RMO - 2015 Paper discussion (Rajasthan Region)

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

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

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.

Q2 Consider 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!

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

how much are you getting?

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

I have a doubt in 5th question.

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.

Answer to Q4

Arrange 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

this the solution of 5th question

This is the solution of 5th question

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

I have done 4 questions correctly

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

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

go here

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

Hey guys I didn't get selected : (

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

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