# 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, 8 months ago

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Are 60 marks enough for selection for INMO?

- 4 years, 8 months ago

More than enough I think. Good luck.

- 4 years, 8 months ago

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

- 4 years, 8 months 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, 8 months ago

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months 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, 8 months 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, 7 months 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, 7 months ago

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

- 4 years, 8 months ago

What's the cutoff bro?

- 4 years, 8 months 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, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

My max is 51.

- 4 years, 8 months ago

Does the cutoff vary with region?

- 4 years, 8 months ago

Also see RMO-2015 Karnataka Region

- 4 years, 8 months ago

- 4 years, 8 months ago

Which 3?

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

But 32 c 3 =4960

- 4 years, 8 months ago

Is the answer 216250, I didn't attempt though

- 4 years, 8 months ago

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

- 4 years, 8 months ago

how much are u getting

- 4 years, 8 months ago

I wrote from Telangana.

- 4 years, 8 months ago

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

- 4 years, 8 months 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, 8 months ago

2 and 3 que were easy..

- 4 years, 8 months ago

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

- 4 years, 8 months ago

Maybe 4 marks. Even I left that case.

- 4 years, 8 months ago

Hope so..

- 4 years, 8 months ago

Max will be 5-6

- 4 years, 8 months ago

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

- 4 years, 8 months ago

When the cutoff and the results will be declared?

- 4 years, 8 months ago

When will the cutoff and results be declared?

- 4 years, 8 months 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, 8 months ago

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

- 4 years, 8 months ago

Which 5 ?

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

I Left it...

- 4 years, 8 months ago

can u plzz provide a solution @Siddhartha Srivastava

- 4 years, 8 months 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, 8 months ago

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

- 4 years, 8 months 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, 8 months ago

also give a solution for Q6

- 4 years, 8 months 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, 8 months ago

I did half , how much can I get?

- 4 years, 8 months 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, 8 months ago

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

- 4 years, 8 months ago

Lol!

- 4 years, 8 months ago

- 4 years, 8 months 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, 8 months ago

so finally there are 8 solutions ?

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months 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, 8 months ago

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

- 4 years, 8 months ago

Yes, I too missed that, any guesses?

- 4 years, 8 months ago

How much marks may be deducted?

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months 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, 8 months 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, 8 months ago

I also gave GMO. How many did you do?

- 4 years, 8 months ago

Can you pleas share the question paper on Brilliant?

- 4 years, 8 months ago

Here Gmo

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months 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, 8 months ago

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

- 4 years, 7 months 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, 8 months ago