Search found 665 matches
- Thu Nov 14, 2013 6:02 pm
- Forum: Algebra
- Topic: F.E. (Z to Z)
- Replies: 5
- Views: 5102
Re: F.E. (Z to Z)
আমিও এইরকম এটা সলুশন সেট পাইছিলাম কিন্তু এই সলুশনের বাগটা খুঁজে পাইতেছি না। কাল টেস্ট শুরু, তাই একটু সময় সংকটে আছি, এজন্যে খালি স্কেচটা দিচছি। একটু আলাদা অ্যাপ্রোচঃ $Q(a) \Rightarrow f(a+1)^3=f(a)^3+3f(a+1)f(a)f(1)+f(1)^3$. $Q(1) \Rightarrow f(2)=2f(1),-f(1)$. C1: $f(2)=2f(1)$. $Q(2) \Rightarrow f(...
- Thu Nov 14, 2013 2:19 pm
- Forum: Algebra
- Topic: F.E. (Z to Z)
- Replies: 5
- Views: 5102
Re: F.E. (Z to Z)
First we look for constant solutions: $c^3-c^3-c^3=3c.c.c \Rightarrow c=0$, which is indeed a solution. Let $P(a,b) \Rightarrow f(a+b)^3-f(a)^3-f(b)^3=3f(a)f(b)f(a+b)$. $P(0,0) \Rightarrow f(0)=0$ $P(a,-a) \Rightarrow f(-a)=-f(a)$. $P(a+1,-a) \Rightarrow f(a+1)^3=f(a)^3+3f(a+1)f(a)f(1)+f(1)^3$. Let'...
- Sun Nov 10, 2013 9:06 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 1989/6
- Replies: 6
- Views: 5089
Re: IMO 1989/6
But $(1,2,4,3)$ has the property $T$. That's not how I defined the injection!
- Sun Nov 10, 2013 5:28 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 1989/6
- Replies: 6
- Views: 5089
Re: IMO 1989/6
Sorry, I meant to write injection . Now I see the bug, fixing $i$, which obviously fixes $j$ removes the bug.
- Sun Nov 10, 2013 1:25 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 1989/6
- Replies: 6
- Views: 5089
Re: IMO 1989/6
Seems rather simple, or I made some trivial mistakes! Could you please check? Suppose $(x_1,...,x_1,...,x_j,...,x_{2n})$ is a permutation without the property $T$ and $|x_i-x_j|=n$ for some $1 \le i \neq j \le 2n$. Consider the following bijection: $(x_1,...x_i,...,x_j,...,x_{2n}) \Rightarrow (x_1,....
- Thu Nov 07, 2013 12:31 pm
- Forum: Geometry
- Topic: Polish MO 2011/2
- Replies: 1
- Views: 2172
Re: Polish MO 2011/2
Let $XYZ$ be that triangle. Let $G,H,I,J,K,L$ be the midpoints of $BD,CD,CE,AE,AF,BF$ respectively. $O,O'$ be the centres of $\odot ABC,\odot XYZ$ respectively. Denote $(P)$ by the degenerate circle centered at $P$ and $l_a,l_b,l_c$ by the perpendicular bisectors of $BC,CA,AB$ respectively. It is we...
- Mon May 20, 2013 12:30 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2013 Problem 4
- Replies: 2
- Views: 5121
Re: APMO 2013 Problem 4
Consider these bijections of $A$ and $B$. $A_a=\{i-a:i \in A\},B_b=\{i+b:i \in B\}$. Note that $|A|=|A_a|,|B|=|B_b|$ Now choose any $i \in A \cup B$. Then $i+a \in A$ or $i-b \in B$. If $i+a \in A$, then $i \in A_a$; else if $i-b \in A$, then $i \in B_b$. So every $i \in A \cup B$ maps to either $A_...
- Sat Apr 20, 2013 8:39 pm
- Forum: Number Theory
- Topic: Shitty Number theory
- Replies: 1
- Views: 2202
Re: Shitty Number theory
From $(2)$ we get $37|b(a+d-1)$. So $37|b$ or $37|a+d-1$. If $37|a+d-1$ we get $ad-bc \equiv ad-(a-a^2) \equiv a(a+d-1) \equiv 0 \pmod{37}$, a contradiction so $37|b$. Similarly from $(3)$ we get $37|c$. So $(1),(4),(5)$ can be reduced to $37|a(a-1),37|d(d-1),ad \equiv 1\pmod{37}$. From the third ex...
- Thu Apr 18, 2013 11:38 pm
- Forum: Geometry
- Topic: Iran TST 2012
- Replies: 4
- Views: 3793
Re: Iran TST 2012
WLOG $\angle MCO=60^{\circ}$. By symmetry we conclude $MA,NB,CO$ are concurrent at a point $D$. Let $P'$ be a point such that $P',M$ are on opposite sides of $CO$ and $\triangle P'CO$ equilateral. It immediately follows that $P' \in \odot OADB$. $\angle OAP'=\angle ODP'=30^{\circ}$. So $\angle MAP'=...
- Wed Apr 17, 2013 11:25 am
- Forum: Number Theory
- Topic: Family Of Functions
- Replies: 5
- Views: 4678
Re: Family Of Functions
Actually it's 2011 N3, the problem Zadid gave that is.*Mahi* wrote:Too much similarity with 2004 N3 in one problem.