Dhaka Secondary 2009/12

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

Dhaka Secondary 2009/12

Unread post by BdMO » Tue Feb 01, 2011 11:51 pm

ABCD is $4 \times 4$ square. $E$ lies on $AB$; $AE=1$. $F$ lies on $AD$ & $AF=AE$. $EFG$ is a right angled triangle where $F$ is the right angle. Find the radius of the circumcirle of the triangle $EFG$.

Shifat
Posts: 53
Joined: Sun Jul 31, 2011 12:21 pm
Location: Dhaka, Bangladesh

Re: Dhaka Secondary 2009/12

Unread post by Shifat » Thu Aug 25, 2011 4:50 pm

the answer is \[\sqrt{5}\]

here is how I got it
$AF=AE=1$
from F, BC er upor FM and from G, AB er upor duita lombo (:P) eke dekhano jay je,
$DG=3$, so $EF=\sqrt{2} $ and $GF= 3.\sqrt{2}$
so $EG= 2.\sqrt{5}$
as $EFG$ right angled, so the midpoint of EG will be the centre of it's circumcircle.
:?: :?:

Ridwan Abrar
Posts: 13
Joined: Mon Aug 05, 2013 8:01 pm

Re: Dhaka Secondary 2009/12

Unread post by Ridwan Abrar » Sat Feb 01, 2014 12:10 am

How did you find out that DG=3?

mdhasib
Posts: 10
Joined: Fri Jul 22, 2016 11:43 am

Re: Dhaka Secondary 2009/12

Unread post by mdhasib » Mon Aug 01, 2016 9:38 am

Look carefully triangle AEF and triangle FGD are similar and AF/DF =1/3 so DG=3AE=3 and then we can use Pythagoras and obtain the value of the hypotenuse and the triangle EFG is right so the so answer would be half the value of the hypotenuse

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