## Search found 64 matches

### Re: FE FE FE

$f(x)f(y+k)=2f(x+(y+k)f(x))=2f(x+yf(x)+\frac{2k}{f(y)}f(x+yf(x))$ [Recall, $f(x)=\frac{2}{f(y)}f(x+yf(x)$] $\Rightarrow f(x)f(y+k) = f(x+yf(x))f(\frac{2k}{f(y)})=\frac{1}{2}f(x)f(y)f(\frac{2k}{f(y)})$ $\Rightarrow 2f(y+k)=f(y)f(\frac{2k}{f(y)})=2f(y+\frac{2k}{f(y)}f(y))=2f(y+2k)$ Inductivly, $f(y+k)...

### FE FE FE

We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:

\[f(x)f(y)=2f(x+yf(x))\]

for all positive real numbers $x$ and $y$.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:

\[f(x)f(y)=2f(x+yf(x))\]

for all positive real numbers $x$ and $y$.

- Mon Apr 09, 2018 9:33 am
- Forum: Geometry
- Topic: China National Olympiad 2018 P4
- Replies:
**2** - Views:
**7974**

### Re: China National Olympiad 2018 P4

China 2018 #4.png Let $X$ and $Y$ be the midpoints of shorter arcs $AD$ and $BC$ in the circles $\circ ADPE$ and $\circ BCPF$ respectively. $Z$ is the intersection point of $XE$ and $YF$. It is obvious that $X, P, Y$ are collinear. Claim 1: $\triangle AXP$ and $\triangle BYP$ are similar. Proof: $\...

- Mon Apr 09, 2018 9:31 am
- Forum: Geometry
- Topic: China National Olympiad 2018 P4
- Replies:
**2** - Views:
**7974**

### China National Olympiad 2018 P4

$ABCD$ is a cyclic quadrilateral whose diagonals intersect at $P$. The circumcircle of $\triangle APD$ meets segment $AB$ at points $A$ and $E$. The circumcircle of $\triangle BPC$ meets segment $AB$ at points $B$ and $F$. Let $I$ and $J$ be the incenters of $\triangle ADE$ and $\triangle BCF$, resp...

- Wed Apr 04, 2018 10:17 pm
- Forum: Number Theory
- Topic: A Conjecture
- Replies:
**1** - Views:
**3769**

### Re: A Conjecture

Doesn't work when $p=127$.

- Tue Apr 03, 2018 12:26 pm
- Forum: Geometry
- Topic: What is the distance?
- Replies:
**3** - Views:
**8045**

### Re: What is the distance?

I'm avoiding the calculations how I got that $PX=2R=65/2$ (By Heron's formula we got the area of the triangle $ABC$, then found out the height from on vertix to oppsite side, and then applied Brahmagupta's theorem.) Using the fact that, $CS/PS=XS/CS$ we get $CS=10$ I think you've made a typo here. ...

- Fri Mar 30, 2018 10:22 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: 2007 number 5 - divisibility
- Replies:
**1** - Views:
**5358**

### Re: 2007 number 5 - divisibility

$4ab-1|(4a^2-1)^2 \Rightarrow 4ab-1|(a-b)^2$. Assume contradiction that $a \neq b$. wlog, $a>b$ $\frac{(a-b)^2}{4ab-1}=k \Rightarrow (a-b)^2=4abk-k$. Let $a$ and $b$ be such a solution such that $a+b$ is minimal. Consider the quadratic equation $$(x-b)^2=4xbk-k$$ $\Rightarrow x^2 - x(2b+4bk)+(b^2+k)...

- Sun Mar 25, 2018 9:29 pm
- Forum: Algebra
- Topic: FE from USAMO 2002
- Replies:
**4** - Views:
**5121**

### Re: FE from USAMO 2002

Let $P(x,y)$ denotes the assertion. $P(0,0) \Rightarrow f(0)=0$ $P(x,0) \Rightarrow f(x^2)=xf(x)$ So, $f(a-b)=f(a)-f(b)$ when $a,b \in \mathbb{R^+}$ It can be proved that this statement is true for negatives too. That is, $f(a-b)=f(a)-f(b)$ for all $a,b \in \mathbb{R}$ Now, $P(x+1,x) \Rightarrow f(2...

- Sun Mar 25, 2018 9:13 pm
- Forum: Algebra
- Topic: FE from USAMO 2002
- Replies:
**4** - Views:
**5121**

### FE from USAMO 2002

Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(x^2 - y^2) = x f(x) - y f(y) \] for all pairs of real numbers $x$ and $y$

- Thu Mar 15, 2018 2:50 pm
- Forum: Algebra
- Topic: Functional equation from Japan MO 2016
- Replies:
**1** - Views:
**3906**

### Re: Functional equation from Japan MO 2016

Let $P(x,y)$ denotes the assertion. $P(0,0) \Rightarrow f(0) = (f(0))^2 \Rightarrow f(0) = 0$ or $1$ Case 1: $f(0)=0$ $P(x,0) \Rightarrow f(-x) = 2x \Rightarrow f(x) = -2x$ which is obviously a solution. Case 2: $ f(0)=1$ $P(x,0) \Rightarrow f(-x) = f(x) + 2x$ ................... (i) $P(x,-y) \Right...