Dhaka Secondary 2011/6

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
Forum rules
Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

Dhaka Secondary 2011/6

Unread post by BdMO » Fri Jan 28, 2011 10:04 pm

The diagonal $AB$ in quadrilateral $ADBC$ bisects the angle $CBD$. $DB$ and $DA$ are tangent to the circumcircle of triangle $ABC$ at points $A$ and $B$. The perimeter of triangle $ABC$ is $20$ and perimeter of triangle $ABD$ is $12$. Find the length of $BD$.

User avatar
Avik Roy
Posts: 156
Joined: Tue Dec 07, 2010 2:07 am

Re: Dhaka Secondary 2011/6

Unread post by Avik Roy » Tue Feb 01, 2011 1:48 am

I'm a bit sad seeing that people are not trying this problem. It's one of my favourite. As a divisional problem, it takes good count of your geometric senses and algebraic capabilities ;)
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

Tahsin24
Posts: 21
Joined: Tue Dec 07, 2010 6:13 pm

Re: Dhaka Secondary 2011/6

Unread post by Tahsin24 » Tue Feb 01, 2011 11:12 am

I've solved it n my ans is (36/11)......I'm sorry that I cant post a full solution because I cant use LATEX yet. I'll try to learn it soon. My procedure was to show tringle ABC and triangle ABD are both isosceles and similar to each other.

User avatar
Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm
Location: Khulna,Bangladesh.

Re: Dhaka Secondary 2011/6

Unread post by Tahmid Hasan » Tue Feb 01, 2011 6:46 pm

i think it's $\frac{36}{11}$
i did something with the ecantor brittangshostho thingyy.
বড় ভালবাসি তোমায়,মা

User avatar
Moon
Site Admin
Posts: 751
Joined: Tue Nov 02, 2010 7:52 pm
Location: Dhaka, Bangladesh
Contact:

Re: Dhaka Secondary 2011/6

Unread post by Moon » Tue Feb 01, 2011 11:36 pm

Tahmid: Please post your solution! :)
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

User avatar
Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm
Location: Khulna,Bangladesh.

Re: Dhaka Secondary 2011/6

Unread post by Tahmid Hasan » Tue Feb 01, 2011 11:38 pm

for that i'll have to draw the pic but i don't know how to post that here.
so if any1 could give a pic i' gladly give the soln
বড় ভালবাসি তোমায়,মা

User avatar
Moon
Site Admin
Posts: 751
Joined: Tue Nov 02, 2010 7:52 pm
Location: Dhaka, Bangladesh
Contact:

Re: Dhaka Secondary 2011/6

Unread post by Moon » Tue Feb 01, 2011 11:55 pm

Posting diagrams is easy. Just create a diagram and attach it with the post :)
(There is an "upload attachment" option below post reply editor...however you must use the "full editor" to get this option.)
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

photon
Posts: 186
Joined: Sat Feb 05, 2011 3:39 pm
Location: dhaka
Contact:

Re: Dhaka Secondary 2011/6

Unread post by photon » Sat Feb 05, 2011 3:55 pm

i find ABD isosceles.how did you get ABC as isosceles?
Try not to become a man of success but rather to become a man of value.-Albert Einstein

photon
Posts: 186
Joined: Sat Feb 05, 2011 3:39 pm
Location: dhaka
Contact:

Re: Dhaka Secondary 2011/6

Unread post by photon » Sat Feb 05, 2011 3:55 pm

how 2 get similarity???
Try not to become a man of success but rather to become a man of value.-Albert Einstein

Tahsin24
Posts: 21
Joined: Tue Dec 07, 2010 6:13 pm

Re: Dhaka Secondary 2011/6

Unread post by Tahsin24 » Mon Feb 07, 2011 12:18 am

$ABD$ and $ABC$ are both isosceles. $\angle BAD=\angle ACB$ (ekanttor brittangshostho kon),
so $\angle ACB=\angle ABD=\angle BAD=\angle ABC$
so in triangle $ABC$ and $ABD$
$\angle ABC=\angle ABD$
$\angle ACB=\angle BAD$
$\angle BAC=\angle ADB$

Post Reply