## BdMO national 2014: junior 8

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
atiab jobayer
Posts: 23
Joined: Thu Dec 19, 2013 7:40 pm

### BdMO national 2014: junior 8

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

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

atiab jobayer
Posts: 23
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### Re: BdMO national 2014: junior 8

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

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

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, $IF^{2}=a^{2}+\frac{a^{2}}{4}=\frac{a^{2}}{5}$.
$\therefore$ $EF^{2}+IE^{2}=IF^{2}$.$\therefore$ $\angle FEI=90^{\circ}$.
"Questions we can't answer are far better than answers we can't question"

badass0
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### Re: BdMO national 2014: junior 8

@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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Joined: Wed Feb 25, 2015 10:07 pm

### Re: BdMO national 2014: junior 8

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

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

apply stewart's theorem in triangle $FVK$ .

badass0
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Joined: Wed Feb 25, 2015 10:07 pm

### Re: BdMO national 2014: junior 8

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

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 $EF^{2}=\frac{5a^{2}} {8}$.
"Questions we can't answer are far better than answers we can't question"

samiul_samin
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Joined: Sat Dec 09, 2017 1:32 pm

### Re: BdMO national 2014: junior 8

It can also be solved without using $Stewart's$ $theorem$.
We can use this diagram!
Screenshot_2019-02-22-22-14-35-1.png (14.93 KiB) Viewed 2777 times