## Search found 40 matches

Sat Aug 05, 2017 11:54 pm
Forum: Number Theory
Topic: Equality and square
Replies: 2
Views: 352

### 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
Topic: IMO 2017 Problem 2
Replies: 3
Views: 1057

### 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: 4294

### 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
Topic: IMO 2001 Problem1
Replies: 2
Views: 237

### 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: 251

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 $b+1>a-1 \Rightarrow a-b<2 \Rightarrow a=b+1$. $b^2-a|a^2+b \Rightarrow b^... Sat Jun 03, 2017 8:54 pm Forum: Geometry Topic: CGMO 2007/5 Replies: 1 Views: 230 ### Re: CGMO 2007/5$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: 276

$\frac {x^2}{2xy^2-y^3+1}=k \in \mathbb{N}$ 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 $(2ky^2-y-1)^2 < X < (2ky^2-y+... Sat May 20, 2017 12:45 am Forum: Geometry Topic: ISL 2006 G3 Replies: 2 Views: 279 ### Re: ISL 2006 G3 Solution with spiral similarity is easy. Here I'm gonna give a different solu. And I chased some angles and lengths.$\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: 294