## Search found 176 matches

Wed Jul 19, 2017 1:17 am
Topic: IMO 2017 Problem 2
Replies: 3
Views: 1939

### Re: IMO 2017 Problem 2

Let $P(x,y)$ denote the FE. Note that if $f$ is a solution then $-f$ is also a solution. Now, $P(2,2)$ imply that $f(f(2)^2)=0$. Let $t=f(2)^2$. Case 1: $t \neq 1$ $P(x,t)$ implies $f(0)+f(x+t)=f(xt)$, and we can find $x$ so that $x+t=xt$. So $f(0)=0$. Now $P(x,0)$ implies $f(x)=0$ for all real $x$....
Wed Jul 19, 2017 12:36 am
Topic: IMO 2017 Problem 2
Replies: 3
Views: 1939

### IMO 2017 Problem 2

Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for any real numbers $x$ and $y:$

$f(f(x)f(y)) + f(x+y) = f(xy).$
Wed Apr 26, 2017 7:39 pm
Forum: Number Theory
Topic: IMO Shortlist 2011 N5
Replies: 4
Views: 1175

### Re: IMO Shortlist 2011 N5

But $f(n)|f(0)$ and $f(0)>0$, doesn't that mean $f$ does achieve a maximal value? So your solution is correct I think.
Mon Apr 24, 2017 1:45 pm
Forum: Geometry
Topic: USA(J)MO 2017 #3
Replies: 6
Views: 1454

### Re: USA(J)MO 2017 #3

Zawadx wrote: There's a typo in the determinant: zero for you~
Edited. Latexing a determinant is a pain in the first place, locating these typos are difficult It was correct in my paper, so if I had submitted, it wouldn't have been a zero, rather a seven.
Sat Apr 22, 2017 6:23 pm
Forum: Geometry
Topic: USA(J)MO 2017 #3
Replies: 6
Views: 1454

### Re: USA(J)MO 2017 #3

We use barycentric coordinates. Let $P\equiv (p:q:r)$. Now, we know that $pq+qr+rp=0$ [The equation of circumcircle for equilateral triangles]. Now, $D\equiv (0:q:r), E\equiv (p:0:r), F\equiv (p:q:0)$. So, the area of $\triangle DEF$ divided by the area of $\triangle ABC$ is: \dfrac{1}{(p+q)(q+r)(...
Sat Apr 22, 2017 3:19 pm
Forum: Number Theory
Topic: USAJMO/USAMO 2017 P1
Replies: 3
Views: 969

### Re: USAJMO/USAMO 2017 P1

aritra barua wrote:When $a,b$ € $N$ and it follows that $a^x$=$b^y$,there exists $t$ € $N$ such that $a$=$t^k$,$b$=$t^q$.This lemma can be quite handy in this problem.
Are you sure that helps? $a$ and $b$ need to be coprime.
Sat Apr 22, 2017 11:24 am
Forum: Algebra
Topic: At last an ineq USAMO '17 #6
Replies: 3
Views: 1070

### Re: At last an ineq USAMO '17 #6

For calculating the tangent, See that the derivative of the function at the point $2$ is $-\dfrac{1}{12}$, and plugging $2$ gives us $\dfrac{1}{12}$. So, the equation of the tangent is $-\dfrac{x}{12}+c$, and it goes through $(2,\dfrac{1}{12})$. This lets us determine the equation.
Sat Apr 22, 2017 11:19 am
Forum: Algebra
Topic: At last an ineq USAMO '17 #6
Replies: 3
Views: 1070

### Re: At last an ineq USAMO '17 #6

Here we use the well known tangent line trick. It is easy to see that the minimum is achieved at $(a,b,c,d)=(0,0,2,2)$ and its cyclic variants. Also, the most troublesome thing in this inequality is the quantities of the form $\dfrac{1}{x^3+4}$. So, we draw a tangent of $\dfrac{1}{x^3+4}$ at the poi...
Thu Apr 20, 2017 9:16 pm
Forum: Social Lounge
Topic: BDMO Forum Mafia #2
Replies: 30
Views: 5362