Search found 8 matches
- Sat Mar 17, 2018 2:28 pm
- Forum: Junior Level
- Topic: BDMO 2016: National Junior/2
- Replies: 5
- Views: 8999
Re: BDMO 2016: National Junior/2
very easy problem
- Thu Mar 15, 2018 12:33 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO national 2016 secondery 7
- Replies: 1
- Views: 2251
BDMO national 2016 secondery 7
see here for full solution
https://artofproblemsolving.com/communi ... _a_problem
https://artofproblemsolving.com/communi ... _a_problem
- Thu Mar 15, 2018 12:31 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2017 P4
- Replies: 4
- Views: 9216
Re: IMO 2017 P4
very easy problem for IMO
- Wed Mar 14, 2018 6:04 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO national 2016, junior 10
- Replies: 7
- Views: 5532
- Tue Mar 13, 2018 9:13 pm
- Forum: National Math Olympiad (BdMO)
- Topic: National BDMO Secondary P8
- Replies: 4
- Views: 4253
Re: National BDMO Secondary P8
[After Tasnood] In the cyclic $BCFE$ quadrileteral,we get $/angleBFC=120$.$/angleDFE=90$ so,$/angleCFD=30$.Now $FD$ bisects $/angleBFC$ then https://artofproblemsolving.com/community/c5h472280p2644107 [it is a well known geometry problem from AIME 2012.So,sad that it is th 8th problem in 2012 nation...
- Sat Mar 03, 2018 4:34 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 junior/4
- Replies: 11
- Views: 7907
Re: BdMO 2017 junior/4
n^3-n=n^2(n-1) Here we have 2 cases. Case 1:n^2 is divisible by 73 and n-1 is divisible by 27. Now n is the form of 73k. We get 73k-1 is congruent to 0(mod27) Or,-(8k+1) is congruent to 0(mod27) Or, 8k is congruent to 26 (mod27) Or, 4k is congruent to 13 (mod27) Or, 4k is congruent to 40 (mod27) Or,...
- Sat Feb 24, 2018 12:48 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL Junior 2016/04
- Replies: 8
- Views: 5940
Re: BDMO NATIONAL Junior 2016/04
hello protaya and prottoy das are the same person.I have forgot the password of protaya das so i am using the prottoy das id.
- Thu Feb 01, 2018 8:14 pm
- Forum: Geometry
- Topic: A hard Geometry Problem
- Replies: 4
- Views: 29688
A hard Geometry Problem
$\triangle ABC$ is an acute angled triangle with orthocenter $H$ and circumcenter $O$. Prove that, $\angle CAH =\angle BAO$.