# RMO 2015 - Delhi Region , Maharashtra Region and Goa Region (Regional Mathematical Olympiad)

1. Two circles $$\Gamma$$ and $$\Sigma$$, with centres $$O$$ and $$O^{'}$$, respectively, are such that $$O^{'}$$ lies on $$\Gamma$$. Let $$A$$ be a point on $$\Sigma$$ and $$M$$ the midpoint of the segment $$AO^{'}$$. If $$B$$ is a point on $$\Sigma$$ different from $$A$$ such that $$AB$$ is parallel to $$OM$$, show that the midpoint of $$AB$$ lies on $$\Gamma$$.

2. Let $$P(x) = x^2 + ax + b$$ be a quadratic polynomial where a and b are real numbers. Suppose $$\left < P(-1)^2 , P(0)^2 , P(1)^2 \right >$$ is an arithmetic progression of integers. Prove that a and b are integers.

3. Show that there are infinitely many triples $$\left ( x , y , z \right )$$ of integers such that $$x^3 + y^4 = z^{31}$$.

4. Suppose 36 objects are placed along a circle at equal distances. In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite?

5. Let $$ABC$$ be a triangle with circumcenter $$\Gamma$$ and incenter $$I$$. Let the internal angle bisectors of $$\angle A , \angle B \text{ and } \angle C$$ meet $$\Gamma$$ in $$A^{'}$$, $$B^{'}$$ and $$C^{'}$$ respectively. Let $$B^{'}C^{'}$$ intersect $$AA^{'}$$ in $$P$$ and $$AC$$ in $$Q$$, and let $$BB^{'}$$ intersect $$AC$$ in $$R$$. Suppose the quadrilateral $$PIRQ$$ is a kite; that is, $$IP=IR$$ and $$QP=QR$$. Prove that $$ABC$$ is an equilateral triangle.

6. Show that there are infinitely many positive real numbers $$a$$ which are not integers such that $$a(a - 3\text{{a}})$$ is an integer. (Here $$\text{{a}}$$ denotes the fractional part of $$a$$. For example $$\text{{1.5}} = 0.5; \text{ {-3.4}} = 0.6$$.)

Note by Rajdeep Dhingra
2 years, 8 months ago

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Q4)We shall use PIE.

Step 1: Number of ways of selecting $$3$$ points from $$36$$ points is $$\binom{36}{3} = 7140$$.

Step 2: Number of ways of selecting $$3$$ adjacent points is $$36$$.

Step 3: Number of ways of selecting $$2$$ adjacent and one not adjacent with them is $$36\times 32 = 1152$$. (Since there are $$32$$ ways to select the non-adjacent point.)

Step 4: Number of ways of selecting two diametrically opposite points are $$18$$ and number of ways of selecting third one not adjacent to both of them are $$30$$ in each case. So total number of ways in this step are $$18 \times 30 = 540$$.

Step 5: Number of ways (what we required) = Total $$-$$ Number of ways of selecting $$3$$ adjacent points $$-$$ Number of ways of selecting $$2$$ adjacent and one not adjacent with them - Number of ways of selecting two diametrically opposite points and selecting third one not adjacent to both of them $$= 7140-36-1152-540=5412$$.

- 2 years, 8 months ago

Even I did something similar to this I think. Don't laugh but at the end I guess I multiplied everything which led to an enormous aolution (face palm). Will I get some marks?

- 2 years, 8 months ago

No u looser

- 1 year, 10 months ago

Please explain me in detail(in general) what is PIE method..

- 2 years, 8 months ago

Thanx a lot!My method was a little different but i got the same answer with n=36.I was right!

- 2 years, 8 months ago

That is right. I took $$36 \times 30$$ instead of $$18 \times \ 30$$ How much do you think I will get?

- 2 years, 8 months ago

NO inequality Question this year :(

- 2 years, 8 months ago

Thank u bhaiya

- 10 months, 2 weeks ago

:(

- 2 years, 6 months ago

Solution to question number 3. Let $x=-m^{4},y=m^{3},z=0$

- 2 years, 8 months ago

Did the same ! Upvoted

- 2 years, 8 months ago

Oh good lord that simple? I really feel bad now.

- 2 years, 8 months ago

Q1) Let $$D$$ be the point of intersection of $$AB$$ with the circle $$\Gamma$$. So, we need to prove that $$D$$ is the mid-point of side $$AB$$ i.e. we have to prove that $$O'D \perp AB$$. Extend $$OD$$ to meet the circle $$\Gamma$$ at $$E$$. Join $$O'E$$. Since, $$DE$$ is diameter and $$O$$ is the center of the circle $$\Gamma$$, it implies that $$DO = OE$$. But $$AM=MB$$. So these ratios are equal and this implies that $$O'E$$ is parallel to $$OM$$ and $$AB$$. Observe that $$ED$$ is diameter. So, $$\angle DO'E = 90^{0}$$ i.e. $$DO' \perp EO'$$. As $$EO' \parallel AB$$, so $$DO' \perp AB$$. Hence proved.

