AMC
AIME
pigeonhole principle
factorial
ratio
geometry
AIME I
\/closed

We are now accepting enrollments for our Special AIME Problem Seminar, a one-weekend class shortly before the AIME in March.  Classes are on March 3 (Sat) and Mar 4 (Sun) from 3:30 PM until 6:30 PM ET (12:30 - 3:30 PM PT).  Each class will have a 30 minute break.

Enrollment information for the Special AIME Problem Seminar can be found [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]here[/url].

[quote="now a ranger"]woah wow why does everything cost hard-earned money?  :( [/quote]
Well Mr. Rusczyk needs to feed his family too.

yes quite true.

Oh wait, it's full time.

Oops.  :blush:

I'll just try to use Asian Persuasion to get my parents to let me sign up sometime this semester.

[quote="Karth"][quote="now a ranger"]I want to enroll
woah wow why does everything cost hard-earned money?  :(
my parents won't let me because of the price..
They are middle class though: My dad makes like 110,000 to 112,000 a yaer. Well that has nothing to do with it. But duno.
Argh  :lol:[/quote]

Actually, the classes are relatively cheap if you compare the cost to those of other online classes that other companies offer (maybe not math, but others). That's what's cool about AoPs... sweet classes with low pay  :D[/quote]
AoPS pwnz all.
About how many problems will be done in this? Will there be message board problems? How many message board problems? I heard from someone that there was a mock last year, what about this year?

[quote="knexpert"]I've taken several practice AIMEs recently and I have received scores of 6-8. Do you think that this seminar will improve my AIME score to a 9 or 10?[/quote]
I don't have that much knowledge about this, but keep in mind two things.
One: It is harder and harder to improve on the AIME the better you get.
Two: This thing is just before the AIME so it's more like a very very effective cram session from what I'm inferring.

Then again, I may be totally wrong.

[quote="now a ranger"]so if you are making 2's and 3's and 4's, it is not as hard to improve your score?[/quote]
It's easier to improve your score by 2 at that point than if you are getting 12s...

He's saying that once you are at a 10+ level, there isn't much more to learn compared with you at a 3-4 level.

oh yeah that's what I thought his message was.

[quote="13375P34K43V312"]I'm a noob, but from what I believe it's that:

Improving from 2-4ish to 7-9ish is a matter of picking up more concepts and gaining more problem experience.
Improving from 10-11ish to 14-15 is a matter of being more creative and noticing cruxes and stuff more quickly.[/quote]

if you're a noob, then I'm a flat retard who doesn't know how to add 1 digit numbers.

Yeah. I'm going to read AoPS 2. I gotta ask my teacher for AoPS 2 though to borrow. I only have AoPS 1.

[quote="13375P34K43V312"]Improving from 10-11ish to 14-15 is a matter of being more creative and noticing cruxes and stuff more quickly.[/quote]

or not making careless mistakes

I can only solve 2-3 on average, and I'm going to ask my teacher to borrow one of the school's AoPS Vol. 2 textbooks.

How should I ask her nicely?

Everyone in Alg. 2 is provided with an AoPS Vol.2 textbook to use for the year, but I'm only in geometry... AoPS 1 is easy

Just to comment:

I would say that the AIME takes two major things:
1) Familiarity with problem solving techniques and general facts.
2) Ability to adapt those techniques to slightly different problems.

[Neither of which is necessarily easy to develop...]

From my point of view, at least half of the AIME problems are twists on the same old techniques.

So basically, you need to have those two things: 1) knowledge of the techniques, 2) ability to apply them when there is a twist.

So you are not really going to learn the second one in a day or two, but if you still don't have the first one, then a class will be helpful. To see what I mean, consider the 2006 AIME I

[url]http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2006[/url]

1. technique: pythagoren theorem...

2. technique: complement [this problem was pretty old...seen in pigeonhole problem from a while ago for a 10 element subest]
new idea: it was a 90 element subset instead...

3. technique: this is a common problem, seen often...writing the digit as a variable...a*10^n+x=29x...

4. technique: number of 0 digits at the end of a factorial
twist: there are 100 factorials now

5. technique: squaring... [but this was from usamts...]

6. technique: 1) average for each place value 2) repeating digits is just multiplying by 1/999

7. technique: side ratio ^2= area ratio for similar triangles
twist: the vertex is not on the parallel lines

8. [i will admit this was pretty unique]

9. technique: 1) log rules, 2) arithmetic series sum, 3) solve linear diophantine in positive intgers
twist: combining the 3

10. [this was a cool problem]

11. [although recursions are a standard technique for these types of problems, this was a unique problem]

12. technique: sum and difference formulas
twist: not obvious to see the factorizations...

13. [cool, but i think i have seen the summation before...]

14. technique: 1) height *base=volume [the rest was brute forcing the lengths and areas] 2) you could do vectors...but that is a bit messy to find them...

15. [this was cool and new]

[quote="calc rulz"]Hmm if I can already solve 8-11ish of the problems in 3 hours (assuming no trivial errors), would reading AoPS Volume 2 help that much?[/quote]

It should help some, but if you're hitting 8-11 now, you should be spending at least half your time (if not more) working on some olympiad problems.

[quote="knexpert"]I've taken several practice AIMEs recently and I have received scores of 6-8. Do you think that this seminar will improve my AIME score to a 9 or 10?[/quote]

I'd expect you'd need more than just 5 hours of class to improve three points on the AIME, but this class is designed for people in the 4-11 range, so it should be right up your alley.

I'm not in the alley.

I make 2's and 3's, unless it's a VERY easy test.

March 4.  (After the first class, before the second.)

[quote="hillarryous"]Because of my state science olympiad tournament, I might have to miss the first hour or so of the March 3 session. Is is still worthwhile to sign up?[/quote]

You'll have access to the transcript after class, so you'll still get to cover everything you miss.

Good luck at the tournament.

[quote="davidyko"][quote="bpms"]Is this class going to include a mock AIME, I've heard that one in the past did?[/quote]
Yes, I was wondering that too. Anyone?[/quote]
He said that they will be using the one from 05.


it's arithmetic mean and geometric mean. the arithmetic mean is always greater than or equal to the geometric mean.

You also need to make sure that $a_1, a_2, a_3, ..., a_n$ are all positive. If they are negative or 0, the inequality does not always hold.

[quote="anirudh"]thanks. also imenat this to go in the getting started section  :D[/quote]

I moved it. Inequalities are NOT getting started. ;) (Other than linear and quadratic)

Please also try the search tool on the top toolbar.

Link: [url]http://www.artofproblemsolving.com/Forum/search.php[/url]

There is a sticky at the top of the intermediate topics forum which is an inequality marathon - there you will find explanations not only on AM-GM but other inequalities and many practice problems.


In the Old Young Chinese Culture We have

$a = r$
$b = j$
$c = k$
etc....

Translate "Thales418" To japanese Math Culture

rofl    :lol:

it may be something like "$7y57893578jfifbaifbibifaibfsdkfwsnwicvw$ or so.

this is chinese math.....

is not Racist Math

nether Russian Math

and other countries


Hi,

Can anybody help out with the following problems: please
Sorry if it in the wrong forum

1. Prove that the following property holds

(a,b)=(a1,b1) if and only if a=a1 and b=b1

2. Are we justified in calling the partial function from X->Y and the total functions from X->Y sets, where X and Y are sets?

3. Show that for any sets X and Y, where Y has atleast 2 elements, that there canot be a 1 to 1 corrospondence between X and the set of functions (X->Y)

4. Let R be a realtion on a set X. We write Rop as the opposite or converse relation where Rop = {(y,x) | (x,y) belongs to R}}
Show that (R U Rop)* is an equivalance realtion and R U (Rop)* need not be an equivalance relation.

Thank you for any help at all

this doesn't look anything like advanced fields to me. I'll move it to a more appropriate forum ...


Let $ a,b,c>0$ such that $ ab\plus{}bc\plus{}ca$≥$ 3$.Prove that
$ \frac{a}{\sqrt{a\plus{}b}}\plus{}\frac{b}{\sqrt{b\plus{}c}}\plus{}\frac{c}{\sqrt{c\plus{}a}}$≥$ \frac{3}{\sqrt2}$

[quote="nguoivn"]It's obviously trues by Honder and Am-Gm  :maybe:[/quote]
Continue with [b]nguoivn's[/b] approach in detailed:
By [b]Holder[/b], we have:
$ (LHS)^2(\sum a(a+b))\ge (a+b+c)^3$
$ \Longleftrightarrow (LHS)^2\ge \frac {(a+b+c)^3}{\sum a(a+b)}$
Hence so as to prove the inequality of the problem, it is enough to prove:
$ 2(a+b+c)^3\ge 9(a^2+b^2+c^2+ab+bc+ca)$, $ (*)$
Denote $ (a+b+c,ab+bc+ca)=(p,q)$
$ (*)\Longleftrightarrow 2p^3\ge 9(p^2-q)$
We have $ q\ge 3\Longrightarrow 9(p^2-q)\le 9(p^2-3)\le 2p^3$ (By $ AM-GM$)
The result is lead as follow.


inequalities
trigonometry
geometry
3D geometry

Let $ ABC$ be an acute-angled triangle. Prove that

\[ \tan\alpha\cdot \tan\beta \cdot \tan\gamma \ge 3\sqrt{3}\]

[quote="Obel1x"]Let $ ABC$ be an acute-angled triangle. Prove that
\[ \tan\alpha\cdot \tan\beta \cdot \tan\gamma \ge 3\sqrt {3}\]
[/quote]
[hide]Knowing that $ \tan\alpha\cdot \tan\beta\cdot \tan\gamma\equal{}\tan\alpha\plus{}\tan\beta\plus{}\tan\gamma$ we now wish to prove $ \tan\gamma\plus{}\tan\beta\plus{}\tan\alpha \geq 3\sqrt{3}$.Also set $ \tan\alpha\cdot \tan\beta\cdot \tan\gamma\equal{}x$. This also means that $ \tan\alpha\plus{}\tan\beta\plus{}\tan\gamma\equal{}x$. Using AM-GM we get that $ \frac{x}{3}\geq \sqrt[3]{x}$. Rearranging this gives us $ x^{3}\geq 27x$. Since we know that $ x$ isn't 0, we have $ x^{2}\geq27$, or $ x\geq3\sqrt{3}$ and now we are done.[/hide]

Hello,

since $ tg(x)$ is convex on $ [0,90]$ we have:

$ tg(x)\plus{}tg(y)\plus{}tg(z)>\equal{}3tg((x\plus{}y\plus{}z)/3)$ by Jensen's inequality.

