OH!.......Parellelograms
- Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
- Location:127.0.0.1
- Contact:
( Inspired from Masum Vaia ) In co-ordinate plane, join points
$[(0,0),(m,0)],[(0,1),(m,1)],...,[(o,n),(m,n)]$ and
$[(0,0],(0,n)], [(1,0),(1,n)...,[(m,0),(m,n)]$.How many parallelograms can you make
joining the intersection points of these lines?
Note:$(0,0),(2,1),(3,3),(1,2)$ is also a parallelogram.So you must count this type of parallelograms too!
$[(0,0),(m,0)],[(0,1),(m,1)],...,[(o,n),(m,n)]$ and
$[(0,0],(0,n)], [(1,0),(1,n)...,[(m,0),(m,n)]$.How many parallelograms can you make
joining the intersection points of these lines?
Note:$(0,0),(2,1),(3,3),(1,2)$ is also a parallelogram.So you must count this type of parallelograms too!
Welcome to BdMO Online Forum. Check out Forum Guides & Rules
- nafistiham
- Posts:829
- Joined:Mon Oct 17, 2011 3:56 pm
- Location:24.758613,90.400161
- Contact:
Re: OH!.......Parellelograms
let me share a hint given by the topic owner [without permission ]
[the topic owner does not want you to read it ]
Last edited by nafistiham on Fri Feb 24, 2012 8:54 pm, edited 1 time in total.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
- Location:127.0.0.1
- Contact:
- nafistiham
- Posts:829
- Joined:Mon Oct 17, 2011 3:56 pm
- Location:24.758613,90.400161
- Contact:
Re: OH!.......Parellelograms
ok.i am hiding that.Phlembac Adib Hasan wrote:Not Fare.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
- Location:127.0.0.1
- Contact:
Re: OH!.......Parellelograms
Ok.Ok.It's alright now.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules
- nafistiham
- Posts:829
- Joined:Mon Oct 17, 2011 3:56 pm
- Location:24.758613,90.400161
- Contact:
Re: OH!.......Parellelograms
my solution was like this
\[\sum_{k=1}^{n-1}k \cdot\sum_{p=1}^{m-1}p\]
adib made it short
\[\sum_{k=1}^{n-1}k \cdot\sum_{p=1}^{m-1}p\]
adib made it short
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
- Location:127.0.0.1
- Contact:
Re: OH!.......Parellelograms
You might do a mistake.Because you are not counting $(0,0),(2,1),(3,3),(1,2)$ type parallelograms,only counting the rectangles.nafistiham vaia wrote:my solution was like this
\[\sum_{k=1}^{n-1}k \cdot\sum_{p=1}^{m-1}p\]
adib made it short
Welcome to BdMO Online Forum. Check out Forum Guides & Rules