Search found 64 matches

by Atonu Roy Chowdhury
Thu Mar 15, 2018 1:27 pm
Forum: Algebra
Topic: Functional equation from Japan MO 2016
Replies: 1
Views: 3895

Functional equation from Japan MO 2016

Find all functions $f: \mathbb{R}\rightarrow \mathbb{R}$
$$f(yf(x)-x)=f(x)f(y)+2x$$
for all $x,y \in \mathbb{R}$
by Atonu Roy Chowdhury
Mon Mar 12, 2018 11:59 am
Forum: Social Lounge
Topic: Chat thread
Replies: 53
Views: 45827

Re: Chat thread

Hello everyone, I'm Atonu from Chittagong.
I request everyone to post problems and solutions in this almost-dead forum.
#Make_BdMO_Forum_Great_Again
by Atonu Roy Chowdhury
Mon Mar 12, 2018 11:56 am
Forum: Higher Secondary Level
Topic: USAMO #1
Replies: 1
Views: 3749

Re: USAMO #1

This problem was posted once. If you need solution, you can check this out.
http://matholympiad.org.bd/forum/viewto ... =26&t=3976
by Atonu Roy Chowdhury
Sun Mar 11, 2018 12:31 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO National Secondary/Higher Secondary 2018/4
Replies: 2
Views: 1509

Re: BdMO National Higher Secondary 2018/4

Lemma: If we draw $x$ lines on plane, we will get maximum $\frac{x(x-1)}{2}$ intersection points. Proof: $2$ lines needed for a intersection point. So, maximum number of intersection point is $xC2=\frac{x(x-1)}{2}$ Here $200$ intersection points. So, $\frac{m(m-1)}{2} \geq 200 \Rightarrow m > 20$. ...
by Atonu Roy Chowdhury
Sat Aug 05, 2017 11:54 pm
Forum: Number Theory
Topic: Equality and square
Replies: 2
Views: 4267

Re: Equality and square

The case of negative integers is quite trivial. Now we'll work with case $a,b > 0$ Lemma 1: $x < 2^x$ Proof: We'll prove it by induction. Base case is solved. Now assume $x<2^x \Rightarrow x+1<2^x+1 \le 2^{x+1}$. Done! Lemma 2: $b^2 < 2^{2b+1}$ Proof: We'll prove it by induction. Base case is solved...
by Atonu Roy Chowdhury
Thu Jul 27, 2017 12:07 am
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2017 Problem 2
Replies: 3
Views: 7402

Re: IMO 2017 Problem 2

NO INJECTIVITY!!! Notice that, if $f(x)$ is a solution, then $-f(x)$ is also a solution. Let $P(x,y)$ denotes the assertion. $P(0,0)$ implies $f(f(0)^2)=0$ . Now if $f(0)=0$ , $P(x,0)$ implies $f(x)=0$ for all $x \in \mathbb{R} $. So, assume $f(0) \neq 0$. Claim 1: If $f(x)=0$, then $ x=1$ Proof: As...
by Atonu Roy Chowdhury
Sun Jun 11, 2017 12:32 am
Forum: Junior Level
Topic: Beginner's Marathon
Replies: 68
Views: 22659

Re: Beginner's Marathon

An easy problem:$\text{Problem 20}$ The fractions $\frac{7x+1}{2}$, $\frac{7x+2}{3}$,......., $\frac{7x+2016}{2017}$ are irreducible.Find all possible values of $x$ such that $x$ is less than or equal to $300$.This Problem was recommended by Zawad bhai in a mock test at CMC. Here's a hint for you g...
by Atonu Roy Chowdhury
Mon Jun 05, 2017 12:02 am
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2001 Problem1
Replies: 2
Views: 5437

Re: IMO 2001 Problem1

Our problem is equivalent to $OP>CP$. Lemma 1: $BP \ge R+PC$ where $R$ is the circumradius of $\triangle ABC$. Proof of lemma: $BP-PC=AB \cos \angle B - AC \cos \angle C = 2R(\sin \angle C \cos \angle B - \sin \angle B \cos \angle C ) \ge R$ because $\angle C- \angle B \ge 30$. Now we back to our p...
by Atonu Roy Chowdhury
Sat Jun 03, 2017 11:37 pm
Forum: Number Theory
Topic: Particular divisibility..
Replies: 1
Views: 1303

Re: Particular divisibility..

Case 1: $a=b>1$ $a^2-a|a^2+a$ or $a^2-a|2a$. So, $a \le 3$. Checking the values, we get $a=2$ and $a=3$. Case 2: $a \neq b$ WLOG $a>b$ $a^2-b|b^2+a \Rightarrow b^2+a \ge a^2-b \Rightarrow b(b+1) \ge a(a-1)$. As $a>b$, we get $b+1>a-1 \Rightarrow a-b<2 \Rightarrow a=b+1$. $b^2-a|a^2+b \Rightarrow b^...
by Atonu Roy Chowdhury
Sat Jun 03, 2017 8:54 pm
Forum: Geometry
Topic: CGMO 2007/5
Replies: 1
Views: 7669

Re: CGMO 2007/5

$M$ is the midpoint of $AC$ and $X$ is the midpoint of $AF$. Proving $\angle DEF = 90$ is equivalent to $MDEF$ cyclic or $\angle MEF = \angle MDF = 30$ It is easy to show $\triangle DFX$ is equilateral. So, $ABDX$ is cyclic. $\angle EFC = \angle BXC = 60+\angle BAD$, so $\angle FEC = 90 - \angle BA...