So $ tg(x)\plus{}tg(y)\plus{}tg(z)>\equal{}3\sqrt{3}$

With AM-GM I got :

$ tg(x)\plus{}tg(y)\plus{}tg(z)>\equal{}3\sqrt[3]{tg(x)*tg(y)*tg(z)}$

We need to show that $ 3\sqrt[3]{tg(x)*tg(y)*tg(z)}>\equal{}3\sqrt[]{3}$

If we cube this expression, we get $ tg(x)tg(y)tg(z)>\equal{}3\sqrt[]{3}$


function
real analysis
real analysis solved

I don't know if this one has been posted before.

$\{f_n(x)\}$ is a sequence of increasing functions on [0,1]. The function sequence converges to $f$ in measure. Show that $f_n(x_0) \to f(x_0)$ where $x_0$ is a continuity of $f$.

If I am not wrong then this problem is quite straightforward :D
 Just use the definitions!!

can you show your proof?
i can only proof that there is a subsequence $\{f_{n_k}(x)\}$ such that $f_{n_k}(x_0) \to f(x_0)$ when $x_0$ is a continuity of $f$.

By the way, the conclusion may not be true if our concerning continuity is $0$ or $1$.

my proof:

 If not, there exist an $\epsilon > 0$ and a subsequence, which we will reenumerate as $\{f_i\}_{i \in \mathbb{N}}$, such that $|f_i(t) - f(t)| > 2\epsilon$ for all $i$

 Since $t$ is a continuity of $f$ then there exist $\delta$ small enough so that $(t-\delta,t+\delta)$ still included in $[0,1]$, and the flunctuation of $f$ on this interval is not greater than $\epsilon$

 Then a straightforward argument show that $m\{(f_i-f) > \epsilon \} > \delta$ for every $i$, hence $\{f_i\}_{i \in \mathbb{N}}$ does not converge to $f$ in measure.

[quote="liyi"]Do you mean the following?

If $f_i(x_0) > f(x_0) + 2\varepsilon$, then for $x\in (x_0, x_0 + \delta)$,
$f_i(x) - f(x) > f_i(x_0) - f(x) > f(x_0) - f(x) + 2\varepsilon > \varepsilon$
and if $f_i < f(x_0) + 2\varepsilon$, $f_i(x) - f(x) > \varepsilon$ for $x \in (x_0-\delta, x_0)$.

and can you provide with a example that the statement is false at $x=0$ and $x=1$?[/quote]

 Yes, you're right!!
 And for the example, consider $f_n(x) = \frac{x}{n}$ for $x <1$ and $f_n(1) = 1 $  for all $n$
 $f(x) = 0$


OK, another problem that I've been unable to solve for awhile now (It sure would be nice if there was a Solution manuel for [u]The Art and Craft of Problem Solving[/u]...).

Consider the operation which transforms the 8-term sequence a1,a2,...,a8 into the new 8-term sequence |a1-a2|,|a2-a3|,...,|a8-a1|. Find all 8-term sequences of integers that have the property that after finitely many applications of this operation, one is left with a sequence, all of whose terms are equal.

By a1 I mean a with subscript 1, etc.

I've come to some basic conclusions, mostly involving the maximum and minimum elements, but I just can't seem to finish it off.

Does a1 through a8 have to be different?

Then a1 through a8 can all be the same number, right?

I have the solution manual for Zeitz. Which problem number and which chapter, etc. is it? I'll look it up to see if it's there for you.

Yes, all the numbers can be the same.

It's problem 3.4.36.

Where were you able to get a solution manual? I found a website that distributes it, but it looked like you had to be a certified instructor to get one, or something like that.

Here is the ``official'' solution:

36. (Sketch due to Kiran Kedlaya) No matter what the starting sequence, eventually all terms become 0. Use induction on the maximum absolute value of the terms, which does not increase after the transformation.

First we show that after 8 steps, all the terms will be even, using the fact
that |x2 −x1| ≡ x2 +x1 (mod 2). (So we can drop the absolute value brackets.) At this point, divide each term by 2, then use the inductive hypothesis.

I got the solution manual by asking Paul Zeitz for it. I guess that's the most direct way of doing things.

Riceboi tweaked the problem a bit and came up with this:

"Given an initial sequence a_1, a_2, ..., a_n of real numbers, we perform a series of steps. At each step, we replace the current sequence x_1, x_2, ..., x_n with |x_1 - a|, |x_2 - a|, ..., |x_n - a| for some a. For each step, the value of a can be different.

Prove that it is always possible to obtain the null sequence consisting of all 0's and determine the minimum number of steps required, regardless of initial sequence, to obtain the null sequence."


If a bird collector wants to buy 100 budgies and wants to spend exactly $ 100. And blue budgies cost$ 10 each, green budgies cost $ 3 each, and yellow budgies cost$ .50 cents each, then how many of each can he buy?

We have that $ 10b\plus{}3g\plus{}.5y\equal{}100 \longmapsto 20b\plus{}6g\plus{}y\equal{}200$ (multiplying by 2).  Subtracting the equation $ b\plus{}g\plus{}y\equal{}100$ from this, we get that:
$ 19b\plus{}5g\equal{}100$.  Since $ b$ and $ g$ are integers, $ 19b$ has to end in a $ 5$ or a $ 0$ so that we can add enough $ g's$ to get to $ 100$.  Obviously, $ 19(5)\equal{}95$ is the desirable choice, so $ \boxed{b\equal{}5}$, and $ g$, therefore, must be $ \boxed{1}$.  Finally, plugging these numbers into our equation $ b\plus{}g\plus{}y\equal{}100$, we get that $ \boxed{y\equal{}94}$.

I don't undertand why you double b and g in the formula?

Because then you get $ 2(.5y)\equal{}y$ in the equation.  From this, we can subtract the second equation, eliminating the $ y$ and making finding the variables a LOT easier.


If $ a$ is a nonzero integer and $ b$ is a positive number such that $ ab^2\equal{}\log _{10}b$, what is the median of the set {$ 0, 1, a, b, 1/b$}?

(a) 0
(b) 1
(c) a
(d) b
(e) 1/b


Prove that if $ x\geq0$ $ y\geq0$,then   $ x^{4}\plus{}y^{4}\plus{}1 \geq x^{3}y\plus{}y^{3}\plus{}1$

_______________________________________
Azerbaijan Land of Fire
$ \Rightarrow\bigstar\Leftarrow$
 :rotfl:

$ x^3(x\minus{}y)\ge y^3(1\minus{}y)$
now if $ y\ge x$ and $ y\le 1$ then lhs is negative  while rhs is  positive :?:

The proposed inequality is not true.

Consider x = y = 0.1

It could be a typo, perhaps $ x^4 \plus{} y^4 \plus{} 1 \geq x^3 y \plus{} y^3 x \plus{} 1$? It seems the problem could be symmetric.
If so:
[hide]
Rearrange to get $ x^4 \minus{} x^3 y \minus{} x y^3 \plus{} y^4 \geq 0$
or $ (x\minus{}y)^2(x^2 \plus{} xy \plus{} y^2)$
$ (x\minus{}y)^2$ is obviously nonnegative since it's a square;
$ (x^2 \plus{} xy \plus{} y^2)$ is obviously nonnegative since x and y are nonnegative and it's all addition.
[/hide]
Either way, what is that extra 1 doing on both sides?

[quote]Either way, what is that extra 1 doing on both sides?[/quote]
it doesnt matter whether its there or not


Sorry if i posted to a wrong place. Can anyone tell me how acient mathematicians found out area of a circle when they do not know calculus? As I know, ancient mathematicians did find out C=2.pi.R by dividing its length with its radius, but I really do not know how they came up with the circular area. Thanks

[quote="surfer"]sorry, I do not know a lot about ancients.
As I remember from my school geometry, circular area can be found with use of regular n-sided polygon area, when n->inf. This has no calculus.
Some info:
http://mathworld.wolfram.com/Circle.html[/quote]
I just thought I had found out something new but it was not.  :blush: Anyway, thank for your link and by the way can you tell me how mathematicians proved e^(ix) = cosx + i.sinx.


Pick's Formula states that A= B/2 + I -1

What does it mean and how can we use it in problems?

I has to do with the area of regions and lattice points.  If you have some shape and you know how many lattice points it covers, you can compute the area of the region.

In a shape with it's vertices on lattice points, Pick's formula states that the area of the shape is $ I\plus{}\frac{P}{2}\minus{}1$
Where P is the number of points on the shape's perimeter and I is the number of points inside the shape.


MATHCOUNTS
AMC
USA(J)MO
USAMO
AMC 8
geometry
analytic geometry

Heed the subtitle.

I got 1st as expected, because my school is kinda, well, lame.

Maybach, I'm from westside! And I beat summai this time (last year I came 7th)!!!

Good job summai though :). (Maybach, remember who summai is?)

And Maybach, how would I vote?

POST EDITED: Sorry everyone :oops:

NO DISCUSSING OF SCORES.  IT SAYS SO ON THE ANNOUNCEMENT.  PLZ EDIT YOUR POST.

Remember that some people haven't taken it yet. I don't even know when it will be for me (probably about a week before Chapter).

Yay I chose the last option.

I never understood the point of discussing results, especially if you aren't discussing the problems; considering if anyone on AoPS knows who you are, they either know you in real life or you are a big shot and likely came in 1st/2nd anyway.

Good luck to all of you though.

I'm in high school, so MC is over for me..  :)  :(  :|

Anyway, when will the test be posted on the MC website? Or is there some other place that puts it on the internet before that?

Mine's not happening until the end of January... I'll probably get in the top four at my school...  :lol:

@SonyWii: After all the Chapter rounds are over...

Deary me I will suck. Probably. Ah well.  :lol:

9th grade now  :(

so...i voted for the last option

I placed third, but there were only like 10 people at my school who took it.

Did you do a countdown at your school for fun? That's what my school did last year.

[quote="eqjj168"]not taking it until Jan 9
and apparently, our "School Round" is for our entire district...[/quote]

What about your chapter then?

I got 2nd, the results came in January 27.
That sucks. :(

The school round was eazy, with all capitals. I think I did well, with all capitals. I hope I am at least 1, with all capitals. It felt eazy, with all capitals. My chapter is on the 28th and I hope I can get 1st, with all capitals. Ahhhhh!!!! Eazy, with all capitals!!!!!!!

I epic failed the School! OH NOZ!
I think I got 6th/didn't make the top 4 :(

[quote="The Hobbit"]AHHHH chapter is THIS saturday!!!! Crazy!!! i is nervous!!![/quote]

...yeah, I agree... !!!
On the bright side I got 1st in the school!!! :yup:
Which I took on Friday!!! :mad:

[quote="bull98"]Now it is in Feb. Chapter is starting.
Can Mathcounts school problems be posted now?[/quote]
[quote="AIME15"]Do not discuss the School or Chapter rounds until they have been posted on the MATHCOUNTS website...this is usually the first week of March. [/quote]

[quote="AdithyaGanesh"]I was 1st at my school.

Are we allowed to mention our score?[/quote]
Don't think so... hmmmmmmmmmm...

