INMO 2017 Practice SET-II (Algebra only)

Hello friends, try these problems and post solutions :

$(1)$ For each positive integer $n$, show that there exists a positive integer $k$ such that

$k = f(x){ (x + 1) }^{ 2n }+ g(x)({ x }^{ 2n }+ 1)$

for some polynomials $f$ , $g$ with integer coefficients, and find the smallest such $k$ as a function of $n$ .

( $2)$ Prove that there exists a polynomial $P(x, y)$ with real coefficients such that $P(x, y) \ge 0$ for all real numbers $x$ and $y$ , which cannot be written as the sum of squares of polynomials with real coefficients.

$(3)$ Let ${ x }_{ 1 } = 2$ and ${ x }_{ n + 1 } = { x }^{ 2 }n - { x }_{ n } + 1$ for $n\ge 1$ .

Prove that

$\large\ 1 - \frac { 1 }{ { { 2 }^{ 2 } }^{ n - 1 } } < \frac { 1 }{ { x }_{ 1 } } + \frac { 1 }{ { x }_{ 2 } } +...+ \frac { 1 }{ { x }_{ n } } < 1 - \frac { 1 }{ { { 2 }^{ 2 } }^{ n } }$ .

$(4)$ Find, with proof, all nonzero polynomials $f(z)$ such that $f({ z }^{ 2 }) + f(z)f(z + 1) = 0$ .

$(5)$ Find the least real number $r$ such that for each triangle with side lengths $a$ , $b$ , $c$ ,

$\large\ \frac { max(a, b, c) }{ \sqrt [ 3 ]{ { a }^{ 3 } + { b }^{ 3 } + { c }^{ 3 } + 3abc } } < r$ .

$(6)$ Let $a$ , $b$ , $c$ be positive real numbers. Prove that

$\large\ { \left( \sum { \frac { a }{ b+c } } \right) }^{ 2016 } \le \left( \sum { \frac { { a }^{ 2016 } }{ { b }^{ 2016 } + { b }^{ 2015 }c } } \right) { \left( \sum { \frac { a }{ c + a } } \right) }^{ 2015 }$ .

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2^{34}

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Easy Math Editor

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Note: you must add a full line of space before and after lists for them to show up correctly
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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

The answer to4 is f(x)=0 &f(x)=(-1)^n+1(x(1-x))^n

himanshu singh - 5 years, 5 months ago

Correct. Can you show the solution?

Priyanshu Mishra - 5 years, 5 months ago

In the he eq put x0 such that f(x0)=0giving that f(x0^2)also equals to zerozero thus by induction there are infinitely many roots of funless x0=0/1 infinitely many roots gives f(x)=0The other case gives after some calc that f(x)=0 for 0,1 only thus the other function is of the formx^m(x-1)^nkeeping in mind that the leading coefficient must be negative keep the above poly in the given funct eq to obtain n=m some thought to the leading coefficient gives the -1at the beginning

I am in desperate need of geometry and combinatorics sets pls hlp

I will post it before INMO.

Is the ans to Q5 2^1/3 i m doubtful

Yes I think you forgot to write it as 1/cube root 2.

Will the other discussions also be present on Brilliant Discussion link as this one

Yes.

I just got the 3 one.its as follows prove by induction that the given expressions equals1-( (x1x2.......xn)^-1) then make the reqd changes in the inequality. By the way the above induction also showds that xn=(x1x2.......xn-1)+1 Now again induction on the transformed ineq completes the sol

Nice, concise solution.

HAVE you solved the problem 6?

Not started yet. But i ve no idea of the first one pls post the sol or give a hint

I cant understand what u say

I suppose that 6 uses the chebyshev's once

This is the proposal so I don't know the answer.

I have solution of Q5 Is answer (-1)^2k +1 {( z(z-1)}^n

kanchan jindal - 2 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$