# KVS 2014 Q7

I was doing the the following problem-
Prove that$\frac { \sqrt { a+b+c } +\sqrt { a } }{ b+c } +\frac { \sqrt { a+b+c } +\sqrt { b } }{ c+a } +\frac { \sqrt { a+b+c } +\sqrt { c } }{ a+b } \ge \frac { 9+3\sqrt { 3 } }{ 2\sqrt { a+b+c } }$.

I normalized this to $a+b+c=1$ and simplified to get-

$\frac { 1 }{ 1-\sqrt { a } } +\frac { 1 }{ 1-\sqrt { b } } +\frac { 1 }{ 1-\sqrt { c } } \ge \frac { 9+3\sqrt { 3 } }{ 2 }$
By Titu's lemma,
$\Rightarrow \quad \frac { 1 }{ 1-\sqrt { a } } +\frac { 1 }{ 1-\sqrt { b } } +\frac { 1 }{ 1-\sqrt { c } } \ge \frac { 9 }{ 3-(\sqrt { a } +\sqrt { b } +\sqrt { c } ) }$
However the $RHS$ is maximized when $\sqrt { a } +\sqrt { b } +\sqrt { c }$ is maximized which is at $\sqrt { 3 }$.
$\Rightarrow \quad \frac { 1 }{ 1-\sqrt { a } } +\frac { 1 }{ 1-\sqrt { b } } +\frac { 1 }{ 1-\sqrt { c } } \ge \frac { 9 }{ 3-(\sqrt { a } +\sqrt { b } +\sqrt { c } ) } \le \frac { 9 }{ 3-\sqrt { 3 } } =\frac { 9+3\sqrt { 3 } }{ 2 }$

What went wrong?

You can view the rest of the problems here

Note by Siddharth G
4 years, 10 months ago

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## Comments

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@Siddharth G :- many congrats for clearing RMO!!!. How many problems did you solve?

- 4 years, 9 months ago

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Thanks! I attempted 5 questions, expected to get 4-4.5. How was your paper? (GMO right?)

- 4 years, 9 months ago

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Read This. I should say, I didn't do as well as you did. (Regardless of the paper's standard)

- 4 years, 9 months ago

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Nice solution!

The second last line should be; "RHS is maximised when √a+ √b+ √c is 'minimised' " as the whole fraction is maximised when its denominator is minimised.

However your deduction is very perfect.

In which class presently you are?

- 4 years, 1 month ago

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11th As for the second last line, $\sqrt{a} +\sqrt{b} +\sqrt{c}$ when maximized, yields $3-(\sqrt{a} +\sqrt{b} +\sqrt{c})$ to be minimized,which implies that the RHS is maximized.

- 4 years, 1 month ago

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Thanks. Now i understood that.

I am also preparing for RMO. So as an elder please guide me that which book will be helpful for me to clear RMO?

Also i get to know that you have cleared RMO , so please tell me which book helped you a lot.

- 4 years, 1 month ago

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A lot of it was luck (including the level of the paper). Regardless I found Challenge and Thrill of Pre-College Mathematics and 'Mathematical Circles' to be good at the basic level. Once you have (sufficiently) completed them go for Problem Solving Strategies(Arthur Engel) and books by Titu Andreescu. And keep trying Olympiad questions all the time.

- 4 years, 1 month ago

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Ok.

Are 'Problem primer for the olympiads' and 'Number theory-problems, structures and examples'(by titu .A.) not fine for RMO/INMO ?

- 4 years, 1 month ago

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Titu is a good book. Even though Problem Primer is recommended by a lot of people, I did not like it that much.

- 4 years, 1 month ago

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Hello Siddharth G,

Please help me in solving this question ( KVS JMO 2015)

A polynomial $f(x)$ with rational coefficients leaves remainder $15$, when divided by $x-3$ and remainder $2x+1$, when divided by $(x-1)^2$. Find the remainder when $f(x)$ is divided by $(x-3)(x-1)^2$.