Someone posted a problem in the Middle School Classroom Math that was in the school. Told him to delete it but he isnt doing it.
[color=red][size=67]Link edited out by mod.  In this circumstance, it is best to contact a moderator of the respective forum (I have done so already).[/size][/color]

[quote="AIME15"]Do not discuss the School or Chapter rounds until they have been posted on the MATHCOUNTS website...this is usually the first week of March. [/quote]

Why??? If Mathcounts people are too busy, do not have time to update their site, then we cannot see the problems?
This is AOPS forum, a place to discuss math problems. It is not a kid of Mathcounts.

I can understand we should not discuss school problems when some schools have not done it yet.
Now all schools are finished the school round. Chapter round has started.
Why we cannot discuss school round problems?

[quote="The Hobbit"]AHHHH chapter is THIS saturday!!!! Crazy!!! i is nervous!!![/quote]
Me too! I'm probably going to be hyperventilating by Friday! :o

[quote="bull98"]Now all schools are finished the school round. Chapter round has started.
Why we cannot discuss school round problems?[/quote]
This is not true. Chapter rounds have to be done any time in February. School rounds can be done any time before chapter. For example, a chapter round could be on Feb. 28, and a school could have its school round on Feb. 27. Not everyone has necessarily taken the school round.

Why are there so many time diferences? It is confusing.  :(

It might be because some areas are more prone to snow? Our Chapter competition is [i]supposed[/i] to be on Saturday, the snow date Sunday, but they're calling for 4-8 inches of snow I think so it could be postponed until after Sunday  :(


function
induction
algebra
functional equation
algebra unsolved

this is really so hard, have you got any sollution?

my solution at the contest is big, but the first ideas are:

prove that f is one bijection and prove that f(-1) using this.
now, use y = -f(x)/x, and a lot of substitutions.

[quote="cyshine"]Find all functions $f\colon \mathbb{R}\to \mathbb{R}$ such that
\[f(xf(y)+f(x)) = 2f(x)+xy \]
for every reals $x,y$.[/quote]
the one answer I got is
\[f(x)=1+x \]
Keep Smiling
Malay

please post your solution, okie? Don't say what it is, say how can you solve it!

Ok,here's my solution:
We recall the problem:
Find all the functional equation $f: R\to R$ satisfying:$f(xf(y)+f(x))=2f(x)+xy$ $\forall x,y \in R$ (*)
From the definition of function $f$,it clears that $f$ is bijection
Put $f(0)=a$,because $f$ is bijection so it exists $b \in R$ such that $f(b)=0$
In (*),Substituting x by b then $f(bf(y))=by$ (1) $\forall y \in R$
                      $\Longrightarrow f(ba)=0$ (do $y=0$)
$\Longrightarrow ba=b$ (because f is bijection)
$\Longrightarrow b=0$ or $a=1$
.If $b=0$ then $f(0)=0$.Put $y=0$ then $f(f(x))=2f(x) \forall x \in R$
Because $f$ is bijection $\Longrightarrow f(x)=2x \forall x \in R$
It clears that $f(x)=2x$ doesn't satisfy (*) (absurd)
So $a=1 \Longrightarrow f(0)=1$
From (1) we have $f(bf(y))=by \forall y \in R$
Put $y$ by $f(by)$ then $f(b^{2}y)=b^{2}f(y) \forall y \in R$
Put $y=0$ then $b^{2}=1 \forall b=-1$ (because $f$ is bijection)
               $\Longrightarrow f(-1)=0$
Because $f(-1)=0$,hence $f(-f(x))=-x \forall x \in R$
Put $y$ by $-1$ then $f(f(x))=2f(x)-x \forall x \in R$ (2)
In (2),substituting $x$ by $-f(x)$:
              $f(f(-f(x)))=2f(-f(x))+f(x)$ $\forall x \in R$
           $\Longleftrightarrow f(-x)=f(x)-2x$ $\forall x \in R$
IN (*),Put $y$ by $-f(y)$ then:
                      $f(f(x)-xy)=2f(x)-xf(y) \forall x,y \in R$ (3)
Substituting $y$ by $\frac{f(x)}{x}$ into (3) then:
           $2f(x)-xf(\frac{f(x)}{x})=1$ (because $f(0)=1$)
$\Longrightarrow xf(\frac{f(x)}{x})=2f(x)-1$ (**)
In (*),Substituting y by $-\frac{f(x)}{x}$ then $f(xf(-\frac{f(x)}{x})+f(x))=f(x) \Longrightarrow xf(-\frac{f(x)}{x})+f(x)=x \forall x \in R$ (because $f$ is bijection)
Because $f(-x)=f(x)-2x \forall x \in R$
${\Longrightarrow f(-\frac{f(x)}{x})=f(\frac{f(x)}{x})-2\frac{f(x)}{x}\forall x \in R\{0}$
Hence: $xf(-\frac{f(x)}{x})+f(x)=x$
      $\Longleftrightarrow xf(\frac{f(x)}{x})=x+f(x)$ (***)
From (**) and (***) $\Longrightarrow f(x)=x+1$
Therefore $f(x)=x+1$ is the only function satisfing (*)

Well, it is not so obvious that f is a bijection.
I have some success proving that f is injective:

[hide]
$f(a)=f(b)$
$af(a)+f(a)=af(b)+f(a)$
$f(af(a)+f(a))=f(af(b)+f(a))$
$a*a+2f(a)=ab+f(a)$
$a*a=ab$
$a=0$ or $a=b$

If $a=0$, then
$f(0)=f(b)$ for any b, and so f is constant, and so
f(0)=xy+2f(0), xy+f(0), false!
[/hide]

Then I have the idea: f=g+1:
g(xg(y)+g(x)+x+1)=xy+2g(x)+1
It is almost good to try g(x)=x.

But I have not success at now  :(

That's clearly correct, Silouan: he showed that $f(x)=2x$ on the image of $f$, which covers all possible values of the variable by surjectivity.

You are right harazi ,thank you pointing out it .

Hmm... my solution goes as follows. First, $f$ is bijective because, plugging $x = 1$, we get $f(f(y)+f(1)) = 2f(1)+x$; $x+2f(1)$ spans all real numbers (so $f$ is surjective) and $f(x) = f(y) \iff f(x)+f(1) = f(y)+f(1) \Leftrightarrow f(f(x)+f(1)) = f(f(y)+f(1))\iff 2f(1)+x = 2f(1)+y \iff x = y$ (so $f$ is injective).

Choose $x \neq 0$ and $y$ such that $2f(x)+xy = f(x) \iff y =-{f(x)\over x}$. By injectivity,
\[xf\left(-{f(x)\over x}\right)+f(x) = x\iff f\left(-{f(x)\over x}\right)+{f(x)\over x}= 1\quad (*)\]

Now, set $x = y = 0$: $f(f(0)) = 2f(0)$. Let $a = f(0)$, so that $f(a) = 2a$. If $a = 0$, setting $y = 0$ we get $f(f(x)) = 2f(x)$ and by surjectivity $f(x) = 2x$ for all $x$. This function doesn't work (just test it!), so $a$ can't be $0$. So $a \neq 0$ and plugging $x = a$ in $(*)$, we obtain (recall that $f(a) = 2a$):
\[f(-2)+2 = 1\iff f(-2) =-1\]

Now set $y = 0$: $f(f(x)-x) = 2(f(x)-x)$. Now we find all real numbers $k$ such that $f(k) = 2k$. If ther is only one possible value for $k$, we are almost done because $f(x)-x = k \iff f9x) = x+k$ and by testing the function, $f(x) = x+1$.

Let $m$ be such that $f(m) = k$ (it exists because $f$ is surjective). Plug $x = m$ and $y = 0$: recalling that $f(0) = a$,
\[f(mf(0)+f(m))=2f(m)\iff f(ma+k) = 2k = f(k)\]

By injectivity,
\[ma+k = k\iff ma = 0 \iff m = 0\]
because we already proved that $a \neq 0$. So there is only one value for $k$, namely, $k = f(0) = a$.
Then, we're done: the only function is $f(x) = x+1$.

