Dhaka Secondary 2009/12
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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.
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.
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 $EFG$.
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.
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.

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Re: Dhaka Secondary 2009/12
How did you find out that DG=3?
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