- 4 years, 1 month ago

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The answer seems to be $2x^2-2x+3$. Though I havent found a good solution yet

- 4 years, 1 month ago

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Your answer is correct.

You can tell me your approach, irrespective of good or bad.

- 4 years, 1 month ago

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I took a cubic which left a remainder $2x+1$ when divided by $(x-1)^2$ ,
$2x^3-4x^2+4x+1 = (x-1)^2(x+2) + 2x+1$. It left a remainder 31 when divided by $x-3$. Note that $(x-1)^2$ leaves a remainder of 4 when divided by $x-3$.
Thus, $2x^3-4x^2+4x+1 - 4(x-1)^2 =2x^3-8x^2+12x-3$ leaves a remainder of $31-4(4)=15$. Hence we find the remainder of $2x^3-8x^2+12x-3$ when divided by $(x-1)^2(x-3)$.

- 4 years, 1 month ago

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Got a better solution from my friend- Clearly, $f(1)=3$ and $f(3)=15$.
Then, $f(x)=q_1(x)(x-3)(x-1)^2+r_1(x)$, where $r_1(x)=ax^2+bx+c$.

Dividing by $(x-1)^2$ and equating remainders,
$2x+1=0+(ax^2+bx+c-a(x-1)^2)$ [Multiplied by a to remove the coeff. of $x^2$]
$\Rightarrow 2x+1=(2a-b)x+(c-a)$ $\Rightarrow 2=2a-b, 1=c-a$
Solving with $f(1)=3=a+b+c -(1)$
(a=2, b=-2, c=3)

- 4 years, 1 month ago

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Your both solutions are pretty.

I have your 2nd solution.

What is the problem that i am not able to understand that why you have multiplied x with q1 ( 2nd line) in the polynomial?

You have taken the polynomial as

$f(x)\quad =\quad { q }_{ 1 }(x)(x-3){ (x-1) }^{ 2 }\quad +\quad { r }_{ 1 }(x)$

I am asking that why you have multiplied x in above polynomial as it is not given?

- 4 years ago

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That is not multiplication. $q_1(x)$ as a whole is a polynomial and in this case the quotient.(ie $q_1 of x$ akin to $f(x)$

- 4 years ago

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Ok. Now i understood that.

Also give me a clue to solve this problem:

Find all real numbers $x$ for which

${ 10 }^{ x }\quad +\quad { 11 }^{ x }\quad +\quad { 12 }^{ x }\quad =\quad { 13 }^{ x }\quad +\quad { 14 }^{ x }$

- 4 years ago

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Try dividing by a suitable number on both sides of the equation.

- 4 years ago

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Ok, got that.

Please tell me that can i apply CAUCHY - SCHWARZ lemma to prove this inequality ? -

Given $a, b, c$ are positive real numbers such that

${ a }^{ 2 }\quad +\quad { b }^{ 2 }\quad +\quad { c }^{ 2 }\quad =\quad 3abc$.

Prove that:

$\large\ \frac { a }{ { b }^{ 2 }{ c }^{ 2 } } +\frac { b }{ { c }^{ 2 }{ a }^{ 2 } } +\frac { c }{ { a }^{ 2 }{ b }^{ 2 } } \ge \quad \frac { 9 }{ a+b+c }$

- 4 years ago

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Make the numerators on the LHS to the fourth power and then apply the lemma.

- 4 years ago

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Did you mean this?

$\large\ \frac { { a }^{ 4 } }{ { a }^{ 3 }{ b }^{ 2 }{ c }^{ 2 } } +\frac { { b }^{ 4 } }{ { b }^{ 3 }{ c }^{ 2 }{ a }^{ 2 } } +\frac { { c }^{ 4 } }{ { c }^{ 3 }{ a }^{ 2 }{ b }^{ 2 } } \ge \frac { 9 }{ a+b+c }$

$\large\ \frac { { a }^{ 4 } }{ { a }^{ 3 }{ b }^{ 2 }{ c }^{ 2 } } +\frac { { b }^{ 4 } }{ { b }^{ 3 }{ c }^{ 2 }{ a }^{ 2 } } +\frac { { c }^{ 4 } }{ { c }^{ 3 }{ a }^{ 2 }{ b }^{ 2 } } \ge \frac { { ({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }) }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }(a+b+c) }$

$\large\ \quad \ge \frac { { 9({ abc }) }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }(a+b+c) }$

$\large\ \quad =\frac { 9 }{ a+b+c }$

- 4 years ago

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Hello Siddharth G,

As you have cleared RMO, can you tell me which book you preferred specially for Geometry and functional equations?

