BdMO national 2014: junior 8

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

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?
math champion

User avatar
Hasibul Haque Himel.
Posts: 2
Joined: Mon Oct 27, 2014 11:24 am
Location: Naogaon

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।

User avatar
atiab jobayer
Posts: 23
Joined: Thu Dec 19, 2013 7:40 pm

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:
math champion

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

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 :)
"Questions we can't answer are far better than answers we can't question"

badass0
Posts: 11
Joined: Wed Feb 25, 2015 10:07 pm

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}
$
Bored of being boring because being bored is boring

badass0
Posts: 11
Joined: Wed Feb 25, 2015 10:07 pm

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}
$
Bored of being boring because being bored is boring

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

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 $FVK$ .

badass0
Posts: 11
Joined: Wed Feb 25, 2015 10:07 pm

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।
Bored of being boring because being bored is boring

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

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

User avatar
samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

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
Screenshot_2019-02-22-22-14-35-1.png (14.93 KiB) Viewed 2476 times

Post Reply