Screenshot_2019-01-10-23-19-29-1.png
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?BdMO regional 2018 set 4 Secondary P 10
Forum rules
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always 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.
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always 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.
-
- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
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 $IHJF$.
$\frac{1}{2}\Big((\frac{1}{3}+\frac{1}{3}+\frac{2}{3})-(\frac{1}{3}+\frac{1}{3}+\frac{1}{3})\Big)=\frac{1}{6}$