BDMO 2018 junior 10

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BDMO 2018 junior 10
$ABC$ 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$ .

 Posts: 17
 Joined: Thu Feb 01, 2018 11:28 am
 Location: Sylhet
Re: BDMO 2018 junior 10
$\angle COD=20$ 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 $306090$ triangle. So,$ \angle AED=\angle AOD=30$ So,$\angle AOC= 60+30=90$.
Re: BDMO 2018 junior 10
I think u need a revision.
Re: BDMO 2018 junior 10
Ummm, Angle AOC =130 Is the correct answer
 samiul_samin
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 Joined: Sat Dec 09, 2017 1:32 pm
Re: BDMO 2018 junior 10
I also found the same answer by solving this problem in a totally different way.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 $306090$ triangle. So,$ \angle AED=\angle AOD=30$ So,$\angle AOC= 60+30=90$.
Can anyone post the full solution?
Re: BDMO 2018 junior 10
@prottoy das,
Post the complete answer.
Post the complete answer.
Wãlkîñg, lõvǐñg, $mīlïñg @nd lìvíñg thě Lîfè

 Posts: 17
 Joined: Thu Feb 01, 2018 11:28 am
 Location: Sylhet
Re: BDMO 2018 junior 10
I mean how it starts. I am little confused because you've written $\angle COD=20$ and again $\angle COD=60$.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 $306090$ triangle. So,$ \angle AED=\angle AOD=30$ So,$\angle AOC= 60+30=90$.
Wãlkîñg, lõvǐñg, $mīlïñg @nd lìvíñg thě Lîfè

 Posts: 17
 Joined: Thu Feb 01, 2018 11:28 am
 Location: Sylhet
Re: BDMO 2018 junior 10
I forgot to write in the problem $D$ and $E$ lies on $AC$ and $AB$.

 Posts: 17
 Joined: Thu Feb 01, 2018 11:28 am
 Location: Sylhet
Re: BDMO 2018 junior 10
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
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