## BdMO National Higher Secondary 2007/9

BdMO
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### 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 $L$.

photon
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### Re: BdMO National Higher Secondary 2007/9

$\sqrt{2}$ is the length of the segments by the mid-points.
So the area is 2...
Try not to become a man of success but rather to become a man of value.-Albert Einstein

Zzzz
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### Re: BdMO National Higher Secondary 2007/9

photon wrote:$\sqrt{2}$ is the length of the segments by the mid-points.
So the area is 2...
How have you got that $L$ will be a square of side length $\sqrt 2$ ?
Every logical solution to a problem has its own beauty.

photon
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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 $\sqrt{2}$
Try not to become a man of success but rather to become a man of value.-Albert Einstein

Zzzz
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### 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
Every logical solution to a problem has its own beauty.

photon
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### Re: BdMO National Higher Secondary 2007/9

oops
Thanks,brother. give some hint?
Try not to become a man of success but rather to become a man of value.-Albert Einstein

Zzzz
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### 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
Every logical solution to a problem has its own beauty.

Hasib
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### Re: BdMO National Higher Secondary 2007/9

is it possible to solve it without calculas?
A man is not finished when he's defeated, he's finished when he quits.

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 $\frac {\pi}{2}$
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

$\frac{1}{0}$