$ f(xf(y)\plus{}f(x))\equal{}2f(x)\plus{}xy$ $ (1)$.
Plugging $ y\equal{}\frac{\minus{}2f(x)}{x}$ in $ (1)$ we get:$ f(xf(y)\plus{}f(x))\equal{}0$. So, there exists such real number $ u$ that $ f(u)\equal{}0$.
Take $ x\equal{}u$ in $ (2)$:
$ f(uf(y))\equal{}uy$ $ (3)$.
Put $ y\equal{}x$ in $ (3)$:
$ f(uf(x))\equal{}ux$ $ (4)$.
If $ f(x)\equal{}f(y)$, $ (3)$ and $ (4)$ gives $ u(x\minus{}y)\equal{}0$.
If $ u\equal{}0$, $ f(0)\equal{}0$.
Take $ y\equal{}0$ in $ (1)$:
$ f(f(x))\equal{}2f(x)$. $ (4)$.
Let's prove that for any real number $ z$ there exists real number $ t$,for which $ f(t)\equal{}z$:Take $ y\equal{}\frac{z\minus{}2f(x)}{x}$ in $ (1)$: $ f(t)\equal{}z$, where $ (z,t)$ are reals, Q.E.D.
Take $ x$ instad of $ f(x)$ in $ (4)$: $ f(x)\equal{}2x$, for any real $ x$, which isn't solution.(by checking it in $ (1)$ we make sure in it).So, $ u$ isn't equal to $ 0$.
So, $ f(x)\equal{}f(y)$ follows $ x\equal{}y$.
Take $ x\equal{}y\equal{}u$ in $ (1)$: $ f(0)\equal{}u^{2}$.
Plugging $ x\equal{}0$ in $ (3)$ we get: $ f(uf(0))\equal{}0\equal{}f(u)$.
So, $ u(f(0)\minus{}1)\equal{}0$. As we have already proved, $ u$ can't be $ 0$. So, $ f(0)\equal{}1$ $ (5)$.
$ u^{2}\equal{}f(0)\equal{}1$ gives $ u\equal{}1$ or $ u\equal{}\minus{}1$.
Let $ u\equal{}1$. Than, plugging $ u\equal{}y\equal{}1$ in $ (1)$ we get: $ f(1)\equal{}2$, which can't be true as $ f(1)\equal{}0$.
So, $ u\equal{}\minus{}1$ and $ f(\minus{}1)\equal{}0$ $ (6)$.
Plugging $ x\equal{}0$ in $ (1)$ we get:
$ f(1)\equal{}2$ $ (7)$.
Let's prove by induction that $ f(n)\equal{}n\plus{}1$ fro any positive integer $ n$.
For $ n\equal{}1$, $ f(1)\equal{}1\plus{}1\equal{}2$.
Let $ f(k)\equal{}k\plus{}1$, for any positive integer $ k$. We have to prove that $ f(k\plus{}1)\equal{}k\plus{}2$.
Take $ x\equal{}k$ and $ y\equal{}\minus{}1$ in $ (1)$ we get: $ f(k\plus{}1)\equal{}k\plus{}2$, Q.E.D.
$ (3)$ gives: $ f(\minus{}f(x))\equal{}\minus{}x$ $ (8)$.
Take $ y\equal{}\minus{}1$ in $ (1)$: $ f(f(x))\equal{}2f(x)\minus{}x$ $ (9)$.
Substitute $ f(x)$ by $ x$: $ f(x)\minus{}f(\minus{}x)\equal{}2x$ $ (10)$.
Let $ n$ be positive integer. $ (10)$ gives: $ f(\minus{}n)\equal{}\minus{}n\plus{}1$.
So, for any integer $ n$, $ f(n)\equal{}n\plus{}1$.
Let's prove that $ f(\frac{x}{y})\equal{}\frac{x}{y}\plus{}1$.
Take $ x\equal{}q$ and $ y\equal{}\frac{p\plus{}2q\minus{}f(q)}{q}$ in $ (1)$. We get: $ f(\frac{p\minus{}2}{q})\equal{}\frac{p\minus{}2}{q}\plus{}1$.
Substituting $ p\minus{}2$ and $ q$ by $ x$ and $ y$ respectively, we get: $ f(\frac{x}{y})\equal{}\frac{x}{y}\plus{}1$,
Q.E.D.
So, for any rational $ n$, $ f(n)\equal{}n\plus{}1$.
Function $ f(x)$ is strictly increasing because it gets each real value exactly once and for any integers $ (m,n)$, where $ m>n$, $ f(m)>f(n)$.
Let, $ f(x)>x\plus{}1$, for any real $ x$:
Take $ x$ irational and $ l$ rational number such that $ x<l<f(x)\minus{}1$.
So, $ x<l\equal{}f(l)\minus{}1<f(x)\minus{}1$. It follows $ f(l)<f(x)$, or $ l<x$. So, $ x<l<x$-clearly contradiction.
So, $ f(x)<\equal{}x\plus{}1$,for any real $ x$ $ (11)$.
Let's prove that $ f(x)>x$, for any real $ x$. Take $ x$ rational number and $ r$ real such that
$ x<r<x\plus{}1$. Obviously, $ x\plus{}1<f(r)<x\plus{}2$. If $ f(r)<\equal{}r$, than $ x\plus{}1<\equal{}r$. So, $ r<x\plus{}1<\equal{}r$-clearly contradiction.
So, $ f(x)>x$, for any real $ x$, Q.E.D.
Let's take $ r$ rational number so that $ f(x)<r<x\plus{}1$. (We let $ f(x)<x\plus{}1$, for any real $ x$).
Than, $ r\equal{}f(r\minus{}1)$.So, $ f(x)<f(r\minus{}1)<x\plus{}1$, or $ x\plus{}1<r$.
So, $ r<x\plus{}1<r$, which is clearly contradiction.
It follows, that $ f(x)>\equal{}x\plus{}1$, but $ f(x)<\equal{}x\plus{}1$. So, $ f(x)\equal{}x\plus{}1$, for any real number $ x$.
By checking it in $ (1)$, we make sure that it's possible answer.

@cyshine:

It's funny, that your solution is EXACTLY the same as the official solution, word for word, and the best is:

WITH THE SAME ERROR, the error is only, that you should replace 0 with -2. The official solution did the same. They wrote 0 instead of -2, and afterwards, they finished the solution correctly.

Best regards, felix

It is quite clear that the function is bijective. We will first show that $ f(\minus{}1) \equal{} 0$.
$ f(xf(y) \plus{} f(x)) \equal{} 2f(x) \plus{} xy \minus{}\minus{}\minus{}\minus{}(*)$
Since $ f$ is bijective, there exists a $ c$ such that $ f(c) \equal{} 0$.
Plugging $ x \equal{} y \equal{}c$ in $ (*)$, we get $ f(0) \equal{} c^2$.
Plugging $ x \equal{} 0$ in $ (*)$, we get $ f(c^2) \equal{} 2c^2$.
Plugging $ x \equal{} c, y \equal{}0$ in $ (*)$, we get $ f(c^3) \equal{} 0$.
Since $ f$ is bijective, $ c^3 \equal{} c \Rightarrow c \in \{\minus{}1, 0, 1\}$.
Case 1: $ c \equal{} 1 \Rightarrow f(1) \equal{} 0$ and $ f(1) \equal{} 2$ which is a contradiction.
Case 2: $ c \equal{} 0 \Rightarrow f(f(x)) \equal{} 2f(x)$. Since $ f$ is a bijection we could replace $ f(x)$ by $ x$ and we have $ f(x) \equal{} 2x$.
Substituting $ f(x) \equal{} 2x$ in $ (*)$ gives us $ 4xy \equal{} xy$ which is a contradiction.
Case 3: $ c \equal{} \minus{}1 \Rightarrow f(\minus{}1) \equal{} 0, f(0) \equal{} 1, f(1) \equal{} 2$.
Plugging $ x \equal{} 1$ in $ (*)$, we get $ f(f(y) \plus{} 2) \equal{} 4 \plus{} y \minus{}\minus{}\minus{}\minus{}\minus{}(**)$.
Plugging $ y \equal{} \minus{}1$ in $ (*)$, we get $ f(f(x)) \equal{} 2f(x) \minus{} x$ (i.e.) $ f(f(y)) \equal{} 2f(y) \minus{} y \minus{}\minus{}\minus{}\minus{}\minus{}(***)$.
Adding (**) and (***), we get $ f(f(y) \plus{}2) \plus{} f(f(y)) \equal{} 2f(y) \plus{} 4$.  Again, since $ f$ is a bijection we could replace $ f(y)$ by $ y$ and we have
$ f(y \plus{} 2) \plus{}f(y) \equal{} 2y \plus{} 4 \minus{}\minus{}\minus{}\minus{}(****)$.
Replacing $ y$ by $ y\plus{}2$, we get
$ f(y \plus{} 4) \plus{}f(y\plus{}2) \equal{} 2y \plus{} 8\minus{}\minus{}\minus{}\minus{}(*****)$.
Applying $ f$ to $ (**)$ and using $ (***)$, we get
$ 2f(f(y) \plus{} 2) \minus{} f(y) \minus{}2 \equal{} f(y \plus{} 4) \Rightarrow 2(y\plus{}4) \minus{} f(y) \minus{} 2 \equal{} f(y\plus{}4) \Rightarrow f(y\plus{}4) \plus{} f(y) \equal{} 2y \plus{} 6 \minus{}\minus{}\minus{}\minus{}\minus{}\minus{}(******)$.
Adding $ (****),(*****)$ and $ (******)$ and dividing by 2, we get
$ f(y \plus{} 4) \plus{}f(y\plus{}2) \plus{}f(y) \equal{} 3y \plus{} 9$.
Subtracting $ (*****)$ from the above equation gives $ f(y) \equal{} y \plus{}1$.

[quote="cyshine"]Hmm... my solution goes as follows. First, $ f$ is bijective because, plugging $ x \equal{} 1$, we get $ f(f(y) \plus{} f(1)) \equal{} 2f(1) \plus{} x$; $ x \plus{} 2f(1)$ spans all real numbers (so $ f$ is surjective)

 $ f(x) \equal{} f(y) \iff f(x) \plus{} f(1) \equal{} f(y) \plus{} f(1) \rightarrow f(f(x) \plus{} f(1)) \equal{} f(f(y) \plus{} f(1))$

$ \iff  2f(1) \plus{} x \equal{} 2f(1) \plus{} y \iff x \equal{} y$ (so $ f$ is injective).

Choose $ x \neq 0$ and $ y$ such that $ 2f(x) \plus{} xy \equal{} f(x) \iff y \equal{} \minus{} {f(x)\over x}$. By injectivity,
\[ xf\left( \minus{} {f(x)\over x}\right) \plus{} f(x) \equal{} x\iff f\left( \minus{} {f(x)\over x}\right) \plus{} {f(x)\over x} \equal{} 1\quad (*)\]
Now, set $ x \equal{} y \equal{} 0$: $ f(f(0)) \equal{} 2f(0)$. Let $ a \equal{} f(0)$, so that $ f(a) \equal{} 2a$. If $ a \equal{} 0$, setting $ y \equal{} 0$ we get $ f(f(x)) \equal{} 2f(x)$ and by surjectivity $ f(x) \equal{} 2x$ for all $ x$. This function doesn't work (just test it!), so $ a$ can't be $ 0$. So $ a \neq 0$ and plugging $ x \equal{} a$ in $ (*)$, we obtain (recall that $ f(a) \equal{} 2a$):
\[ f( \minus{} 2) \plus{} 2 \equal{} 1\iff f( \minus{} 2) \equal{} \minus{} 1\]
Now set $ y \equal{} \minus{} 2$: $ f(f(x) \minus{} x) \equal{} 2(f(x) \minus{} x)$. Now we find all real numbers $ k$ such that $ f(k) \equal{} 2k$. If ther is only one possible value for $ k$, we are almost done because $ f(x) \minus{} x \equal{} k \iff f(x) \equal{} x \plus{} k$ and by testing the function, $ f(x) \equal{} x \plus{} 1$.

Let $ m$ be such that $ f(m) \equal{} k$ (m exists because $ f$ is surjective). Plug $ x \equal{} m$ and $ y \equal{} 0$: recalling that $ f(0) \equal{} a$,
\[ f(mf(0) \plus{} f(m)) \equal{} 2f(m)\iff f(ma \plus{} k) \equal{} 2k \equal{} f(k)\]
By injectivity,
\[ ma \plus{} k \equal{} k\iff ma \equal{} 0 \iff m \equal{} 0\]
because we already proved that $ a \neq 0$. So there is only one value for $ k$, namely, $ k \equal{} f(0) \equal{} a$.
Then, we're done: the only function is $ f(x) \equal{} x \plus{} 1$.[/quote]

[i]  A very good solution , Cyshine , congratulation :lol:

 I have edited some mistakes in order to be easy to understand[/i] :lol:

It's easy to see that $f$ is bijective(we just need to fix $x$). Let $f(0)=c,$ if $c=0,$ taking $y=0$ yields $f(f(x))=2f(x),$ i.e. $f(x)=2x$ follows from bijective, which is clearly absurd, so $f(0)=c\ne 0,$ and $f(c)=2c.$ Assume there's a $y_0\in \mathbb{R}$ for which $f(y_0)=-1,$ observe that
\[2c=f(-c+2c)=4c+y_0c \implies (2+y_0)c=0\implies y_0=-2\]
by taking $x=c, y=y_0$ and $c\ne 0.$ Now take $x=-2,y=-2$ to get $f(1)=-2+4=2,$ setting $x=1$ we get
\[f(f(y)+2)=y+4\qquad (\spadesuit)\]
for all $y\in \mathbb{R}.$ Assume there's an $a\in\mathbb{R}$ for which $f(a)=1,$ we claim that $a=0.$ Taking $(x,y)=(a,0)$ yields that $f(ac+1)=2\implies a=0$ since $c\ne 0.$ Now setting $y=0$ in our initial assertion yields
\[f(x+f(x))=2f(x).\qquad (\bigstar)\]
Taking $x=f(y)+2$ in $(\bigstar)$ gives
\[\begin{aligned}
f(f(y)+2+f(f(y)+2))=f((f(y)+y+4)+2)&=(2y+4)+4\qquad &(\text{follows by} (\spadesuit)) \\
\implies f((2y+2)+2)&=(f(y)+y)+4\qquad &(\text{follows by} (\spadesuit))\\
\implies f(f(y)+y)=2y+2&=2f(y) \qquad &(\text{also repeat $(\spadesuit)$ and combine $(\bigstar)$})
\end{aligned}\]
from which we conclude that $\boxed{f(x)=x+1,}$ which is the solution indeed. :)

$x=1, y=r-2f(1)$ gives $f(stuff)=r$, thus surjectivity. $f(a)=f(b)$ gives (by $x=1, y=a$ and $x=1, y=b$) $a=b$, thus injectivity and bijectivity.