- 4 years ago

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I am terrible at geometry, so I would not recommend anything for that. As for functional equations, the key is to do a lot of problems after learning the basic techniques. I have not found a 'perfect' book, but the good ones are by BJ Venkatachala, Christopher G Small and Problem solving strategies by Arthur Engel.

- 4 years ago

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Ok. Also i want to ask you that who guided you the most for RMO ?

Any professor or you did by yourself or anyone else?

Also without geometry how can one dream for winning RMO or further olympiads? It is the whole and sole .

- 4 years ago

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Did it by myself. I have to make-do without geometry. I am doing geometry (in Challenge and thrill) but it seems futile.

- 4 years ago

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O.k.

Please help me to factorise this one:

$\large\ { x }^{ 2 }+4{ y }^{ 2 }-2xy-2x-4y-8=0$

- 4 years ago

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well this can be simplified by completing the squares.

- 3 years, 10 months ago

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OK but please explain it here.

- 3 years, 10 months ago

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Multiply both sides by 2 and try completing the square method.

- 3 years, 8 months ago

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Hello Siddharth G,

Appeared in RMO 2015? How was it? What is your expected score?

- 3 years, 10 months ago

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4 Questions. How was yours?

- 3 years, 10 months ago

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Actually I am in K.V JNU so i cannot appear in RMO directly. I have to qualify JMO, which i gave this year but unfortunately did not make it. I will try next year( class 11th).

By the way, how was your preparation this year?

Did you study something more this year than what you studied last year for RMO 2014?

- 3 years, 10 months ago

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Nothing special, just deeper into the usual topics.

- 3 years, 10 months ago

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Was the paper harder than last year?

- 3 years, 10 months ago

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A bit, but not harder than average.

- 3 years, 10 months ago

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So, are you happy with what you did in the exam?

What is the your probability of clearing RMO?

- 3 years, 10 months ago

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I prefer not to think about it. Waiting for the results.

- 3 years, 10 months ago

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Nice.

As you have qualified RMO 2014, can you tell me that is the INMO camps conducted in delhi or no camp is conducted?

- 3 years, 10 months ago

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Sorry, didnt see the comment. An INMO camp is conducted int IIT Delhi for RMO awardees

- 3 years, 10 months ago

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For how many days?

Which topics were taught to you?

If possible, can you send me some notes of that camp to my e-mail?

- 3 years, 10 months ago

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I can send the assignments, but my notes are terrible.

- 3 years, 10 months ago

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Thanks for the initiation.

Have you solved this RMO question in the exam? :

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

- 3 years, 10 months ago

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Put z=0 and $x=-a^4, y=a^3$ for some natural a.

- 3 years, 10 months ago

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How putting of these variables clicked to your mind?

Before getting this, what steps you did?

- 3 years, 10 months ago

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The exponent 31 seemed awkward and misplaced. Thus it seemed that it had no effective value.

- 3 years, 10 months ago

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Can't we use the FMID- variant in this question?

- 3 years, 10 months ago

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FMID is usually used for proving that no solutions exist. However, ther are infinitely many non-trivial solutions as well.

- 3 years, 10 months ago

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Ok.

I was doing this question:

Find all positive integers $n$ such that $3^{n - 1} + 5^{n - 1}$ divides $3^n + 5^n$.

