**Time - 3hrs.**

**Instructions :**

Calculators (in any form) and protector are not allowed.

Rulers and compasses are allowed

All questions carry equal marks. Maximum marks : 102.

Let

**ABC**be a triangle. Let**B'**and**C'**denote respectively the reflection of**B**and**C**in the internal angle bisected of \(\angle A \). Show that the**ABC**and**AB'C'**have the same incentre.Let \(P(x)=x^{2}+ax+b \) be a quadratic polynomial with real coefficients. Suppose there are real numbers \(s \neq t\) such that \(P(s) =t\) and \(P(t) =s\). Prove that \(b-st\) is a root of the equation \(x^{2}+ax+b-st=0\)

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

Suppose

**32**objects placed along a circle at equal distances. In how many ways can**3**objects be chosen from among them so that no two of three chosen objects are adjacent nor diametrically opposite ?Two circles \(T\) and \(\sum\) in the plane intersect at two distinct points

**A**and**B**, and the center of \(\sum\) lies on \(T\). Let points C and D on \(T\) and \(\sum\) respectively such that**C**,**B**and**D**are co-linear. Let point**E**be on \(\sum\) such that**DE**is parallel to**AC**. Show that AE=AB.Find all real numbers \(a\) such that \(4<a<5\) and \(a(a-3\{a\}) \) is an integer.

{Here \(\{a \} \) represents fractional part of \(a\)}.

## Comments

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TopNewestAnswer 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 · 12 months agoLog in to reply

– Anik Mandal · 11 months, 2 weeks ago

Why are solutions like \((2,-1,-1)\) not included?Log in to reply

– Anik Mandal · 11 months, 1 week ago

Maybe it's not but how do we know that these are the only solutions?Log in to reply

This gives that the equation will have a solution when \(a-b=0 \) or \(a+b=-c\). That's how we get to know about the solutions we are gonna get. – Shubhendra Singh · 11 months, 1 week ago

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– Anik Mandal · 11 months, 1 week ago

It means all the solutions of \(a+b+c = 0\) are solutions?Log in to reply

– Shubhendra Singh · 11 months, 1 week ago

No, this is just a condition between a, b, c that will help us in getting the solutions.Log in to reply

– Shubhendra Singh · 11 months, 2 weeks ago

I think it's not a solution, please check it again.Log in to reply

– Adarsh Kumar · 11 months, 2 weeks ago

What do you think the cut off would be.of GMO 2015 paper?Log in to reply

– Shubhendra Singh · 11 months, 2 weeks ago

I think it's gonna be around 50-55, the results could be declared any time as the result of rmo have been declared. How much are you accepting??Log in to reply

– Kushagra Sahni · 11 months, 1 week ago

Did you get selected?Log in to reply

– Shubhendra Singh · 11 months, 1 week ago

Yes 8-)Log in to reply

– Kushagra Sahni · 11 months, 1 week ago

What About Dev Sharma?Log in to reply

– Dev Sharma · 11 months, 1 week ago

nope and u?Log in to reply

– Chirayu Bhardwaj · 8 months, 1 week ago

Dev you are in which class ??Log in to reply

– Kushagra Sahni · 11 months, 1 week ago

I also didn't get selected. I did 4 but my writing was horrible.Log in to reply

– Dev Sharma · 11 months, 1 week ago

Would it be sahni at place of juneja!!Log in to reply

– Kushagra Sahni · 11 months, 1 week ago

I wish there was, Next year there will be Sahni and no Juneja. Interestingly my mother's surname was Juneja before marriage.Log in to reply

– Harsh Shrivastava · 11 months, 1 week ago

How many marks did you got? And Congrats!Log in to reply

– Shubhendra Singh · 11 months, 1 week ago

I don't know how much I have scored, how did you get to know about your marks.Log in to reply

– Harsh Shrivastava · 11 months, 1 week ago

Its written in the result list of my state.Log in to reply

– Shubhendra Singh · 11 months, 1 week ago

There's nothing like this in rajasthan's list, even the cut-off is not mentioned.Log in to reply

