## 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

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

### Re: Dhaka Secondary 2009/12

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

How did you find out that DG=3?

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

### Re: Dhaka Secondary 2009/12

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