Sanzeed, most of the theorems of inequality are not applicable in all general cases.You have to be careful before applying them.and finding out the case.SANZEED wrote:Mahi vai,I did not understand the thing you wrote for me,would you please make it clear by examples?
BOMC-2012 Test Day 1 Problem 2
- nafistiham
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\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- Tahmid Hasan
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Re: BOMC-2012 Test Day 1 Problem 2
It's a little late to post the solution but my method was different form others and quite algebric and brute force
so here it goes
we can write this equation into a quadratic equation with variable $a$.
\[ca^2+(b^2-3bc)a+bc^2=0\]
$\Leftrightarrow$ \[a= \frac {3bc-b^2 \pm \sqrt {(b^2-3bc)^2-4bc^3}}{2c}\].....(1)
the expression under the square root can be simplified into
$(b-c)^2(b^2-4bc)$
which means $b^2-4bc$ is the square of an integer.Let it be $t$.
so $b^2-4bc=t^2$
or,$c= \frac {b^2-t^2}{4b}$.....(2)
plugging this value in (1) and simplifying we get
\[a= \frac {-b^3-3bt^2 \pm (3b^2t+t^3)}{8bc}\]
or,$abc=(\frac {-b \pm t}{2})^3$
so it suffices to prove that $-b \pm t$ is even.
from (2) we get $4 \mid b^2-t^2$,which means $b,t$ are both even or odd which satisfies our above condition.[Showed]
so here it goes
we can write this equation into a quadratic equation with variable $a$.
\[ca^2+(b^2-3bc)a+bc^2=0\]
$\Leftrightarrow$ \[a= \frac {3bc-b^2 \pm \sqrt {(b^2-3bc)^2-4bc^3}}{2c}\].....(1)
the expression under the square root can be simplified into
$(b-c)^2(b^2-4bc)$
which means $b^2-4bc$ is the square of an integer.Let it be $t$.
so $b^2-4bc=t^2$
or,$c= \frac {b^2-t^2}{4b}$.....(2)
plugging this value in (1) and simplifying we get
\[a= \frac {-b^3-3bt^2 \pm (3b^2t+t^3)}{8bc}\]
or,$abc=(\frac {-b \pm t}{2})^3$
so it suffices to prove that $-b \pm t$ is even.
from (2) we get $4 \mid b^2-t^2$,which means $b,t$ are both even or odd which satisfies our above condition.[Showed]
বড় ভালবাসি তোমায়,মা
Re: BOMC-2012 Test Day 1 Problem 2
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
- Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
Re: BOMC-2012 Test Day 1 Problem 2
Yes,just like I mentioned I've faced the problem before.*Mahi* wrote:Hmmm http://www.artofproblemsolving.com/Foru ... 3#p1857643
বড় ভালবাসি তোমায়,মা
Re: BOMC-2012 Test Day 1 Problem 2
I just shared the link
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi