## BDMO 2018 junior 10

prottoy das
Posts: 17
Joined: Thu Feb 01, 2018 11:28 am
Location: Sylhet

### 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\$ .

prottoy das
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 \$30-60-90\$ triangle. So,\$ \angle AED=\angle AOD=30\$ So,\$\angle AOC= 60+30=90\$.

ankon dey
Posts: 3
Joined: Wed Nov 22, 2017 2:16 pm

### Re: BDMO 2018 junior 10

I think u need a revision.

Arifa
Posts: 7
Joined: Wed Nov 22, 2017 8:09 pm

### Re: BDMO 2018 junior 10

Ummm, Angle AOC =130 Is the correct answer

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### Re: BDMO 2018 junior 10

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
Posts: 35
Joined: Sat Dec 15, 2018 5:19 pm
Location: Munshigonj, Dhaka

### Re: BDMO 2018 junior 10

@prottoy das,
Wãlkîñg, lõvǐñg, \$mīlïñg @nd lìvíñg thě Lîfè

prottoy das
Posts: 17
Joined: Thu Feb 01, 2018 11:28 am
Location: Sylhet

### Re: BDMO 2018 junior 10

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

NABILA
Posts: 35
Joined: Sat Dec 15, 2018 5:19 pm
Location: Munshigonj, Dhaka

### Re: BDMO 2018 junior 10

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
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\$.

prottoy das
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