BdMO national 2014: junior 8
- atiab jobayer
- Posts:23
- Joined:Thu Dec 19, 2013 7:40 pm
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
- Joined:Thu Dec 19, 2013 7:40 pm
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.
math champion
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 $ $\angle FEI=90^{\circ}$.
$IE^{2}+EF^{2}=\frac{5a^{2}}{4}$.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"
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}
$
$EF^2$ 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}
$
Bored of being boring because being bored is boring
Re: BdMO national 2014: junior 8
$EF^2$ এর মান কিভাবে $ \frac{5a^2}{8}$ হয়. আমারতো $EF^2$ মান এইরকম আসছেঃtanmoy wrote: By Stewart's theorem,we get: $EF^{2}=\frac{5a^{2}}{8}$
$
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}
$
Bored of being boring because being bored is boring
Re: BdMO national 2014: junior 8
badass0 wrote:
$EF^2$ এর মান কিভাবে $ \frac{5a^2}{8}$ হয়.
apply stewart's theorem in triangle $FVK$ .
Re: BdMO national 2014: junior 8
Stewart's Theorem apply করেও তো আমার একই জিনিস আসছে।Tahmid wrote: apply stewart's theorem in triangle $FVK$ .
Bored of being boring because being bored is boring
Re: BdMO national 2014: junior 8
What is $O$ BTW,apply $\text{Stewart's Theorem}$ carefully.It gives $EF^{2}=\frac{5a^{2}} {8}$.badass0 wrote:$EF^2$ এর মান কিভাবে $ \frac{5a^2}{8}$ হয়. আমারতো $EF^2$ মান এইরকম আসছেঃtanmoy wrote: By Stewart's theorem,we get: $EF^{2}=\frac{5a^{2}}{8}$
$
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}
$
"Questions we can't answer are far better than answers we can't question"
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Re: BdMO national 2014: junior 8
It can also be solved without using $Stewart's$ $theorem$.
We can use this diagram!
We can use this diagram!