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# 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 $$4<a<5$$ and $$a(a-3\{a\})$$ is an integer.
{Here $$\{a \}$$ represents fractional part of $$a$$}.

Note by Shubhendra Singh
12 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 $(a,b,c)=(1,1,0),(-1,-1,0).$. · 12 months ago

Why are solutions like $$(2,-1,-1)$$ not included? · 11 months, 2 weeks ago

Maybe it's not but how do we know that these are the only solutions? · 11 months, 1 week 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. · 11 months, 1 week ago

It means all the solutions of $$a+b+c = 0$$ are solutions? · 11 months, 1 week ago

No, this is just a condition between a, b, c that will help us in getting the solutions. · 11 months, 1 week ago

I think it's not a solution, please check it again. · 11 months, 2 weeks ago

What do you think the cut off would be.of GMO 2015 paper? · 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?? · 11 months, 2 weeks ago

Did you get selected? · 11 months, 1 week ago

Yes 8-) · 11 months, 1 week ago

What About Dev Sharma? · 11 months, 1 week ago

nope and u? · 11 months, 1 week ago

Dev you are in which class ?? · 8 months, 1 week ago

I also didn't get selected. I did 4 but my writing was horrible. · 11 months, 1 week ago

Would it be sahni at place of juneja!! · 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. · 11 months, 1 week ago

How many marks did you got? And Congrats! · 11 months, 1 week ago

I don't know how much I have scored, how did you get to know about your marks. · 11 months, 1 week ago

Its written in the result list of my state. · 11 months, 1 week ago

There's nothing like this in rajasthan's list, even the cut-off is not mentioned. · 11 months, 1 week ago

Are you selected for INMO @Harsh Shrivastava · 11 months, 1 week ago

Luckily yes. · 11 months, 1 week ago

Why luckily ? you are intelligent bro . Unfortunately , our principal doesn't allow us for RMO.@Harsh Shrivastava · 11 months, 1 week ago

Why? · 11 months, 1 week ago

Dont know . H e thinks we should focus on our school math and so , · 11 months, 1 week ago

Congrats Harsh, 52 is a good score. All the best for INMO · 11 months, 1 week ago

Thanks! · 11 months, 1 week ago

Meet you at INMO training camp on 14th-15th. · 11 months, 1 week ago

What's this?? · 11 months, 1 week ago

Camp for RMO qualified from Chhattisgarh · 11 months, 1 week ago

Oh you have also qualified? Congo!

Btw where's training camp? · 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 list · 11 months, 1 week ago

How do u know a=b from the equation a+b+c=o · 12 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 $$a+b+c=0$$ · 12 months ago

I am also not sure at how you arrive at that conclusion · 12 months ago

That's the best way. · 12 months ago

Did the same !!! Nice and standard solution btw... · 12 months ago

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

Sorry to say that, substitute a = 4 + 1 / 2 + sqrt 6 · 12 months ago

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

$$4 + \frac{1}{\sqrt{6} + 2} = 4 + \frac{\sqrt{6} - 2}{2} = 3 + \frac{\sqrt{6}}{2}$$ · 12 months ago

ya · 12 months ago

i am sorry. couldnt understand. that is >5 · 12 months ago

Hey ! 4 + 1/ something is never greater than 5 and that something is > 1 · 12 months ago

 i am unable to understand what you are saying. · 12 months ago

How did you get the RHS in the second line ? · 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. · 12 months ago

What answer were you getting for Q.4 · 12 months ago

sorry man, my combinatorics is weak(very) · 12 months ago

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

Similar Approach ... Nice and elegant solution. · 12 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! · 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?? · 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. · 12 months ago

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

How much u r getting?? · 12 months ago

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

Don't get sad, work on your weaknesses and try again. · 11 months, 1 week ago

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

There are 33 selections from rajasthan, and luckily I'm one of them. · 11 months, 2 weeks ago

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

go here · 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. · 12 months ago

See this · 12 months ago

Wow. Did you get it right in the exam as well? · 12 months ago

Nah the solutions are written by my FIITJEE sir,not me. · 12 months ago

How many did you solve in the exam? · 12 months ago

I screwed up the paper.( only 2 solutions are perfectly correct rest contain some flaws :( ) · 12 months ago

I got that right check out what I have done in my solution below. · 11 months, 4 weeks ago

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

I have done 4 questions correctly · 12 months ago

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

45-55 · 11 months, 3 weeks ago

This is the solution of 5th question

· 12 months ago

this the solution of 5th question

· 12 months ago

Yup did the same way but rather with a more tedious approach · 12 months ago

I too did the same way · 12 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 · 12 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. · 12 months ago

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

Comment deleted 12 months ago

There are five solutions: $$3+\sqrt{x}$$ where x={1.5,2,2.5,3,3.5} · 12 months ago

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

how much are you getting? · 12 months ago

Did you appear for RMO Dev ? · 12 months ago

Yeah · 12 months ago

How much will you score ? And how many questions did you attempt? All? · 12 months ago

Well I was able to do all of them, let's see how the marking is done · 12 months ago

Can you please tell what was your answer for the 4th question? A very similar problem had appeared in my region. · 12 months ago

I got 3616 · 12 months ago

i was also getting something like this, btw in which class you are? · 12 months ago

11th · 12 months ago

In 11 and 15 yrs old ? Did you put wrong age while signing up? · 12 months ago

Happy Birthday Dev! · 12 months ago

Thanks · 12 months ago

I got 2015!! what was your approach? · 12 months ago

First tell me yours · 12 months ago

What was your approach · 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 :( · 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. · 12 months ago

How did you solve for the condition that none of the two are adjacent or diametrically opposite ? · 12 months ago

Pheww ^_^ got all except the geometry problems... and 4) · 12 months ago

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

That's obvious : $$17\times5$$ = 85(lol) · 12 months ago

Yeah that's very funny. 85 of course. · 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. · 12 months ago

Are you serious? Which region? · 12 months ago