# 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
5 years, 6 months ago

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## Comments

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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 $= 7140-36-1152-540=5412$.

- 5 years, 6 months ago

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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?

- 5 years, 6 months ago

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No u looser

- 4 years, 8 months ago

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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?

- 5 years, 6 months ago

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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!

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Did the same ! Upvoted

- 5 years, 6 months ago

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NO inequality Question this year :(

- 5 years, 6 months ago

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:(

- 5 years, 4 months ago

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Thank u bhaiya

- 3 years, 8 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)

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Yaah!! That is also a solution.

- 5 years, 6 months 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.

- 5 years, 6 months ago

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Smae paper was for maharashtra and goa

- 5 years, 6 months ago

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How do we do q2?

- 5 years, 6 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.

- 5 years, 6 months ago

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

- 5 years, 5 months ago

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Can you please write the full solution ?

- 5 years, 5 months ago

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Any News about the result?

- 5 years, 5 months ago

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Not till now.

- 5 years, 5 months ago

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

- 5 years, 6 months ago

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How many did you solve Rajdeep?

- 5 years, 6 months ago

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4

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Not solved. Burnt an hour on it.

- 5 years, 6 months ago

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I also spent an hour on it and got 6275.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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What is the answer of question 6

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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3616 in rajasthan region 32 objects were taken

- 5 years, 6 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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Yup, I know.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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Ans of 4th is 5412

- 5 years, 6 months ago

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Q 4) 5376 (INCLUSION AND EXCLUSION PRINCIPLE)

- 5 years, 6 months ago

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according to me answer to the fourth is 5412

- 5 years, 6 months ago

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Do you study at FIITJEE?

- 5 years, 6 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!

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Well Aryan r u selected for GMO?

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Answer to question 6. Let $a=m+\frac{b}{c}$ where $m$ is any integer and $0 . 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.

- 5 years, 6 months ago

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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.

- 5 years, 6 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...

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Please Inform me what is PIE method?

- 5 years, 6 months ago

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- 5 years, 6 months ago

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The marks are out guys.

- 5 years, 6 months ago

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what marks ??

- 5 years, 6 months ago

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THE RESULTS( for delhi region ) ARE OUT GUYS !

- 5 years, 5 months ago

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

- 5 years, 5 months ago

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Why didn't you give RMO this year??

- 4 years, 8 months ago

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I will give from Rajasthan this year.

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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Yup

- 4 years, 8 months ago

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Best of luck.

- 4 years, 8 months ago

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Hey you left FIITJEE?

- 4 years, 8 months ago

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Sorry was asking @Rajdeep Dhingra

- 4 years, 8 months ago

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Me = yes

- 4 years, 8 months ago

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Oh where do u go now?

- 4 years, 8 months ago

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Allen

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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He is Allen Jaipur I think. I am in kota

- 4 years, 8 months ago

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So u k him?

- 4 years, 8 months ago

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Is FIITJEE South Delhi better or Allen.

- 4 years, 8 months ago

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@rajdeep das Allen I guess

- 4 years, 8 months ago

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Same doubt

- 4 years, 8 months ago

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U are from FIITJEE?

- 4 years, 8 months ago

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Yeah I am

- 4 years, 8 months ago

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Both have there pro and cons

- 4 years, 8 months ago

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Yaah but still...?

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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Same to you :)

- 4 years, 8 months ago

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Congrats @rohit kumar

- 5 years, 5 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.

- 5 years, 6 months ago

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Are you sure it is 4800?

- 5 years, 6 months ago

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Well solutions aren't out but I guess so.

- 5 years, 6 months ago

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I got 6275

- 5 years, 6 months ago

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

- 5 years, 6 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

- 5 years, 6 months ago

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Yes i did the same

- 5 years, 6 months ago

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Hey @Rajdeep Dhingra how was ur paper ...

- 5 years, 6 months ago

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How was your?

- 5 years, 6 months ago

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screwed up

- 5 years, 6 months ago

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Same here.

- 5 years, 6 months ago

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How much are u getting I may get around 60

- 5 years, 6 months ago

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40 - 45 :(

- 5 years, 6 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

- 5 years, 6 months ago

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Please solve 5th.

- 5 years, 6 months ago

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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

- 5 years, 6 months ago

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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).

- 5 years, 6 months ago

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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)

- 5 years, 6 months 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 ?

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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results out.sorry to say but u are not selected

- 5 years, 5 months ago

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Did you get selected?

- 5 years, 5 months ago

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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.

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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

- 5 years, 6 months 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.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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Where can we get full detailed solutions for this paper

- 5 years, 5 months ago

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Any idea on how to solve 2 and 5 ?

- 5 years, 6 months ago

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5 was very very lengthy at least for me.

- 5 years, 6 months ago

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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 ?

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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6-7, because half the work was done.

- 5 years, 6 months ago

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Our sir said we would get mininmum 10

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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