Try to draw the diagram and then read this. Though it looks big it is easy.

- 2 years, 8 months ago

1. I will just give my rough idea.
Clearly $$(1,0,1)$$ is a solution.

Let $$(x,y,z)$$ be a solution. Then $$2^{124} x, 2^{93} y , 2^{12} z$$ is also a solution. So there are infinitely many solutions.(By induction)

- 2 years, 8 months ago

I put z = 0, and then x^3 +y^4 = 0 obviously has infinite solutions.

- 2 years, 8 months ago

Yaah!! That is also a solution.

- 2 years, 8 months ago

Please also post the solutions for my region(Karnataka). I have posted the paper.

- 2 years, 8 months ago

Q3 can be solved by modular arithmetic.(Chinese remainder therom)

- 2 years, 7 months ago

Can you please write the full solution ?

- 2 years, 7 months ago

- 2 years, 7 months ago

Not till now.

- 2 years, 7 months ago

For question 6, take $$[a]$$ to be an odd positive integer, and the fractional part to be 0.5. That does the job.

- 2 years, 8 months ago

How do we do q2?

- 2 years, 8 months ago

Smae paper was for maharashtra and goa

- 2 years, 8 months ago

THE RESULTS( for delhi region ) ARE OUT GUYS !

- 2 years, 7 months ago

I am not able to view the results can someone send me ?

- 2 years, 7 months ago

Why didn't you give RMO this year??

- 1 year, 10 months ago

I will give from Rajasthan this year.

- 1 year, 10 months ago

Oh thats why I didn't see your name. So you have gone to Kota?

- 1 year, 10 months ago

Yup

- 1 year, 10 months ago

Best of luck.

- 1 year, 10 months ago

Same to you :)

- 1 year, 10 months ago

Hey you left FIITJEE?

- 1 year, 10 months ago

Me = yes

- 1 year, 10 months ago

Oh where do u go now?

- 1 year, 10 months ago

Allen

- 1 year, 10 months ago

Is FIITJEE South Delhi better or Allen.

- 1 year, 10 months ago

Both have there pro and cons

- 1 year, 10 months ago

Yaah but still...?

- 1 year, 10 months ago

I really can't say anything. It depends on your need.

- 1 year, 10 months ago

Same doubt

- 1 year, 10 months ago

U are from FIITJEE?

- 1 year, 10 months ago

Yeah I am

- 1 year, 10 months ago

@rajdeep das Allen I guess

- 1 year, 10 months ago

All the best .Is there Any Somender Jha sir there? You top there too?

- 1 year, 10 months ago

He is Allen Jaipur I think. I am in kota

- 1 year, 10 months ago

So u k him?

- 1 year, 10 months ago

- 1 year, 10 months ago

Congrats @rohit kumar

- 2 years, 7 months ago

Where can we get full detailed solutions for this paper

- 2 years, 8 months ago

The marks are out guys.

- 2 years, 8 months ago

what marks ??

- 2 years, 8 months ago

Please Inform me what is PIE method?

- 2 years, 8 months ago

- 2 years, 8 months ago

well..can anyone tell me that have i solved this question : prove that the roots of the equation x^3 - 3x^2 - 1 = 0 are never rational. correctly? my solution is like this:-

well i approached like this:- let the roots be a,b,c a+b+c=3 ab+bc+ac=0 abc=1 assuming roots to be rational..i took a=p1/q1,b as p2/q2 and c as p3/q3

so i got p1/q1+p2/q2+p3/q3=3---------eq.1 p1p2/q1q2 + p2p3/q2/q3 + p1p3/q1q3 = 0-----eq.2 and p1p2p3=q1q2q3 -----------eq-3

proceeding with equation 1

i got after expanding and replacing q1q2q3 by p1p2p3...

reciprocal of eq.2=3 (after three steps of monotonous algebraic expansion)

taking eq2 as x+y+z = 0 and then its reciprocal from above as 1/x+x/y+1/z = 3

by A.M-G.M we know that (x+y+z)(1/x+1/y+1/z)>=9

but here we are getting it as 3*0=0

therefore by contradiction roots can't be rational...

- 2 years, 8 months ago

How do you know that $$x,y,z$$ are positive (you applied AM-GM)?

