## Search found 244 matches

Tue Jan 01, 2013 6:28 pm
Forum: Algebra
Topic: Nice Ineq
Replies: 2
Views: 1302

### Nice Ineq

Let $a,b,c$ be positive reals so that $ab+bc+ca=abc$
prove that ,
$\displaystyle \frac{a^4+b^4}{ab(a^3+b^3)} +\frac{b^4+c^4}{bc(b^3+c^3)}+\frac{c^4+a^4}{ca(c^3+a^3)} \geq 1$
Sun Dec 30, 2012 4:52 pm
Forum: Algebra
Topic: f(2x-f(x))=x
Replies: 0
Views: 856

### f(2x-f(x))=x

Find all function $f: \mathbb R \rightarrow \mathbb R$ so that ,
$f(2x-f(x))=x \forall x \in \mathbb R$
Sun Dec 30, 2012 3:29 pm
Forum: Number Theory
Topic: Iran-NMO-2010-1
Replies: 13
Views: 4242

### Re: Iran-NMO-2010-1

আওপ্স , আমি তাইলে ভুল ছিলাম । @ আদিব ।
(মাহি , সরি )
Sun Dec 30, 2012 12:39 pm
Forum: Number Theory
Topic: Iran-NMO-2010-1
Replies: 13
Views: 4242

### Re: Iran-NMO-2010-1

আচ্ছা এইসব জিনিস কি প্রফ কইরা ইউস করা লাগে ??
Sun Dec 30, 2012 12:03 pm
Forum: Geometry
Topic: Prove that $CH \bot AB$
Replies: 4
Views: 1790

### Re: Prove that $CH \bot AB$

$D,H$ are isogonal Conjugates
Sun Dec 30, 2012 10:45 am
Forum: Number Theory
Topic: Iran-NMO-2010-1
Replies: 13
Views: 4242

### Re: Iran-NMO-2010-1

পীথাগোরিয়ান ট্রিপল এর ক্ষেত্রে , Primitive ট্রিপল থাকে একটা । বাট ,

$(a-b)^2+(ab+1)^2=(a+b)^2+(ab-1)^2$
Tue Dec 25, 2012 11:28 am
Forum: Geometry
Topic: CMO 2012
Replies: 8
Views: 3096

### Re: CMO 2012

*Mahi* wrote: if we can prove that $c_1, c_2$ have equal radius, then we are done.
yesss. I also used this idea .
Sun Dec 23, 2012 4:56 pm
Forum: Geometry
Topic: CMO 2012
Replies: 8
Views: 3096

### Re: CMO 2012

I also solved this in Tahmid's way .. But at first I accidentally proved something else :p In the $\triangle ABC$ , $\angle A$ is biggest. On the circumcircle of $\triangle ABC$ , let $D$ be the midpoint of arc$ABC$ and $E$ be the midpoint of arc$ACB$ . The circle $c_1$ passes through $A,D$ and is t...
Tue Dec 18, 2012 11:04 am
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 57015

### Re: Secondary and Higher Secondary Marathon

Complete Rectengle $ABCE$ and $ACBG$ . $GB \cap CE=F$ . Note that $AX=AY,CY=CF,\angle XAY=\angle YCF$ . So $X,Y,F$ are collinear & $XF$ the bisector of $\angle EFG$ . Let the circumcircle of$\triangle CFB$ intersect $XF$ at $P'$ . Now $P'B$ being equal to $P'C$ imply $P'=P$ .And \$\angle BAC=\angle B...
Tue Dec 18, 2012 10:51 am
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 57015

### Re: Secondary and Higher Secondary Marathon

Sol 17

Check it out
The killer construction.PNG (45.51 KiB) Viewed 3210 times