## Search found 110 matches

Mon Mar 02, 2015 12:44 am
Forum: Number Theory
Topic: Some GCD Problems
Replies: 6
Views: 2603

### Re: Some GCD Problems

tanmoy wrote:.$a^{2^{n}}+1$ is a divisor of $a^{2^{m}}-1$.
how ?  $a=3$ ; $n=2$ ; $m=3$ breaks it .
Sun Mar 01, 2015 3:41 pm
Forum: Number Theory
Topic: Some GCD Problems
Replies: 6
Views: 2603

### Re: Some GCD Problems

just apply $euclidean$ $algorithm$. nothing else.
Sat Feb 28, 2015 2:39 am
Forum: Geometry
Topic: Bulgaria 1996
Replies: 4
Views: 2233

let , $k_{1},k_{2}$ touches $k$ at points $X,Y$ respectively. and $l\cap AB=Z$ so, $AX,BY,CZ$ are concurrent at the orthocentre of triangle $ABC$ by ceva's theorem we have $\frac{AZ}{ZB}=\frac{r_{1}}{r_{2}}$ now let, $AO_{1}\cap BC=R$ ; $BO_{2}\cap AC=S$ then , $\frac{CS}{SA}\cdot \frac{AZ}{ZB}\cdot... Sat Feb 28, 2015 1:21 am Forum: Geometry Topic: Determine the angles Replies: 1 Views: 1193 ### Re: Determine the angles$\angle A=90\angle B=22.5\angle C=67.5$Thu Feb 26, 2015 7:44 pm Forum: Geometry Topic: Sine and Cosine Replies: 2 Views: 1589 ### Re: Sine and Cosine very cool need to prove$sin \angle C=2sin \angle APC cos \angle APC$find the value of$cos \angle APC$by cosine rule in triangle$APC$then use stewart's theorem to find$AP^{2}$. plugging them into$2sin \angle APC cos \angle APC$and just manipulate . Wed Feb 25, 2015 11:24 pm Forum: Geometry Topic: USAMO 2009/5 Replies: 4 Views: 3489 ### Re: USAMO 2009/5 i have solved this . but my solution is too large main part of my solution is to prove$DX$=$CY$where$X=QR\cap w ; Y=PS\cap w$Wed Feb 25, 2015 10:55 am Forum: Geometry Topic: USAMO 1999/6 Replies: 2 Views: 1592 ### Re: USAMO 1999/6 tanmoy , did you solve this ? ... if not , then try to show that$E$is the touch point of the excircle of triangle$ADC$with side$DC$Tue Feb 24, 2015 8:07 pm Forum: Geometry Topic: Balkan MO 2005 Replies: 3 Views: 1857 ### Re: Balkan MO 2005 well, i have proved the first part in differrent way . here it is , let$BX\cap AC={X}'$and the incircle touches$BC$at$F$now , apply menelus's$ \frac{AE}{E{X}'}\frac{{X}'X}{XB}\frac{BD}{DA}=1 \Leftrightarrow \frac{{X}'X}{XB}=\frac{E{X}'}{BD}=\frac{E{X}'}{BF}$as ,$\frac{{X}'X}{XB}=\frac{CX}{C...
Tue Feb 24, 2015 12:08 pm
Forum: Geometry
Topic: Balkan MO 2005
Replies: 3
Views: 1857