BdMO National Higher Secondary 2007/9
BdMO National Higher Secondary 2007/9
Problem 9:
A square has sides of length $2$. Let $S$ is the set of all line segments that have length $2$ and whose endpoints are on adjacent side of the square. Say $L$ is the set of the midpoints of all segments in $S$. Find out the area enclosed by $L$.
A square has sides of length $2$. Let $S$ is the set of all line segments that have length $2$ and whose endpoints are on adjacent side of the square. Say $L$ is the set of the midpoints of all segments in $S$. Find out the area enclosed by $L$.
Re: BdMO National Higher Secondary 2007/9
\[\sqrt{2}\] is the length of the segments by the midpoints.
So the area is 2...
So the area is 2...
Try not to become a man of success but rather to become a man of value.Albert Einstein
Re: BdMO National Higher Secondary 2007/9
How have you got that $L$ will be a square of side length $\sqrt 2$ ?photon wrote:\[\sqrt{2}\] is the length of the segments by the midpoints.
So the area is 2...
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Re: BdMO National Higher Secondary 2007/9
I just used pythagorean theorem.
Here the value of 2 sides of right angle is 1(for each right angled triangle)
So hipotenuse is \[\sqrt{2}\]
Here the value of 2 sides of right angle is 1(for each right angled triangle)
So hipotenuse is \[\sqrt{2}\]
Try not to become a man of success but rather to become a man of value.Albert Einstein
Re: BdMO National Higher Secondary 2007/9
How can you be sure that $L$ is a square? See the image below. The Blue lines are some segments that have length $2$ and whose endpoints are on adjacent side of the square. $S$ is the set of all such segments. Join the midpoints of the segments and find out the shape of the area enclosed by the midpoints
 Attachments

 untitled.JPG (13.96 KiB) Viewed 3921 times
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Re: BdMO National Higher Secondary 2007/9
oops
Thanks,brother. give some hint?
Thanks,brother. give some hint?
Try not to become a man of success but rather to become a man of value.Albert Einstein
Re: BdMO National Higher Secondary 2007/9
draw segments and guess the shape. try to prove your assumption... try it yourself, it will be fun
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Re: BdMO National Higher Secondary 2007/9
is it possible to solve it without calculas?
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 nafistiham
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Re: BdMO National Higher Secondary 2007/9
i think it'll be a circle whose radius is $\frac {1}{\sqrt 2}$
so,the area will be $\frac {\pi}{2}$
so,the area will be $\frac {\pi}{2}$
\[\sum_{k=0}^{n1}e^{\frac{2 \pi i k}{n}}=0\]
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