Search found 461 matches
- Thu May 05, 2011 12:11 am
- Forum: Geometry
- Topic: Triangles and Medians
- Replies: 2
- Views: 2398
Re: Triangles and Medians
First let's use a trick :D Trick: Just try it on a equilateral triangle and use the theorem that the ratio of area of two similar triangle is equal to the ratio of area of squares of similar sides.( This trick is not my own way , i get it from a bhaiia) Now the direct proof :mrgreen: Direct solution...
- Wed May 04, 2011 11:33 pm
- Forum: Geometry
- Topic: A QUESTION ABOUT PROPERTIES OF TRIANGLE
- Replies: 3
- Views: 2898
Re: A QUESTION ABOUT PROPERTIES OF TRIANGLE
You are right that orthocenter $H$ of a triangle is isogonal conjugated of circumcenter $O$ of that triangle, but that doesn't mean that $HO$ must pass through incenter $I$.
- Tue Mar 15, 2011 7:10 pm
- Forum: Geometry
- Topic: Bangladesh IMO TST 1: 2011/ Geometry-1 (P 2)
- Replies: 1
- Views: 2413
Re: Bangladesh IMO TST 1: 2011/ Geometry-1 (P 2)
I find a solution. Please reply if i am right or wrong. :) Solution: Draw three perpandicular $HE,HF,HG$ from $H$ to respectively on $AB,AC,PQ$ . $AE=AHsin<EHA=AHsinB$ $AF=AHsin<AHF=AHsinC$ So,$AE/AF=sinB/sinC$ $BPQC$ is cyclic. So,$<BPQ=180-<BCQ=<C$ $EPGH$ is cyclic and $PH=2R$ $therefore EG=HPsinC...
- Sun Mar 13, 2011 7:01 pm
- Forum: Geometry
- Topic: Bangladesh IMO TST 1: 2011/ Geometry-2 (P 2)
- Replies: 2
- Views: 2815
Re: Bangladesh IMO TST 1: 2011/ Geometry-2 (P 2)
What is $J$?
- Sun Mar 13, 2011 6:43 pm
- Forum: Geometry
- Topic: Geometry Marathon v1.0
- Replies: 23
- Views: 16508
Re: Geometry Marathon v1.0
Draw three parallel lines $DE,MN,PQ$ through the inercenter of $\Delta ABC.$ If $S$ is the inercenter of $\Delta ABC$ then, \[\angle SBD= \angle SBM, \quad \angle MSB= \angle BSD\] So $MSBD$ is a rombos. Therefore,$SD=BD$ In this way, $SP=PC$ $\therefore SP+SD+PD=a$ In this way , we can prove that, ...
- Thu Feb 24, 2011 8:11 pm
- Forum: Combinatorics
- Topic: Find the number of surjective functions
- Replies: 3
- Views: 3581
Re: Find the number of surjective functions
There are \[n^{m}\] ways to make any function. There are \[\binom{n}{1}\left ( n-1 \right )^{m}\] ways to make a function not using one element from co-domain. There are \[\binom{n}{2}\left ( n-2 \right )^{m}\]ways to make a function not using two elements from co-domain. In this way, . . . There ar...
- Wed Feb 16, 2011 12:04 pm
- Forum: National Math Camp
- Topic: Bangladesh National Math Camp 2011
- Replies: 13
- Views: 9857
Re: Bangladesh National Math Camp 2011
Bhaia, i have never got any chance to attend camp.But this year i am a divisional champion and a national champion too. I really want to attend camp. Can i get chance to camp? And my request is "Please,please and please :Start camp after S.S.C. practical ." :cry: Otherwise my risky attempt (attendin...
- Wed Feb 16, 2011 11:37 am
- Forum: Combinatorics
- Topic: Find the number of surjective functions
- Replies: 3
- Views: 3581
Re: Find the number of surjective functions
Bhaia, I don't know whether my ans. is right or wrong;But i use exclusion-inclusion to solve it.
And the ans. is:
\[n^{m}-\binom{n}{1}\left ( n-1 \right )^{m}+\binom{n}{2}(n-2)^{m}-...............+(-1)^{n-1}n\]
Please reply.
And the ans. is:
\[n^{m}-\binom{n}{1}\left ( n-1 \right )^{m}+\binom{n}{2}(n-2)^{m}-...............+(-1)^{n-1}n\]
Please reply.
- Wed Jan 26, 2011 1:54 pm
- Forum: Secondary Level
- Topic: Find the Product of the terms of a sequence
- Replies: 6
- Views: 4503
Re: Find the Product of the terms of a sequence
I think it has a better solution.\[x_{1}x_{2}=2;x_{3}x_{4}=4....
\therefore x_{1}x_{2}...x_{8}=2*4*6*8=384
\]
Am i right?
\therefore x_{1}x_{2}...x_{8}=2*4*6*8=384
\]
Am i right?
- Wed Jan 26, 2011 1:36 pm
- Forum: Secondary Level
- Topic: সেট থিওরীঃ পাওয়ার সেট
- Replies: 2
- Views: 2769
Re: সেট থিওরীঃ পাওয়ার সেট
Ans: {@} (Sorry, i cannot use the sign of subset)