RMO - 2015 Paper discussion (Rajasthan Region)

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.

1. 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.

2. 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 $x^{2}+ax+b-st=0$

3. Find all integers $a, b, c$ such that $a^{2}=bc+1 \ \\\, \\\ b^{2}=ca +1$

4. 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 ?

5. 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.

6. Find all real numbers $a$ such that $4 and $a(a-3\{a\})$ is an integer.
{Here $\{a \}$ represents fractional part of $a$}.

Note by Shubhendra Singh
5 years, 4 months ago

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

- 5 years, 4 months ago

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

- 5 years, 4 months ago

That's the best way.

- 5 years, 4 months ago

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

- 5 years, 4 months ago

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

- 5 years, 4 months ago

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 $a+b+c=0$

- 5 years, 4 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

What do you think the cut off would be.of GMO 2015 paper?

- 5 years, 3 months 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??

- 5 years, 3 months ago

Did you get selected?

- 5 years, 3 months ago

Yes 8-)

- 5 years, 3 months ago

How many marks did you got? And Congrats!

- 5 years, 3 months ago

Congrats Harsh, 52 is a good score. All the best for INMO

- 5 years, 3 months ago

Thanks!

- 5 years, 3 months ago

Meet you at INMO training camp on 14th-15th.

- 5 years, 3 months ago

Oh you have also qualified? Congo!

Btw where's training camp?

- 5 years, 3 months 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 list

- 5 years, 3 months ago

What's this??

- 5 years, 3 months ago

Camp for RMO qualified from Chhattisgarh

- 5 years, 3 months ago

- 5 years, 3 months ago

nope and u?

- 5 years, 3 months ago

I also didn't get selected. I did 4 but my writing was horrible.

- 5 years, 3 months ago

Would it be sahni at place of juneja!!

- 5 years, 3 months ago

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

- 5 years, 3 months ago

Dev you are in which class ??

- 5 years ago

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

- 5 years, 3 months ago

Its written in the result list of my state.

- 5 years, 3 months ago

Are you selected for INMO @Harsh Shrivastava

- 5 years, 3 months ago

Luckily yes.

- 5 years, 3 months ago

Why luckily ? you are intelligent bro . Unfortunately , our principal doesn't allow us for RMO.@Harsh Shrivastava

- 5 years, 3 months ago

Why?

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

By subtracting the given equations we get $a^{2}-b^{2}=-c(a-b)$ $\Rightarrow (a+b)(a-b) =-c(a-b)$

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.

- 5 years, 3 months ago

It means all the solutions of $a+b+c = 0$ are solutions?

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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 put values n=11,12,13,14,15 to find $a=\dfrac{6+\sqrt{36-2n}}{2}=3+\dfrac{\sqrt{14}}{2},3+\dfrac{\sqrt{12}}{2},3+\dfrac{\sqrt{10}}{2},3+\dfrac{\sqrt{8}}{2},3+\dfrac{\sqrt{6}}{2}$

- 5 years, 4 months ago

Perfect, I too got the same answer.

- 5 years, 4 months ago

What answer were you getting for Q.4

- 5 years, 4 months ago

sorry man, my combinatorics is weak(very)

- 5 years, 4 months ago

How did you get the RHS in the second line ?

- 5 years, 4 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.

- 5 years, 4 months ago

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

- 5 years, 4 months ago

i am sorry. couldnt understand. that is >5

- 5 years, 4 months ago

Hey ! 4 + 1/ something is never greater than 5 and that something is > 1

- 5 years, 4 months ago

 i am unable to understand what you are saying.

- 5 years, 4 months ago

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

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

- 5 years, 4 months ago

ya

- 5 years, 4 months ago

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

- 3 years, 7 months ago

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

- 3 years, 7 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.

- 5 years, 4 months ago

Similar Approach ... Nice and elegant solution.

- 5 years, 4 months ago

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!

- 5 years, 4 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??

- 5 years, 4 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.

- 5 years, 4 months ago

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

- 5 years, 4 months ago

How much u r getting??

- 5 years, 4 months ago

how much are you getting?

- 5 years, 4 months ago

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

- 5 years, 4 months ago

- 5 years, 4 months ago

I got 3616

- 5 years, 4 months ago

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

- 5 years, 4 months ago

11th

- 5 years, 4 months ago

In 11 and 15 yrs old ? Did you put wrong age while signing up?

- 5 years, 4 months ago

Happy Birthday Dev!

- 5 years, 4 months ago

Thanks

- 5 years, 4 months ago

I got 2015!! what was your approach?

- 5 years, 4 months ago

First tell me yours

- 5 years, 4 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 :(

- 5 years, 4 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.

- 5 years, 4 months ago

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

- 5 years, 4 months ago

- 5 years, 4 months ago

Did you appear for RMO Dev ?

- 5 years, 4 months ago

Yeah

- 5 years, 4 months ago

How much will you score ? And how many questions did you attempt? All?

- 5 years, 4 months ago

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

- 5 years, 4 months ago

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

- 5 years, 4 months ago

I have a doubt in 5th question.

- 5 years, 4 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.

- 5 years, 4 months ago

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

- 5 years, 4 months ago

this the solution of 5th question

- 5 years, 4 months ago

I too did the same way

- 5 years, 4 months ago

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

- 5 years, 4 months ago

This is the solution of 5th question

- 5 years, 4 months ago

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

- 5 years, 4 months ago

45-55

- 5 years, 4 months ago

I have done 4 questions correctly

- 5 years, 4 months ago

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

- 5 years, 4 months ago

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

- 5 years, 4 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.

- 5 years, 4 months ago

- 5 years, 4 months ago

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

- 5 years, 4 months ago

Nah the solutions are written by my FIITJEE sir,not me.

- 5 years, 4 months ago

How many did you solve in the exam?

- 5 years, 4 months ago

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

- 5 years, 4 months ago

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

- 5 years, 4 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

Hey guys I didn't get selected : (

- 5 years, 3 months ago

- 5 years, 3 months ago

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

- 5 years, 4 months ago

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

- 5 years, 4 months ago

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

- 5 years, 4 months ago

Yeah that's very funny. 85 of course.

- 5 years, 4 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.

- 5 years, 4 months ago

Are you serious? Which region?

- 5 years, 4 months ago

Rajasthan

- 5 years, 4 months ago