Now let $z, u$ be the reals such that $f(z)=0, f(u)=1$. $x=z, y=u$ gives $0=zu$.
Case1: $z=0 \Leftrightarrow f(0)=0$:
$x=f^{-1}(k), y=0$ gives $f(k)=2k$, which clearly doesn't work.

Case2: $u=0 \Leftrightarrow f(0)=1$:
$x=z$ gives $f(zf(y))=zy$ (*).

$y=z$ gives $f(f(x))=2f(x)+xz\overbrace{=}^{(*)} 2f(x)+f(zf(x))$ and $k=f(x)$ makes it $f(k)=2k+f(zk)$ (#).
$x=0$ gives $f(1)=2$
$x=y=z$ gives $1=z^2$, but $z\neq 1$, so $z=-1$, and (#) becomes $f(k)=2k+f(-k)$.
$x=-1, y=1$ gives $f(-2)=-1$.
$y=-2$ gives $f(f(x)-x)=2f(x)-2x\Leftrightarrow$(#) $2f(x)-2x+f(-f(x)+x)=2f(x)-2x \Leftrightarrow f(-f(x)+x)=0 \Leftrightarrow -f(x)+x=-1 \Leftrightarrow f(x)=x+1$. This works

[quote=cyshine]Find all functions $f\colon \mathbb{R}\to \mathbb{R}$ such that
\[f(xf(y)+f(x)) = 2f(x)+xy\]
for every reals $x,y$.[/quote]

Suppose $f(a)=(b)$. Fix $x=1$ and plug $y=a$ and $y=b$ to prove injectivity. Now fix $x=1$ and vary $y \in \mathbb{R}$; thus $f$ is surjective. Pick $y_0 \in \mathbb{R}$ with $f(y_0)=0$ and plug $y=y_0$; then $f(f(x))=2f(x)+xy_0$ for all $x \in \mathbb{R}$. If $y_0=0$ then $f(t)=2t$ for all $t \in \mathbb{R}$ which is not a solution. Pick $t_0$ with $f(t_0)=1$ and plug $y=t_0, x=y_0$; then $f(y_0f(t_0))=2f(y_0)+y_0t_0$ hence $t_0=0$ and $f(0)=1$. Now $f(f(0))=2f(0)$ hence $f(1)=2$. Also $f(f(y_0))=2f(y_0)+y_0^2$ so $y_0 \in \{\pm 1\}$ but $f(1) \ne 0$ hence $y_0=-1$ and $f(-1)=0$. Consequently, $f(f(x))=2f(x)-x$ for all $x \in \mathbb{R}$.

Plug $x=1$ to get $f(f(y)+2)=y+4$ for all $y \in \mathbb{R}$. Now $y=f(f(y-4)+2)$ for all $y$; so $f(y)=f(f(f(y)-4)+2)$ for all $y$ and injectivity applies; so $=f(f(y)-4)=y-2=f(f(y-6)+2)$ for all $y$. Thus, $f(y)=f(y-6)+6$ for all $y \in \mathbb{R}$.

[b]Claim.[/b] $f(x)-f(0)$ is additive.

(Proof) Fix $a,b \in \mathbb{R}$ with $b \ne 0$. We claim that $f(a+2b)=2f(a+b)-f(a)$. Indeed set $x=\tfrac{b}{6}$ and $y=f^{-1}\left(\frac{a-f\left(\frac{b}{6}\right)}{\left(\frac{b}{6}\right)}\right)$ then $$f(a)=f(xf(y)+f(x))=2f(x)+xy$$ $$f(a+b)=f(xf(y)+f(x)+6x)=f(xf(y+6)+f(x))=2f(x)+x(y+6)$$ $$f(a+2b)=f(xf(y)+f(x)+12x)=f(xf(y+12)+f(x))=2f(x)+x(y+12)$$ hence proving the claim.

Now we plug $a=0$ to get $f(2b)=2f(b)-f(0)$ for all $b \in \mathbb{R}$ (write-up issue: deal with $b=0$ separately). Hence $f(2a+2b)=2f(a+b)-f(0)$ as long as $a \ne -b$. Therefore, for all $x,y \in \mathbb{R}$ we have $f(x+y)=f(x)+f(y)-f(0)$. $\blacksquare$

Thus, $f(xf(y))+2f(x)-x-1=2f(x)+xy \implies f(xf(y))=xy+x+1$ for all $x,y \in \mathbb{R}$. Plug $y=1$ and $t=2x$ then $f(t)=t+1$ for all $t \in \mathbb{R}$. This also satisfies the equation hence it is the only solution! $\blacksquare$

A quick glance says this is hopefully different from the above?

[hide = Solution]
We claim that $f(x) \equiv x+1$ is the only solution. It is easily verified that this indeed works, we will now show it is unique.

Let $P(x, y)$ denote the assertion that $f(xf(y)+f(x)) = 2f(x) + xy.$

Firstly, if $f(a) = f(b)$, then $P(\pi, a) - P(\pi,b)$ gives that $\pi a = \pi b$, so that
$$f \; \text{is} \; \text{injective}.$$
By $P(1, y)$, we get that $2f(1) + y \in \text{Range}(f) \;  \forall y \in \mathbb{R}$, which gives that
$$f \; \text{is} \; \text{surjective},$$ which combined with the above gives that
$$f \; \text{is} \; \text{bijective}.$$
Hence, it makes sense to let $f(t) = 0$ for some $t \in \mathbb{R}.$ Then, $P(x, t)$ gives that
$$f(f(x)) = 2f(x) + tx. \qquad (1)$$
$P(t, y)$ gives that
$$f(tf(y)) = ty. \qquad (2)$$
Letting $y = 0$ in $(2)$ gives us that $f(tf(0)) = 0$, which together with injectivity gives us that $t = tf(0)$. Therefore, we either have that $t = 0$ or $f(0) = 1.$ If $t = 0$, then $(1)$ gives us that $f(f(x)) = 2f(x),$ which together with surjectivity implies that $f(x) \equiv 2x \; \forall x \in \mathbb{R},$ which can be easily checked to not work.

Therefore, we know that $f(0) = 1.$ $P(0, 0)$ now implies that $f(1) = 2.$ Hence, $P(1, y)$ gives us that $f(f(y)+2) = 4+y.$ Therefore, since $f(f(-2)+2) = 4 + (-2) = 2 = f(1),$ we have from injectivity that $f(-2) + 2 = 1 \Rightarrow f(-2) = -1.$ Therefore, $P(x, -2)$ gives us that
$$f(f(x) - x) = 2(f(x)-x) \; \forall x \in \mathbb{R}. \qquad (3)$$
However, remember that from $(1)$ we know that $f(x) = 2x + t f^{-1}(x),$ where $f^{-1}$ is the inverse function of $f$ (exists since $f$ is bijective). Therefore, if we plug in $x = f(a) - a$ into this equation, we have that $2(f(a) - a) =^{(3)} f(f(a) - a) = 2(f(a) - a) + t f^{-1} (f(a) - a),$ which implies since $t \neq 0$ that
$$f^{-1}(f(a) -a) = 0 \Leftrightarrow f(a) - a = f(0) \; \forall a \in \mathbb{R}. \qquad (4).$$
With $(4)$ we are nearly done, we now have that $f(x) \equiv x + c \; \forall x \in \mathbb{R}$ for some $c \in \mathbb{R}.$ Plugging this into $P(x, y)$ then easily yields our only solution $\boxed{f(x) \equiv x+1},$ as claimed.

$\square$

[/hide]

Clearly, $f$ is bijective.
[hide=finding f(-1) and f(2)]
Let $a=f(-1)$
$$P(-1,x) \implies f(a-f(x))=2a-x$$
$$P(-1,a-f(x)) \implies f(x-a)=a+f(x)$$
$$P(x,y)-P(x,y-a) \implies f(xf(y)+f(x)+ax)=f(xf(y)+f(x))-ax \implies f(y+ax)=f(y)-ax \implies a=0$$
$$P(-1,-1) \implies f(0)=1$$
$$P(0,0) \implies f(1)=2$$
$$P(1,-1) \implies f(2)=3$$
[/hide]
Plug-in $y=-1$ to see that $f(f(x))=2f(x)-x$
$$P(2,x) \implies f(2f(x)+3)=6+2x$$
$$f(xf(y)+f(x))=2f(x)+xy \implies f(f(xf(y)+f(x)))=f(2f(x)+xy) \implies f(2f(x)+xy)=3f(x)+2xy-xf(y) \implies$$
$$\stackrel{ y\mapsto \frac{3}{x}}{ \Longrightarrow}f(2f(x)+3)=3f(x)+6-xf\left(\frac{3}{x}\right)= 6+2x \ \ \ (*)$$
Now let $x \mapsto 3x$ in $(*)$ to see that $f(3x)=xf\left(\frac{1}{x}\right)+2x$ and let $x\mapsto \frac{1}{x}$ to see that $3xf\left(\frac{1}{x}\right)=2+f(3x)$ and now $f(x)=x+1$ trivially follows.

Denote the assertion as $P(x,y)$.

$P(0,0)$ gives $f(f(0))=2f(0)$.

If $f(y_1)=f(y_2)$, $P(x,y_1)$ and $P(x,y_2)$ for $x\neq 0$ gives $y_1=y_2$, so $f$ is injective. Also, varying $y$ for fixed $x$ allows RHS to achieve all real numbers, so $f$ is bijective.

