Page 1 of 1

BdMO national 2014: junior 8

Posted: Mon Dec 01, 2014 11:21 am
by atiab jobayer
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?

Re: BdMO national 2014: junior 8

Posted: Mon Dec 01, 2014 11:45 am
by Hasibul Haque Himel.
ভাই আপনে প্রশ্ন দিয়েছেন ৮ নুম্বের। কিন্ত চিত্র দিয়েছেন ৭ নাম্বারের।

Re: BdMO national 2014: junior 8

Posted: Thu Dec 04, 2014 10:30 am
by atiab jobayer
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:

Re: BdMO national 2014: junior 8

Posted: Thu Dec 04, 2014 10:21 pm
by tanmoy
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 :)

Re: BdMO national 2014: junior 8

Posted: Tue Mar 03, 2015 10:51 pm
by badass0
@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}
$

Re: BdMO national 2014: junior 8

Posted: Tue Mar 03, 2015 10:56 pm
by badass0
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}
$

Re: BdMO national 2014: junior 8

Posted: Wed Mar 04, 2015 12:02 am
by Tahmid
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

Posted: Wed Mar 04, 2015 12:41 am
by badass0
Tahmid wrote: apply stewart's theorem in triangle $FVK$ .
Stewart's Theorem apply করেও তো আমার একই জিনিস আসছে।

Re: BdMO national 2014: junior 8

Posted: Wed Mar 04, 2015 4:40 pm
by tanmoy
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 $EF^{2}=\frac{5a^{2}} {8}$.

Re: BdMO national 2014: junior 8

Posted: Fri Feb 22, 2019 10:18 pm
by samiul_samin
It can also be solved without using $Stewart's$ $theorem$.
We can use this diagram!
Screenshot_2019-02-22-22-14-35-1.png
Screenshot_2019-02-22-22-14-35-1.png (14.93 KiB) Viewed 2481 times