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BdMO Online Forum • Search

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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: 64

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: 70

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: 70

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: 672

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: 182

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: 866

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: 417

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: 63
Views: 2133

Re: BDMO Forum Mafia #1

The reason behind voting me is not strong. If I was mafia then I would try to be more active in the thread and tried to suspect anyone with a logic. I didn't do so. At first I voted Jahin as all of you had voted him. I don't want to be killed indeed. "Morite Chahina Ami Sundoro Forume" Aga...
by Ananya Promi
Fri Mar 31, 2017 5:53 pm
 
Forum: Social Lounge
Topic: BDMO Forum Mafia #1
Replies: 50
Views: 1684

Re: BDMO Forum Mafia #1

VOTE: (ahmedittihad)

No strong logic but I think he is a mafia...
by Ananya Promi
Fri Mar 31, 2017 2:53 pm
 
Forum: Social Lounge
Topic: BDMO Forum Mafia #1
Replies: 50
Views: 1684
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