The solution in the book was:

Note that $\large\ 1 < \frac { { 3 }^{ n } + { 5 }^{ n } }{ { 3 }^{ n - 1 } + { 5 }^{ n - 1 } } <5$, so we can have only $\large\ \frac { { 3 }^{ n } + { 5 }^{ n } }{ { 3 }^{ n - 1 } + { 5 }^{ n - 1 } } \in \{ 2,3,4\}$
cases, which are easily checked.

Can you explain me why the solution wrote $\large\ 1 < \frac { { 3 }^{ n } + { 5 }^{ n } }{ { 3 }^{ n - 1 } + { 5 }^{ n - 1 } } <5$ above ?

I am not understanding how that expression came.

- 3 years, 10 months ago

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The numerator is clearly greater than the denominator. For the second part of the inequality note that $3^n<3^{n-1} \cdot 5$

- 3 years, 10 months ago

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But why the solution bounded the fractional part between 1 and 5?

Why it chose 1 and 5 only?

- 3 years, 10 months ago

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1 is just a lower bound by noticing that the numerator is greater than the denominator. For 5, it might help to see that a n tends to infinity, the expression tends to 5

- 3 years, 10 months ago

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Ok, i got it.

Are the RMO results out?

- 3 years, 10 months ago

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no

- 3 years, 10 months ago

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Which questions you left in RMO?

Were they geometry ones?

- 3 years, 10 months ago

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Did 1,3,4,6.

- 3 years, 10 months ago

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Means you did this also:

Show that there are infinitely many positive real numbers $a$ which are not integers such that $a(a-3\text{{a}})$ is an integer.

Which method or theorem you applied?

- 3 years, 10 months ago

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My method is too long. You should check out some other solutions. Eg A family of infinite solutions is odd+0.5.

- 3 years, 10 months ago

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Can you help me to draw the diagram for question 1?

I am unable to draw it even after trying 5 times?

Please elaborate the steps.

- 3 years, 10 months ago

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http://postimg.org/image/cafnb0sc5/

- 3 years, 10 months ago

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Thanks for the image. Can you please now elaborate the solution here.?

- 3 years, 10 months ago

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http://www.isical.ac.in/~rmo/rmo(underscore)2015(underscore)wb(underscore)questions(underscore)and(underscore)solutions(underscore)u(underscore)1.pdf. Replace (underscore) by _

- 3 years, 10 months ago

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Sorry, i found that.

- 3 years, 10 months ago

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You can do problems for INMO 2016 from the set " INMO 2016 PRACTICE SET-1" posted by me . It consists only of 6 number theory problems.

- 3 years, 10 months ago

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thanks!

- 3 years, 10 months ago

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You can also post solutions there if you have any.

- 3 years, 10 months ago

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Congratulations Siddharth for RMO 2015 winner.

Hope that you qualify INMO 2016 also, with flying colours.

- 3 years, 10 months ago

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Thanks! I hope so too, but it seems unlikely.

- 3 years, 10 months ago

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It seems unlikely (seems likely by me) , but with some little efforts it is also possible.

- 3 years, 10 months ago

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Help me in this NMTC question:

$f(x)$ is a fifth degree polynomial. It is given that $f(x) + 1$ is divisible by $(x - 1)^3$ and $f(x) - 1$ is divisible by $(x + 1)^3$. Find $f(x)$.

- 3 years, 9 months ago

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Let $f(x)=ax^5+bx^4+cx^3+dx^2+ex+f$. Note that $f(1)-1=f(-1)+1=f'(1)=f'(-1)=f''(1)=f''(-1)=0$. You have six variables and six equations.

- 3 years, 9 months ago

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What is meant by $f '(1)$ and $f ''(1)$?

Integration or differentiation?

Thanks, but I am unable to understand your solution.

- 3 years, 9 months ago

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differentiation

- 3 years, 9 months ago

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How can i differentiate $f '(1)$ and $f ''(1)$ as $f(1) = a + b + c + d + e + f$?

According to my knowledge we can differentiate a function only w.r.t some variable , but here are 6 variables.

Could you elaborate the differentiation here?