- 2 years, 8 months ago

the above problem was asked in JHARKHAND RMO 2015...

- 2 years, 8 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.

- 2 years, 8 months ago

But its given 4<a<5 why didn't you consider that thing and they asked to find all integers a I also got infinite solutions but in a quadratic form.

- 2 years, 8 months ago

Also second can be done by first applying condition of AP then making some cases like a is a multiple of 4 or not , b is a perfect square or not. i made a total of 6 cases and the result was proved!

- 2 years, 8 months ago

prakhar you must try my newly posted question .they are of gmo .

- 2 years, 8 months ago

i will give them a try for sure . try my faulty pendulum its good

- 2 years, 8 months ago

Well Aryan r u selected for GMO?

- 2 years, 8 months ago

yes.selection for GMO is made only on basis of a screening test conducted in school itself.

- 2 years, 8 months ago

according to me answer to the fourth is 5412

- 2 years, 8 months ago

Do you study at FIITJEE?

- 2 years, 8 months ago

Q 4) 5376 (INCLUSION AND EXCLUSION PRINCIPLE)

- 2 years, 8 months ago

Ans of 4th is 5412

- 2 years, 8 months ago

What should be the cut off guys? Can solving 4 questions completely be enough?

- 2 years, 8 months ago

Hello Kushagra,

If you want , you can try this set "INMO 2016 PRACTICE SET-1 (NUMBER THEORY ONLY)" posted by me.

It has $$6$$ problems on Number theory for INMO 2016. You can post solutions also.

Convey this message to all your friends.

- 2 years, 8 months ago

The first question can be done by using basic coordinate geometry.

- 2 years, 8 months ago

Yup, I know.

- 2 years, 8 months ago

question 3 is very confusing if x,y,z can be 0 or not because if x or y= 0 we can find the general solution easily . I wrote (x , y , z ) = (0, k^31 ,k^4 ) is a solution where k is any integer is my solution correct .
Yes there can be numerous general solutions but any one can do the job I think

- 2 years, 8 months ago

Take z=0 and y=m^3 amd x=-(m^4)

- 2 years, 8 months ago

What is the answer to question 4 if 32 objects are there?

- 2 years, 8 months ago

3616 in rajasthan region 32 objects were taken

- 2 years, 8 months ago

What is the answer of question 6

- 2 years, 8 months ago

Comment deleted Dec 06, 2015

Someone was saying that 6-(3)^1/2 is also a solution i.e 6 solutions

- 2 years, 8 months ago

$$6-\sqrt{3}=4+f \Rightarrow f=2-\sqrt{3}$$ putting these values gives final result $$6-\sqrt{3}(2\sqrt{3})$$ which is not integer.

- 2 years, 8 months ago

Comment deleted Dec 06, 2015

Sorry in our region there was a restriction of 4<a<5

- 2 years, 8 months ago

Well in their region (UP ) they had to find all solutions between 4 and 5. You can see Raven Herd's post for the same.

- 2 years, 8 months ago

Please post solutions for 3rd and 4th. If possible for rest also.

- 2 years, 8 months ago

How many did you solve Rajdeep?

- 2 years, 8 months ago

4

What about you ? Did you solve the 3rd and 4th ? If yes pls post solution.

- 2 years, 8 months ago

4th would be 62832+62932=10944 First choice from 36 Next from 32 the last one from either 28 or 29. All divided by 6 because these are combinations

- 2 years, 8 months ago

For 4th I've posted a solution in my note you can check it out.

- 2 years, 8 months ago

Me too 4. What did you get in Q.4

- 2 years, 8 months ago

Not solved. Burnt an hour on it.

- 2 years, 8 months ago

I also spent an hour on it and got 6275.

- 2 years, 8 months ago

The combi question was common to many states just the number was changed.

- 2 years, 8 months ago

Hello, everybody. I'm new in here. Like all of you I love maths and certainly I intend to have a great amount of knowledge pertaining to every topic, be it calculus(fav), number theory, geometry.. Etc. Anyway, besides all of this I'm immensely impressed by every individual who's there in this community reason being that at very young age you all know so much. Hats off!! Since I'm in 12th I'm may be good in what is being taught to me and that is just CBSE. Somehow I grab questions from other books as well but it's no good. As it is I'm not good in everything. Above all of this one thing that confuses me is that gow do you all study maths of higher level and how do you manage studying your academics and this ? I dont go to fiitjee or institutions like these. So is it possible for meto attain such knowlodge like you without going to such institutions ? If yes then how? Thanks. :) I wish to give RMO next year. I hope you'll all help me with this.

- 2 years, 8 months ago

You are in std XII. RMO can be given only by students from std XI and below.

