Same herePhlembac Adib Hasan wrote:My solution is same as Najif, so there is no need to post it.
Search found 244 matches
- Wed Oct 17, 2012 9:48 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO MOCK5 (iii)
- Replies: 7
- Views: 5453
Re: IMO MOCK5 (iii)
- Sat Oct 13, 2012 7:14 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO MOCK5 (iii)
- Replies: 7
- Views: 5453
Re: IMO MOCK5 (iii)
Solution
- Sat Oct 13, 2012 7:03 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO MOCK5 (iii)
- Replies: 7
- Views: 5453
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...
- Sat Oct 13, 2012 6:54 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO MOCK5 (V)
- Replies: 1
- Views: 2749
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...
- Sat Oct 13, 2012 6:49 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO MOCK5 (V)
- Replies: 1
- Views: 2749
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
$BC^2=4PF.AK$
meets ray $BC$ at $P$. Points $K$ and $F$ lie on line $PA$ such that $MF⊥BC$ and $MK⊥PA$. Prove that
$BC^2=4PF.AK$
- Fri Oct 12, 2012 2:21 pm
- Forum: Secondary Level
- Topic: $3^{x}+4^{y}=5^{z}$
- Replies: 1
- Views: 2147
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,...
- Mon Sep 24, 2012 5:30 pm
- Forum: Algebra
- Topic: Functional Equation
- Replies: 1
- Views: 2577
Functional Equation
Find all $f:R \to R$ so that
$f(x^3+y^3)=xf(x^2)+yf(y^2) $ for all $x,y \in R$
$f(x^3+y^3)=xf(x^2)+yf(y^2) $ for all $x,y \in R$
- Mon Sep 24, 2012 4:28 pm
- Forum: Algebra
- Topic: How's that FE?
- Replies: 9
- Views: 5289
Re: How's that FE?
Umm what about this bro ?
Thanku Tahmid Vai ... <3
- Mon Sep 24, 2012 4:05 pm
- Forum: Geometry
- Topic: Hexagon-quest
- Replies: 1
- Views: 1852
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 ...
- Sat Sep 22, 2012 4:24 pm
- Forum: Number Theory
- Topic: Prime
- Replies: 2
- Views: 2319
Re: Prime
what if $13|x$ .....?