BdMO regional 2018 set 4 Secondary P 10

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samiul_samin
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BdMO regional 2018 set 4 Secondary P 10

Unread post by samiul_samin » Thu Jan 10, 2019 11:26 pm

Screenshot_2019-01-10-23-19-29-1.png
Screenshot_2019-01-10-23-19-29-1.png (10.96 KiB) Viewed 763 times
In rectangle $ABCD$, $AB=1$, $BC=2$, $E, F, G$ is the midpoint of $BC, CD, AD$ respectively.$ H$ is the midpoint of $GE$. What is the area of the unshaded region?

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Tasnood
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Re: BdMO regional 2018 set 4 Secondary P 10

Unread post by Tasnood » Wed Jan 16, 2019 11:18 pm

I set $D$ as $(0,0)$ and worked with analytic geometry.
$AF\rightarrow y=-4x+2$
$FB\rightarrow y=4x-2$
$DH\rightarrow y=2x$
$HC\rightarrow y=-2x-2$
Then i called $I$ the intersection of $AF$ and $DH$ and $J$ the intersection of $FB$ and $HC$.
Hence, $I=(\frac{1}{3},\frac{2}{3})$ and $J=(\frac{2}{3},\frac{2}{3})$.
Now its easy to determine the area of the polygon $IHJF$.
$\frac{1}{2}\Big((\frac{1}{3}+\frac{1}{3}+\frac{2}{3})-(\frac{1}{3}+\frac{1}{3}+\frac{1}{3})\Big)=\frac{1}{6}$

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