@Harsh Shrivastava – Chinmay Sangawadekar · 11 months, 1 week ago

Are you selected for INMOLog in to reply

– Harsh Shrivastava · 11 months, 1 week ago

Luckily yes.Log in to reply

@Harsh Shrivastava – Chinmay Sangawadekar · 11 months, 1 week ago

Why luckily ? you are intelligent bro . Unfortunately , our principal doesn't allow us for RMO.Log in to reply

– Harsh Shrivastava · 11 months, 1 week ago

Why?Log in to reply

– Chinmay Sangawadekar · 11 months, 1 week ago

Dont know . H e thinks we should focus on our school math and so ,Log in to reply

– Mayank Chaturvedi · 11 months, 1 week ago

Congrats Harsh, 52 is a good score. All the best for INMOLog in to reply

– Harsh Shrivastava · 11 months, 1 week ago

Thanks!Log in to reply

– Mayank Chaturvedi · 11 months, 1 week ago

Meet you at INMO training camp on 14th-15th.Log in to reply

– Shubhendra Singh · 11 months, 1 week ago

What's this??Log in to reply

– Mayank Chaturvedi · 11 months, 1 week ago

Camp for RMO qualified from ChhattisgarhLog in to reply

Btw where's training camp? – Harsh Shrivastava · 11 months, 1 week ago

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– Mayank Chaturvedi · 11 months, 1 week ago

B.C.S.Govt. P.G. College, Dhamtari C.G on January 14-15, 2016. All the details are written on top of the result listLog in to reply

– Ashish Sacheti · 12 months ago

How do u know a=b from the equation a+b+c=oLog in to reply

\((a-b)(a+b)=-c (a-b) \)

Thus we have two cases:

Case 1:\((a-b)=0\) or \(a=b\)

Case 2:If \(a\not=b\) , we cancel out \((a-b)\) from botb sides to obtain \(a+b=-c\) or \(a+b+c=0\) – Sualeh Asif · 12 months ago

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– Mardokay Mosazghi · 12 months ago

I am also not sure at how you arrive at that conclusionLog in to reply

– Shubhendra Singh · 12 months ago

That's the best way.Log in to reply

– Chinmay Sangawadekar · 12 months ago

Did the same !!! Nice and standard solution btw...Log in to reply

6)\[\{a\}=a-4\] \[a(a-3a+12)=-2a^2+12a\] \[-2a^2+12a=n\] \[(a-3)^2=\dfrac{-n+18}{2}\] \[(4-3)^2<(a-3)^2<(5-3)^2\] \[1<\dfrac{-n+18}{2}<4\] \[10<n<16\] put values n=11,12,13,14,15 to find \[a=\dfrac{6+\sqrt{36-2n}}{2}=3+\dfrac{\sqrt{14}}{2},3+\dfrac{\sqrt{12}}{2},3+\dfrac{\sqrt{10}}{2},3+\dfrac{\sqrt{8}}{2},3+\dfrac{\sqrt{6}}{2}\] – Aareyan Manzoor · 12 months ago

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– Easha Manideep D · 12 months ago

Sorry to say that, substitute a = 4 + 1 / 2 + sqrt 6Log in to reply

\( 4 + \frac{1}{\sqrt{6} + 2} = 4 + \frac{\sqrt{6} - 2}{2} = 3 + \frac{\sqrt{6}}{2} \) – Siddhartha Srivastava · 12 months ago

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– Easha Manideep D · 12 months ago

yaLog in to reply

– Aareyan Manzoor · 12 months ago

i am sorry. couldnt understand. that is >5Log in to reply

– Easha Manideep D · 12 months ago

Hey ! 4 + 1/ something is never greater than 5 and that something is > 1Log in to reply

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– Vishal Yadav · 12 months ago

How did you get the RHS in the second line ?Log in to reply

– Aareyan Manzoor · 12 months ago

second line: we deduced \(\{a\}=a-4\). we substitute this value into the given expression to find \[a(a-3\{a\})=a(a-3(a-4))=a(a-3a+12)=a(-2a+12)=-2a^2+12\]. hope this helps.Log in to reply

– Shubhendra Singh · 12 months ago

What answer were you getting for Q.4Log in to reply

– Aareyan Manzoor · 12 months ago

sorry man, my combinatorics is weak(very)Log in to reply

– Shubhendra Singh · 12 months ago

Perfect, I too got the same answer.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 · 12 months ago

