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## BdMO regional 2018 set 4 Secondary P 10

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- samiul_samin
**Posts:**1007**Joined:**Sat Dec 09, 2017 1:32 pm

### BdMO regional 2018 set 4 Secondary P 10

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?

### Re: BdMO regional 2018 set 4 Secondary P 10

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 $IHJF$.

$\frac{1}{2}\Big((\frac{1}{3}+\frac{1}{3}+\frac{2}{3})-(\frac{1}{3}+\frac{1}{3}+\frac{1}{3})\Big)=\frac{1}{6}$

$AF\rightarrow y=-4x+2$

$FB\rightarrow y=4x-2$

$DH\rightarrow y=2x$

$HC\rightarrow y=-2x-2$

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 $IHJF$.

$\frac{1}{2}\Big((\frac{1}{3}+\frac{1}{3}+\frac{2}{3})-(\frac{1}{3}+\frac{1}{3}+\frac{1}{3})\Big)=\frac{1}{6}$