BdMO National Higher Secondary 2011/8

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BdMO
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BdMO National Higher Secondary 2011/8

Unread post by BdMO » Sat Feb 12, 2011 4:50 pm

Problem 8:
$ABC$ is a right angled triangle with $\angle A = 90^{\circ}$ and $D$ be the midpoint of $BC$. A point $F$ is chosen on $AB$. $CA$ and $DF$ meet at $G$ and $GB \parallel AD$. $CF$ and $AD$ meet at $O$ and $AF = FO$. $GO$ meets $BC$ at $R$. Find the sides of $ABC$ if the area of $GDR$ is $\dfrac{2}{\sqrt{15}}$

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FahimFerdous
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Re: BdMO National Higher Secondary 2011/8

Unread post by FahimFerdous » Mon Feb 14, 2011 12:22 am

Yay, I solved it. I can't post the solution right now, coz I don't have time now. But I'll post it tomorrow. I did many unnecessary things during proving it. So, I've to cut down the unnecessary portions and arrange it and post it. Right I'm telling the answer I've got. The answer is BC=4(2/5)^1/2, AC=2 and AB=6/(15)^1/2.

Actually, I did many fast calculations. There might be mistakes. But I believe my proccess is right. Forgive me if there's any mistake. :-)
Your hot head might dominate your good heart!

Mehfuj Zahir
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Re: BdMO National Higher Secondary 2011/8

Unread post by Mehfuj Zahir » Mon Feb 14, 2011 12:33 am

Fahim you are right............and i am senior to you

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Re: BdMO National Higher Secondary 2011/8

Unread post by Mehfuj Zahir » Mon Feb 14, 2011 1:09 am

As D is the mid point of BC then AD=BD=CD.GB andAD is parallel.SO that<DAB=<ABG
AD=BD,<ABC=<ABG.Now we can prove that the triangle ABC and GBA is equal.Let the extention of CF meets BC at I.So CI,GD,AB is a median of the triangle GBC.
Now by Cevas theorem in triangle GDC GF.DR.CA=FD.RC.AG
We know that GF/FD=2/1 and CA=AG
So DR=2RC
now the area of triangle GDC is 6.(15)^(1/2)
so area of triangle ABC is twice of GBC which we can express as AB.AC
AO and GI is parallel so CO=OI=2FI as the triangle AOF and GFI is similiar.
now using AF=FO ,AB=3OF,the value of AB.ac and using pythagoreous in triangle ABC and ACF find the desired answer.

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FahimFerdous
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Re: BdMO National Higher Secondary 2011/8

Unread post by FahimFerdous » Mon Feb 14, 2011 10:02 am

Oh, thanx, Mehfuz vaia. Your solution is same to mine. Thanx for writing it and saving me from the trouble of writing it from MOBILE PHONE! :-)
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abir91
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Re: BdMO National Higher Secondary 2011/8

Unread post by abir91 » Mon Feb 14, 2011 1:37 pm

If it is troublesome, then why do you bother writing the solutions? :P
Abir

Have you read the Forum Rules and Guidelines?

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TIUrmi
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Re: BdMO National Higher Secondary 2011/8

Unread post by TIUrmi » Mon Feb 14, 2011 6:07 pm

Two reasons:

1. To let know that he solved it.
2. To check if it's correct. Though I guess most of them do not submit solutions that they are confused with. So, point 1 is the only reason in most cases.. xD
"Go down deep enough into anything and you will find mathematics." ~Dean Schlicter

Mehfuj Zahir
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Re: BdMO National Higher Secondary 2011/8

Unread post by Mehfuj Zahir » Mon Feb 14, 2011 10:00 pm

TIURMI what do you mean by this?

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FahimFerdous
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Re: BdMO National Higher Secondary 2011/8

Unread post by FahimFerdous » Mon Feb 14, 2011 10:20 pm

@Urmi apu: you're quite right. But some people get a little excited after a solving a problem they thought they couldn't solve (people like me) and want others to know that they solved it and if they're wrong, then they want to be corrected by others. Actually I dnt knw if there's any other, or is it just me? :-)
Your hot head might dominate your good heart!

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FahimFerdous
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Re: BdMO National Higher Secondary 2011/8

Unread post by FahimFerdous » Mon Feb 14, 2011 10:23 pm

@Urmi apu: But again, I write 'Is my solution correct? :-s' everytime (like you) I post a solution. Actually, I can never be sure whether I solved a problem correctly. I always think that I can't solve any problem smartly like other people. I dnt knw why is that. :-s
Your hot head might dominate your good heart!

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