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– Chinmay Sangawadekar · 12 months ago

Similar Approach ... Nice and elegant solution.Log in to reply

Q2Consider the polynomial \(Q(x)=x^2+ax+b-st\) When \(Q(x)=0\), we realise that if \(b-st\) is a root then the other root is 1 by Vieta's formulas. So the sum of the roots is \((b-st+1)\). We can factor it to \(Q(x)=(x-(b-st))(x-1).\) Now realise that this means \[a=-(b-st+1) \implies st=a+b+1\]. Thus if we can prove this we are done. Now move to the other pat of the information provided. Realise that \[P(s)-t=P(t)-s, \\ s^2 -t^2 =a(t-s) +(t-s) \\ a=-1-s-t\] We can cancel \(t-s\) out since \(t\not=s\). Substitute \( a=-1-s-t\) back into \[P(s)-t=0 \implies b=st+s+t \\ \therefore a+b+1=st\] And we are done! – Sualeh Asif · 12 months agoLog in to reply

– Mohit Gupta · 12 months ago

In this question instead of proving b-st as the root it is sufficient to prove that 1 is the root of the eqn. Isn't it??Log in to reply

– Sualeh Asif · 12 months ago

Well using vietas formulas you see that the multipleof the two roots is \(b-st\). Hence the roots should be 1 and \(b-st\). Thus if you prove 1 is a root of the equation \(b-st\) is automatically a root.Log in to reply

As per the level of paper how much one need to score for clear this level – Aakash Khandelwal · 12 months ago

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– Devansh Shah · 12 months ago

How much u r getting??Log in to reply

Hey guys I didn't get selected : ( – Easha Manideep D · 11 months, 1 week ago

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– Shubhendra Singh · 11 months, 1 week ago

Don't get sad, work on your weaknesses and try again.Log in to reply

Who got selected for rmo and how much scores....? – Shubham Gupta · 11 months, 2 weeks ago

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– Shubhendra Singh · 11 months, 2 weeks ago

There are 33 selections from rajasthan, and luckily I'm one of them.Log in to reply

everybody....the answer to the combinatorics question is here:

go here – Vaibhav Prasad · 12 months ago

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– Svatejas Shivakumar · 12 months ago

Where did you get the answers from? I think that your answer is way too small. The answer for 28 object has been posted in a website(resonance) and that answer is bigger than the one you posted.Log in to reply

See this – Harsh Shrivastava · 12 months ago

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– Svatejas Shivakumar · 12 months ago

Wow. Did you get it right in the exam as well?Log in to reply

– Harsh Shrivastava · 12 months ago

Nah the solutions are written by my FIITJEE sir,not me.Log in to reply

– Svatejas Shivakumar · 12 months ago

How many did you solve in the exam?Log in to reply

– Harsh Shrivastava · 12 months ago

I screwed up the paper.( only 2 solutions are perfectly correct rest contain some flaws :( )Log in to reply

– Shubhendra Singh · 11 months, 4 weeks ago

I got that right check out what I have done in my solution below.Log in to reply

I attmpted 5 my logic is correct but I commited calculation mistake in 2 problems – Satyam Mani · 12 months ago

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I have done 4 questions correctly – Ashutosh Kaul · 12 months ago

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The same questions were asked in Jharkand . What should be the expected cut off ? – Wasif Jawad Hussain · 12 months ago

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– Aditya Kr · 11 months, 3 weeks ago

45-55Log in to reply

This is the solution of 5th question

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this the solution of 5th question

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– Mohit Gupta · 12 months ago

Yup did the same way but rather with a more tedious approachLog in to reply

– Shubhendra Singh · 12 months ago

I too did the same wayLog in to reply

Answer to

Q4Arrange all things in a straight line \(a_{1},a_{2}...….a_{32}\)

See the following arrangement

\(P \\ \boxed{a} \\ Q \\ \boxed{b} \\ R \\ \boxed{c} \\ S \)

Here a, b, c are any three things P, Q, R, S are no of things left in their middle and aside.

