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BdMO National Higher Secondary 2007/9
Posted: Sun Feb 06, 2011 10:20 pm
by BdMO
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 $L$.
Re: BdMO National Higher Secondary 2007/9
Posted: Mon Feb 07, 2011 9:07 am
by photon
\[\sqrt{2}\] is the length of the segments by the mid-points.
So the area is 2...
Re: BdMO National Higher Secondary 2007/9
Posted: Mon Feb 07, 2011 10:33 am
by Zzzz
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 $\sqrt 2$ ?
Re: BdMO National Higher Secondary 2007/9
Posted: Mon Feb 07, 2011 12:55 pm
by photon
I just used pythagorean theorem.
Here the value of 2 sides of right angle is 1(for each right angled triangle)
So hipotenuse is \[\sqrt{2}\]
Re: BdMO National Higher Secondary 2007/9
Posted: Mon Feb 07, 2011 1:13 pm
by Zzzz
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
Re: BdMO National Higher Secondary 2007/9
Posted: Mon Feb 07, 2011 2:19 pm
by photon
oops
Thanks,brother.
give some hint?
Re: BdMO National Higher Secondary 2007/9
Posted: Mon Feb 07, 2011 3:05 pm
by Zzzz
draw segments and guess the shape. try to prove your assumption... try it yourself, it will be fun
Re: BdMO National Higher Secondary 2007/9
Posted: Mon Feb 07, 2011 7:00 pm
by Hasib
is it possible to solve it without calculas?
Re: BdMO National Higher Secondary 2007/9
Posted: Mon Jan 09, 2012 10:29 pm
by nafistiham
i think it'll be a circle whose radius is $\frac {1}{\sqrt 2}$
so,the area will be $\frac {\pi}{2}$
Re: BdMO National Higher Secondary 2007/9
Posted: Tue Jan 10, 2012 6:26 pm
by Nadim Ul Abrar