- 3 years, 9 months ago

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differentiate f(x) and then substitute x=1 into f '(x).

- 3 years, 9 months ago

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Understood but what is differentiation of a constant? 1 or 0?

- 3 years, 9 months ago

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1. Do you have problem solving stategies? This method is (kind of) mentioned in the second page of polynomials.

- 3 years, 9 months ago

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I saw that but no such problem is there. On second page of polynomials there are points numbered 4, 5, 6 but i am unable to find. Can you tell the point number?

- 3 years, 9 months ago

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not a problem. See point 4

- 3 years, 9 months ago

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What is meant by multiplicity?

- 3 years, 9 months ago

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A root $a$ of $f(x)$ has multiplicity $m$ if $(x-a)^m | f(x)$ and $(x-a)^{m+1}$ does not divide f(x).

- 3 years, 9 months ago

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I think i should leave it. I am not of that level to understand.

Sorry for your inconvenience.

- 3 years, 9 months ago

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The whole concept might be a bit difficult without a grasp of differentiation. But you can still solve the problem, if you know how to differentiate a polynomial.

- 3 years, 9 months ago

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I solved six equations but i am getting all the values to be 0?!!

Can you tell me how to start?

Also thanks a lot for sending INMO 2015 notes.

- 3 years, 9 months ago

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a=9/4, c=-5/4 rest all are zero

- 3 years, 9 months ago

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But a=-3/8)\ \(c=5/4 (e=-15/8)\ and all rest 0 is the correct answer.

- 3 years, 9 months ago

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Indeed. I made a mistake in my calculations. This is the system you got right? $a+b+c+d+e+f=1 \\ -a+b-c+d-e+f=-1 \\ 5a+4b+3c+2d+e=0 \\ 5a-4b+3c-2d+e=0 \\ 20a+12b+6c+2d=0 \\-20a+12b-6c+2d=0$

- 3 years, 9 months ago

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Yes that it is.

I have found another solution without calculus(working half an hour on it). It is completely algebraic.

Here it is:

Let $f(x) = (x - 1)^3(a{x}^2 + bx + c) - 1$. $...(1)$

As $(x + 1)^3|f(x) - 1$ we have

$f(x) - 1 = a{x}^5 + x^4(b - 3a) + x^3(c - 3b + 3a) + x^2(3b - 3c - a) + x(3c - b) - c - 2$.

and the RHS is divisible by $(x + 1)^3$ , so by dividing RHS by $(x + 1)^3$

We have the remainder $(-38a - 6c +18b)x^2 + (-48a + 16b)x - 2c - 18a +6b - 2 = 0$

So, we have three simultaneous equations viz:

$38a + 6c -18b = 0$ ; $48a - 16b = 0$ ; $2c + 18a - 6b + 2 = 0$

which on solving together gives $a = \frac {-3}{8}$ ; $b = \frac {-9}{8}$ ; $c = -1$.

Putting these values in $(1)$ and doing some tedious calculations we get

$f(x) = \boxed{\frac { -3 }{ 8 } { x }^{ 5 }+ \frac { 5 }{ 4 } { x }^{ 3 } - \frac { 15 }{ 8 } x}$ which is the correct answer.

Can you please rate my solution?

- 3 years, 9 months ago

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I received assignments of Functional equations and geometry. Are there any assignments on NUMBER THEORY?

- 3 years, 9 months ago

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No. Though I have a collection of Number Theory Questions from Mumbai RMO.

- 3 years, 9 months ago

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Ok. Have you received any e-mail regarding INMOTC 2016?

- 3 years, 9 months ago

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No.

- 3 years, 9 months ago

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Explain the meaning of this question:

The arithmetic mean of a pair wise distinct prime numbers is 27. Determine the biggest prime among them.

- 3 years, 9 months ago

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The question is terribly framed. I am not sure but I think it means the AM of two different primes is 27. Fins thw maximal possible value of one of the primes.

- 3 years, 9 months ago

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Yes, it seemed like that to me also.

