RMO-2015 for Rajasthan region was held on 06-12-2015, Sunday between 1300hrs and 1600hrs IST.

Hi guys this is the paper of RMO-2015 that I have given from Ajmer,Rajasthan.

Please do post solutions and enjoy.

\(1\) Let \(ABC\) be a triangle. Let \(B'\) and \(C'\) denote respectively the reflection of \(B\) and \(C\) in the internal bisector of \(\angle A\). Show that the triangle \(ABC\) and \(AB'C'\) have same incentre.

\(2\) Let \(P(x)= x^{2}+ax+b\) be a quadratic polynomial with real coefficients. Suppose there are real numbers \(s≠t\) such that \(P(s)= t\) and \(P(t)= s\). Prove that \(b-st\) is a root of equation \(x^{2}+ax+b-st=0\).

\(3\) Find all integers \(a,b,c\) such that \(a^{2}=bc+1, b^{2}=ca+1\).

\(4\) Suppose 32 objects are placed along a circle at equal distances. In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite?

\(5\) Two circles \(G_{1}\) and \(G_{2}\) in thea plane intersect at two points \(A\) and \(B\), and the centre of \(G_{2}\) lies on \(G_{1}\). Let \(C\) and \(D\) be on \(G_{1}\) and \(G_{2}\), respectively, such that \(C\),\(B\) and \(D\) are collinear. Let \(E\) on \(G_{2}\) be such that \(DE\) is parallel to \(AC\). Show that \(AE=AB\).

\(6\) Find all real numbers \(a\) such that \(4<a<5\) and \(a(a-3\){\(a\)}\()\) is an integer. (Here \({a}\) denotes fractional part of \(a\). For example {\(1.5\)}= \(0.5\); {\(-3.4\)}=\(0.6\))

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## Comments

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TopNewestAre 60 marks enough for selection for INMO?

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More than enough I think. Good luck.

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What about 45-50?

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I think the cut off will be high......This year paper was very easy.

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When the cutoff and the results will be declared?

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2 and 3 que were easy..

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I left the case a=b in third one and got 6 solutions ...how much would be penalised??

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Can you please tell what extra solutions (excluding 6) will the case a=b give:)

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Maybe 4 marks. Even I left that case.

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I have done three but I am sure about only two. Please do post your solutions.

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Which 3?

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2,3 and 4 , however I am not sure about 4. I got its answer as 21696. Please solve that one.

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Also see RMO-2015 Karnataka Region

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What's the cutoff bro?

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Swapnil, I don't think they would've declared the cutoff yet. Though it is generally around 3 questions out of 6.

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It is not number of questions but total marks scored I think.... How much are u guyz scoring?

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Can it be lower than it, as our teacher was telling it is 2 here?

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Can we use coordinate geometry in q1 to prove that both triangles have same incentre

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well..can anyone tell me that have i solved this question : prove that the roots of the equation x^3 - 3x^2 - 1 = 0 are never rational. correctly? my solution is like this:-

well i approached like this:- let the roots be a,b,c a+b+c=3 ab+bc+ac=0 abc=1 assuming roots to be rational..i took a=p1/q1,b as p2/q2 and c as p3/q3

so i got p1/q1+p2/q2+p3/q3=3---------eq.1 p1p2/q1q2 + p2p3/q2/q3 + p1p3/q1q3 = 0-----eq.2 and p1p2p3=q1q2q3 -----------eq-3

proceeding with equation 1

i got after expanding and replacing q1q2q3 by p1p2p3...

reciprocal of eq.2=3 (after three steps of monotonous algebraic expansion)

taking eq2 as x+y+z = 0 and then its reciprocal from above as 1/x+x/y+1/z = 3

by A.M-G.M we know that (x+y+z)(1/x+1/y+1/z)>=9

but here we are getting it as 3*0=0

therefore by contradiction roots can't be rational... THIS APPEARED IN JHARKHAND RMO 2015.

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I gave CBSE Group Mathematics Olympiad today instead of rmo, the first and last questions are same as in rmo! Second level of both these exams is INMO

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I also gave GMO. How many did you do?

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I m sure about 3 questions.... 2 are wrong......and in one I have a doubt

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I made a silly mistake in the third question and I didn't do no.5, rest was ok

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Can you pleas share the question paper on Brilliant?

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Gmo

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Solution to question 1 The incenter of triangle ABC will lie on the angle bisector of angle A itself. Now when we reflect the triangle ABC to AB'C' about the internal angle bisector of A then we also reflect its incenter, circumcenter etc. But the reflection of incenter will be the original incenter itself as it lies on the mirror. It can be also easily proved that B' and C' lie on AC and AB respectively. We join B and B'. Let the point of intersection of BB' and internal angle bisector be D. Then AD =AD and BD=B'D and angles ADB and ADB' are 90 each.

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Answer to question 3.\[a^{2}-b^{2}=c(b-a)\] This implies \[a+b+c=0 ...or... a=b\]

Case 1\[a+b+c=0\] Solutions are \[(a,b,c)=(1,0,-1),(-1,0,1),(0,1,-1),(0,-1,1),(1,1,0),(1,-1,0),(-1,1,0),(-1,-1,0).\]Case 2\[a=b \] Therefore \[(a,b,c)=(1,1,0),(-1,-1,0).\].Log in to reply

How do you get the solutions in Case I? Why can't there be more solutions?

