## Dhaka Junior 2011/9

Problem for Junior Group from Divisional Mathematical Olympiad will be solved here.
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### Dhaka Junior 2011/9

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

Answer is 13

ahmedulkavi
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### Re: Dhaka Junior 2011/9

How can we get the answer ?

Mehfuj Zahir
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### Re: Dhaka Junior 2011/9

BE^2=AD^2+DE^2

jkisor
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### Re: Dhaka Junior 2011/9 May be BE=DC.
(BOE and DOC is similar)
DC^2=OC^2-OD^2=2.5^2-1.5^2=4

nafistiham
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### Re: Dhaka Junior 2011/9

$BD=2$
$DE=DO+OE=DO+DO=AC=3$
$BE^2=BD^2+DE^2=13$
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

Shafin
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### Re: Dhaka Junior 2011/9

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

nafistiham
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### Re: Dhaka Junior 2011/9

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.  here is what you wanted. there is something called 'extension' :p
junior 2011-9.png (18.07 KiB) Viewed 2218 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.