Dhaka Junior 2011/9

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Moon
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Dhaka Junior 2011/9

Unread post by Moon » Wed Feb 02, 2011 8:27 am

9. $ABC$ is a triangle with $AB = 4$, $BC = 5$ and $AC = 3$. $O$ is the midpoint of $BC$. A line from $O$ parallel to $AC$ meets $AB$ at $D$. $E$ is on $DO$ so that $DO = OE$ and $D$ and $E$ are on opposite sides of $BC$. Find $BE^2$.
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Mehfuj Zahir
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Re: Dhaka Junior 2011/9

Unread post by Mehfuj Zahir » Wed Feb 02, 2011 11:31 am

Answer is 13

ahmedulkavi
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Re: Dhaka Junior 2011/9

Unread post by ahmedulkavi » Wed Feb 02, 2011 1:55 pm

How can we get the answer [13]?

Mehfuj Zahir
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Re: Dhaka Junior 2011/9

Unread post by Mehfuj Zahir » Wed Feb 02, 2011 2:47 pm

BE^2=AD^2+DE^2

jkisor
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Re: Dhaka Junior 2011/9

Unread post by jkisor » Wed Jul 11, 2012 8:56 am

:?: May be BE=DC.
(BOE and DOC is similar)
DC^2=OC^2-OD^2=2.5^2-1.5^2=4

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Re: Dhaka Junior 2011/9

Unread post by nafistiham » Wed Aug 29, 2012 11:48 pm

$BD=2$
$DE=DO+OE=DO+DO=AC=3$
$BE^2=BD^2+DE^2=13$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Shafin
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Re: Dhaka Junior 2011/9

Unread post by Shafin » Thu Jan 10, 2013 5:10 pm

nafistiham wrote:$BD=2$
$DE=DO+OE=DO+DO=AC=3$
$BE^2=BD^2+DE^2=13$
Can u draw it on a paper and then upload the image with positions of all e points? I cant understand the position of e

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Re: Dhaka Junior 2011/9

Unread post by nafistiham » Fri Jan 25, 2013 12:56 pm

Shafin wrote:
nafistiham wrote:$BD=2$
$DE=DO+OE=DO+DO=AC=3$
$BE^2=BD^2+DE^2=13$
Can u draw it on a paper and then upload the image with positions of all e points? I cant understand the position of e
Sorry for being so late. :oops: :oops:
here is what you wanted.
junior 2011-9.png
there is something called 'extension' :p
junior 2011-9.png (18.07KiB)Viewed 12228 times
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
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nafistiham@gmail

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