Dhaka Secondary 2009/8

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Secondary 2009/8

Unread post by BdMO » Tue Feb 01, 2011 11:49 pm

For triangle $ABC$ $ \angle C$ is $90^{\circ}$. $\angle BAC$ is $30^{\circ}$ & $AB$ is $1$cm. $D$ is a point within $ABC$ so that angle $\angle BDC$ is $90^{\circ}$ & $\angle ACD = \angle DBA$. $AB$ & $CD$ meets at $E$. Find $AE$.

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Tahmid Hasan
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Re: Dhaka Secondary 2009/8

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

ans is $\frac{1}{2}$. i just used similiar triangle properties and general trigonometric means.:P
বড় ভালবাসি তোমায়,মা

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Moon
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Re: Dhaka Secondary 2009/8

Unread post by Moon » Wed Feb 02, 2011 12:02 am

^By outline I did not mean that you should just post what you used to solved the problems (it is more or less apparent that some Euclidean geometry and trig can solve almost any problem at this level). You better post what you used to solve what.

It is also best for everyone if you post complete solution. Remember, in National we need to write complete solutions. :)
"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.

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leonardo shawon
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Re: Dhaka Secondary 2009/8

Unread post by leonardo shawon » Wed Feb 02, 2011 12:04 am

tahmid, full sollution or at least a little hint.. Please
Ibtehaz Shawon
BRAC University.

long way to go .....

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Tahmid Hasan
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Re: Dhaka Secondary 2009/8

Unread post by Tahmid Hasan » Wed Feb 02, 2011 12:05 am

sorry i'll try from next time and post almost full but not formal solns
বড় ভালবাসি তোমায়,মা

Shapnil
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Re: Dhaka Secondary 2009/8

Unread post by Shapnil » Thu Mar 22, 2012 8:20 pm

Plz give the full solution or a hint

Ridwan Abrar
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Re: Dhaka Secondary 2009/8

Unread post by Ridwan Abrar » Fri Jan 31, 2014 9:11 pm

in triangle ABC, cos 30=AC/1
ANGLE DCB+ ANGLE DBC=90
ANGLE ACD+ANGLE DCB=90
GIVEN, ANGLE ACD=ANGLE DBA
SO, ANGLE ACD=ANGLE DBA
ANGLE DBC=ANGLE DBA=60/2=30
sin(120)/AC=sin(30)/AE
AE=0.5

mdhasib
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Re: Dhaka Secondary 2009/8

Unread post by mdhasib » Fri Jul 22, 2016 12:47 pm

We can easily show that triangle FCD and triangle EDB are similar . So FD/DE = CD/BD = FC/BE . Again we can see that triangle FCD and triangle BCD are similar . Again FD/CD = CD/BD = FC/BC .If we merge the conclusions we get FD/CD = FC/BC = FD/DE = CD/BD = FC/BE. We have FC/BC = FC/BE hence BC=BE. Now we can apply trigonometry and get the value of BE then i think we all know what to do :D :roll: :mrgreen:

[Q.E.D]

SYED ASHFAQ TASIN
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Re: Dhaka Secondary 2009/8

Unread post by SYED ASHFAQ TASIN » Mon Sep 05, 2016 3:55 pm

please insert picture

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