RMO-2015-Rajasthan

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.

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Comments

Are 60 marks enough for selection for INMO?

Samarth Agarwal - 5 years, 6 months ago

More than enough I think. Good luck.

A Former Brilliant Member - 5 years, 6 months ago

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

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

What about 45-50?

Swapnil Das - 5 years, 6 months ago

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

@A Former Brilliant Member – May your words be true, how much are you getting?

@Swapnil Das – 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.

@A Former Brilliant Member – 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.

Shrihari B - 5 years, 6 months ago

@Shrihari B – 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?

A Former Brilliant Member - 5 years, 5 months ago

@Swapnil Das – Dont you think Algebra questions were easy this time...

What's the cutoff bro?

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

Mehul Arora - 5 years, 6 months ago

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

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

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

Harsh Shrivastava - 5 years, 6 months ago

@Harsh Shrivastava – My max is 51.

@Harsh Shrivastava – Does the cutoff vary with region?

Also see RMO-2015 Karnataka Region

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

Akshay Yadav - 5 years, 6 months ago

Which 3?

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

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

@A Former Brilliant Member – I've posted the solution of Q4. in my note

Shubhendra Singh - 5 years, 6 months ago

@Akshay Yadav – But 32 c 3 =4960

Adarsh Kumar - 5 years, 6 months ago

@Adarsh Kumar – Is the answer 216250, I didn't attempt though

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

@Adarsh Kumar – how much are u getting

Dev Sharma - 5 years, 6 months ago

@Dev Sharma – I wrote from Telangana.

@Adarsh Kumar – It is not that straight as you think. Try it with more time.

@Akshay Yadav – 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.

2 and 3 que were easy..

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

Maybe 4 marks. Even I left that case.

@A Former Brilliant Member – Hope so..

@Samarth Agarwal – Max will be 5-6

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

Siddharth Singh - 5 years, 6 months ago

When the cutoff and the results will be declared?

When will the cutoff and results be declared?

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?

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

Siddhartha Srivastava - 5 years, 6 months ago

Which 5 ?

@Harsh Shrivastava – $3 + \sqrt{\frac{k}{2}}$ for $3 \leq k \leq 7$

@Siddhartha Srivastava – Hmm, I m wrong.BTW is there step marking?

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

@Samarth Agarwal – I Left it...

@Siddhartha Srivastava – can u plzz provide a solution @Siddhartha Srivastava

Vaibhav Prasad - 5 years, 6 months ago

@Vaibhav Prasad – 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 $4 < a < 5$ , $2 < k < 8$ .

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

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 - 5 years, 6 months ago

also give a solution for Q6

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

I did half , how much can I get?

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.

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

Lol!

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

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

so finally there are 8 solutions ?

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

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

@A Former Brilliant Member – 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?

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

@A Former Brilliant Member – Yes, I too missed that, any guesses?

@Swapnil Das – How much marks may be deducted?

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

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

@Swapnil Das – Depends on your region. Check when the results of your region came out last year.

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 - 5 years, 6 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

Manisha Garg - 5 years, 6 months ago

I also gave GMO. How many did you do?

Aditya Chauhan - 5 years, 6 months ago

Can you pleas share the question paper on Brilliant?

@Akshay Yadav – Here Gmo

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

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

Devansh Shah - 5 years, 6 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.

Gyanendra Prakash - 5 years, 6 months ago

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

Mayank Jha - 5 years, 5 months ago

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.

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$