Now, suppose that $f(z)=0$ (note $z$ is well defined due to bijectivity.) Then, $P(z,y)$ gives $f(zf(y))=zy$. $y=z$ into this gives $f(0)=z^2$, and $y=0$ gives $f(zf(0))=0$, which by injectivity means $f(0)=1$. Thus, $z^2=1\implies z=\pm 1$. If $z=1$, then we get that $f(f(x))=x$, so $f(f(0))=0$. However, we know that $f(f(0))=2f(0)=2$, which is a contradiction. Thus, $z=-1$.

Now, we have $f(-f(y))=-y$. As $f(1)=2$, we get $f(-2)=-1$. Also, $P(1,-3)$ gives $f(f(-3)+2)=1=f(0)$, so injectivity yields $f(-3)=-2$.

Now, $P(x,-2)$ gives $f(f(x)-x)=2(f(x)-x)$. If we have a real number $k$ such that $f(k)=2k$, $P(k,y)$ gives $f(k(f(y)+2))=k(y+4)$. Substituting $y=-3$ gives $f(0)=k\implies k=1$. Thus, we must have $f(x)-x=1\forall x\implies f(x)=x+1$. This satisfies the initial assertion, so our only solution is $f(x)=x+1$.

Hard but not so hard. ;)
[hide=Solution]We claim that only $f \equiv y+1$ is our solution. It's easy to check that it works. We now show this is the only one.

 Let $P(x,y)$ denote the assertion. Note that in $P(1,y)$ varying $y$ accordingly yields surjectivity. Suppose if possible $f(u)=f(v)$ for $u \neq v$. Then $P(x,u)$ and $P(x,v)$ combined implies, for non-zero $x,$ that $u=v$ which is a contradiction. We conclude that $f$ is bijective.

Since $f$ is bijective, we know $\exists a$ such that $f(a)=0$. Note that $P(a,y)$ implies $f(af(y))=ay~(\clubsuit)$ and $P(af(y),a)$ combined with previous result implies $$f(ay)=f(f(af(y)))=2f(af(y))+a^2f(y)=2ay+a^2f(y)$$In addition, $y=0$ yields $f(0)=0$ or $a=\pm 1$. We deal with this in cases as follows:

[list]
[*]If $f(0)=0$ then $f(a)=0=f(0)$ implying $a=0$ by injectivity. Now $P(x,0)$ yields $f(f(x))=2f(x)$ but then since $f$ is bijective we conclude $f(x)=2x$ for all $x \in \mathbb{R}$ which is not a solution to our original assertion.
[*]If $a=1$ then $f(1)=0$. Now $P(1,y)$ yields $f(f(y))=y$ but on other hand, by $P(x,1)$ we have $f(f(x))=2f(x)+x$ which then, combined with previous result, implies $f \equiv 0$ which is not a solution.
[*]If $a=-1$ then $f(-1)=0$. Now $P(x,-1)$ implies $f(f(x))=2f(x)-x~(\spadesuit)$. Here, in $(\spadesuit),$ $x=-1$ yields $f(0)=1$. Then $P(0,y)$ yields $f(1)=2$. From $P(1,y),$ we have $f(f(y)+2)=4+y~(\bigstar)$. Taking $f$ both sides in previous equation yields, with $(\clubsuit)$ combined that $$f(4+y)=2f(f(y)+2)-(f(y)+2)=2(4+y)-f(y)-2=6+2y-f(y)$$By $y \to f(y)$ shift in previous equation implies, combined with $(\clubsuit),$ that $f(4+f(y))=6+y$. By $y \to f(y)+2$ in $(\bigstar)$ implies, again combined with $(\clubsuit),$ that $$f(y)+6=f(6+y)=f(f(4+f(y)))=2f(4+f(y))-4-f(y)=2(6+y)-4-f(y)$$Simplying this equation yields $f(y)=y+1$ for all $y$.
[/list]
Summing up all cases, we have $f \equiv y+1$ as our only solution, as required. $\blacksquare$

Decently difficult.

Storage:

Let $P(x,y)$ denote the original expression.

[b]Claim 1[/b]: $f$ is bijective.
[b]Proof[/b]: Surjectivity is obvious. If $f(a)=f(b)$, the $P(x,a)$ and $P(x,b)$ for some non-zero $x$ gives $a=b$. $\blacksquare$

Let $u$ be the real number satisfying $f(u)=0$. If $u=0$, then $P(x,0)$ $\implies$ $f(f(x))=2f(x)$ $\implies$ $f(x)=2x$ by bijectivity, which is clearly not a solution. Therefore $u \neq 0$.

[b]Claim 2[/b]: $f(0)=1$ and $f(1)=2$.
[b]Proof[/b]: $P(u,0)$ $\implies$ $f(uf(0))=0=f(u)$ $\implies$ $f(0)=1$ since $u \neq 0$. Now $P(0,0)$ gives $f(1)=2$. $\blacksquare$

[b]Claim 3[/b]: $f(x+6)=f(x)+6$ for all $x$.
[b]Proof[/b]: $P(1,x)$ $\implies$ $f(f(x)+2)=x+4$ $$\implies \ f(x+6)=f(f(f(x)+2)+2)=f(x)+6$$
as required. $\blacksquare$

For $x \neq 0$, choose $y$ such that $f(y)=-\dfrac{f(x)}{x}$. Then from $P(x,y)$, $P(x,y+6)$ and Claim 3 we get $$f(x(f(y)+6)+f(x))=2f(x)+xy+6x=f(xf(y)+f(x))+6x$$ $$\implies \ f(6x)=6x+1$$
for all $x \neq 0$. This, combined with $f(0)=1$ gives the solution $$\boxed{f(x)=x+1 \ \ \forall x \in \mathbb{R}}$$

[hide = Redacted]Here is my solution.
Let $P(x,y)$ be the assertion to the initial eqn.
First lets solve for polynomial $f(x)'s$.
Assuming $deg(f(x)) = d$ $$P(0,x) \implies f(f(0)) = 2f(0)$$
$P(x,0) \implies f(xf(0) + f(x)) = 2f(x) \implies d \cdot$ max{$1,d$}$= d \implies d = 0 \text{ or } 1$
For $d = 0$, let $f(x) = c \implies c = 2c + xy$ which is clearly impossible.

For $d = 1$, let $f(x) = ax + b$

$P(x,x) \implies f(f(x) (x+1)) = 2f(x) + x^2$
$$\implies a((ax+b)(x+1)) + b = a^2 x^2 + x(a^2 + ab) + ab + b = x^2 + 2ax + 2b$$
$$a^2 = 1, a^2 + ab = 2a, ab + b = 2b \implies f(x) = x + 1$$

Now all that is left to show is that this is the only solution.

$P(1,x) \implies f(f(x) + f(1)) = 2f(1) + x$ as $f(1)$ has to be constant, $2f(1) + x = m \ \forall \ m  \in \ \mathbb{R}$ thus $f(x)$ is surjective.
$f(x) + 1= f(y) +1 \implies 2f(1) + x = 2f(1) + y \implies x = y$ thus $f(x)$ is Injective.
Therefore, $f(x)$ has to be a polynomial. So we are done.
[/hide]

[quote=quirtt]... thus $f(x)$ is Injective.
Therefore, $f(x)$ has to be a polynomial. So we are done.[/quote]
Could you kindly explain us how the bijectivity of $f(x)$ implies that it must be a polynomial ?
Thanks a lot for any precision.

[quote=quirtt]How to prove that it can only be a polynomial?[/quote]
Nearly all the posts in this thread end proving $f(x)=x+1$
And since $x+1$ is a polynomial, nearly all the posts in this thread are proofs that solution must be a plynomial.

Tough FE. Denote the assertion by $P(x,y)$. Fixing $x$ and letting $y$ vary over $\mathbb{R}$ it follows that $f$ is surjective.

Now let $c$ be any real such that $f(c)=0$. $P(x,c) \implies f(f(x))=2f(x)+cx$. If $c=0$, then $f(f(x))=2f(x)$. Since $f$ is surjective it follows that $f(r)=2r$ for all reals $r$. It is easy to check that this solution fails the original equation, so $c \neq 0$.

Suppose there exist reals $a,b$ such that $f(a)=f(b)$. Then
\[ 2f(a)+ca=f(f(a))=f(f(b))=2f(b)=cb \implies a=b \]
so $f$ is injective. Note $c$ is unique by injectivity.

Take any reals $x_1,y_1$. We have $f(x_1f(y_1)+f(x_1))=2f(x_1)+x_1y_1$. But $P(y_1, x_1) \implies f(y_1f(x_1)+f(y_1))=2f(y_1)+x_1y_1$. This holds for all $x_1, y_1$ real so we have
\[ f(xf(y)+f(x))-2f(x)=f(yf(x)+f(y))-2f(y) \]
for all reals $x,y$. Taking $y=0, x=c$ in the above and using $f(f(x))=2f(x)+cx$ it follows that
\[ f(cf(0))=f(f(0)) -2f(0) = 0 =f(c) \implies cf(0)=c \]
by injectivity. $c \neq 0$, so $f(0)=1$.

Now
\[ P(0,0) \implies f(1)=f(f(0))=2f(0)=2 \]
\[ P(c,c) \implies 1=f(0)=f(f(c))=2f(c)+c^2=c^2 \implies c= \pm 1\]
If $c=1$ then $2=f(1)=0$, contradiction. So $c = -1$, and the above rewrites as $f(f(x))=2f(x)-x$. This means
\[ f(2)=f(f(1)) = 2f(1)-1 = 3 \]
Next
\[ P(-1, x) \implies f(-f(x))=-x = f(f(x))-2f(x) \]
Since $f$ is surjective, it follows that
\[ f(-r) = f(r)-2r \]
for all reals $r$. Then
\[ P(1,y) \implies f(f(y)+2) = y+4 \]
\[ P(-1, y) \implies f(-f(y))=-y \implies -y = f(-f(y))=f(f(y))-2f(y) \]
\[ \implies f(f(y)+2)=2f(y)-f(f(y))+4 \]
Since $f$ is surjective, it follows that for all reals $r$,
\[ f(r+2)+f(r)=2r+4 \]
Now
\[ P(2,y) \implies f(2f(y)+3)=6+2y = 6+2(2f(y)-f(f(y))) \implies f(2f(y)+3)+2f(f(y)) = 4f(y)+6 \]
By surjectivity it follows that for all reals $r$,
\[ f(2r+3)+2f(r)=4r+6 \]
This holds for all reals $r$, so
\[ r \rightarrow r - \frac{1}{2} \implies f(2r+2)+2f(r-\frac{1}{2})=4r+4 \]
Now since $f(r+2)+f(r)=2r+4$ for all reals $r$,
\[ r \rightarrow 2r \implies f(2r+2)+f(2r) = 4r+4 \]
Injectivity of $f$ implies for all reals $r$ we have
\[ f(2r) = 2f(r-\frac{1}{2}) \implies f(2r+1)=2f(r) \]
taking $r-\frac{1}{2} \rightarrow r$, since these relations hold for all reals $r$.

