BdMO National Secondary 2007#6
-
samiul_samin
- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
What is the area bounded by the region $|x+y|+|x-y|=4$ ?Where $x$ and $y$ are real numbers.
-
samiul_samin
- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
Re: BdMO National Secondary 2007#6
I found a solution from $AoPS$.
Solution:
there can be $4$ cases:
$|x+y| + |x-y| = 4$
A. $x+y+x-y = 4 \Rightarrow x=2$
B. $x+y-x+y = 4 \Rightarrow y=2$
C. $-x-y +x-y= 4 \Rightarrow y=-2$
D. $-x-y-x+y=4 \Rightarrow x=-2$
The area bounded by 4 graphs $x=2, x=-2, y=2, y=-2$ will be square with the side of $4$ so the area will be $4\times4=16.$
Solution:
there can be $4$ cases:
$|x+y| + |x-y| = 4$
A. $x+y+x-y = 4 \Rightarrow x=2$
B. $x+y-x+y = 4 \Rightarrow y=2$
C. $-x-y +x-y= 4 \Rightarrow y=-2$
D. $-x-y-x+y=4 \Rightarrow x=-2$
The area bounded by 4 graphs $x=2, x=-2, y=2, y=-2$ will be square with the side of $4$ so the area will be $4\times4=16.$