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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
1 year 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 \(= 7140-36-1152-540=5412\). Surya Prakash · 12 months ago

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@Surya Prakash 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? Svatejas Shivakumar · 12 months ago

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@Svatejas Shivakumar No u looser Manav Jaral · 1 month, 4 weeks ago

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@Surya Prakash Please explain me in detail(in general) what is PIE method.. Dhaval Pandya · 12 months ago

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@Surya Prakash Thanx a lot!My method was a little different but i got the same answer with n=36.I was right! Adarsh Kumar · 12 months ago

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@Surya Prakash That is right. I took \( 36 \times 30 \) instead of \( 18 \times \ 30 \) How much do you think I will get? Kunal Verma · 12 months ago

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Solution to question number 3. Let \[x=-m^{4},y=m^{3},z=0\] Shivam Jadhav · 12 months ago

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@Shivam Jadhav Did the same ! Upvoted Rohit Kumar · 12 months ago

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@Shivam Jadhav Oh good lord that simple? I really feel bad now. Kunal Verma · 12 months ago

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NO inequality Question this year :( Raghav Rathi · 12 months ago

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@Raghav Rathi :( Samanvay Vajpayee · 10 months, 1 week ago

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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. Surya Prakash · 12 months ago

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  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) Surya Prakash · 12 months ago

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@Surya Prakash I put z = 0, and then x^3 +y^4 = 0 obviously has infinite solutions. Rohit Kumar · 12 months ago

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@Rohit Kumar Yaah!! That is also a solution. Surya Prakash · 12 months ago

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@Surya Prakash Please also post the solutions for my region(Karnataka). I have posted the paper. Svatejas Shivakumar · 12 months ago

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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. Shourya Pandey · 12 months ago

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Smae paper was for maharashtra and goa Tejas Khairnar · 12 months ago

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THE RESULTS( for delhi region ) ARE OUT GUYS ! Rohit Kumar · 11 months, 2 weeks ago

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@Rohit Kumar I am not able to view the results can someone send me ? Rajdeep Dhingra · 11 months, 2 weeks ago

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@Rajdeep Dhingra Why didn't you give RMO this year?? Rajdeep Das · 2 months ago

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@Rajdeep Das I will give from Rajasthan this year. Rajdeep Dhingra · 2 months ago

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@Rajdeep Dhingra Oh thats why I didn't see your name. So you have gone to Kota? Rajdeep Das · 2 months ago

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@Rajdeep Das Yup Rajdeep Dhingra · 2 months ago

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@Rajdeep Dhingra Best of luck. Rajdeep Das · 2 months ago

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@Rajdeep Das Same to you :) Rajdeep Dhingra · 2 months ago

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@Rajdeep Das Hey you left FIITJEE? Kaustubh Miglani · 2 months ago

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@Kaustubh Miglani Me = yes Rajdeep Dhingra · 2 months ago

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@Rajdeep Dhingra Oh where do u go now? Kaustubh Miglani · 2 months ago

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@Kaustubh Miglani Allen Rajdeep Dhingra · 2 months ago

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@Rajdeep Dhingra Is FIITJEE South Delhi better or Allen. Rajdeep Das · 2 months ago

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@Rajdeep Das Both have there pro and cons Rajdeep Dhingra · 2 months ago

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@Rajdeep Dhingra Yaah but still...? Rajdeep Das · 2 months ago

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@Rajdeep Das I really can't say anything. It depends on your need. Rajdeep Dhingra · 2 months ago

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@Rajdeep Das Same doubt Kaustubh Miglani · 2 months ago

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@Kaustubh Miglani U are from FIITJEE? Rajdeep Das · 2 months ago

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@Rajdeep Das Yeah I am Kaustubh Miglani · 2 months ago

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@Rajdeep Das @rajdeep das Allen I guess Rajdeep Das · 2 months ago

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@Rajdeep Dhingra All the best .Is there Any Somender Jha sir there? You top there too? Kaustubh Miglani · 2 months ago

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@Kaustubh Miglani He is Allen Jaipur I think. I am in kota Rajdeep Dhingra · 2 months ago

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@Rajdeep Dhingra So u k him? Kaustubh Miglani · 2 months ago

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@Kaustubh Miglani Sorry was asking @Rajdeep Dhingra Kaustubh Miglani · 2 months ago

