BdMO National Higher Secondary 2007/9

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BdMO
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BdMO National Higher Secondary 2007/9

Unread post by BdMO » Sun Feb 06, 2011 10:20 pm

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 $L$.

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Re: BdMO National Higher Secondary 2007/9

Unread post by photon » Mon Feb 07, 2011 9:07 am

\[\sqrt{2}\] 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

Unread post by Zzzz » Mon Feb 07, 2011 10:33 am

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 $\sqrt 2$ ?
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Re: BdMO National Higher Secondary 2007/9

Unread post by photon » Mon Feb 07, 2011 12:55 pm

I just used pythagorean theorem.
Here the value of 2 sides of right angle is 1(for each right angled triangle)
So hipotenuse is \[\sqrt{2}\] :P
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Re: BdMO National Higher Secondary 2007/9

Unread post by Zzzz » Mon Feb 07, 2011 1:13 pm

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 ;)
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Re: BdMO National Higher Secondary 2007/9

Unread post by photon » Mon Feb 07, 2011 2:19 pm

oops
Thanks,brother. :) give some hint?
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Re: BdMO National Higher Secondary 2007/9

Unread post by Zzzz » Mon Feb 07, 2011 3:05 pm

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

Unread post by Hasib » Mon Feb 07, 2011 7:00 pm

is it possible to solve it without calculas?
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Re: BdMO National Higher Secondary 2007/9

Unread post by nafistiham » Mon Jan 09, 2012 10:29 pm

i think it'll be a circle whose radius is $\frac {1}{\sqrt 2}$
so,the area will be $\frac {\pi}{2}$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Nadim Ul Abrar
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Re: BdMO National Higher Secondary 2007/9

Unread post by Nadim Ul Abrar » Tue Jan 10, 2012 6:26 pm

Hint
সমকোনী ত্রিভূজের অতিভূজ এর মধ্যবিন্দু হতে ,অতিভূজ এর বিপরীত শী‍‍‍‌‌র্ষ এর দূরত্ব ত্রিভূজের পরিবৃত্তের ব্যাসার্ধ এর সমান ।
Solution
$2^2-\pi .4(\frac{{(\frac{2}{2})^2}}{4})=4-\pi$
$\frac{1}{0}$

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