BOMC-2012 Test Day 1 Problem 2

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*Mahi*
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BOMC-2012 Test Day 1 Problem 2

Unread post by *Mahi* » Wed Apr 11, 2012 11:06 pm

Show that if a, b, c be integers satisfying \[\frac ab +\frac bc +\frac ca = 3\] then $abc$ is cube of an integer.
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Nadim Ul Abrar
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by Nadim Ul Abrar » Thu Apr 12, 2012 1:17 pm

Hiks.. hiks ... I made a silly misteke in this problem ..... Ammmmmaaaaa.....
*Mahi* thanks ,
$\frac{1}{0}$

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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by SANZEED » Thu Apr 12, 2012 5:27 pm

I am again suggesting that it can be solved by rearrangement inequality,I found it later.(My $70$th post :P :P :P )
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by Phlembac Adib Hasan » Thu Apr 12, 2012 5:55 pm

SANZEED wrote:I am again suggesting that it can be solved by rearrangement inequality,I found it later.(My $70$th post :P :P :P )
Serious crime.They all are not necessarily positive integers.If they are, it directly comes from AM-GM that $a=b=c$.
I did not give a complete solution of this one because of the crises of time.I heard about the problem set at noon in the PMS class.As I had practicals I sat to take the exam at 4 and stopped at 11.That's the reason.But I solved this one later.Everything will be discussed in the next PMS class.And look what I have got today
http://answers.yahoo.com/question/index ... 119AAfBOXz
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by *Mahi* » Thu Apr 12, 2012 7:10 pm

SANZEED wrote:I am again suggesting that it can be solved by rearrangement inequality,I found it later.(My $70$th post :P :P :P )
Phlembac Adib Hasan wrote:
SANZEED wrote:I am again suggesting that it can be solved by rearrangement inequality,I found it later.(My $70$th post :P :P :P )
Serious crime.They all are not necessarily positive integers.If they are, it directly comes from AM-GM that $a=b=c$.
In moments like this, sometimes I think, "What should I do? :? "
Sanzeed, you are wrong, I posted about that yesterday, that you can not WLOG assume $a \geq b \geq c$ here, which is necessary to prove it by rearrangement inequality, because the given equation is cyclic and not symmetric for $a,b,c$.

And Adib, wikipedia says you do not need positive numbers for rearrangement inequality, so I think you are wrong too.

By the way, where did you get the Yahoo answers link? I posted my solutions in general topic on the exam solutions, and I was wondering why the $3$ was not necessary.
(And I also submitted my solutions at 3pm of the first day, if you want to know ;) )
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by Tahmid Hasan » Fri Apr 13, 2012 6:20 am

*Mahi* wrote:By the way, where did you get the Yahoo answers link?
,I just googled 'a/b+b/c+c/a perfect cube' and the first result showed me that link. ;)
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by Phlembac Adib Hasan » Sat Apr 14, 2012 9:20 am

*Mahi* wrote:
SANZEED wrote:I am again suggesting that it can be solved by rearrangement inequality,I found it later.(My $70$th post :P :P :P )
Phlembac Adib Hasan wrote:
SANZEED wrote:I am again suggesting that it can be solved by rearrangement inequality,I found it later.(My $70$th post :P :P :P )
Serious crime.They all are not necessarily positive integers.If they are, it directly comes from AM-GM that $a=b=c$.
In moments like this, sometimes I think, "What should I do? :? "
Sanzeed, you are wrong, I posted about that yesterday, that you can not WLOG assume $a \geq b \geq c$ here, which is necessary to prove it by rearrangement inequality, because the given equation is cyclic and not symmetric for $a,b,c$.

And Adib, wikipedia says you do not need positive numbers for rearrangement inequality, so I think you are wrong too.

By the way, where did you get the Yahoo answers link? I posted my solutions in general topic on the exam solutions, and I was wondering why the $3$ was not necessary.
(And I also submitted my solutions at 3pm of the first day, if you want to know ;) )
Re-arrangement inequality-র বিষয়টি clear করার জন্য ধন্যবাদ।আসলেই জানা ছিল না।আর পরশু google search দিয়ে ঐটা পাইসি।আর শোনেন নায়েল ভাইয়াও জানেন যে আমি দুই number-এর complete solution দেই নাই।পরে যদিও পুরা solve করসি, exam-এর মাঝে নিজে যেটুকু পারসি ততটুকুই post করসি।আমার intention মোটেও ঐ রকম না যে কোন website থেকে টুকলিফাই করে post করব।যদি আমি ঐটা exam-এর মাঝেও পাইতাম তবুও কখনও post করতাম না।
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by Tahmid Hasan » Sat Apr 14, 2012 11:14 am

Actually I have seen this problem before while I was solving AoPS NT marathon problems last year when I opened a mathlinks account.This one was quite difficult to me and I gave up after some time,then I took a hint(make the expression into a quadratic equation) and finally solved it in an algebric way.(And of course,I informed Nayel vaiya about it).There is a good collection of olympiad level problems there,here's the link:http://www.artofproblemsolving.com/Foru ... start=1140
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by SANZEED » Sat Apr 14, 2012 6:09 pm

Mahi vai,I did not understand the thing you wrote for me,would you please make it clear by examples? :? :? :oops:
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Re: BOMC-2012 Test Day 1 Problem 2

Unread post by *Mahi* » Sat Apr 14, 2012 7:05 pm

I said that, if you assumed $a \geq b \geq c$, then you are wrong. Because the given equation is cyclic, not symmetric. You can find the contradiction by setting $c \geq b \geq a$ or something like that.
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