**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 \(a\)}.

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## 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 \[(a,b,c)=(1,1,0),(-1,-1,0).\].Log in to reply

Why are solutions like \((2,-1,-1)\) not included?

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

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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.

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I think it's not a solution, please check it again.

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@Harsh Shrivastava

Are you selected for INMOLog in to reply

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@Harsh Shrivastava

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

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Btw where's training camp?

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How do u know a=b from the equation a+b+c=o

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Well here is how he arrived at the conclusion.

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

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 \(a+b+c=0\)

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I am also not sure at how you arrive at that conclusion

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That's the best way.

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Did the same !!! Nice and standard solution btw...

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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 \[a=\dfrac{6+\sqrt{36-2n}}{2}=3+\dfrac{\sqrt{14}}{2},3+\dfrac{\sqrt{12}}{2},3+\dfrac{\sqrt{10}}{2},3+\dfrac{\sqrt{8}}{2},3+\dfrac{\sqrt{6}}{2}\]

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From where did (a-3) come.....I think the way I be done is quite different.....And I think the answer is 6-√2,6-√3

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Sorry.... I got it.....

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

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Do you mean \( 4 + \frac{1}{\sqrt{6} + 2} \)

\( 4 + \frac{1}{\sqrt{6} + 2} = 4 + \frac{\sqrt{6} - 2}{2} = 3 + \frac{\sqrt{6}}{2} \)

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i am sorry. couldnt understand. that is >5

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How did you get the RHS in the second line ?

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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.

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What answer were you getting for Q.4

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sorry man, my combinatorics is weak(very)

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Perfect, I too got the same answer.

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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.

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Similar Approach ... Nice and elegant solution.

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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!Log in to reply

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??

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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.

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As per the level of paper how much one need to score for clear this level

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How much u r getting??

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Hey guys I didn't get selected : (

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Don't get sad, work on your weaknesses and try again.

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Who got selected for rmo and how much scores....?

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

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

go here

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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.

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See this

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

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I attmpted 5 my logic is correct but I commited calculation mistake in 2 problems

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I have done 4 questions correctly

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The same questions were asked in Jharkand . What should be the expected cut off ?

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45-55

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This is the solution of 5th question

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this the solution of 5th question

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Yup did the same way but rather with a more tedious approach

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I too did the same way

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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

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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.

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I have a doubt in 5th question.

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Comment deleted Dec 06, 2015

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

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[@Nihar Mahajan , [@Swapnil Das , did you appear for RMO ?

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how much are you getting?

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Did you appear for RMO Dev ?

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Yeah

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

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

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Pheww ^_^ got all except the geometry problems... and 4)

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I got 5 questions correct expect that 4 th one. How much u all have done

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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

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That's obvious : \(17\times5\) = 85(lol)

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Yeah that's very funny. 85 of course.

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