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(I got the idea to use Harmonic Division in this problem from Nadim Ul Abrar vai after the exam. This problem actually made me learn Harmonic Division) Let $BR\cap \omega=E', CE'\cap AR=S$. Now by the first lemma of Zhao, $AC$ is a symmedian of $\triangle ABD$. Thus $(C,A;D,B)=-1$. Thus $E'(C,A;D,B)... Fri May 10, 2013 12:59 am Forum: Asian Pacific Math Olympiad (APMO) Topic: APMO 2013 Problem 5 Replies: 1 Views: 1184 ### Re: APMO 2013 Problem 3 (This is a solution I saw later in the math camp.) Let$\displaystyle\sum_{i=1}^{k}a_i=A,\displaystyle\sum_{i=1}^{k}b_i=B$. Also let$X_i=X_1+(i-1)d$. Since$X_1,X_2$are integers,$d=X_2-X_1$is also an integer. Then,$\displaystyle\sum_{i=1}^{k}(a_in+b_i+1)>\displaystyle\sum_{i=1}^{k}\lfloor a_in+...
Fri May 10, 2013 12:15 am

Forum: Asian Pacific Math Olympiad (APMO)
Topic: APMO 2013 Problem 3
Replies: 2
Views: 1169

It's easy to check that $n$ can't be a perfect square. Then let $n=m^{2}+r$, where $1\leq r\leq 2m$. Then, the expression becomes, $\displaystyle\frac{(m^{2}+r)^{2}+1}{m^{2}+2}=\displaystyle\frac{m^{2}(m^{2}+2)+2(r-1)(m^{2}+2)+r^{2}-4r+4+1}{m^{2}+2}$ $=m^{2}+2(r-1)+\displaystyle\frac{(r-2)^{2}+1}{m^... Thu May 09, 2013 11:47 pm Forum: Asian Pacific Math Olympiad (APMO) Topic: APMO 2013 Problem 2 Replies: 1 Views: 813 ### China TST 2013 Day 5 Problem 2 The circumcircle of triangle ABC has centre$O$.$P$is the midpoint of$\widehat{BAC}$and$QP$is the diameter. Let$I$be the incentre of$\triangle ABC$and let$D$be the intersection of$PI$and$BC$. The circumcircle of$\triangle AID$and the extension of$PA$meet at$F$. The point$E$lies... Wed May 01, 2013 10:33 pm Forum: Geometry Topic: China TST 2013 Day 5 Problem 2 Replies: 0 Views: 647 ### Re: Cool Number Theory There are primes as powers,primes as divisors-I can't think of sth else than Fermat' Little Theorem! :twisted: Let us assume,by symmetry,that$p\leq q$. Now$p|(5^p-2^p)$or$p|(5^q-2^q)$. By Fermat's theorem,if a prime$p$divides$5^p-2^p$,then$5^p-2^p\equiv 5-2\equiv 3\equiv 0\Rightarrow p=3$. A... Sun Apr 14, 2013 12:16 pm Forum: Number Theory Topic: Cool Number Theory Replies: 1 Views: 526 ### Re: 2002 Czech-Polish-Slovak My solution's quite the same. Sat Apr 13, 2013 11:59 pm Forum: Number Theory Topic: 2002 Czech-Polish-Slovak Replies: 4 Views: 767 ### BdMC-Not So Easy II problem 5 Find all pairs of$(a,b)$such that for all$n\in \mathbb{N}$,$a\left \lfloor bn \right \rfloor=b\left \lfloor an \right \rfloor$Thu Apr 11, 2013 11:42 pm Forum: Number Theory Topic: BdMC-Not So Easy II problem 5 Replies: 1 Views: 514 ### m and n Let$m,n$be positive integers. Prove that$(2^m-1)^2|(2^n-1)$if and only if$m(2^m-1)|n$. Source:Russia 1997 Wed Apr 10, 2013 11:49 pm Forum: Number Theory Topic: m and n Replies: 4 Views: 1230 ### Yammy...GCD Find all triples$(m,n,l)$of positive integers such that$m+n=gcd(m,n)^2,m+l= gcd(m,l)^2,n+l=gcd(n,l)^2$. Source:$1997$Russian Mathematical Olympiad Mon Apr 08, 2013 11:52 pm Forum: Number Theory Topic: Yammy...GCD Replies: 0 Views: 354 ### Re: f(f(n))=3n (I saw a similar problem in IMO math's functional equation tutorial that asked for the value of$f(2006)$. I have used the same procedure here.) We claim first that, :arrow:$f(2\cdot 3^{n})=3^{n+1}$:arrow:$f(3^{n})=2\cdot 3^{n}$for$n=0,1,2,....$We use induction. For$n=0,f(...
Sat Apr 06, 2013 11:53 pm

Forum: Algebra
Topic: f(f(n))=3n
Replies: 1
Views: 445
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