Which inequality will you apply to solve this ?:

a, b, c are real numbers such that their sum is $0$ an sum of their square is 1. Find the maximum value of $a^2.b^2.c^2$.

- 3 years, 9 months ago

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Which point number?

- 3 years, 9 months ago

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Hey, I remember this. I have solved this one in some NMTC set on brilliant just using simple differential.

- 3 years, 8 months ago

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Hello Siddharth,

how you drew the image of the rmo question on the website which you sent me?

Please tell the website where I can draw the diagrams.

- 3 years, 9 months ago

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http://web.geogebra.org/. Though it is preferable that you you should at least try to draw it on your own first. PS Sorry for the late response. Couldnt solve your previous problem. Do you have a solution?

- 3 years, 9 months ago

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I have drawn a number of diagrams in geogebra but could not save it on PC so that I can post problems in BRILLIANT. This is my problem. Please help me that how can I save it in PC.

No I don't have solution to that inequality .

Please tell me how to save the diagrams of geogebra in desktop as soon as possible.

- 3 years, 9 months ago

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Under menu-- export--png

- 3 years, 9 months ago

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Not understood properly.

Should I need to login on the geogebra?

- 3 years, 9 months ago

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I dont think so. On the top right corner, (its like an equivalence sign with horizontal bars), go to export and select png. this works on the web version. An alternate method is to take a screenshot with web apps like Nimbus.

- 3 years, 9 months ago

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Thanks for the assist. I will try it and hope that it will work.

By the way how's your preparation for INMO? Was INMOTC conducted at IIT?

If possible, please send me the assignments. You can take as many days you can.

- 3 years, 9 months ago

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No INMOTC this year.

- 3 years, 9 months ago

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What? So bad.

IIT professors don't take it initiative. They are just regional coordinators for name.

In KV, 20 DAYS camp is conducted but rarely a student is selected in INMO, where students from delhi region attend no camp, yet so many qualify. GREAT IRONY!!

Also thanks for helping me to post geometric figures. The method worked.

How was your INMO? Can you send me the paper?

Also, clear my doubts of 2 questions i have posted - "Help residues mod 7 continents".

Please post solutions also.

- 3 years, 9 months ago

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http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYS85LzlkNDNhMWMzNGM3ZDdmNWQ4ZGE3ZDI2NjIxYzg2MThkNjQ2MWM1LmpwZw==&rn=U2Nhbl8yMDE2MDExNy5qcGc= It was okay. Asfor the camp, last time we had one professional olympiad teacher (he gave the notes), one prof from IIT and the rest were past INMO awardees, who studied in IITD. This time the problem(I think) was mainly due to lack of time.

- 3 years, 9 months ago

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OK. How many questions you solved in INMO?

- 3 years, 9 months ago

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Suppose i have opened NIMBUS on one side and on the webpage i have opened the geogebra, where i have drawn the image. Then how can i take the snapshot of the image by nimbus?

- 3 years, 9 months ago

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I solved ~3 questions. In nimbus select "selected area".

- 3 years, 9 months ago

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WHERE IS ITS LOCATION IN NIMBUS?

- 3 years, 9 months ago

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You are using nimbus screenshot (in chrome) right?

- 3 years, 9 months ago

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YES.

- 3 years, 9 months ago

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https://whatsonmypc.files.wordpress.com/2014/10/nimbus.jpeg?w=393&h=494

- 3 years, 9 months ago

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Thanks so much. I understood.

Can you tell me what is meant by the test "SMT" conducted in all DPSs? I got to know about it from my fiitjee friend who studies at DPS RKP.

- 3 years, 9 months ago

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DPS conducts an annual exam for class nine students in partnership with Sof. It is in PCMB.

- 3 years, 9 months ago

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Ok.

Have you solved question 1 of INMO?

Also help me in these questions:

- 3 years, 9 months ago

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Did Q1 with coordinate geometry. Already tried, couldnt do it.

- 3 years, 9 months ago

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No problem. You keep trying and when you get solution , do convey me.

