Search found 64 matches

by Atonu Roy Chowdhury
Fri Jun 02, 2017 6:48 am
Forum: Number Theory
Topic: A strange divisibility
Replies: 1
Views: 1297

Re: A strange divisibility

$\frac {x^2}{2xy^2-y^3+1}=k \in \mathbb{N}$ If $y=1$ , we can easily find $x=2k$ . Now, let $y>1$. Then, $\frac {x^2}{2xy^2-y^3+1}=k$ gives us the quadratic equation $x^2-2ky^2x+k(y^3-1)=0$ . Here the discriminant $X=4k^2y^4-4ky^3+4k$ is a perfect square . Notice that $(2ky^2-y-1)^2 < X < (2ky^2-y+...
by Atonu Roy Chowdhury
Sat May 20, 2017 12:45 am
Forum: Geometry
Topic: ISL 2006 G3
Replies: 2
Views: 7697

Re: ISL 2006 G3

Solution with spiral similarity is easy. Here I'm gonna give a different solu. And I chased some angles and lengths. $\triangle AED$ and $\triangle ABC$ are similar and so $AC.AD=AB.AE$ which along with $\angle BAD = \angle CAE$ implies $\triangle ABD$ and $\triangle ACE$ are similar and so $ABCP$,$...
by Atonu Roy Chowdhury
Sat May 20, 2017 12:07 am
Forum: Algebra
Topic: Inequality with abc = 1
Replies: 3
Views: 4873

Re: Inequality with abc = 1

My solution. Same as Asif Bhai's. $(x-1+\frac{1}{y})(y-1+\frac{1}{z})=xy-x+\frac{x}{z}-y+1-\frac{1}{z}+1-\frac{1}{y}+\frac{1}{yz}=\frac{x}{z}+2-(y+\frac{1}{y})\le \frac{x}{z}$ because $y+\frac{1}{y} \ge 2$ Similarly we get, $(y-1+\frac{1}{z})(z-1+\frac{1}{x}) \le \frac{y}{x}$ and $(z-1+\frac{1}{x})(...
by Atonu Roy Chowdhury
Thu May 11, 2017 10:37 pm
Forum: Number Theory
Topic: Divisibility with a and b
Replies: 2
Views: 1544

Re: Divisibility with a and b

$\frac{b^3-1}{ab-1}=x$ Firstly notice that if $(a,b)$ is a solution, then $(b,a)$ is also a solution. So, we may assume wlog $a \ge b$ . If $b=1$, $x=0$ . So, we got a solution $(a,1)$ where $a>1$ . Now consider the case $b \ge 2$ . $x>0$. Notice that $x \equiv 1 (mod b)$ . So, $x=nb+1$ . As $a \ge...
by Atonu Roy Chowdhury
Thu May 04, 2017 8:05 pm
Forum: Algebra
Topic: FE ^_^
Replies: 1
Views: 3917

Re: FE ^_^

Multiplying $f(x)$ we get $f(x)^3 = f(x)f(y)f(z)[1+f(x^y)-f(y^z)]$ and two similar expressions for $f(y)^3$ and $f(z)^3$ . Adding them, we get $ f(x)^3+ f(y)^3+ f(z)^3 = 3f(x)f(y)f(z)$ . Equality case of AM-GM! So, $f(x) = c$ where $c \in \mathbb{R}^+$
by Atonu Roy Chowdhury
Thu May 04, 2017 7:59 pm
Forum: Algebra
Topic: FE ^_^
Replies: 1
Views: 3917

FE ^_^

Find all $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all distinct $x,y,z$
$f(x)^2-f(y)f(z)=f(x^y)f(y)f(z)[f(y^z)-f(z^x)]$
by Atonu Roy Chowdhury
Mon May 01, 2017 9:45 pm
Forum: Number Theory
Topic: IMO Shortlist 2011 N5
Replies: 4
Views: 2526

Re: IMO Shortlist 2011 N5

$f(m-n)|f(m)-f(n)$ $n=0$ implies $f(m)|f(0)$ $m=0$ and $n=m$ implies $f(-m)|f(m)$ . Similarly we get $f(m)|f(-m)$ . So, $f(m)=f(-m)$ $n=-n$ implies $f(m+n)|f(m)-f(n)$ from which we get $ f(m+n) \le f(n)-f(m)$ Again, $m=m+n$ and $n=m$ implies $f(n)|f(m+n)-f(m)$ . Assume $f(m) \neq f(m+n)$ . Here we'...
by Atonu Roy Chowdhury
Thu Apr 27, 2017 12:12 am
Forum: Junior Level
Topic: 14x14 grid
Replies: 1
Views: 1480

Re: 14x14 grid

Put $1$ and $0$ alternately in that grid. Then in any $T$ shape, $2$ configuration is possible: #1 three $1$'s and one $0$'s #2 three $0$'s and one $1$'s In both configuration, the sum of the numbers in that $T$ is odd. The grid is $14 x 14$. So, we need an odd number of $T$ - tiles. The initial su...
by Atonu Roy Chowdhury
Wed Apr 26, 2017 12:13 am
Forum: Number Theory
Topic: JBMO:NT
Replies: 1
Views: 1286

Re: JBMO:NT

It is obvious that $x \ge 1$ . Case 1: $y=0$ Then, $2^x - 1 = 5^z . 7^w$ . Subcase 1.1: $z > 0$ So, $2^x \equiv 1 (mod 5) $ . So, $ 4|x$. $2^{4k} - 1 = 16^k - 1$ which is divisible by $3$ . But $5^z . 7^w$ is not divisible by $3$. . Subcase 1.2: $z=0$ So, $2^x - 7^w = 1$ $x=1 \rightarrow w = 0$ $x=2...
by Atonu Roy Chowdhury
Mon Apr 24, 2017 11:26 am
Forum: Geometry
Topic: USA(J)MO 2017 #3
Replies: 6
Views: 10232

Re: USA(J)MO 2017 #3

WLOG $P$ lies on the shorter arc $BC$ . So, $[DEF]=[AEF]-[ABC]-[BDF]-[CDE]$ $\angle BAD = \alpha $ Use Sine Law to find $BD$,$DC$,$BF$,$CE$ in terms of $a$ and sine of $\alpha$ and $60-\alpha$, where $a$ is the length of the sides of $\triangle ABC$ . Then we'll use these lengths to find $[AEF]$,$[B...