Math problem 2 (need solution)

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ataher.sams
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Math problem 2 (need solution)

Unread post by ataher.sams » Thu Dec 01, 2011 7:56 pm

There are three points in a plane[somotol-a].One can draw as many parallelograms as possible keeping those three points as the three vertices of the parallelogram.Find the difference between the area of parallelogram having the largest perimeter possible and the parallelogram having the minimum permeter possible.

I have done it. But I am not sure abt it... So anyone please give me the answer..... :? :?
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Re: Math problem 2 (need solution)

Unread post by *Mahi* » Fri Dec 02, 2011 12:49 am

Try posting your own solution. Then you'll know whether that's right or wrong.
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Re: Math problem 2 (need solution)

Unread post by Corei13 » Fri Dec 02, 2011 12:51 pm

ataher.sams wrote:One can draw as many parallelograms as possible keeping those three points as the three vertices of the parallelogram
How ?
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ataher.sams
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Re: Math problem 2 (need solution)

Unread post by ataher.sams » Fri Dec 02, 2011 1:50 pm

I have to draw it... But how can I draw it here????
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Re: Math problem 2 (need solution)

Unread post by Corei13 » Fri Dec 02, 2011 2:22 pm

Not actually, you can draw only 3 parallelogram with 3 given vertexes. So if the given points are $A, B,C$ then the answer of your problem is $2\left(\max\{AB,BC,CA\}-\min\{AB,BC,CA\}\right)$

By the way, if you want to post something drawn, you can draw it in your PC and then submit it as a attachment.
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Re: Math problem 2 (need solution)

Unread post by *Mahi* » Fri Dec 02, 2011 2:41 pm

Corei13 wrote:Not actually, you can draw only 3 parallelogram with 3 given vertexes. So if the given points are $A, B,C$ then the answer of your problem is $2\left(\max\{AB,BC,CA\}-\min\{AB,BC,CA\}\right)$
@Corei13:
Notice that all 3 parallelograms have the same area ;)
@ataher.sams:
You can see this topic.
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Re: Math problem 2 (need solution)

Unread post by Labib » Fri Dec 02, 2011 2:51 pm

ডিজে...
সে যে কয়টাই আকুক, সবগুলাই একটার উপর আরেকটা সমাপতিত হবে... ;)
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Re: Math problem 2 (need solution)

Unread post by nafistiham » Fri Dec 02, 2011 3:12 pm

any three vertices of a parallellogram make a triangle having half the area of it.and, the other vertex is such a point that creates another triangle having a common side with the previous one.these triangles are congruent.so the answer is $0$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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ataher.sams
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Re: Math problem 2 (need solution)

Unread post by ataher.sams » Fri Dec 02, 2011 9:16 pm

Yes... My ans also was "0" ...
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Re: Math problem 2 (need solution)

Unread post by Corei13 » Fri Dec 02, 2011 10:09 pm

I misread as it asked the difference between maximum and minimum perimeter. Actually $2\left(\max\{AB,BC,CA\}-\min\{AB,BC,CA\}\right)$ doesn't express an area here!
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