## BdMO National 2012: Higher Secondary, Secondary 05

Moon
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### BdMO National 2012: Higher Secondary, Secondary 05

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

learn how to write equations, and don't forget to read Forum Guide and Rules.

photon
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### Re: BdMO National 2012: Higher Secondary, Secondary 05

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'. $RR'M,QQ'N,GMN$ are congruent.so GM=MR' and GN=NQ'.
thus we get that fraction dividing area...
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nafistiham
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### Re: BdMO National 2012: Higher Secondary, Secondary 05

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.

Labib
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### Re: BdMO National 2012: Higher Secondary, Secondary 05

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

*Mahi*
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### Re: BdMO National 2012: Higher Secondary, Secondary 05

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 $(1,1)$.

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

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 $(GNM)=\frac{GST}{4}=\frac{GSPT}{8}=\frac{(A)}{8}$