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Hi, do you have any idea when cutoff will be declared?

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I mentioned that a,b,c belongs to set 1,-1,0 rather than mentioning solutions, how much can I get?

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You should get full marks if you mentioned both the cases.

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so finally there are 8 solutions ?

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Please tell the answer to 4th one...It is most doubtful....

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Actually cutoff was already declared for RMO for their respective classes. 2 questions for class 9, 3 for 10 and 4 for 11.This is in odisha.

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What about for class 8? Hope it is 1 question :P

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Lol!

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Answer to question 2.\[P(s)=s^{2}+as+b=t,P(t)=t^{2}+at+b=s\] \[P(s)-P(t)=s^{2}-t^{2}+a(s-t)=t-s\] \[P(s)-P(t)=s+t+a+1=0\] \[P(s)P(t)=(st)^{2}+ast^{2}+bt^{2}+bs^{2}+bas+b^{2}+as^{2}t+a^{2}st+abt=st\] \[(st)^{2}+b^{2}-st=-(ast^{2}+as^{2}t+a^{2}st+bt^{2}+abt+bs^{2}+bas)\] \[(st)^{2}+b^{2}-st=-(ast(a+s+t)+bt(a+t)+bs(a+s))\] \[(st)^{2}+b^{2}-st=(ast(1)+bt(1+s)+bs(1+t))\] \[(st)^{2}+b^{2}-st=(ast+b(s+t)\] \[(st)^{2}+b^{2}-st=(ast-b(1+a)\] \[(st)^{2}+b^{2}-st-ast+b+ab=0\] But \[P(b-st)-st=(st)^{2}+b^{2}-st-ast+b+ab\] Therefore \[P(b-st)-st=0\] Hence proved.

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I did half , how much can I get?

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also give a solution for Q6

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\(Solution\quad to\quad Ques\quad 6:\\ let\quad a=4+f\\ 0<f<1\\ f\neq 0\quad so\quad that\quad a>4.\\ (4+f)(4+f-3f)=Integer\\ (4+f)(4-2f)=Int.\\ 16-4f-2{ f }^{ 2 }=Ingeter\\ 2{ f }^{ 2 }+4f=k\\ where\quad k\quad is\quad integer.\\ 2{ f }^{ 2 }+4f-k=0\\ BY\quad Shir\quad Dharacharya\quad method:\\ f=\frac { -4\pm \sqrt { 16+8k } }{ 4 } \\ f>0\\ \therefore \quad f=\frac { -4+\sqrt { 16+8k } }{ 4 } \\ 0<f<1\\ 0<\frac { -4+\sqrt { 16+8k } }{ 4 } <1\\ 0<-4+\sqrt { 16+8k } <4\\ \\ 4<\sqrt { 16+8k } <8\\ 16<16+8k<64\\ 0<8k<48\\ 0<k<6\\ \therefore \quad k=\{ 1,2,3,4,5\} \\ On\quad solving\quad a=3+\sqrt { x } \\ where\quad x=\{ 1.5,2,2.5,3,3.5\} \)

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What is the answer to 4th....I am getting 2015!!!

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Hey guyz in question no. 6 , I made a very very silly mistake. I took 3 + sqrt(3) less than 4 ( : P) and hence concluded that only one value exists.I have showed all other steps, so how much would I get in that question?

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There are five values. How did you get only one value? And what do you mean by showed other steps?

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Which 5 ?

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@Siddhartha Srivastava

can u plzz provide a solutionLog in to reply

Therefore \( a(a - 3\{a\}) = a(12 - 2a) = -2(a - 3)^2 +36 \)

If \( -2(a - 3)^2 +36 \) is an integer, so is \( -2(a-3)^2 \) and so is \( 2(a-3)^2 \). The reverse is also true.

Therefore \( 2(a-3)^2 = k \implies a = 3 + \sqrt{\frac{k}{2}} \). Since \( 4 < a < 5 \), \( 2 < k < 8 \).

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When will the cutoff and results be declared?

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Answer to question 6. Let \[a=m+\frac{b}{c}\] where \(m\) is any integer and \[0<b<c\] . Then \[a(a-3{a})=( m+\frac{b}{c})( m-2\frac{b}{c}) \] \[m^{2}-\frac{bm}{c}+\frac{2b^{2}}{c^{2}}\]. \[m^{2}-\frac{2b^{2}-bcm}{c^{2}}\] Now, \(m\) is an integer . Let's consider \[ \frac{2b^{2}-bcm}{c^{2}}=k\] where \(k\) is an integer . After solving we get \[\frac{b}{c}=\frac{m+_{-}\sqrt{m^{2}+8k}}{4}\].....(I) But \[\frac{b}{c}<1\]....(II) Now putting value of \(\frac{b}{c}\) from (I) to (II). We get \[m+k<2\] Therefore there are infinitely many integers \(m,k\) such that \[m+k<2\]. Hence proved.

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