Jharkhand RMO 2015

This time RMO of jharkhand was more difficult than last year .. what do you all reckon what will be cut off for this year?..

#RMO #JHARKHAND

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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

one question was: prove that the roots of the equation x^3-3x^2-1 = 0 are never rational.

Gyanendra Prakash - 5 years, 6 months ago

By rational root theorem, the only possible rational roots of the above polynomial could be $\pm1$ which are not in this case. Therefore, the roots are Irrational of Imaginary.

Either it is not a RMO question or my approach is not complete.

Vishal Yadav - 5 years, 6 months ago

this is surely from RMO (jharkhand RMO 2015)...but i have solved this problem integrating 3 concepts : theory of equations, inequalities and then final proof by contradiction

Gyanendra Prakash - 5 years, 6 months ago

@Gyanendra Prakash – 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

procceding with equation 1

i got after expanding and replacing q1q2q3 by p1p2p3...

recoiprocal 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

therfore by contradiction roots can't be rational...

i suppose it's corrcet..

Gyanendra Prakash - 5 years, 6 months ago

@Gyanendra Prakash – so does my solution deserve any credit?

Gyanendra Prakash - 5 years, 6 months ago

@Gyanendra Prakash – I don't know, but if it is a RMO Question it should be solved somewhat the way you did it. I can't actually understand your solution, should use LaTex and you left the most part of your solution that is the steps with which you solved.

Vishal Yadav - 5 years, 6 months ago

@Vishal Yadav – ok..i will improve it

Gyanendra Prakash - 5 years, 6 months ago

In a group of seven persons every person is asked to write the sum of ages of all the other six persons.if all the seven sums form a six element set {63,64,65,66,68,70}..then find the individual ages assuming ages are whole numbers..

Gyanendra Prakash - 5 years, 6 months ago

find all triplets such that (a,b,c) of positive integers such that a<=b<=c: 3 does not divide b.c and (1+1/a)(2+1/b)(3+1/c)=13

Gyanendra Prakash - 5 years, 6 months ago

last time cut off was 45. by the way how much are you getting this time?

Aditya Kr - 5 years, 6 months ago

If the roots of x^3+px^2+qx+r=0 are all real and +ve...prove pq-9r>=0 and q^3-27r^2>=0

Gyanendra Prakash - 5 years, 6 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}`	$\boxed{123}$