Also you share this problem with your friends of school and FiitJee.

- 3 years, 9 months ago

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This time a greater number of students will qualify INMO from the KVS region also. I believe 2 or 3.

- 3 years, 8 months ago

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How do you know that?

- 3 years, 8 months ago

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I qualified JMO and KVS RMO this year and appeared for INMO via KVS region, so I met them at INMOTC and also after the exam, I asked some, who I know can qualify.

- 3 years, 8 months ago

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Can you please send me the KVS RMO paper 2015? I need that as i will appear in JMO this year also.

Also is it possible for you to send me the notes of INMOTC conducted in Delhi and Hyderabad?

- 3 years, 8 months ago

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As per the notes are concerned, I am not that good at taking down notes. However, I can send you the two KVS RMO papers within this week if you give me your e-mail ID. I will send after 2 or 3 days coz we are having ASL tomorrow and the day after. BTW, which class are you in? And did you appear for JMO this year?

- 3 years, 8 months ago

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Ok , please send the papers on this e-mail id- priyanshu_2feb@rediffmail.com.

I am presently in class 10. Yes, i appeared in JMO in 2015 but could not qualify as i scored just $48$ marks out of 100.

If you cannot send the notes, then can you atleast send me the assignments provided during the camp?

You can take as many days you want.

- 3 years, 8 months ago

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How did you get to know your marks?? The excel sheet with everyone's marks is not yet out, only those who qualified have got to know their marks.

- 3 years, 8 months ago

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Actually i asked the JMO coordinator, Mr G.S. LAWANIA personally. He told me. Otherwise i was also not knowing the marks.

By the way, which camp was good - DELHI or HYDERABAD?

- 3 years, 8 months ago

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Hyderabad one was better I think

- 3 years, 8 months ago

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I doubt. If you qualified JMO, then your name must be there in the pdf format list and the marks are given there.

- 3 years, 8 months ago

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It's there, see clearly

- 3 years, 8 months ago

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When are you sending the KVS RMO papers?

- 3 years, 8 months ago

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There is a list of selected students for INMO in KVS's website now, my name is there at 16th serial no. :3

- 3 years, 8 months ago

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That's not the merit list of JMO qualifiers. Its KVS RMO qualifiers's list for INMO 2016.

This is the list of JMO 2015 qualifiers:

- 3 years, 8 months ago

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What is there at 31 no.?? See clearly :3

- 3 years, 8 months ago

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Yes that's your name, so why are you asking your marks of JMO as it is already given there?

- 3 years, 8 months ago

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No, I was not asking mine. I was just telling as ur name is not there, how did you come to know your marks?? One friend of mine also rang up Lawania sir, but he didn't tell the marks!!

- 3 years, 8 months ago

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Actually one maths teacher of my school requested him to tell the my marks of JMO. As he checked my JMO answer sheet, he knew the marks and told orally. He was also impressed by my solutions. He checked my copy 5 times so that i may be selected for INMOTC but my fate was not that good.

By the way, in which class are you?

- 3 years, 8 months ago

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In 10th

- 3 years, 8 months ago

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Can you help me in this question? :

Find all positive integers $(a, b)$ such that $\large\ \frac { \sqrt { 2 } + \sqrt { a } }{ \sqrt { 3 } + \sqrt { b } }$ is a rational number.

- 3 years, 8 months ago

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For integers, $(3,2)$ is the only solution. Just rationalize the numerator and the denominator, and then note that the radical part of each term of the numerator should be 6. So, fix $a = 2l^2$ and $b = 2k^2$ and then simplify to get $(1+l)(1-k) = 0$. And hence you get the only solution as $(3,2)$

- 3 years, 8 months ago

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However, if it is asked to find solutions not only in integers but also in rational numbers, then we can generalize a solution to get infinitely many solutions

- 3 years, 8 months ago

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Ok.

When will you send the assignments to my mail?

- 3 years, 10 months ago

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Can you please send me your notes and assignments on - kuleshvandan8@gmail.com

- 3 years, 1 month ago

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