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

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 {\(1.5\)}= \(0.5\); {\(-3.4\)}=\(0.6\))

Please do reshare and post your views about the paper.

## Comments

Sort by:

TopNewestAre 60 marks enough for selection for INMO? – Samarth Agarwal · 9 months, 4 weeks ago

Log in to reply

– Svatejas Shivakumar · 9 months, 4 weeks ago

More than enough I think. Good luck.Log in to reply

– Swapnil Das · 9 months, 4 weeks ago

What about 45-50?Log in to reply

– Chinmay Sangawadekar · 9 months, 4 weeks ago

Dont you think Algebra questions were easy this time...Log in to reply

– Svatejas Shivakumar · 9 months, 4 weeks ago

I think that too should be enough. I may be wrong as this is the first time I am writing RMO.!!Log in to reply

– Swapnil Das · 9 months, 4 weeks ago

May your words be true, how much are you getting?Log in to reply

– Svatejas Shivakumar · 9 months, 4 weeks ago

Don't know. Hopefully around 40-50 (Did many silly mistakes :( . Not a great score but this year there is some special eligibility for students of class 8 for those who don't come in top 30 or meet the required cut off so some hopes are there. Good luck to you and everyone who wrote RMO.Log in to reply

– Shrihari B · 9 months, 2 weeks ago

Are you in class 8 ? If you are in class 8 right now I bet you will be in the Indian team for IMO by the time you reach your 12th standard.Log in to reply

– Svatejas Shivakumar · 9 months, 2 weeks ago

Yes I am in 8th. Thanks for your well wishes. Wish you too all the best for getting selected in the Indian team for IMO!! Can you please tell me how you prepared for RMO?Log in to reply

– Samarth Agarwal · 9 months, 4 weeks ago

I think the cut off will be high......This year paper was very easy.Log in to reply

– Svatejas Shivakumar · 9 months, 4 weeks ago

Can you you guess what will be the cut off for my region (Karnataka)? I have posted the paper. Please check it.Log in to reply

When the cutoff and the results will be declared? – Siddharth Singh · 9 months, 4 weeks ago

Log in to reply

2 and 3 que were easy.. – Dev Sharma · 9 months, 4 weeks ago

Log in to reply

– Samarth Agarwal · 9 months, 4 weeks ago

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

– Siddharth Singh · 9 months, 4 weeks ago

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

– Svatejas Shivakumar · 9 months, 4 weeks ago

Maybe 4 marks. Even I left that case.Log in to reply

– Samarth Agarwal · 9 months, 4 weeks ago

Hope so..Log in to reply

– Svatejas Shivakumar · 9 months, 4 weeks ago

Max will be 5-6Log in to reply

I have done three but I am sure about only two. Please do post your solutions. – Akshay Yadav · 9 months, 4 weeks ago

Log in to reply

– Harsh Shrivastava · 9 months, 4 weeks ago

Which 3?Log in to reply

– Akshay Yadav · 9 months, 4 weeks ago

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

– Svatejas Shivakumar · 9 months, 4 weeks ago

Can you please tell how you got question no.4? Even I had a very similar question in my region.Log in to reply

– Shubhendra Singh · 9 months, 4 weeks ago

I've posted the solution of Q4. in my noteLog in to reply

– Adarsh Kumar · 9 months, 4 weeks ago

But 32 c 3 =4960Log in to reply

– Akshay Yadav · 9 months, 4 weeks ago

It is not that straight as you think. Try it with more time.Log in to reply

– Adarsh Kumar · 9 months, 4 weeks ago

Dude,i was trying to tell you that your answer couldn't possibly be correct.I know that the answer isn't 32 c 3.Log in to reply

– Swapnil Das · 9 months, 4 weeks ago

Is the answer 216250, I didn't attempt thoughLog in to reply

– Adarsh Kumar · 9 months, 4 weeks ago

I am sorry Swapnil but i don't think so as the number is larger than 32 c 3.Log in to reply

– Dev Sharma · 9 months, 4 weeks ago

how much are u gettingLog in to reply

– Adarsh Kumar · 9 months, 4 weeks ago

I wrote from Telangana.Log in to reply

Also see RMO-2015 Karnataka Region – Svatejas Shivakumar · 9 months, 4 weeks ago

Log in to reply

What's the cutoff bro? – Swapnil Das · 9 months, 4 weeks ago

Log in to reply

– Mehul Arora · 9 months, 4 weeks ago

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

– Harsh Shrivastava · 9 months, 4 weeks ago

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

– Swapnil Das · 9 months, 4 weeks ago

Does the cutoff vary with region?Log in to reply

– Swapnil Das · 9 months, 4 weeks ago

My max is 51.Log in to reply

– Swapnil Das · 9 months, 4 weeks ago

Can it be lower than it, as our teacher was telling it is 2 here?Log in to reply

– Mehul Arora · 9 months, 4 weeks ago

Yes it coukd be. Harsh, I'm not scoring that well. I have done 1 for sure.Log in to reply

