Junior Divisional 2013/1

Problem for Junior Group from Divisional Mathematical Olympiad will be solved here.
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barnik
Posts: 13
Joined: Wed Dec 03, 2014 3:37 pm

Junior Divisional 2013/1

Unread post by barnik » Wed Dec 10, 2014 5:41 pm

$O$ is the midpoint of $AB$ and $N$ is the midpoint of $AC$.The ratio $AD:AB=2:5$ and the ratio $AF:AC=2:5$. The area of triangle $ABC$ is $50\text m^2$. What is the difference between the area of $ADPF$ and triangle $PON$?
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tanmoy
Posts: 282
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

Re: Junior Divisional 2013/1

Unread post by tanmoy » Thu Dec 11, 2014 2:58 pm

$(ABC)=50m^{2}.\therefore (AON)=\frac{50}{4}=\frac{25}{2}m^{2}.(ADF)=50\times \frac{4}{25}=8$
$(DPF)=\frac{16}{25}(PON).(DPO)=(FPN)=\frac{4}{5}(PON).$
$(DONF)=\frac{25}{2}-8=\frac{9}{2}$
$\therefore \frac{16}{25}(PON)+\frac{4}{5}(PON)+\frac{4}{5}(PON)+(PON)=\frac{9}{2}$
$\therefore (PON)=\frac{25}{18}.\therefore (DPF)=\frac{8}{9}$
$\therefore (ADPF)=8+\frac{8}{9}=\frac{80}{9}m^{2}.
\therefore (ADPF)-(PON)=\frac{80}{9}-\frac{25}{18}=\frac{15}{2}m^{2}$ :)
"Questions we can't answer are far better than answers we can't question"

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