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@Rohit Kumar Congrats @rohit kumar Harsh Shrivastava · 11 months, 2 weeks ago

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Q3 can be solved by modular arithmetic.(Chinese remainder therom) Sameer Pimparkhede · 11 months, 2 weeks ago

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@Sameer Pimparkhede Can you please write the full solution ? Rajdeep Dhingra · 11 months, 2 weeks ago

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@Rajdeep Dhingra Any News about the result? Kushagra Sahni · 11 months, 2 weeks ago

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@Kushagra Sahni Not till now. Rajdeep Dhingra · 11 months, 2 weeks ago

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Where can we get full detailed solutions for this paper Ganesh Ayyappan · 11 months, 3 weeks ago

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The marks are out guys. Pawan Dogra · 12 months ago

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@Pawan Dogra what marks ?? Vaibhav Prasad · 12 months ago

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Please Inform me what is PIE method? Dhaval Pandya · 12 months ago

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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... Gyanendra Prakash · 12 months ago

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@Gyanendra Prakash How do you know that \(x,y,z\) are positive (you applied AM-GM)? Shourya Pandey · 11 months, 4 weeks ago

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@Gyanendra Prakash the above problem was asked in JHARKHAND RMO 2015... Gyanendra Prakash · 12 months ago

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How do we do q2? Shaurya Raj Singh · 12 months ago

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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. Shivam Jadhav · 12 months ago

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@Shivam Jadhav 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. Easha Manideep D · 12 months ago

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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! Prakhar Bindal · 12 months ago

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@Prakhar Bindal prakhar you must try my newly posted question .they are of gmo . Aryan Goyat · 12 months ago

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@Aryan Goyat i will give them a try for sure . try my faulty pendulum its good Prakhar Bindal · 12 months ago

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@Aryan Goyat Well Aryan r u selected for GMO? Yash Kumar · 12 months ago

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@Yash Kumar yes.selection for GMO is made only on basis of a screening test conducted in school itself. Aryan Goyat · 12 months ago

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according to me answer to the fourth is 5412 Sankush Gupta · 12 months ago

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@Sankush Gupta Do you study at FIITJEE? Priyanshu Mishra · 11 months, 3 weeks ago

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Q 4) 5376 (INCLUSION AND EXCLUSION PRINCIPLE) Atharva Sarage · 12 months ago

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Ans of 4th is 5412 Tejas Khairnar · 12 months ago

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What should be the cut off guys? Can solving 4 questions completely be enough? Kushagra Sahni · 12 months ago

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@Kushagra Sahni 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. Priyanshu Mishra · 11 months, 3 weeks ago

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The first question can be done by using basic coordinate geometry. Aditya Kumar · 12 months ago

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@Aditya Kumar Yup, I know. Rajdeep Dhingra · 12 months ago

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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 Sauditya Yo Yo · 12 months ago

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@Sauditya Yo Yo Take z=0 and y=m^3 amd x=-(m^4) Tejas Khairnar · 12 months ago

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What is the answer to question 4 if 32 objects are there? Samarth Agarwal · 12 months ago

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@Samarth Agarwal 3616 in rajasthan region 32 objects were taken Shubhendra Singh · 12 months ago

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What is the answer of question 6 Samarth Agarwal · 12 months ago

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@Shubhendra Singh Someone was saying that 6-(3)^1/2 is also a solution i.e 6 solutions Samarth Agarwal · 12 months ago

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@Samarth Agarwal \(6-\sqrt{3}=4+f \Rightarrow f=2-\sqrt{3}\) putting these values gives final result \(6-\sqrt{3}(2\sqrt{3}) \) which is not integer. Shubhendra Singh · 12 months ago

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@Rajdeep Dhingra Sorry in our region there was a restriction of 4<a<5 Samarth Agarwal · 12 months ago

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@Rajdeep Dhingra 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. Kunal Verma · 12 months ago

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@Sharky Kesa@Surya Prakash

Please post solutions for 3rd and 4th. If possible for rest also. Rajdeep Dhingra · 12 months ago

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How many did you solve Rajdeep? Kushagra Sahni · 1 year ago

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@Kushagra Sahni 4

What about you ? Did you solve the 3rd and 4th ? If yes pls post solution. Rajdeep Dhingra · 12 months ago

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@Rajdeep Dhingra 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 Kunal Jain · 12 months ago

