Search found 81 matches
- Sun Aug 07, 2016 11:56 am
- Forum: Number Theory
- Topic: IMO Shortlist 2012 N1
- Replies: 7
- Views: 5437
Re: IMO Shortlist 2012 N1
OK. I have approached it like this. Let, $d= gcd(m,n).$ If, $d > 1$, if $d|x$ & $ d|y $ then $d|x^2+kxy+y^2 $ for all k.So,though the multiple of d satisfies condition, but $A \ne \mathbb{Z}$. here it is easy to see, $ d=1$ . Now,We let,the set $A$ is admissible containing m,n.So,if $ x^2 \in A$ , ...
- Sun Aug 07, 2016 12:00 am
- Forum: Number Theory
- Topic: IMO Shortlist 2012 N1
- Replies: 7
- Views: 5437
Re: IMO Shortlist 2012 N1
OK. I have approached it like this. Let, $d= gcd(m,n).$ If, $d > 1$, if $d|x$ & $ d|y $ then $d|x^2+kxy+y^2 $ for all k.So,though the multiple of d satisfies condition, but $A \ne \mathbb{Z}$. here it is easy to see, $ d=1$ . Now,We let,the set $A$ is admissible containing m,n.So,if $ x^2 \in A$ , t...
- Sat Aug 06, 2016 1:59 pm
- Forum: Social Lounge
- Topic: Chat thread
- Replies: 53
- Views: 77889
Re: Chat thread
Well, I am Nahin Munkar (not moon-car !). I also live in Dhaka . I am highly interested in literature & mathematics as well. I am also fond of reading,kidding,writing,fighting and so on....
It's really interesting place . May $BDMO$ $FORUM$ live long.
It's really interesting place . May $BDMO$ $FORUM$ live long.
- Fri Aug 05, 2016 4:13 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2016 #4
- Replies: 2
- Views: 7161
Re: APMO 2016 #4
I think, the answer is $57$.(may be). According to condition, we can make a cycle of $57$ cities where it is always possible to have a reach-connection between any two cities using at most $28$ flights . So,any $2$ of those $57$ cities cannot situate in the same group. So, at least $57$ groups are n...
- Thu Aug 04, 2016 9:19 pm
- Forum: Combinatorics
- Topic: Tennis League (USAMO 1989)
- Replies: 3
- Views: 3307
Re: Tennis League (USAMO 1989)
14 two-person games make $14*2=28$ place to participate. Now,under condition,if all participate in this game just one time,there will be 8 place to play more as there r $20$ players of this tournament . if the players play in those 8 position,there will be 8 games where any player plays more than on...
- Wed Aug 03, 2016 10:27 pm
- Forum: Geometry
- Topic: Heros of 71 (Geometry)
- Replies: 7
- Views: 5684
Re: Heros of 71 (Geometry)
kazi zareer actually wants the total manipulation of this result. (He already makes me mad in the class..)
- Sat May 28, 2016 1:12 am
- Forum: Combinatorics
- Topic: Combi identity proof
- Replies: 4
- Views: 3810
Re: Combi identity proof
Use symmetric identity that, $n C r = n C (n-r)$. Then u get at R.H.S.: $(2n) C (n+1)=(2n) C (n-1)$; [1st & last term same.] Then, use another summation identity: $nCr+nC(r+1)=(n+1)C(r+1)$ .Then after some calculation u will get the form to prove: $(2n+2)C(n+1)=(2)*{{(2n+1)C(n+1)}}$.It's just easy s...
- Sat May 28, 2016 12:47 am
- Forum: Geometry
- Topic: Easy Problem in geo, (cyclic, angle chasing)
- Replies: 6
- Views: 4823
Re: Easy Problem in geo, (cyclic, angle chasing)
Here,we can show $AQYR$ & $XQDP$ is cyclic. So,By easy angle chasing, we can show $\angle RQF$ =$\angle FQP$.That's the proof.[here, RG cuts AC at Y & EP cuts BD at X. It's a simple proof.Try to yourself.]. Proved. :) I have already solved it and given the main clues of my solution. ;) Yes,dude.I s...
- Sat May 28, 2016 12:41 am
- Forum: Geometry
- Topic: China TST 2016 P1
- Replies: 3
- Views: 3414
Re: China TST 2016 P1
Yeah, solved.Kazi_Zareer wrote:Did you solve it?nahin munkar wrote:Use $ homotety$.It may be a useful way to get the solution.
- Sun May 22, 2016 5:54 pm
- Forum: Geometry
- Topic: Easy Problem in geo, (cyclic, angle chasing)
- Replies: 6
- Views: 4823
Re: Easy Problem in geo, (cyclic, angle chasing)
Well done.It's another good approach in projective way.