VOTE : ThamimZahin
Reason : Same as ahmedittihad bro.
Search found 66 matches
- Tue Mar 28, 2017 8:09 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #1
- Replies: 52
- Views: 57337
- Mon Mar 27, 2017 9:57 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies: 68
- Views: 44450
Re: Beginner's Marathon
I'm posting a new problem. $\text {Problem 4}$ Let $ABC$ be an acute triangle with $D,E,F$ the feet of the altitudes lying on $BC, CA,AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P$. The lines $BP$ and $DF$ meet at point $Q$. Prove that $AP =AQ$. (Source...
- Mon Mar 27, 2017 5:04 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #1
- Replies: 52
- Views: 57337
Re: BDMO Forum Mafia #1
Looks like I'm the first player
- Thu Mar 23, 2017 9:05 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia
- Replies: 5
- Views: 4450
Re: BDMO Forum Mafia
Sounds fun!..I am already in.
- Mon Feb 27, 2017 7:14 pm
- Forum: Number Theory
- Topic: Pretty Diophantine Equation
- Replies: 2
- Views: 2991
Re: Pretty Diophantine Equation
Note that the equation can be rewritten as $m_2^2 + 8n_3^2 = 41n_3$, where $x = 7\times 7\times 2\times m_2$ and $y = 7\times 7\times 2\times 2\times n_3$, which gives the only solution when $(m_2, n_3) = (6,4) \Rightarrow (x,y) = (588,784)$
- Mon Feb 13, 2017 12:12 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 junior/4
- Replies: 11
- Views: 7898
Re: BdMO 2017 junior/4
I also messed up part(b) during actual exam. However, I was able to find an answer using brute force then.
I got $n = 730$.
Am I correct?..
I got $n = 730$.
Am I correct?..
- Sun Feb 12, 2017 9:32 pm
- Forum: National Math Olympiad (BdMO)
- Topic: National BDMO 2017 : Junior 8
- Replies: 8
- Views: 4932
Re: National BDMO 2017 : Junior 8
Take point E on $BC$ such that $BD = DE = EC$. Join $O, E$. Now, by $SAS$ theorem, $\bigtriangleup OCE$ is congruent to $\bigtriangleup OBD$. SO, $OD = OE =DE$ and $\bigtriangleup ODE$ is an equilateral triangle. So, $\angle OBD = \angle BOD = 30$. Again, $\angle BAD + \angle ABO + \angle OBD = 60 ...
- Fri Feb 10, 2017 9:56 pm
- Forum: National Math Olympiad (BdMO)
- Topic: National BDMO 2017 : Junior 8
- Replies: 8
- Views: 4932
National BDMO 2017 : Junior 8
In $\bigtriangleup ABC$, the perpendicular bisector of $AB$ and $AC$ meet at $O$. $AO$ meets $BC$ at $D$. Now, $OD$ = $BD$ = $\dfrac {1}{3}BC$. Find the angles of $\bigtriangleup ABC$.
- Sun Feb 05, 2017 7:04 pm
- Forum: Junior Level
- Topic: Find $x$
- Replies: 4
- Views: 3829
Re: Find $x$
$x = 11n + 7$ for any non-negative integer n.
This can be proved very easily. $x \equiv {0,1,2,3,4,5,6,7,8,9,10}$. Now checking the values for $x(x-3) mod 11$ gives the only solution when $x \equiv 7(mod 11)$.
This can be proved very easily. $x \equiv {0,1,2,3,4,5,6,7,8,9,10}$. Now checking the values for $x(x-3) mod 11$ gives the only solution when $x \equiv 7(mod 11)$.
- Fri Feb 03, 2017 4:54 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary: Problem Collection(2016)
- Replies: 17
- Views: 11087
Re: BdMO National Secondary: Problem Collection(2016)
Another solution to Problem 6(a) It can be easily proved that $C, O, I$ are collinear. Let $\angle DBO = \angle a, \angle IDO = \angle b, \angle IDB = \angle c, \angle BOD = \angle d$. $O$ is the circumcenter of $\bigtriangleup ABC$. So, $\angle OCB = \angle OBC = \angle CID = \angle a$. So, $\angle...