BdMO National 2012: Higher Secondary, Secondary 05

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
User avatar
Moon
Site Admin
Posts: 751
Joined: Tue Nov 02, 2010 7:52 pm
Location: Dhaka, Bangladesh
Contact:

BdMO National 2012: Higher Secondary, Secondary 05

Unread post by Moon » Sat Feb 11, 2012 11:20 pm

Problem:
In triangle $ABC$, medians $AD$ and $CF$ intersect at point $G$. $P$ is an arbitrary point on $AC$. $PQ$ & $PR$ are parallel to $AD$ & $CF$ respectively. $PQ$ intersects $BC$ at $Q$ and $PR$ intersects $AB$ at $R$. If $QR$ intersects $AD$ at $M$ & $CF$ at $N$, then prove that area of triangle $GMN$ is $\frac{(A)}{8}$ where $(A)$ = area enclosed by $PQ, PR, AD, CF$.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

photon
Posts: 186
Joined: Sat Feb 05, 2011 3:39 pm
Location: dhaka
Contact:

Re: BdMO National 2012: Higher Secondary, Secondary 05

Unread post by photon » Wed Feb 15, 2012 7:37 am

that area is a parallelogam.with similarity we can show, $2RM==MQ$ and $2NQ=RN$.which implies $RM=MN=NQ$.so let PR and PQ intersect AD and CF at R' and Q'. :arrow: $RR'M,QQ'N,GMN$ are congruent.so GM=MR' and GN=NQ'.
thus we get that fraction dividing area...
Try not to become a man of success but rather to become a man of value.-Albert Einstein

User avatar
nafistiham
Posts: 829
Joined: Mon Oct 17, 2011 3:56 pm
Location: 24.758613,90.400161
Contact:

Re: BdMO National 2012: Higher Secondary, Secondary 05

Unread post by nafistiham » Wed Feb 15, 2012 2:51 pm

2012 sec 5.JPG
2012 sec 5.JPG (38.54 KiB) Viewed 2113 times
the only thing is needed that
\[DG=\frac {AG}{2}\]
\[FG=\frac {CG}{2}\]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

User avatar
Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm
Location: Dhaka, Bangladesh.
Contact:

Re: BdMO National 2012: Higher Secondary, Secondary 05

Unread post by Labib » Fri Feb 24, 2012 3:03 pm

Tiham, I wanted a detailed solution.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.


"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

User avatar
*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
Location: 23.786228,90.354974
Contact:

Re: BdMO National 2012: Higher Secondary, Secondary 05

Unread post by *Mahi* » Fri Feb 24, 2012 3:22 pm

Let $PR \cap GC =X$ and $AG \cap QN =Y$. Then try to prove $\triangle RXN,\triangle GNM,\triangle MQY$ are homothetic. Then it follows that for some $x,y$; $xRX+yRX=2RX$ and $2QY=\frac 1yQY +\frac xyQY$ and the only nonzero solution for $x,y$ is $(1,1)$.
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

User avatar
asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm
Location: Sylhet,Bangladesh

Re: BdMO National 2012: Higher Secondary, Secondary 05

Unread post by asif e elahi » Sun Feb 02, 2014 1:03 pm

Let $PQ\cap AG=S,PR\cap CG=F$. So $PSGT$ is a paraleogram. As $\bigtriangleup AQP\sim \bigtriangleup AFC$
$\frac{1}{2}=\frac{CG}{GF}=\frac{PS}{QS}$. Again $PR\parallel MS$. So $\frac{QM}{MR}= \frac{QS}{PS}=2$. So $MR=2QM$.Similarly $QN=2NR$.this implies $QM=MN=NR$.
$\frac{GM}{SM}=\frac{NM}{QM}=1$.$M$ is the midpoint of $GS$ and $N$ is the midpoint of $GT$.So $(GNM)=\frac{GST}{4}=\frac{GSPT}{8}=\frac{(A)}{8}$

Post Reply