Here \(P+Q+R+S =29 \) where \(P, S \geq 0 \& Q, R \geq 1\)

So the no. of solutions of this equation are \(^{27+4-1}C_{3}=4060\). Now here we had taken them in a line.

When arranged in a circle \(a_{1}\) and \(a_{32}\) can't be together so we will remove the cases in which both of them are together, there are 28 cases to be removed.

So now we get 4060-28 = 4032 cases

Now we have to remove the cases of diametrically opposite things.

Here two things \(a_{p}; a_{q} \) p>q are diametrically opposite if \(p-q=16\) we will get 16 pairs with 26 arrangements corresponding to every pair so total cases are 26×16=416

Finally the answer comes out to be 4032-416=3616 – Shubhendra Singh · 12 months ago

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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. – Shivam Jadhav · 12 months ago

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I have a doubt in 5th question. – Chinmay Sangawadekar · 12 months ago

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– Samarth Agarwal · 12 months ago

There are five solutions: \(3+\sqrt{x}\) where x={1.5,2,2.5,3,3.5}Log in to reply

[@Nihar Mahajan , [@Swapnil Das , did you appear for RMO ? – Chinmay Sangawadekar · 12 months ago

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how much are you getting? – Dev Sharma · 12 months ago

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– Chinmay Sangawadekar · 12 months ago

Did you appear for RMO Dev ?Log in to reply

– Dev Sharma · 12 months ago

YeahLog in to reply

– Chinmay Sangawadekar · 12 months ago

How much will you score ? And how many questions did you attempt? All?Log in to reply

– Shubhendra Singh · 12 months ago

Well I was able to do all of them, let's see how the marking is doneLog in to reply

– Svatejas Shivakumar · 12 months ago

Can you please tell what was your answer for the 4th question? A very similar problem had appeared in my region.Log in to reply

– Shubhendra Singh · 12 months ago

I got 3616Log in to reply

– Dev Sharma · 12 months ago

i was also getting something like this, btw in which class you are?Log in to reply

– Shubhendra Singh · 12 months ago

11thLog in to reply

– Chinmay Sangawadekar · 12 months ago

In 11 and 15 yrs old ? Did you put wrong age while signing up?Log in to reply

– Kushagra Sahni · 12 months ago

Happy Birthday Dev!Log in to reply

– Dev Sharma · 12 months ago

ThanksLog in to reply

– Samarth Agarwal · 12 months ago

I got 2015!! what was your approach?Log in to reply

– Shubhendra Singh · 12 months ago

First tell me yoursLog in to reply

– Samarth Agarwal · 12 months ago

What was your approachLog in to reply

– Samarth Agarwal · 12 months ago

A weird one: There are 30+29+28+...+1 ways to choose the restricted objects and total ways I think should be (32C3)/2 as I studied in a book that for circur permutation we need to divide by two... so 2015 :(Log in to reply

– Shubhendra Singh · 12 months ago

Well I had a long one but here are results that I got. No of ways of selecting 3 such that no 2 are adjacent \(^{30}C_{3}- 28=4032\) and the most of arrangements containing diametrically opposite were \(26 ×16\). Finally getting 4032-416=3616.Log in to reply

– Vishal Yadav · 12 months ago

How did you solve for the condition that none of the two are adjacent or diametrically opposite ?Log in to reply

– Chinmay Sangawadekar · 12 months ago

Pheww ^_^ got all except the geometry problems... and 4)Log in to reply

I got 5 questions correct expect that 4 th one. How much u all have done – Shubham Gupta · 12 months ago

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What is the expected cutoff. I did 5 correct questions and am in class 10. Can I know the max marks they give if the solution is completely correct – Shubham Gupta · 12 months ago

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– Vishal Yadav · 12 months ago

That's obvious : \(17\times5\) = 85(lol)Log in to reply

– Kushagra Sahni · 12 months ago

Yeah that's very funny. 85 of course.Log in to reply

– Shubham Gupta · 12 months ago

I mean that have u appeared for RMO last year and know that how much lenient checking they do. I have heard that they give max 14 marks in full correct solutions.Log in to reply

– Svatejas Shivakumar · 12 months ago

Are you serious? Which region?Log in to reply

– Shubham Gupta · 12 months ago

RajasthanLog in to reply