BOMC-2012 Test Day 1 Problem 2

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nafistiham
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by nafistiham » Sat Apr 14, 2012 9:45 pm

SANZEED wrote:Mahi vai,I did not understand the thing you wrote for me,would you please make it clear by examples? :? :? :oops:
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. :)
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Tahmid Hasan
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by Tahmid Hasan » Thu Apr 19, 2012 11:37 am

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]
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by *Mahi* » Thu Apr 19, 2012 12:46 pm

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Tahmid Hasan
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by Tahmid Hasan » Thu Apr 19, 2012 12:53 pm

Yes,just like I mentioned I've faced the problem before.
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by *Mahi* » Thu Apr 19, 2012 2:44 pm

I just shared the link :)
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