Search found 21 matches
- Tue Dec 27, 2011 12:53 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2009/5
- Replies: 5
- Views: 4598
Re: BdMO National Higher Secondary 2009/5
let\[O_{1}=\]circumcenter of \[\triangle AMC\]\[X=O_{1}P\cap AB,Y=BA\cap CO_{1}\]\[O_{1}AXM\]is a rhombus. \[\Rightarrow AE=EM\Rightarrow AP=PM\Rightarrow \angle PYA=\angle PCA=\angle PAM=\angle PCM\]
- Thu Dec 15, 2011 9:10 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2007
- Replies: 5
- Views: 5408
APMO 2007
Given \[\sqrt{x}+\sqrt{y}+\sqrt{z}=1\] for all positive real \[x,y,z\]Prove \[\frac{x^2+yz}{\sqrt{2x^2(y+z)}} + \frac{y^2+zx}{\sqrt{2y^2(z+x)}} + \frac{z^2+xy}{\sqrt{2z^2(x+y)}} \geq1\]
- Sun Nov 13, 2011 4:04 pm
- Forum: Number Theory
- Topic: IMO LONGLISTED PROBLEM 1974
- Replies: 1
- Views: 2170
Re: IMO LONGLISTED PROBLEM 1974
we've to prove \[2^{147}=1(mod 343)\]
we know \[2^{9}=169(mod343)\Rightarrow 2^{144}=43(mod343)\]
and \[2^{3}=351(mod343)\]
multiplying both this we get the desired result.
we know \[2^{9}=169(mod343)\Rightarrow 2^{144}=43(mod343)\]
and \[2^{3}=351(mod343)\]
multiplying both this we get the desired result.
- Tue Nov 08, 2011 9:14 pm
- Forum: Algebra
- Topic: Quadratic function
- Replies: 0
- Views: 1886
Quadratic function
could somebody please explain me why a discriminant has to be negative for a quadratic function to be positive?? when a quadratic function is positive, does it refer that it's value is positive or the signs are all positive??
- Tue Nov 08, 2011 11:55 am
- Forum: Physics
- Topic: রিলেটিবিটি
- Replies: 8
- Views: 7330
Re: রিলেটিবিটি
I think so. Cuz the speed of light is absolute irrespective of all spectators and objects of arbitrary speed.
- Mon Nov 07, 2011 9:17 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Problems Involving Triangles
- Replies: 8
- Views: 5824
Re: Problems Involving Triangles
number 3. \[AP,BP,CP \] are concurrent at \[P\]
By definition, Reflection of P across the midpoint of BC lies on AP. So all those reflective lines are concurrent at P.
By definition, Reflection of P across the midpoint of BC lies on AP. So all those reflective lines are concurrent at P.
- Mon Nov 07, 2011 9:07 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Problems Involving Triangles
- Replies: 8
- Views: 5824
Re: Problems Involving Triangles
Number 2. \[13/2,5\]
@tiham, rookie mistake
@tiham, rookie mistake
- Mon Nov 07, 2011 1:03 pm
- Forum: National Math Camp
- Topic: Exercise-1.15(new book) (BOMC-2011)
- Replies: 10
- Views: 8045
Re: Exercise-1.15(new book) (BOMC-2011)
could somebody please explain me why a discriminant has to be negative for a quadratic function to be positive?? when a quadratic function is positive, does it refer that it's value is positive or the signs are all positive??
- Mon Nov 07, 2011 6:42 am
- Forum: National Math Camp
- Topic: Exercise-1.14(new book) (BOMC-2011)
- Replies: 16
- Views: 10052
Re: Exercise-1.14(new book) (BOMC-2011)
\[\left \lfloor \sqrt{\left ( 4n^2+n \right )} \right \rfloor=2n\]
which leads us to \[n< \left ( n+\left ( 1/16 \right ) \right )\]and its obvious.
which leads us to \[n< \left ( n+\left ( 1/16 \right ) \right )\]and its obvious.
- Mon Nov 07, 2011 6:35 am
- Forum: National Math Camp
- Topic: Exercise-1.15(new book) (BOMC-2011)
- Replies: 10
- Views: 8045
Re: Exercise-1.15(new book) (BOMC-2011)
oww. so silly of me. abc=1 condition was missed by me. Now i understand sourov's approach. Bt is my process right??