## Search found 40 matches

- Sat Aug 05, 2017 11:54 pm
- Forum: Number Theory
- Topic: Equality and square
- Replies:
**2** - Views:
**325**

### 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...

- Thu Jul 27, 2017 12:07 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2017 Problem 2
- Replies:
**3** - Views:
**994**

### 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...

- Sun Jun 11, 2017 12:32 am
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**4048**

### 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...

- Mon Jun 05, 2017 12:02 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2001 Problem1
- Replies:
**2** - Views:
**219**

### 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...

- Sat Jun 03, 2017 11:37 pm
- Forum: Number Theory
- Topic: Particular divisibility..
- Replies:
**1** - Views:
**220**

### 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^...

- Sat Jun 03, 2017 8:54 pm
- Forum: Geometry
- Topic: CGMO 2007/5
- Replies:
**1** - Views:
**194**

### 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...

- Fri Jun 02, 2017 6:48 am
- Forum: Number Theory
- Topic: A strange divisibility
- Replies:
**1** - Views:
**258**

### 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+...

- Sat May 20, 2017 12:45 am
- Forum: Geometry
- Topic: ISL 2006 G3
- Replies:
**2** - Views:
**230**

### 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 $ABCP$,$...

- Sat May 20, 2017 12:07 am
- Forum: Algebra
- Topic: Inequality with abc = 1
- Replies:
**3** - Views:
**263**

### 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})(...

- Thu May 11, 2017 10:37 pm
- Forum: Number Theory
- Topic: Divisibility with a and b
- Replies:
**2** - Views:
**224**

### 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...