## Search found 8 matches

Sat Mar 17, 2018 2:28 pm
Forum: Junior Level
Topic: BDMO 2016: National Junior/2
Replies: 5
Views: 6686

### 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: 1233

### BDMO national 2016 secondery 7

Thu Mar 15, 2018 12:31 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2017 P4
Replies: 4
Views: 7109

### 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: 3332

### Re: BDMO national 2016, junior 10

Tue Mar 13, 2018 9:13 pm
Forum: National Math Olympiad (BdMO)
Topic: National BDMO Secondary P8
Replies: 4
Views: 2533

### 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: 4492

### 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: 3405

### 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: 27654

### A hard Geometry Problem

\$\triangle ABC\$ is an acute angled triangle with orthocenter \$H\$ and circumcenter \$O\$. Prove that, \$\angle CAH =\angle BAO\$.