We are almost there. $f(2r+3)+2f(r)=4r+6$ holds for all reals $r$, so
\[ r \rightarrow r-1 \implies f(2r+1)+2f(r-1)=4r+2 \]
\[ f(2r+1)=2f(r) \implies f(r)+f(r-1)=2r+1 \]
This holds for all reals $r$ so
\[ r \rightarrow r+1 \implies f(r+1)+f(r)=2r+3 \]
\[ r \rightarrow r+2 \implies f(r+2)+f(r+1)=2r+5 \]
\[ \implies f(r+2)-f(r) = 2 \]
This holds for all reals $r$. We also have $f(r+2)+f(r)=2r+4$ for all reals $r$, so subtracting yields
\[ 2f(r)=2r+2 \implies f(r)=r+1 \]
for all reals $r$. It is easy to check that this solution works, so the only function is $f(x) \equiv x+1$, finishing the proof.

Let $P(x,y)$ be the assertion $f(xf(y)+f(x)) = 2f(x)+xy$.

$P(1,x)\Rightarrow f(f(x)+f(1))=2f(1)+x$
So $f$ is bijective.
We have that $\exists k,n:f(k)=0,f(n)=1$ by surjectivity.
$P(k,n)\Rightarrow 0=kn$
$P(n,k)\Rightarrow f(1)=2$
Since $kn=0$, we now have two cases to consider.

[b]Case 1:[/b] $k=0\Rightarrow f(0)=0$
$P(x,0)\Rightarrow f(f(x))=2f(x)$ so by surjectivity we have $f(x)=2x$, which doesn't work.

[b]Case 2:[/b] $n=0\Rightarrow f(0)=1$
$P(k,k)\Rightarrow 1=k^2$, and since $f(1)=2$, we have $k\ne1$, so $k=-1\Rightarrow f(-1)=0$.

$P(-2,-2)\Rightarrow f(-f(-2))=2f(-2)+4$
$P(-1,-2)\Rightarrow f(-f(-2))=2$
So $f(-2)=-1$.
$P(-1,x)\Rightarrow f(-f(x))=-x$

The next two lines are the crucial step.
$P(-2,-f(x))\Rightarrow f(2x-1)=2f(x)-2$
$P\left(x,-\frac2x\right)\Rightarrow f\left(xf\left(-\frac2x\right)+f(x)\right)=2f(x)-2$

So $2x-1=xf\left(-\frac2x\right)+f(x)\forall x\ne0$ after using injectivity.
Now this is routine, since $-\frac2x$ is involutive. Taking $x\mapsto-\frac2x$ and multiplying by $-x$ yields:
$$4+x=2f(x)-xf\left(-\frac2x\right)$$
Adding, we get $3f(x)=3x+3\forall x\ne0$, hence the unique solution $\boxed{f(x)=x+1}$ which works. $\square$

Tricky problem.
As in previous posts let $P(x,y)$ be the assertion $f(xf(y)+f(x)) = 2f(x)+xy$.
$P(1,y): f(f(y)+f(1))=2f(1)-y$, so f is obviously bijection. Now if $f$ is bijective, we can find such $t$ that $f(t)=0$. :
$P(t,0): f(tf(0)+f(t))=2f(t)+0\Rightarrow f(tf(0))=0$. But $f(t)=0$ and f is a bijection so if $f(tf(0))=f(t)=0$ then $tf(0)=t\Rightarrow t=0$ or $f(0)=1$. It`s equivalent to two cases $f(0)=0$ and $f(0)=1$.
[b]First Case: f(0)=0[/b]. Then plugging $P(x,0): f(x*f(0)+f(x))=2f(x)+0\Rightarrow f(f(x))=2f(x)$. Now by surjectivity we can find such $t$ for every $y$ in R that $f(t)=y$. So for every $y$: $f(f(t))=f(y)=2f(t)=2y$ so [u]$f(y)=2y$[/u]. Plugging this function into the equation we get that the function doesn`t fit.
[b]Second Case: f(0)=1[/b]. Now $P(x,0): f(x+f(x))=2f(x)          \bigstar$.
Plugging $P(x+f(x),y_0)$ where $y_0=\frac {2xy} {x+f(x)}$, we get :
$f((x+f(x))f(y_0)+f(x+f(x))=f((x+f(x))f(y_0)+2f(x))=2f(x+f(x))+(x+f(x))y_0=4f(x)+(x+f(x))y_0$, so $f(x+f(x))(f(y_0)+2f(x))=4f(x)+(x+f(x))y_0=4f(x)+(x+f(x))\frac {2xy}{x+f(x)}=4f(x)+2xy=2f(xf(y)+f(x))$. But from $\bigstar$ we get that $2f(xf(y)+f(x))=f(xf(y)+f(x)+f(xf(y)+f(x)))=f(xf(y)+f(x)+2f(x)+xy)=f(3f(x)+xy+xf(y))$ so $f(x+f(x))(f(y_0)+2f(x))=f(3f(x)+xy+xf(y))$, by injectivity we get that $(x+f(x))f(y_0)+2f(x)=3f(x)+xy+xf(y)\  -     \spadesuit$. Now plugging $x=t$ where $t$ is such number that $f(t)=0$ we get that that $y_0=\frac{2ty}{t+0}=2y$, so $\spadesuit$ becomes $tf(2y)=ty+tf(y)\Rightarrow y+f(y)=f(2y)$ for every $y$. Plugging $P(\frac{x}{2},2y):\ f(\frac{x}{2}f(2y)+f(\frac{x}{2}))=2f(\frac{x}{2})+xy=2(f(x)-\frac{x}{2})+xy=2f(x)+x(y-1)=f(xf(y-1)+f(x))$,
so $f(\frac{x}{2}f(2y)+f(\frac{x}{2}))=f(xf(y-1)+f(x))$, by injectivity $\frac{x}{2}f(2y)+f(\frac{x}{2})=xf(y-1)+f(x)\Rightarrow \frac{x}{2}(f(y)+y)+f(x)-\frac{x}{2}=xf(y-1)+f(x)\Rightarrow \frac{x}{2}(f(y)+y)-\frac{x}{2}=xf(y-1)\Rightarrow {f(y)+y-1}=2f(y-1)$ Setting $y=0$ into the last equation: $f(0)+0-1=2f(-1)\Rightarrow 1-1=2f(-1)\Rightarrow f(-1)=0$. Plugging $P(x,-1): f(f(x))=2f(x)-x$. But now from $f(y)+y-1=2f(y-1)\Rightarrow f(y)=2f(y-1)-(y-1)=f(f(y-1))$, that`s why by injection we get that $y=f(y-1)$ for every $y$, so [u]$f(x)=x+1$[/u], which is correct. $\blacksquare$

Woah woah woah really hard (ok maybe not too much but yea it's hard and nice :D).
As usual let $P(x,y)$ denote the given assertion.
[hide=Bijectivity]
Fixing $x \neq 0$ gives us that $f$ is surjective. Thus there exists $a$ such that $f(a)=0$ (and let's define this specific $a$ for convenience).
$\textbf{Case: } f(0)=0$.
$P(x,0) : f(f(x))=2f(x) \implies f(t)=2t$ but this does not satisfy our original equation hence $f(0) \neq 0$.
$P(a,t) : f(af(t))=at$.
If $f(x)=f(y)$
$$\implies af(x)=af(y) \implies f(af(x))=f(af(y)) \implies ax = ay \implies x=y.$$
Thus $f$ is bijective.
[/hide]
[hide=Some required values of f]
There exists $b$ such that $f(b)=1$.
$P(a,b) : f(a)=ab =0 \implies b=0 \implies \boxed{f(0)=1}.$
$P(0,a) : f(f(0))=2 \implies \boxed{f(1)=2}.$
$P(-1,-1) : f(0) = 2f(-1)+1 \implies \boxed{f(-1)=0}.$
[/hide]
Now to finish
$P(x,-1) : f(f(x))=2f(x)-x$.
$P(1,x) : f(f(x)+2)=x+4$.
Adding both we obtain $f(t)+f(t+2)=2t + 4$.
$$\implies f(t-2) + f(t)=2t \implies f(t+2)-f(t-2)=4 \implies f(t+4)=f(t)+4.$$
$$f(f(t)+2)=t+4 \implies f(f(f(t)+2)+2)= f(t)+6 \implies f(t+6) = f(t)+6.$$
$$\implies f(t+4)+f(t+6) = 2f(t)+10 \implies f(t+4) + f(t+4 +2) =2f(t)+10$$
$$\implies 2(t+4) + 4 = 2f(t) + 10 \implies f(t)=t+1.$$
And we are done :D

Beautiful functional equation. It is not actually as difficult as previously suggested. Here is a relatively new solution which almost overuses that $f$ is a bijection. Solved (the first half) with [b]L567[/b].
[hide=Answer]
$f(x) = x+1$ for all $x \in \mathbb{R}$ is the only solution to the functional equation.[/hide]
Let $P(x,y)$ denote the assertion to the above functional equation. \\
[hide=Routine Work]
$P(1,x):$ $$f(f(x)+f(1)) = 2f(1)+x$$ meaning that $f$ is bijective. \\
Let $a \in \mathbb{R}$ such that $f(a) = 0$. \\
$P(a,0):$ gives that $af(0) = a$ meaning that $a=0$ or $f(0) = 1$. \\
$\textbf{Case 1)}$ $a=0$ \\
We get that $f(0)=0$ and from $P(x,0):$ $$f(f(x)) = 2f(x) $$ using that $f$ is bijective we have that $f(x) = 2x$ for all $x\in \mathbb{R}$ which is not a solution. \\
$\textbf{Case 2)}$ $f(0) = 1$ \\
Notice therefore that $f(1) = f(f(0)) = 2f(0) = 2$. \\
$P(a,a)$ gives that $1 = f(0) = a^2$. \\
Therefore $f(-1) = 0$ as $a \in -1,1$ and $f(1) \neq 0$. \\[/hide]
[hide=Surjection Masterclass]
Going back to $P(1,x)$ notice that $$f(f(x)+2) = x+4 $$
Also
$P(x,-1):$ $$f(f(x)) = 2f(x) -x$$
Therefore $$f(f(x)) = 2f(x) - x = 2f(x) - f(f(x)+2) +4$$
Using the fact that $f$ is surjective $f(x) \rightarrow$ gives us that $$f(x) + f(x+2) = 2x+4$$
Also $P(-1,x):$ $$f(-f(x)) = -x = 4-f(f(x)+2) $$
Once again taking $f(x) \rightarrow x$ we have that $$4 = f(-x)+f(x+2)$$
Combining this with the fact that $f(x) + f(x+2) = 2x+4$, we have that $$f(x) = f(-x) +2x$$
Then $$2x+4 = f(x)+f(x+2) = f(x) + f(-x-2)+2x+4$$
$\implies$ $$f(x)+f(-x-2) = 0$$
Now we continue with some magic to finish off the problem. \\
$P(-1,f(x)):$
$$f(-2f(x)+x) = f(-f(f(x)) = -f(x) $$
where we use that $f(f(x)) = 2f(x)-x$ by $P(x,-1)$ which gives us that $$f(-2f(x)+x) + f(x) = 0 = f(x) + f(-2-x)$$
Consequently, we have that $f(-2f(x)+x) = f(-2-x)$ and therefore $f(x) = x+1$ for all $x \in \mathbb{R}$ using the fact that $f$ is injective. $\blacksquare$[/hide]

[quote=cyshine]Find all functions $f\colon \mathbb{R}\to \mathbb{R}$ such that
\[f(xf(y)+f(x)) = 2f(x)+xy\]
for every reals $x,y$.[/quote]

Hopefully correct solution.

