[phpBB Debug] PHP Warning: in file [ROOT]/includes/bbcode.php on line 122: include(/home/shoeb/public_html/www.matholympiad.org.bd/forum/includes/phpbb-latex.php) [function.include]: failed to open stream: No such file or directory
[phpBB Debug] PHP Warning: in file [ROOT]/includes/bbcode.php on line 122: include() [function.include]: Failed opening '/home/shoeb/public_html/www.matholympiad.org.bd/forum/includes/phpbb-latex.php' for inclusion (include_path='.:/opt/php53/lib/php')
[phpBB Debug] PHP Warning: in file [ROOT]/includes/functions.php on line 4786: Cannot modify header information - headers already sent by (output started at [ROOT]/includes/functions.php:3887)
[phpBB Debug] PHP Warning: in file [ROOT]/includes/functions.php on line 4788: Cannot modify header information - headers already sent by (output started at [ROOT]/includes/functions.php:3887)
[phpBB Debug] PHP Warning: in file [ROOT]/includes/functions.php on line 4789: Cannot modify header information - headers already sent by (output started at [ROOT]/includes/functions.php:3887)
[phpBB Debug] PHP Warning: in file [ROOT]/includes/functions.php on line 4790: Cannot modify header information - headers already sent by (output started at [ROOT]/includes/functions.php:3887)
BdMO Online Forum • Search

Search found 15 matches

Return to advanced search

Re: IMO 2017 P4

We get $TA$ parallel to $KR$ because $\angle{ATS}=\angle{SJK}=\angle{SRK}$ We extend $KS$ to $P$ where $KS$ intersects $TA$ at $P$ Now, It's easy to prove that $TPRK$ is a rombus So, $\angle{TPK}=\angle{PKR}$ Again, $\angle{ARS}=\angle{SKR}$ So, $\angle{TPK}=\angle{ARS}$ So, $APSR$ is cyclic. $\angl...
by Ananya Promi
Sat Jul 29, 2017 12:38 pm
 
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2017 P4
Replies: 1
Views: 96

IMO 2017 P4

Let R and S be dierent points on a circle Ω such that RS is not a diameter. Let be the tangent line to Ω at R. Point T is such that S is the midpoint of the line segment RT. Point J is chosen on the shorter arc RS of Ω so that the circumcircle Γ of triangle JST intersects at two distinct points. Let...
by Ananya Promi
Sat Jul 29, 2017 12:29 pm
 
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2017 P4
Replies: 1
Views: 96

CGMO 2007/5

Point $D$ lies inside triangle $ABC$ such that $$\angle{DAC}=\angle{DCA}=30^o$$ and $$\angle{DBA}=60^o$$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF=2FC$. Prove that $DE$ I $EF$.
by Ananya Promi
Sat Jun 03, 2017 3:21 pm
 
Forum: Geometry
Topic: CGMO 2007/5
Replies: 1
Views: 84

Re: IMO 2001 Problem1

$$\angle{A}+\angle{COP} <90° \Rightarrow \angle{COP} <90°-\angle{COD}$$ $$\Rightarrow \angle{COP} <\angle{OCP} \Rightarrow CP<PO$$ is enough to prove. $\angle C -\angle B \geq 30°$ as both $(\angle{C}-\angle{B})$ and 30° are less than $90°$, $$\begin{align*}&\sin(\angle{C}-\angle{B})\geq \sin30°...
by Ananya Promi
Fri Jun 02, 2017 3:04 pm
 
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2001 Problem1
Replies: 2
Views: 100

IMO 2001 Problem1

Consider an acute angled triangle $ABC$. Let $P$ be the foot of the altitude of triangle $ABC$ issuing from the vertex $A$, and let $O$ be the circumvented of triangle $ABC$. Assume that $$\angle{C}\geq\angle{B}+30°$$. Prove that $$\angle{A}+\angle{COP}<90°$$
by Ananya Promi
Fri Jun 02, 2017 3:01 pm
 
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2001 Problem1
Replies: 2
Views: 100

Re: IMO 2007 Problem 4

Let's draw $RX$ perpendicular on $PK$ which intersects thevcircumcirle of triangle $ABC$ at $M$ $RX$ is parallel to $BC$ So, $BRMC$ is a trapizoid. So, $BR$=$MC$ Again, $BR$=$AR$ is known as $CR$ in the angle bisector of angle $ACB$ So, $AR$=$MC$ and it makes $ARCM$ cyclic trapizoid Then $MR$=$b$ $X...
by Ananya Promi
Fri Jun 02, 2017 1:59 pm
 
Forum: Geometry
Topic: IMO 2007 Problem 4
Replies: 4
Views: 713

Re: BDMO 2017 National round Secondary 2

Here, $AD$ is perpendicular on $BC$ Hence, $BE=CE$ We can show that $$\triangle$$$BDE$ and $$\triangle$$$CFE$ are congruent. So, $FE=ED=OF=\frac{1}{3}OD=\frac{1}{3}OB$ In the right angle triangle $OBE$, $OE^2+BE^2=OB^2$ or,$4OF^2+5=9OF^2$ or,$OF=1=DE$ in rigth angle triangle $CED$, $CE^2+DE^2=CD^2$ ...
by Ananya Promi
Thu Apr 27, 2017 4:31 pm
 
Forum: National Math Olympiad (BdMO)
Topic: BDMO 2017 National round Secondary 2
Replies: 1
Views: 243

Re: BDMO Forum Mafia #2

I am also in
by Ananya Promi
Wed Apr 12, 2017 10:42 am
 
Forum: Social Lounge
Topic: BDMO Forum Mafia #2
Replies: 27
Views: 1112

Re: The Gonit IshChool Project - Beta FE Class

I will be able to dedicate about 10 hours per week.
My experience is Adib vai's class.
by Ananya Promi
Sun Apr 02, 2017 2:31 pm
 
Forum: National Math Camp
Topic: The Gonit IshChool Project - Beta FE Class
Replies: 8
Views: 535

Re: Beginner's Marathon

Another solution to P6: As $EM$ is perpendicular on $AC$ we can say by perpendicular lemma, $CE^2 + AM^2 = AE^2 + CM^2$ or, $CM^2 - AM^2 = CE^2 - AE^2$ Again, $MF$ is perpendicular on $BC$ so, $CF^2 + BM^2 = CM^2 + BF^2$ or, $CM^2 - BM^2 = CF^2 - BF^2 = CM^2 - AM^2$ so, $CE^2 - AE^2 = CF^2 - BF^2$ ...
by Ananya Promi
Fri Mar 31, 2017 6:28 pm
 
Forum: Junior Level
Topic: Beginner's Marathon
Replies: 68
Views: 2802
Next

Return to advanced search

cron