INMO 2017 Practice SET-II (Algebra only)

Hello friends, try these problems and post solutions :

(1)(1) For each positive integer nn, show that there exists a positive integer kk such that

k=f(x)(x+1)2n+g(x)(x2n+1)k = f(x){ (x + 1) }^{ 2n }+ g(x)({ x }^{ 2n }+ 1)

for some polynomials ff, gg with integer coefficients, and find the smallest such kk as a function of nn.


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


(3)(3) Let x1=2{ x }_{ 1 } = 2 and xn+1=x2nxn+1{ x }_{ n + 1 } = { x }^{ 2 }n - { x }_{ n } + 1 for n1n\ge 1.

Prove that

 1122n1<1x1+1x2+...+1xn<1122n\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)(4) Find, with proof, all nonzero polynomials f(z)f(z) such that f(z2)+f(z)f(z+1)=0f({ z }^{ 2 }) + f(z)f(z + 1) = 0.


(5)(5) Find the least real number rr such that for each triangle with side lengths aa, bb, cc,

 max(a,b,c)a3+b3+c3+3abc3<r\large\ \frac { max(a, b, c) }{ \sqrt [ 3 ]{ { a }^{ 3 } + { b }^{ 3 } + { c }^{ 3 } + 3abc } } < r.


(6)(6) Let aa, bb, cc be positive real numbers. Prove that

 (ab+c)2016(a2016b2016+b2015c)(ac+a)2015\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 }.

Note by Priyanshu Mishra
3 years, 10 months ago

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1 vote

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

himanshu singh - 3 years, 10 months ago

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Correct. Can you show the solution?

Priyanshu Mishra - 3 years, 10 months ago

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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

himanshu singh - 3 years, 10 months ago

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

himanshu singh - 3 years, 10 months ago

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I will post it before INMO.

Priyanshu Mishra - 3 years, 10 months ago

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

himanshu singh - 3 years, 10 months ago

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

Priyanshu Mishra - 3 years, 10 months ago

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

himanshu singh - 3 years, 10 months ago

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Yes.

Priyanshu Mishra - 3 years, 10 months ago

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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

himanshu singh - 3 years, 10 months ago

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Nice, concise solution.

HAVE you solved the problem 6?

Priyanshu Mishra - 3 years, 10 months ago

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

himanshu singh - 3 years, 10 months ago

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I cant understand what u say

himanshu singh - 3 years, 10 months ago

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I suppose that 6 uses the chebyshev's once

himanshu singh - 3 years, 10 months ago

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

Priyanshu Mishra - 3 years, 10 months ago

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

kanchan jindal - 1 year, 4 months ago

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