Can we use coordinate geometry in q1 to prove that both triangles have same incentre – Mayank Jha · 9 months, 1 week ago

Log in to reply

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. – Gyanendra Prakash · 9 months, 4 weeks ago

Log in to reply

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 – Manisha Garg · 9 months, 4 weeks ago

Log in to reply

– Aditya Chauhan · 9 months, 4 weeks ago

I also gave GMO. How many did you do?Log in to reply

– Devansh Shah · 9 months, 4 weeks ago

I m sure about 3 questions.... 2 are wrong......and in one I have a doubtLog in to reply

– Manisha Garg · 9 months, 4 weeks ago

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

– Akshay Yadav · 9 months, 4 weeks ago

Can you pleas share the question paper on Brilliant?Log in to reply

Gmo – Aditya Chauhan · 9 months, 4 weeks ago

HereLog in to reply

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. – Pranav Rao · 9 months, 4 weeks ago

Log in to reply

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 \[(a,b,c)=(1,1,0),(-1,-1,0).\]. – Shivam Jadhav · 9 months, 4 weeks agoLog in to reply

– Siddhartha Srivastava · 9 months, 4 weeks ago

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

– Swapnil Das · 9 months, 4 weeks ago

Hi, do you have any idea when cutoff will be declared?Log in to reply

– Siddhartha Srivastava · 9 months, 4 weeks ago

Depends on your region. Check when the results of your region came out last year.Log in to reply

– Swapnil Das · 9 months, 4 weeks ago

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

– Svatejas Shivakumar · 9 months, 4 weeks ago

You should get full marks if you mentioned both the cases.Log in to reply

– Siddharth Singh · 9 months, 4 weeks ago

I did a blunder,I mentioned the case a=b ,but wrote since in a*a=ac+1 L.H.S is divisible by \(a\) but not the R.H.S(forgot about 1).how much marks will be deducted?Log in to reply

– Svatejas Shivakumar · 9 months, 4 weeks ago

Maybe 3-4 marks. Do you know how much marks will be deducted for missing the case a=b? Good luck.Log in to reply

– Swapnil Das · 9 months, 4 weeks ago

Yes, I too missed that, any guesses?Log in to reply

– Swapnil Das · 9 months, 4 weeks ago

How much marks may be deducted?Log in to reply

– Vaibhav Prasad · 9 months, 4 weeks ago

so finally there are 8 solutions ?Log in to reply

Please tell the answer to 4th one...It is most doubtful.... – Samarth Agarwal · 9 months, 4 weeks ago

Log in to reply

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. – Swapnil Das · 9 months, 4 weeks ago

Log in to reply

– Svatejas Shivakumar · 9 months, 4 weeks ago

What about for class 8? Hope it is 1 question :PLog in to reply

– Swapnil Das · 9 months, 4 weeks ago

Lol!Log in to reply

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. – Shivam Jadhav · 9 months, 4 weeks ago

Log in to reply

– Swapnil Das · 9 months, 4 weeks ago

I did half , how much can I get?Log in to reply

– Vaibhav Prasad · 9 months, 4 weeks ago

also give a solution for Q6Log in to reply

– Samarth Agarwal · 9 months, 4 weeks ago

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

What is the answer to 4th....I am getting 2015!!! – Samarth Agarwal · 9 months, 4 weeks ago

Log in to reply

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? – Harsh Shrivastava · 9 months, 4 weeks ago

Log in to reply

– Siddhartha Srivastava · 9 months, 4 weeks ago

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

– Harsh Shrivastava · 9 months, 4 weeks ago

Which 5 ?Log in to reply

– Siddhartha Srivastava · 9 months, 4 weeks ago

\( 3 + \sqrt{\frac{k}{2}} \) for \( 3 \leq k \leq 7 \)Log in to reply

– Harsh Shrivastava · 9 months, 4 weeks ago

Hmm, I m wrong.BTW is there step marking?Log in to reply

– Samarth Agarwal · 9 months, 4 weeks ago

I think so....Please tell the answer to 4th one I got 2015 ways!Log in to reply

– Harsh Shrivastava · 9 months, 4 weeks ago

I Left it...Log in to reply

@Siddhartha Srivastava – Vaibhav Prasad · 9 months, 4 weeks ago

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 \( 4 < a < 5 \), \( 2 < k < 8 \). – Siddhartha Srivastava · 9 months, 4 weeks ago

Log in to reply

When will the cutoff and results be declared? – Swapnil Das · 9 months, 4 weeks ago

Log in to reply

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. – Shivam Jadhav · 9 months, 4 weeks ago

Log in to reply