Search found 57 matches
- Wed Feb 22, 2017 2:41 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL JUNIOR 2012/03
- Replies: 4
- Views: 3577
Re: BDMO NATIONAL JUNIOR 2012/03
Let,the total days needed by Tanvir be M1+E1,where M1 denotes the total number of mornings and E1 denotes the total number of evenings.Now,without rain,M1+E1=29.........(1),again with rain,M1+E1=11.........(2);By adding (1) and (2)..we get M1+E1=20;therefore,the total number of days to climb the mou...
- Wed Feb 22, 2017 2:21 pm
- Forum: National Math Olympiad (BdMO)
- Topic: How two altitudes determine the third
- Replies: 3
- Views: 3259
Re: How two altitudes determine the third
Let,ABC denote a right angled triangle.Let,the sides be 2014,1 and y units.Then the area of the triangle will be 1007 sq units.If we construct another altitude x in length,then the area will be xy/2 units.Then,we apply an equation,xy/2=1007;then x results in 2014/√(2014^2+1) units which denotes the ...
- Fri Feb 17, 2017 9:14 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 junior/10
- Replies: 2
- Views: 2714
Re: BdMO 2017 junior/10
Let,the maximum number of combinations be x....which follows the form ab+ab+....+ab...where a=2,b=-1;Now,when there are 100 terms in a sequence,the numbers of combinations will be maximum when number of a=number of b.....because here distinct terms will be applicable and 2 terms are followed....In t...
- Thu Feb 16, 2017 8:51 pm
- Forum: National Math Olympiad (BdMO)
- Topic: National BDMO 2017 : Junior 8
- Replies: 8
- Views: 4939
Re: National BDMO 2017 : Junior 8
You are right,Thamim...I also feel like kicking myself for not paying much attention to this problem as I was working on other problems...or else I could have attained a bit more than being 2nd runners up......Now,This problem does need only a bit of algebraic calculations..We may let angle BAO to b...
- Sun Jan 29, 2017 8:22 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL JUNIOR 2015/5
- Replies: 4
- Views: 3877
Re: BDMO NATIONAL JUNIOR 2015/5
Hmmm....you are right...I forgot to consider those cases
- Sun Jan 29, 2017 12:00 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL JUNIOR 2015/5
- Replies: 4
- Views: 3877
Re: BDMO NATIONAL JUNIOR 2015/5
Suppose,there are m girls and n boys.Then,1 boy makes m handshakes.For this reason,n boys make in total mn/2 handshakes.Again,1 girl makes m+n-1 handshakes,so m girls make(m^2+mn-m)/2 handshakes.Now,establish the condition,mn/2+(m^2+mn-m)/2=40.....then you may derive m(m+2n-1)=80,then a little check...
- Sat Jan 28, 2017 9:53 pm
- Forum: Junior Level
- Topic: BDMO NATIONAL Junior 2016/03
- Replies: 4
- Views: 4171
Re: BDMO NATIONAL Junior 2016/03
I think it should have been 2*2013+2012+2011......+3+2+1,because I did not consider the adjacent points in the 1st case
- Sat Jan 28, 2017 7:49 pm
- Forum: Junior Level
- Topic: BDMO NATIONAL Junior 2016/03
- Replies: 4
- Views: 4171
Re: BDMO NATIONAL Junior 2016/03
I am wrong indeed......but as I am still in class 8,I don't understand what triangulation of polygon is....
- Sat Jan 28, 2017 1:49 pm
- Forum: Junior Level
- Topic: COMBINATORICS!!!
- Replies: 4
- Views: 3778
Re: COMBINATORICS!!!
Correct!My solution is more or less as yours too.....
- Sat Jan 28, 2017 1:33 pm
- Forum: Junior Level
- Topic: BDMO NATIONAL Junior 2016/03
- Replies: 4
- Views: 4171
Re: BDMO NATIONAL Junior 2016/03
If we mark our total diagonals as M,we find M=C(2016,2)-2016.Then our required diagonals will be less than the total number of diagonals.By cutting out a pattern.....we follow that for an N sided polygon,the required diagonals as per the question is 2n+(n-1)+(n-2).........+2+1...So,the required numb...