## Search found 665 matches

Fri Feb 21, 2014 11:02 am
Forum: Algebra
Topic: FE from real to real
Replies: 2
Views: 1874

I don't have access to a pc currently, so just a sketch. Let $P(x,y)$ be the given equation. $P(x,0)$ implies $f(0)=0,f(1)=1$. $P(xy,y)-P(x,1)$ gives $f(xy)=f(x)f(y)$. So write $P(x,y)$ as $\frac {f(x+y)}{f(x-y)}=\frac {x+y}{x-y}$. Now let this be $Q(x,y)$. Use $P(2,1)P(3,1),P(4,1)P(5,1),f(6)=f(2)f(... Thu Feb 20, 2014 7:35 pm Forum: Number Theory Topic: prime divisors Replies: 5 Views: 3090 ### Re: prime divisors Yeah, and it directly kills SL-2006-N5. And this very problem(and a little গুতানি from Masum vai ) motivated me to learn about cyclotomic polynomials. Tue Feb 18, 2014 9:30 pm Forum: Number Theory Topic: prime divisors Replies: 5 Views: 3090 ### Re: prime divisors By Zsigmondy's theorem$p^p-1$has a prime divisor not dividing$p-1$[Check the exceptions, none works]. Let it be$q$. Let$d=ord_qp$. So$p^d \equiv 1 \pmod q$. So$d|p \Rightarrow d=1,p$. If$d=1$, then$q|p-1$, a contradiction! So$d=p$. Now$d|\phi (q) \Rightarrow p|q-1 \Rightarrow q=kp+1$, don... Tue Feb 11, 2014 2:27 pm Forum: National Math Olympiad (BdMO) Topic: warm-up problems for national BdMO'14 Replies: 25 Views: 8402 ### Re: warm-up problems for national BdMO'14$[6]$( APMO'08 )$\Gamma$is the circum-circle of$\triangle ABC$. A circle passing through$A$and$C$meets$BC$and$BA$at$D$and$E$respectively.$AD$and$CE$intersect$\Gamma$again at$G$and$H$. The tangents of$\Gamma$at$A$and$C$meet$DE$at$L$and$M$. Prove that, the intersect... Thu Feb 06, 2014 7:39 pm Forum: Combinatorics Topic: rotating a colored square Replies: 5 Views: 3221 ### Re: rotating a colored square So let me get this straight$\mathbb B \rightarrow \mathbb B=\mathbb B, \mathbb B \rightarrow \mathbb W=\mathbb B, \mathbb W \rightarrow \mathbb B=\mathbb B, \mathbb W \rightarrow \mathbb W=\mathbb W$. WLOG we assume the square is rotated anti-clockwise. For simplicity, we assume the centre of a squ... Tue Feb 04, 2014 5:04 pm Forum: Number Theory Topic: positive cube =$p^2-p-1$Replies: 7 Views: 3190 ### Re: positive cube =$p^2-p-1$Similar to a Balkan problem which was in 2012 TST too! Let$n^3=p^2-p-1 \Rightarrow (n+1)(n^2-n+1)=p(p-1)$. If$p|n+1$, then$n+1 \ge p \Rightarrow n(n+1) \ge p(p-1)=(n+1)(n^2-n+1) \Rightarrow n \ge n^2-n+1 \Rightarrow n=1,2.$Plugging them in the original equation we find one solution$(n,p)=(1,2)$... Tue Jan 28, 2014 6:12 pm Forum: Geometry Topic: Geometry problem concerning duality Replies: 2 Views: 2059 ### Re: Geometry problem concerning duality Mine involves duality as well. (Pascal is a result of dual projective axioms so far as I know :? ) Let$AC \cap BD=G, AC \cap PQ=G', AP \cap CQ=X, AQ \cap CP=Y$. Applying Pascal's theorem on$ADCQMP$, we get$AD \cap MQ=F, DC \cap MP=E, AP \cap CQ=X$are collinear. Similarly applying Pascal's theore... Wed Jan 22, 2014 9:16 pm Forum: Junior Level Topic: MCQ highest Replies: 3 Views: 2849 ### Re: MCQ highest Labib wrote: I'm posting a solution posted by a brilliant member named Patrick Corn. Lord pco Tue Jan 07, 2014 5:53 pm Forum: Number Theory Topic: Nice In a Tricky Way! Replies: 2 Views: 1985 ### Re: Nice In a Tricky Way! Am I missing something? Clearly$i \neq j$. So let$i<j$. Now by some algebraic manipulation we get $\frac{\binom nj}{\binom ni}=\frac{\binom{n-i}{n-j}}{\binom ji}$ Since$LHS\$ is irreducible, comparing denominators we have $\binom ni|\binom ji \Rightarrow \binom ni \leq \binom ji$ which is a co...
Fri Nov 15, 2013 7:43 pm
Topic: IMO 2013, Day 1-P3
Replies: 3
Views: 3409

### Re: IMO 2013, Day 1-P3

King Jakaria wrote:Should the converse necessarily been true? That is, given ABC right-angled, should necessarily the center of A1B1C1 lie on the circumcircle of ABC?
Yes.