# 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 and $a(a-3${$a$}$)$ is an integer. (Here ${a}$ denotes fractional part of $a$. For example {$1.5$}= $0.5$; {$-3.4$}=$0.6$) 4 years, 1 month ago

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Are 60 marks enough for selection for INMO?

- 4 years, 1 month ago

More than enough I think. Good luck.

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

- 4 years, 1 month ago

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

- 4 years, 1 month ago

May your words be true, how much are you getting?

- 4 years, 1 month 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.

- 4 years, 1 month 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.

- 4 years, 1 month 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?

- 4 years, 1 month ago

Dont you think Algebra questions were easy this time...

- 4 years, 1 month ago

What's the cutoff bro?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

My max is 51.

- 4 years, 1 month ago

Does the cutoff vary with region?

- 4 years, 1 month ago

Also see RMO-2015 Karnataka Region

- 4 years, 1 month ago

- 4 years, 1 month ago

Which 3?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

I've posted the solution of Q4. in my note

- 4 years, 1 month ago

But 32 c 3 =4960

- 4 years, 1 month ago

Is the answer 216250, I didn't attempt though

- 4 years, 1 month ago

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

- 4 years, 1 month ago

how much are u getting

- 4 years, 1 month ago

I wrote from Telangana.

- 4 years, 1 month ago

It is not that straight as you think. Try it with more time.

- 4 years, 1 month 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.

- 4 years, 1 month ago

2 and 3 que were easy..

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Maybe 4 marks. Even I left that case.

- 4 years, 1 month ago

Hope so..

- 4 years, 1 month ago

Max will be 5-6

- 4 years, 1 month ago

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

- 4 years, 1 month ago

When the cutoff and the results will be declared?

- 4 years, 1 month ago

When will the cutoff and results be declared?

- 4 years, 1 month ago

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?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Which 5 ?

- 4 years, 1 month ago

$3 + \sqrt{\frac{k}{2}}$ for $3 \leq k \leq 7$

- 4 years, 1 month ago

Hmm, I m wrong.BTW is there step marking?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

I Left it...

- 4 years, 1 month ago

can u plzz provide a solution @Siddhartha Srivastava

- 4 years, 1 month ago

Since $4 < a < 5$, we have $a = 4 + \{ a \}$.

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 $4 < a < 5$, $2 < k < 8$.

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

also give a solution for Q6

- 4 years, 1 month ago

$Solution\quad to\quad Ques\quad 6:\\ let\quad a=4+f\\ 04.\\ (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

- 4 years, 1 month ago

I did half , how much can I get?

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Lol!

- 4 years, 1 month ago

- 4 years, 1 month ago

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 $(a,b,c)=(1,1,0),(-1,-1,0).$.

- 4 years, 1 month ago

so finally there are 8 solutions ?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month 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?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Yes, I too missed that, any guesses?

- 4 years, 1 month ago

How much marks may be deducted?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Depends on your region. Check when the results of your region came out last year.

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

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

- 4 years, 1 month ago

I also gave GMO. How many did you do?

- 4 years, 1 month ago

Can you pleas share the question paper on Brilliant?

- 4 years, 1 month ago

Here Gmo

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

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

- 4 years ago

Answer to question 6. Let $a=m+\frac{b}{c}$ where $m$ is any integer and $0 . 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.

- 4 years, 1 month ago