Search found 244 matches

by Nadim Ul Abrar
Wed Oct 17, 2012 9:48 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO MOCK5 (iii)
Replies: 7
Views: 3394

Re: IMO MOCK5 (iii)

Phlembac Adib Hasan wrote:My solution is same as Najif, so there is no need to post it.
Same here :)
by Nadim Ul Abrar
Sat Oct 13, 2012 7:14 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO MOCK5 (iii)
Replies: 7
Views: 3394

Re: IMO MOCK5 (iii)

Solution
MOCK 5 (iii).JPG
MOCK 5 (iii).JPG (37.21 KiB) Viewed 3392 times
by Nadim Ul Abrar
Sat Oct 13, 2012 7:03 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO MOCK5 (iii)
Replies: 7
Views: 3394

IMO MOCK5 (iii)

Let $O$ and $H$ be the circumcenter and orthocenter of acute $△ABC$. The bisector of $\angle BAC$ meets the circumcircle $Γ$ of $△ABC$ at $D$. Let $E$ be the mirror image of $D$ with respect to line $BC$. Let $F$ be on $Γ$ such that $DF$ is a diameter. Let lines $AE$ and $FH$ meet at $G$. Let $M$ be...
by Nadim Ul Abrar
Sat Oct 13, 2012 6:54 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO MOCK5 (V)
Replies: 1
Views: 1711

Re: IMO MOCK5 (V)

My solution : MOCK 5 (v).JPG Let the circumcircle of the triangle $ABC$ intersect the line $MF$ at the point $F'$ (where $F',F$ lie on the same side of $BC$) and $BC \cap AF'=P'$ Now as $M$ is the midpoint of $BC$ and $MF'? BC$ , $\angle BF'M=\angle CF'M=\frac{1}{2} \angle BF'C=\frac{A}{2}$ So $\ang...
by Nadim Ul Abrar
Sat Oct 13, 2012 6:49 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO MOCK5 (V)
Replies: 1
Views: 1711

IMO MOCK5 (V)

In acute $△ABC$, $AB>AC$. Let $M$ be the midpoint of $BC$. The exterior angle bisector of $\angle BAC$
meets ray $BC$ at $P$. Points $K$ and $F$ lie on line $PA$ such that $MF⊥BC$ and $MK⊥PA$. Prove that
$BC^2=4PF.AK$
by Nadim Ul Abrar
Fri Oct 12, 2012 2:21 pm
Forum: Secondary Level
Topic: $3^{x}+4^{y}=5^{z}$
Replies: 1
Views: 1354

Re: $3^{x}+4^{y}=5^{z}$

Case 1 : $z$ is non negative Then $x,y$ will also be non negative Sub Case 1 : $x=0$ Then $4^y+1=5^z$ . As we know $v_5(4^y+1)=1+v_5(y)$ ... $(i)$ let $y=5^k.a$ Using $(i)$ $4^{5^k.a}+1=5^{k+1}$ that imply the only possible value of $k$ is $0$ , and $a=1$ So Solution for this case is $(x,y,z)=(1,0,...
by Nadim Ul Abrar
Mon Sep 24, 2012 5:30 pm
Forum: Algebra
Topic: Functional Equation
Replies: 1
Views: 1684

Functional Equation

Find all $f:R \to R$ so that
$f(x^3+y^3)=xf(x^2)+yf(y^2) $ for all $x,y \in R$
by Nadim Ul Abrar
Mon Sep 24, 2012 4:28 pm
Forum: Algebra
Topic: How's that FE?
Replies: 9
Views: 3141

Re: How's that FE?

Umm what about this bro ?
Let $g(x)=f(x)-\frac{x(x-1)}{2}$
Now $g:Q \to Q$
and $g(x+y)=g(x)+g(y)$

Using Cauchy , $g(x)=ax=f(x)-\frac{x(x-1)}{2}$
So $f(x)=\frac{x(x-1)}{2}+ax$
Thanku Tahmid Vai ... <3
by Nadim Ul Abrar
Mon Sep 24, 2012 4:05 pm
Forum: Geometry
Topic: Hexagon-quest
Replies: 1
Views: 1185

Re: Hexagon-quest

Solution . (If $p'$ is reflection of point $p$ over segment $xy$ the we will write that statement thus $p'=R(p,xy)$) HEX 1.JPG $G \in AB$ , $G'=R(G,BC)$ Now $(IH+HG)_{min}=IG'$ . $G''=R(G',CD)$ So $(IH+HG+IJ)_{min}=(IJ+IG')_{min}=JG"$ .. Thus $GM'=h_{min}$ . So Its enough to prove that $GM'=3 \sqrt ...
by Nadim Ul Abrar
Sat Sep 22, 2012 4:24 pm
Forum: Number Theory
Topic: Prime
Replies: 2
Views: 1493

Re: Prime

what if $13|x$ .....?