BDMO 2018 junior 10

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prottoy das
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BDMO 2018 junior 10

Unread post by prottoy das » Tue Mar 20, 2018 2:51 pm

$ABC$ is an equilateral triangle. $D$ and $E$ is such a point that, $AD:CD=1:2$ and $BE:AE=1:2$. If $O$ is the intersection point of $BD$ and $CE$ find $ \angle AOC$ .

prottoy das
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Re: BDMO 2018 junior 10

Unread post by prottoy das » Tue Mar 20, 2018 2:59 pm

$\angle COD=20$ and $\angle BOE=40$ So $\angle COD=60$. Now $\angle EAD=\angle COD$. so, $ADOE$ is a cyclic quadrilateral. Join $D,E$. Here $AE={2/3}AB$ and $AD$ is half of $AE$ and their intersected angle is $60$. SO, $AED$ is a $30-60-90$ triangle. So,$ \angle AED=\angle AOD=30$ So,$\angle AOC= 60+30=90$.

ankon dey
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Re: BDMO 2018 junior 10

Unread post by ankon dey » Thu Mar 22, 2018 10:38 am

I think u need a revision.

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Arifa
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Re: BDMO 2018 junior 10

Unread post by Arifa » Wed Dec 19, 2018 10:21 pm

Ummm, Angle AOC =130 Is the correct answer :)

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samiul_samin
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Re: BDMO 2018 junior 10

Unread post by samiul_samin » Fri Jan 11, 2019 1:00 pm

prottoy das wrote:
Tue Mar 20, 2018 2:59 pm
$\angle COD=20$ and $\angle BOE=40$ So $\angle COD=60$. Now $\angle EAD=\angle COD$. so, $ADOE$ is a cyclic quadrilateral. Join $D,E$. Here $AE={2/3}AB$ and $AD$ is half of $AE$ and their intersected angle is $60$. SO, $AED$ is a $30-60-90$ triangle. So,$ \angle AED=\angle AOD=30$ So,$\angle AOC= 60+30=90$.
I also found the same answer by solving this problem in a totally different way.
Can anyone post the full solution?

NABILA
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Re: BDMO 2018 junior 10

Unread post by NABILA » Tue Jan 15, 2019 7:51 pm

@prottoy das,
Post the complete answer.
Wãlkîñg, lõvǐñg, $mīlïñg @nd lìvíñg thě Lîfè

prottoy das
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Re: BDMO 2018 junior 10

Unread post by prottoy das » Wed Jan 16, 2019 11:17 pm

NABILA wrote:
Tue Jan 15, 2019 7:51 pm
@prottoy das,
Post the complete answer.
I posted the complete answer!!!!!!!!!! Did you mean to explain it?

NABILA
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Re: BDMO 2018 junior 10

Unread post by NABILA » Thu Jan 17, 2019 2:51 pm

prottoy das wrote:
Tue Mar 20, 2018 2:59 pm
$\angle COD=20$ and $\angle BOE=40$ So $\angle COD=60$. Now $\angle EAD=\angle COD$. so, $ADOE$ is a cyclic quadrilateral. Join $D,E$. Here $AE={2/3}AB$ and $AD$ is half of $AE$ and their intersected angle is $60$. SO, $AED$ is a $30-60-90$ triangle. So,$ \angle AED=\angle AOD=30$ So,$\angle AOC= 60+30=90$.
I mean how it starts. I am little confused because you've written $\angle COD=20$ and again $\angle COD=60$.
Wãlkîñg, lõvǐñg, $mīlïñg @nd lìvíñg thě Lîfè

prottoy das
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Re: BDMO 2018 junior 10

Unread post by prottoy das » Thu Jan 17, 2019 7:22 pm

I forgot to write in the problem $D$ and $E$ lies on $AC$ and $AB$.

prottoy das
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Re: BDMO 2018 junior 10

Unread post by prottoy das » Thu Jan 17, 2019 7:33 pm

And the first line of my solution has a typing mistake.
Correct:
As, $\triangle ABC$ is equilateral, So,$\angle CBD:\angle ABD=2:1$. So, similarly you can find $\angle BCE$. As, $\angle COD$ is the exterior angle of $\triangle BCO$, so, $\angle COD=\angle CBD+\angle BCE$. Then you will get $\angle COD=\angle EAD=60^\circ$
Then everything is okay

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