur Engel

himanshu singh - 3 years, 9 months ago

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If q=a2+b2ab+1q=\dfrac{a^2+b^2}{ab+1} where a,b,qZa,b,q \in Z, prove that qq is a perfect square. (very hard)

A Former Brilliant Member - 3 years, 9 months ago

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Counterexample : a=4,b=3 a = 4, b = 3 .

Karthik Venkata - 3 years, 9 months ago

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q must be an integer

A Former Brilliant Member - 3 years, 9 months ago

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@A Former Brilliant Member Then please edit the question.

Karthik Venkata - 3 years, 9 months ago

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@Karthik Venkata Sorry I put Q for Z

A Former Brilliant Member - 3 years, 9 months ago

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This problem must be read as this expression is a perfect square if it is an integer

Bimit Mandal - 1 year, 12 months ago

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Prove that the equation x2+y2+z2=2xyzx^2+y^2+z^2=2xyz has no integral solution except x=y=z=0x=y=z=0.

Extension Prove that the equation x2+y2+z2=kxyzx^2+y^2+z^2=kxyz has infinitely many solutions only for k=1k=1 and k=3k=3.

A Former Brilliant Member - 3 years, 9 months ago

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A simple use of fermat s infinite descent

himanshu singh - 3 years, 9 months ago

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Same problem is in MATHEMATICAL OLYMPIAD CHALLENGES by TITU ANDREESCU.

Priyanshu Mishra - 3 years, 9 months ago

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If nNn \in N and 4n+2n+14^{n}+2^{n}+1 is prime, prove that nn is a power of 33.

A Former Brilliant Member - 3 years, 9 months ago

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

Department 8 - 3 years, 9 months ago

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Thanks. I will take a look at it.

A Former Brilliant Member - 3 years, 9 months 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 - 3 years, 8 months ago

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It seems that u ve done a typographical error the ans seems trivial

himanshu singh - 3 years, 8 months ago

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there is no error

Hemant Kumae - 3 years, 8 months ago

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