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 \(\Gamma\).

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.

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

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?

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.

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 \(\text{{1.5}} = 0.5; \text{ {-3.4}} = 0.6\).)

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## Comments

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TopNewestQ4)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 \(= 7140-36-1152-540=5412\).Log in to reply

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?

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No u looser

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

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

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

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NO inequality Question this year :(

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Thank u bhaiya

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:(

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

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Did the same ! Upvoted

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

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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.Log in to reply

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)

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

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Yaah!! That is also a solution.

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

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

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Can you please write the full solution ?

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Any News about the result?

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

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How do we do q2?

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Smae paper was for maharashtra and goa

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THE RESULTS( for delhi region ) ARE OUT GUYS !

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

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Why didn't you give RMO this year??

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@rajdeep das Allen I guess

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@Rajdeep Dhingra

Sorry was askingLog in to reply

Congrats @rohit kumar

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The marks are out guys.

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what marks ??

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Please Inform me what is PIE method?

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Principle of Inclusion and Exclusion (PIE)

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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...

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

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

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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.

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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.

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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!

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

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

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Well Aryan r u selected for GMO?

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according to me answer to the fourth is 5412

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Do you study at FIITJEE?

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Q 4) 5376 (INCLUSION AND EXCLUSION PRINCIPLE)

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Ans of 4th is 5412

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

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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.

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

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Yup, I know.

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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

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

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

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3616 in rajasthan region 32 objects were taken

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What is the answer of question 6

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Comment deleted Dec 06, 2015

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

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Comment deleted Dec 06, 2015

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

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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.

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@Sharky Kesa@Surya Prakash

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

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How many did you solve Rajdeep?

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4

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

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4th would be 6

2832+62932=10944 First choice from 36 Next from 32 the last one from either 28 or 29. All divided by 6 because these are combinationsLog in to reply

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

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

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

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Where can we get full detailed solutions for this paper

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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.

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

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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.

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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 ?

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

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results out.sorry to say but u are not selected

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Did you get selected?

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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.

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

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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

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Please solve 5th.

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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

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

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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)

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Hey @Rajdeep Dhingra how was ur paper ...

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How was your?

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screwed up

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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

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Yes i did the same

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Comment deleted Dec 06, 2015

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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 \( 2k \ - \ 0.5 \)

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Comment deleted Dec 06, 2015

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4 Answer \( 4800 \)

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

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Are you sure it is 4800?

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Well solutions aren't out but I guess so.

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Any idea on how to solve 2 and 5 ?

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5 was very very lengthy at least for me.

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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 ?

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