Junior Divisional 2013/1

Problem for Junior Group from Divisional Mathematical Olympiad will be solved here.
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Please don't post problems (by starting a topic) in the "Junior: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
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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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Re: Junior Divisional 2013/1

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

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

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