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 $PQ, PR, AD, CF$.
BdMO National 2012: Higher Secondary, Secondary 05
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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...
$RR'M,QQ'N,GMN$ 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 (38.54KiB)Viewed 5114 times
\[DG=\frac {AG}{2}\]
\[FG=\frac {CG}{2}\]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: BdMO National 2012: Higher Secondary, Secondary 05
Tiham, I wanted a detailed solution.
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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 $(1,1)$.
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Re: BdMO National 2012: Higher Secondary, Secondary 05
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 $(GNM)=\frac{GST}{4}=\frac{GSPT}{8}=\frac{(A)}{8}$
$\frac{1}{2}=\frac{CG}{GF}=\frac{PS}{QS}$. 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 $(GNM)=\frac{GST}{4}=\frac{GSPT}{8}=\frac{(A)}{8}$