BdMO national 2014: junior 8

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atiab jobayer
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BdMO national 2014: junior 8

Unread post by atiab jobayer » Mon Dec 01, 2014 11:21 am

AVIK is a square. The point E is taken on VK in such a way that 3VE=EK. F is the midpoint of AK. What is the value of <FEI?
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Hasibul Haque Himel.
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Re: BdMO national 2014: junior 8

Unread post by Hasibul Haque Himel. » Mon Dec 01, 2014 11:45 am

ভাই আপনে প্রশ্ন দিয়েছেন ৮ নুম্বের। কিন্ত চিত্র দিয়েছেন ৭ নাম্বারের।

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atiab jobayer
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Re: BdMO national 2014: junior 8

Unread post by atiab jobayer » Thu Dec 04, 2014 10:30 am

Hasibul Haque Himel. wrote:ভাই আপনে প্রশ্ন দিয়েছেন ৮ নুম্বের। কিন্ত চিত্র দিয়েছেন ৭ নাম্বারের।


Where have you find picture? :roll: there is no picture as attachment. If you can do the solution. :evil:
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tanmoy
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Re: BdMO national 2014: junior 8

Unread post by tanmoy » Thu Dec 04, 2014 10:21 pm

Join $AE$ and $IF$.Suppose,the length of the sides of the square is $a$. By Stewart's theorem,we get:$IE^{2}=\frac{5a^{2}}{8}$ and also $EF^{2}=\frac{5a^{2}}{8}$.
$IE^{2}+EF^{2}=\frac{5a^{2}}{4}$.Again, $IF^{2}=a^{2}+\frac{a^{2}}{4}=\frac{a^{2}}{5}$.
$\therefore $ $EF^{2}+IE^{2}=IF^{2}$.$\therefore $ $\angle FEI=90^{\circ}$. :D :)
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badass0
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Re: BdMO national 2014: junior 8

Unread post by badass0 » Tue Mar 03, 2015 10:51 pm

@tonmoy
$EF^2$ এর মান কিভাবে $ \frac{5a^2}{8}$ হয়. আমারতো $EF^2$ মান এইরকম আসছেঃ
$
EF^2=OF^2-OE^2
=(\frac{a}{2}) ^2 - (\frac{a \sqrt{2}}{4}) ^2
=\frac{a^2}{4} - \frac{2a^2}{16}
=\frac{a^2}{8}
$
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badass0
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Re: BdMO national 2014: junior 8

Unread post by badass0 » Tue Mar 03, 2015 10:56 pm

tanmoy wrote: By Stewart's theorem,we get: $EF^{2}=\frac{5a^{2}}{8}$
$EF^2$ এর মান কিভাবে $ \frac{5a^2}{8}$ হয়. আমারতো $EF^2$ মান এইরকম আসছেঃ
$
EF^2=OF^2-OE^2
=(\frac{a}{2}) ^2 - (\frac{a \sqrt{2}}{4}) ^2
=\frac{a^2}{4} - \frac{2a^2}{16}
=\frac{a^2}{8}
$
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Tahmid
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Re: BdMO national 2014: junior 8

Unread post by Tahmid » Wed Mar 04, 2015 12:02 am

badass0 wrote:
$EF^2$ এর মান কিভাবে $ \frac{5a^2}{8}$ হয়.

apply stewart's theorem in triangle $FVK$ .

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Re: BdMO national 2014: junior 8

Unread post by badass0 » Wed Mar 04, 2015 12:41 am

Tahmid wrote: apply stewart's theorem in triangle $FVK$ .
Stewart's Theorem apply করেও তো আমার একই জিনিস আসছে।
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tanmoy
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Re: BdMO national 2014: junior 8

Unread post by tanmoy » Wed Mar 04, 2015 4:40 pm

badass0 wrote:
tanmoy wrote: By Stewart's theorem,we get: $EF^{2}=\frac{5a^{2}}{8}$
$EF^2$ এর মান কিভাবে $ \frac{5a^2}{8}$ হয়. আমারতো $EF^2$ মান এইরকম আসছেঃ
$
EF^2=OF^2-OE^2
=(\frac{a}{2}) ^2 - (\frac{a \sqrt{2}}{4}) ^2
=\frac{a^2}{4} - \frac{2a^2}{16}
=\frac{a^2}{8}
$
What is $O$ :?:BTW,apply $\text{Stewart's Theorem}$ carefully.It gives $EF^{2}=\frac{5a^{2}} {8}$.
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samiul_samin
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Re: BdMO national 2014: junior 8

Unread post by samiul_samin » Fri Feb 22, 2019 10:18 pm

It can also be solved without using $Stewart's$ $theorem$.
We can use this diagram!
Screenshot_2019-02-22-22-14-35-1.png

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