## 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:
**2961**

### 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:
**3659**

### 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:
**1410**

### 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:
**1162**

### 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:
**4138**

### 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:
**3723**

### 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:
**3723**

### 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:
**5226**

### 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:
**6804**

### 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:
**5226**

### 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??