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 mid-points.
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 mid-points.
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
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- untitled.JPG (13.96KiB)Viewed 6885 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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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 $\frac {\pi}{2}$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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