Search found 185 matches
- Tue May 31, 2016 4:53 pm
- Forum: Algebra
- Topic: Inequality with xyz=1
- Replies: 1
- Views: 2958
Re: Inequality with xyz=1
Let \begin{align*} \dfrac{1}{2x-1}=a\\ \dfrac{1}{2y-1}=b\\ \dfrac{1}{2z-1}=c \end{align*} So \begin{align*} x=\dfrac{a+1}{2a}\\ y=\dfrac{b+1}{2b}\\ z=\dfrac{c+1}{2c} \end{align*} Now $xyz=1$ implies $(a+1)(b+1)(c+1)=8abc$ So \begin{align*} 8abc & = \prod\limits_{cyc}^{}(a+1)\\ & \geq \prod\limits_{c...
- Wed Apr 27, 2016 4:17 pm
- Forum: Geometry
- Topic: feet of perpendicular and midpoint
- Replies: 1
- Views: 2540
Re: feet of perpendicular and midpoint
It will be equal to $0$ if you take directed segment, otherwise not.
Hint
Hint
- Mon Feb 15, 2016 8:48 am
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO national higher secondery -prb 07
- Replies: 2
- Views: 4534
Re: BDMO national higher secondery -prb 07
Assume triangles $COB,COA,AOB$ have area $p,q,r$ respectively. Then express every ratio in terms of $p,q,r$.
- Fri Feb 05, 2016 11:29 pm
- Forum: Primary Level
- Topic: Divisional Math Olympiad, Dhaka-2016,primary, ques. 7
- Replies: 10
- Views: 16848
Re: Divisional Math Olympiad, Dhaka-2016,primary, ques. 7
Because $x$ has the only value $99$.ahsaf wrote:I didn't understand why are we taking $99$ to prove the problem
- Wed Feb 03, 2016 12:36 pm
- Forum: Junior Level
- Topic: Easy Chess Tournament Problem
- Replies: 1
- Views: 2690
Re: Easy Chess Tournament Problem
Let $A,B,C$ be the set of first $3$,middle $4$ and last $5$ players respectively according to their rank and $U$ is the union set.Let $f(X,Y)$ denote the total score gained by the players of set $X$ against players of set $Y$.Easy to see that every match has total outcome of point $1$. now $f(A,U)=f...
- Sun Jan 31, 2016 4:03 pm
- Forum: Primary Level
- Topic: Divisional Math Olympiad, Dhaka-2016,primary, ques. 7
- Replies: 10
- Views: 16848
- Mon Jan 25, 2016 10:46 pm
- Forum: Secondary Level
- Topic: 2014-national
- Replies: 15
- Views: 11758
Re: 2014-national
Can you please explain your solution??seemanta001 wrote:The solution is:
For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.
- Mon Jan 25, 2016 10:32 pm
- Forum: Geometry
- Topic: A BEAUTIFUL GEO
- Replies: 1
- Views: 2481
- Mon Jan 25, 2016 9:33 pm
- Forum: Divisional Math Olympiad
- Topic: Dhaka regional higher secondary/8
- Replies: 3
- Views: 3985
Re: Dhaka regional higher secondary/8
Prove that if $5 \nmid x$, then $f(x)=0$ using the first 2 conditions.
- Mon Jan 25, 2016 9:04 pm
- Forum: Divisional Math Olympiad
- Topic: Dhaka regional 2015 secondary/4
- Replies: 2
- Views: 3170