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# 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
9 months, 3 weeks 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 $$= 7140-36-1152-540=5412$$. · 9 months, 3 weeks 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? · 9 months, 3 weeks ago

Please explain me in detail(in general) what is PIE method.. · 9 months, 3 weeks ago

Thanx a lot!My method was a little different but i got the same answer with n=36.I was right! · 9 months, 3 weeks ago

That is right. I took $$36 \times 30$$ instead of $$18 \times \ 30$$ How much do you think I will get? · 9 months, 3 weeks ago

Solution to question number 3. Let $x=-m^{4},y=m^{3},z=0$ · 9 months, 3 weeks ago

Did the same ! Upvoted · 9 months, 3 weeks ago

Oh good lord that simple? I really feel bad now. · 9 months, 3 weeks ago

NO inequality Question this year :( · 9 months, 2 weeks ago

:( · 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. · 9 months, 3 weeks 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) · 9 months, 3 weeks ago

I put z = 0, and then x^3 +y^4 = 0 obviously has infinite solutions. · 9 months, 3 weeks ago

Yaah!! That is also a solution. · 9 months, 3 weeks ago

Please also post the solutions for my region(Karnataka). I have posted the paper. · 9 months, 3 weeks ago

For question 6, take $$[a]$$ to be an odd positive integer, and the fractional part to be 0.5. That does the job. · 9 months, 2 weeks ago

Smae paper was for maharashtra and goa · 9 months, 3 weeks ago

THE RESULTS( for delhi region ) ARE OUT GUYS ! · 9 months ago

I am not able to view the results can someone send me ? · 9 months ago

Congrats @rohit kumar · 9 months ago

Q3 can be solved by modular arithmetic.(Chinese remainder therom) · 9 months ago

Can you please write the full solution ? · 9 months ago

Any News about the result? · 9 months ago

Not till now. · 9 months ago

Where can we get full detailed solutions for this paper · 9 months, 1 week ago

The marks are out guys. · 9 months, 2 weeks ago

what marks ?? · 9 months, 2 weeks 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. · 9 months, 3 weeks ago

You are in std XII. RMO can be given only by students from std XI and below. · 9 months, 2 weeks 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. · 9 months, 2 weeks ago

Please Inform me what is PIE method? · 9 months, 3 weeks 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... · 9 months, 3 weeks ago

How do you know that $$x,y,z$$ are positive (you applied AM-GM)? · 9 months, 2 weeks ago

the above problem was asked in JHARKHAND RMO 2015... · 9 months, 3 weeks ago

How do we do q2? · 9 months, 3 weeks 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. · 9 months, 3 weeks 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. · 9 months, 3 weeks 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! · 9 months, 3 weeks ago

prakhar you must try my newly posted question .they are of gmo . · 9 months, 3 weeks ago

i will give them a try for sure . try my faulty pendulum its good · 9 months, 3 weeks ago

Well Aryan r u selected for GMO? · 9 months, 3 weeks ago

yes.selection for GMO is made only on basis of a screening test conducted in school itself. · 9 months, 3 weeks ago

according to me answer to the fourth is 5412 · 9 months, 3 weeks ago

Do you study at FIITJEE? · 9 months, 2 weeks ago

Q 4) 5376 (INCLUSION AND EXCLUSION PRINCIPLE) · 9 months, 3 weeks ago

Ans of 4th is 5412 · 9 months, 3 weeks ago

What should be the cut off guys? Can solving 4 questions completely be enough? · 9 months, 3 weeks 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. · 9 months, 1 week ago

The first question can be done by using basic coordinate geometry. · 9 months, 3 weeks ago

Yup, I know. · 9 months, 3 weeks 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 · 9 months, 3 weeks ago

Take z=0 and y=m^3 amd x=-(m^4) · 9 months, 3 weeks ago

What is the answer to question 4 if 32 objects are there? · 9 months, 3 weeks ago

3616 in rajasthan region 32 objects were taken · 9 months, 3 weeks ago

What is the answer of question 6 · 9 months, 3 weeks ago

Comment deleted 9 months ago

Someone was saying that 6-(3)^1/2 is also a solution i.e 6 solutions · 9 months, 3 weeks 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. · 9 months, 3 weeks ago

Comment deleted 9 months ago

Sorry in our region there was a restriction of 4<a<5 · 9 months, 3 weeks 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. · 9 months, 3 weeks ago

Please post solutions for 3rd and 4th. If possible for rest also. · 9 months, 3 weeks ago

How many did you solve Rajdeep? · 9 months, 3 weeks ago

4

What about you ? Did you solve the 3rd and 4th ? If yes pls post solution. · 9 months, 3 weeks 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 · 9 months, 3 weeks ago

For 4th I've posted a solution in my note you can check it out. · 9 months, 3 weeks ago

Me too 4. What did you get in Q.4 · 9 months, 3 weeks ago

Not solved. Burnt an hour on it. · 9 months, 3 weeks ago

I also spent an hour on it and got 6275. · 9 months, 3 weeks ago

The combi question was common to many states just the number was changed. · 9 months, 3 weeks 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 ? · 9 months, 3 weeks ago

Our sir said it must be 60 +-2 , i m from mprakssh academy · 9 months, 3 weeks ago

results out.sorry to say but u are not selected · 9 months ago

Did you get selected? · 9 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. · 9 months, 3 weeks ago

But the question does not say so . It only asks us to prove that there exist infinite no of non integer real solutions. · 9 months, 3 weeks ago

SORRY ! In my Region TELANGANA they asked to find all possible solutions. · 9 months, 2 weeks 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 · 9 months, 3 weeks ago

Please solve 5th. · 9 months, 3 weeks 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 · 9 months, 3 weeks 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). · 9 months, 3 weeks 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) · 9 months, 3 weeks ago

Hey @Rajdeep Dhingra how was ur paper ... · 9 months, 3 weeks ago

How was your? · 9 months, 3 weeks ago

screwed up · 9 months, 3 weeks ago

Same here. · 9 months, 3 weeks ago

How much are u getting I may get around 60 · 9 months, 3 weeks ago

40 - 45 :( · 9 months, 3 weeks 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 · 9 months, 3 weeks ago

Yes i did the same · 9 months, 3 weeks ago

Comment deleted 9 months ago

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$$ · 9 months, 3 weeks ago

Comment deleted 9 months ago

It is not required to produce all answers. It is required to prove that there are infinitely many which this one general solution does. · 9 months, 3 weeks ago

4 Answer $$4800$$

Not sure about the 3rd one. No one in my centre even got the wind of it. · 9 months, 3 weeks ago

Are you sure it is 4800? · 9 months, 3 weeks ago

Well solutions aren't out but I guess so. · 9 months, 3 weeks ago

I got 6275 · 9 months, 3 weeks ago

Well nobody knows you maybe correct. Let's just wait for the solution. Anyways all I did was 36C3 - 36C1 - 36x30-36x34 · 9 months, 3 weeks ago

Any idea on how to solve 2 and 5 ? · 9 months, 3 weeks ago

5 was very very lengthy at least for me. · 9 months, 3 weeks 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 ? · 9 months, 3 weeks ago

How many marks u would expect , i did the same thing in the problem · 9 months, 3 weeks ago

6-7, because half the work was done. · 9 months, 3 weeks ago

Our sir said we would get mininmum 10 · 9 months, 3 weeks ago

Let's hope so. Did you ask the solution by the way ? · 9 months, 3 weeks ago