- 2 years, 8 months ago

Okay. Thanks, I was not aware of this. Anyway, my motive is to study maths of higher level and solve questions. For that I can download RMO question paper. But other than this, there is something else that I asked in my comment. Could you please read it and answer that as well ? I'd appreciate that.

- 2 years, 8 months ago

Question 6

Take {a} to be 0.5 and the integer part of a to be an odd integer .Then there are an infinite no of solutions.

What do you expect the cutoff to be ?

- 2 years, 8 months ago

Our sir said it must be 60 +-2 , i m from mprakssh academy

- 2 years, 8 months ago

results out.sorry to say but u are not selected

- 2 years, 7 months ago

Did you get selected?

- 2 years, 7 months ago

But in the question its given a lies between 4 and 5 and you can choose integer part of a as only 4 not any thing else.

- 2 years, 8 months ago

But the question does not say so . It only asks us to prove that there exist infinite no of non integer real solutions.

- 2 years, 8 months ago

SORRY ! In my Region TELANGANA they asked to find all possible solutions.

- 2 years, 8 months ago

Well i did not appeared in exam but for 4th problem i am getting 5412 . Can be very easily done using exclusion inclusion or bijection

- 2 years, 8 months ago

- 2 years, 8 months ago

Finally i am able to solve it. !

:) best problem of geometry in rmo in last 10 years! .

Biggest hint - Join AB' . try proving AQP is congruent to QRB' . Try it.

its beautiful

Caution: Its Not at all lengthy . just chase some angles and triangles

- 2 years, 8 months ago

Yeah, got it now. Although I didn't do exactly what you told, but I came up with a proof.

Can you give me hint for the second question too ?(I'm so shameless).

- 2 years, 8 months ago

No Problem . but congrats i am sure you will qualify for INMO . According to my estimation cutoff will no way be more than 60. so your are well above.

Its a kind of problem for which we have to make cases (as in typical problems of number theory) .

first apply condition of arithemetic progression by simply dumping values .

then apply condition for the roots of a quadratic to be integers .

Next Just Make Cases for a and b . (u will realise it once you reach till this extent)

- 2 years, 8 months ago

Hey @Rajdeep Dhingra how was ur paper ...

- 2 years, 8 months ago

How was your?

- 2 years, 8 months ago

screwed up

- 2 years, 8 months ago

Same here.

- 2 years, 8 months ago

How much are u getting I may get around 60

- 2 years, 8 months ago

40 - 45 :(

- 2 years, 8 months ago

solution to 6 ) all no of form 2m+ 3/2 , where m is any integer is a solution see that the fractional part would be 1/2 and then the rest is trivial

- 2 years, 8 months ago

Yes i did the same

- 2 years, 8 months ago

Comment deleted Dec 06, 2015

We have to get the given expression mod $$1$$ = 0. It is trivial that we need to create a factor of $$10$$ in the decimal point = $$2 \times 5 \ ). Take one case where fractional part of \( a \ = \ 0.2$$ and the other as $$0.5$$. $$0.2$$ clearly doesn't work and it is easy to notice that for odd numbers + $$0.5$$ always works as we get an even number for the expression inside the bracket. Hence the general solution $$2k \ - \ 0.5$$

- 2 years, 8 months ago

Comment deleted Dec 06, 2015

It is not required to produce all answers. It is required to prove that there are infinitely many which this one general solution does.

- 2 years, 8 months ago

4 Answer $$4800$$

Not sure about the 3rd one. No one in my centre even got the wind of it.

- 2 years, 8 months ago

Are you sure it is 4800?

- 2 years, 8 months ago

I got 6275

- 2 years, 8 months ago

Well nobody knows you maybe correct. Let's just wait for the solution. Anyways all I did was 36C3 - 36C1 - 36x30-36x34

- 2 years, 8 months ago

Well solutions aren't out but I guess so.

- 2 years, 8 months ago

Any idea on how to solve 2 and 5 ?

- 2 years, 8 months ago

5 was very very lengthy at least for me.

- 2 years, 8 months ago

Just give me an idea how you proceeded after proving AB = BC. It was the only significant progress I made. By the way, what about others from DPS ?

- 2 years, 8 months ago

How many marks u would expect , i did the same thing in the problem

- 2 years, 8 months ago

6-7, because half the work was done.

- 2 years, 8 months ago

Our sir said we would get mininmum 10

- 2 years, 8 months ago

Let's hope so. Did you ask the solution by the way ?

- 2 years, 8 months ago

I actually did not. Just try and prove PIRQ to be a cyclic quadrilateral. Aman did 3. That kid did 3. Don't know about the others.

- 2 years, 8 months ago