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@Rajdeep Dhingra For 4th I've posted a solution in my note you can check it out. Shubhendra Singh · 12 months ago

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@Rajdeep Dhingra Me too 4. What did you get in Q.4 Kushagra Sahni · 12 months ago

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@Kushagra Sahni Not solved. Burnt an hour on it. Rajdeep Dhingra · 12 months ago

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@Rajdeep Dhingra I also spent an hour on it and got 6275. Kushagra Sahni · 12 months ago

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The combi question was common to many states just the number was changed. Adarsh Kumar · 1 year ago

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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. Tushar Jawalia · 12 months ago

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@Tushar Jawalia You are in std XII. RMO can be given only by students from std XI and below. Shourya Pandey · 11 months, 4 weeks ago

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@Shourya Pandey 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. Tushar Jawalia · 11 months, 4 weeks ago

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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 ? Utsav Bhardwaj · 12 months ago

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@Utsav Bhardwaj Our sir said it must be 60 +-2 , i m from mprakssh academy Tejas Khairnar · 12 months ago

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@Utsav Bhardwaj results out.sorry to say but u are not selected Kaustubh Miglani · 11 months, 2 weeks ago

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@Kaustubh Miglani Did you get selected? Kushagra Sahni · 11 months, 2 weeks ago

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@Utsav Bhardwaj 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. Easha Manideep D · 12 months ago

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@Easha Manideep D But the question does not say so . It only asks us to prove that there exist infinite no of non integer real solutions. Utsav Bhardwaj · 12 months ago

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@Utsav Bhardwaj SORRY ! In my Region TELANGANA they asked to find all possible solutions. Easha Manideep D · 12 months ago

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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 Prakhar Bindal · 12 months ago

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@Prakhar Bindal Please solve 5th. Rohit Kumar · 12 months ago

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@Rohit Kumar 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 Prakhar Bindal · 12 months ago

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@Prakhar Bindal 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). Rohit Kumar · 12 months ago

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@Rohit Kumar 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) Prakhar Bindal · 12 months ago

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Hey @Rajdeep Dhingra how was ur paper ... Vaibhav Prasad · 12 months ago

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@Vaibhav Prasad How was your? Harsh Shrivastava · 12 months ago

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@Harsh Shrivastava screwed up Vaibhav Prasad · 12 months ago

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@Vaibhav Prasad Same here. Harsh Shrivastava · 12 months ago

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@Harsh Shrivastava How much are u getting I may get around 60 Samarth Agarwal · 12 months ago

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@Samarth Agarwal 40 - 45 :( Harsh Shrivastava · 12 months ago

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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 Sauditya Yo Yo · 12 months ago

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@Sauditya Yo Yo Yes i did the same Tejas Khairnar · 12 months ago

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@Rajdeep Dhingra 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 \( 2k \ - \ 0.5 \) Kunal Verma · 12 months ago

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@Rajdeep Dhingra It is not required to produce all answers. It is required to prove that there are infinitely many which this one general solution does. Kunal Verma · 12 months ago

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4 Answer \( 4800 \)

Not sure about the 3rd one. No one in my centre even got the wind of it. Kunal Verma · 12 months ago

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@Kunal Verma Are you sure it is 4800? Kushagra Sahni · 12 months ago

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@Kushagra Sahni Well solutions aren't out but I guess so. Kunal Verma · 12 months ago

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@Kunal Verma I got 6275 Kushagra Sahni · 12 months ago

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@Kushagra Sahni Well nobody knows you maybe correct. Let's just wait for the solution. Anyways all I did was 36C3 - 36C1 - 36x30-36x34 Kunal Verma · 12 months ago

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Any idea on how to solve 2 and 5 ? Rohit Kumar · 12 months ago

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@Rohit Kumar 5 was very very lengthy at least for me. Kunal Verma · 12 months ago

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@Kunal Verma 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 ? Rohit Kumar · 12 months ago

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@Rohit Kumar How many marks u would expect , i did the same thing in the problem Tejas Khairnar · 12 months ago

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@Tejas Khairnar 6-7, because half the work was done. Rohit Kumar · 12 months ago

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@Rohit Kumar Our sir said we would get mininmum 10 Tejas Khairnar · 12 months ago

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@Tejas Khairnar Let's hope so. Did you ask the solution by the way ? Rohit Kumar · 12 months ago

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@Rohit Kumar 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. Kunal Verma · 12 months ago

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