Let $P(x, y)$ be the assertion. $P(1, y)$ implies $f$ is bijective. If $f(0) = 0$, then $P(x, 0) \implies f(f(x)) = 2f(x)$ which by surjectivity means that $f(x) = 2x$ for all reals $x$ which does not work. Therefore, $f(0) \neq 0$. Now, $P(f^{-1}(0), f^{-1}(1)) \implies 0 = f^{-1}(0) \cdot f^{-1}(1) \implies f^{-1}(1) = 0 \implies f(0) = 1$. $P(0, 0) \implies f(f(0)) = 2f(0) \implies f(1) = 2$. If $f(x) = 0$, then $P(x, k) \implies f(xf(k)) = xk$. Since $x \neq 0$, $k = \dfrac{1}{x}$ here implies that $f(xf(k)) = 1 \implies xf(k) = 0$. This means $x = 0$ or $f(k) = 0$, but the previous is impossible which means that $f(k) = 0$. But this forces $k = x$ or $\dfrac{1}{x} = x \implies x \in \{ -1, 1\}$, but since $f(1) = 2$, we have that $f(-1) = 0$. If $f(y) = -1$, then $P(1, x) \implies f(1) = 2 = 2f(1) + y = 4 + y$, which means that $y = -2$. Now $P(x, -2) \implies  f(f(x) -x) = 2f(x) - 2x = 2(f(x)-x)$, now plugging $c_x = f(x) - x$ gives that $f(c_x) = 2c_x$ and $P(f^{-1}(c_x), 0) \implies f(f^{-1}(c_x) + c_x) = 2c_x = f(c_x) \implies f^{-1}(c_x) = 0 \implies c_x = 1$ for all reals $x$, implying that $\boxed{f(x) = x + 1}$ for all reals $x$.

Here is my solution is was really difficult for me atleast not very hard but still its hard i solved with [b][color=#f00]Fafnir[/color][/b] and we are done hope its correct. :maybe:
[hide=Proving Bijectivity]
[b][color=#f00]Claim:[/color][/b] $f$ is bijective function.
Let $P(x,y)$ is the assertion of the following functional equation\\
we have\\
Lets suppose $f(a)=f(b)$ so we have \\
$P(x,a)$
\[\implies f(xf(a)+f(x))=2f(x)+ax\]
$P(x,b)$
\[\implies f(xf(b)+f(x)=2f(x)+bx\]
since $f(xf(a)+f(x))=f(xf(b)+f(x))$ because $f(a)=f(b)$ so we can say that
\[\implies 2f(x)+ax=2f(x)+bx\implies ax=bx\implies a=b\]
so we can say that $f$ is injective function now\\
$P(1,y)$
\[\implies f(f(y)+f(1))=2f(1)+y\]
so we can see here that $f$ is surjective so hence $f$ is injective and surjective it is bijective.$\square$\\[/hide]
[hide=Finding important values]
Lets suppose for some $k$ we have $f(k)=0$ then putting\\
$P(k,0)$\[\implies f(kf(0))=0\implies f(kf(0))=f(k)\]we know that $f$ is injective so we can say that
\[\implies kf(0)=k\implies f(0)=1\]
now putting\\
$P(0,k)$\[\implies f(f(0))=2f(0)\implies f(1)=2\]
now putting\\
$P(k,k)$\[\implies f(kf(k)+f(k))=2f(k)+k^2\implies f(0)=k^2=1\]
now $k=\pm 1$ so for $k=1$ we know that $f(1)=2$ so its clear that $f(-1)=0$
[/hide]
[hide=playing with equations]
Putting $P(x,-1)$
	\[\implies f(xf(-1)+f(x))=2f(x)-x\implies f(f(x))=2f(x)-x------(1)\]
Putting $P(1,x)$
	\[f(f(x)+f(1))=2f(1)+x\implies f(f(x)+2)=4+x-------(2)\]
by using (2) we can write this as
\[\implies f(f(x))=2f(x)-f(f(x)+2)+4\]
since we know that $f$ is surjective
replacing $f(x)\rightarrow x$ we have
	\[\implies f(x)+f(x+2)=2x+4-------(3)\]
now putting $P(-1,x)$\[\implies f(-f(x))=-x=4-f(f(x)+2)\implies f(-f(x))=4-f(f(x)+2)\]
once again replacing $f(x)\rightarrow x$
	\[\implies f(-x)=4-f(x+2)\implies f(-x)+f(x+2)=4-------(4)\]
substracting eq(3)-eq(2) we have
	\[\implies f(x)=f(-x)+2x-------(5)\]
replacing $x\rightarrow x+2$ we get
\[\implies f(x+2)=f(-x-2)+2x+4\]adding $f(x)$ on both side we have
	\[f(x)+f(x+2)=f(x)+f(-x-2)+2x+4\implies f(x)+f(-x-2)=0-------(6)\]
now putting $P(-1,f(f(x)))$
\[\implies f(-f(f(x))=-f(x)\]
from eq(1) we can write this as
\[\implies f(-2f(x)+x)=-f(x)\implies f(-2f(x)+x)+f(x)=0\]
from eq(6) we can write this as
\[\implies f(-2f(x)+x)=f(-2-x)\implies -2f(x)+x=-2-x\implies f(x)=x+1\]
we get $f(x)=x+1$ $\forall x\in\mathbb{R}$ because since we know that $f$ is injective.  :D $\blacksquare$[/hide]

Really difficult FE.

The solution is very long(with a lot of substitutions), so we will split it into a couple of main parts.

[color=#f00]1. Bijective[/color]: Fixing $x$ shows that $f$ is surjective. Pick a $b$ such that $f(b)=0$. $P(x,b)$ gives that $f(f(x))=2f(x)+bx$, and $P(0,b)$ gives that $f(f(0))=2f(0)$. If $b=0,$ we have that $f(f(x))=2f(x),$ and by surjectivity, that $f(x)=2x\forall x\in \mathbb{R}$. Plugging this into the original equation, it evidently does not work, so we assume that $b\neq 0$. Then, from $f(f(x))=2f(x)+bx,$ we have that $f$ is injective. This establishes that $f$ is bijective.

[color=#f00]2. Establish that [/color]$f(0)=1$: To give some motivation, originally, I actually wanted to show that $f(0)=0,$ so I let $a$ be such that $f(a)=0$. There are two ways to go from here:

1) From $P(a,y)$ we get that $f(af(y))=ay$. Pick a $y$ such that $f(y)=1$. Then this gives that $0=ay\implies a=0$ or $y=0$(i.e. either $f(0)=0$ or $f(0)=1$). But we already established a contradiction when $f(0)=0,$ so this means $f(0)=1$.

2) $P(a,a)$ gives that $f(0)=a^2,$ so $f(f(a))=a^2.$ Follow the same steps in 1) to establish that $f(af(y))=ay$, then setting $y=f(a)$ gives that $f(a^3)=0$. But $f$ is injective, so we have that $a^3=a,$ so $a=\pm 1$. I am pretty sure you could do something with this here, but I did not use this solution path.

[color=#f00]Finishing the problem and making a TON of substitutions[/color] From here, we get from $P(-1,-1)$ that $f(-1)=0$, and then $P(x,-1)$ gives that $f(f(x))=2f(x)-x$. From this, it is easy to get by induction that $f(x)=x+1\forall x\in \mathbb{Z}$. From here, my finish was identical to @nhat, so I will not post it.

Quite a good problem  :)
We claim the only answer is $\fbox{f(x)= x+1}$
[b]Claim[/b]:-$f$ is bijective
Let $P(x,y)$ be our assertion
$P(1,y)$ $\rightarrow$ $f(f(y)+f(1))=2f(1)+y$ which immediatey gives away surjectivity.
For injectivity,
let $f(a)=f(b)$
$P(1,a)$ $\rightarrow$ $f(f(a)+f(1))=2f(1)+a$
$P(1,b)$ $\rightarrow$ $f(f(b)+f(1))=2f(1)+b$
from here $a=b$ and hence $f$ is indeed bijective.
[b]Claim[/b]:- $f(-1)=0$
Let $f(\alpha)=0$ for some $\alpha \in \mathbb{R}$
$P(\alpha,\alpha)$ $\Rightarrow$ $f(0)=\alpha^2$
$P(\alpha,o)$ $\Rightarrow$ $f(\alpha^3)=0=f(\alpha)$
As $f$ is injective therefore $\alpha^3=\alpha$ $\Rightarrow$  $\alpha=-1,0,1$
[b]Case 1:-[/b] if $\alpha=0$
We get that $f(0)=0$ and from $P(x,0):$
$$f(f(x)) = 2f(x) $$
using that $f$ is bijective we have that $f(x) = 2x$ for all $x\in \mathbb{R}$ which is not a solution.
[b]Case 2:-[/b]If $\alpha=1$
From the above equations we get that $f(0)=1$ and $f(1)=0$
$P(0,0)$ $\rightarrow$ $f(1)=2$ which is a contradiction.
So, $\alpha=-1$
From here $f(0)=1$ and $f(-1)=0$ which concludes our claim.
[b]Claim:-[/b] $f$ is linear
$P(1,x)$ $\rightarrow$ $f(f(x)+2) = x+4 $
Also,$P(x,-1)$ $\rightarrow$ $f(f(x)) = 2f(x) -x$
Adding this two we get, $f(f(x))+f(f(x)+2)=2f(x)+4$
Taking $f(x) \rightarrow x$ we get $f(x)+f(x+2)=2x+4$
$P(-1,x)$ $\rightarrow$ $f(-f(x))=-x=f(f(x)-2f(x)$
=$\Rightarrow$  $f(-x)=f(x)-2x$
Combining the equations we get,
$f(-x)+f(x+2)=4$
$x \rightarrow x+2$
$$f(-x-2)-f(x+2)=-2x+4$$
$$f(-x)+f(x+2)=4$$
$f(-x)+f(-x-2)=-2x=f(-x)+f(x)$
$\Rightarrow f(-x-2)=-f(x)$
$P(-1,f(x))$  $\rightarrow$ $f(-2f(x)+x)+f(x)=f(x)+f(-x-2)$
=$f(-2f(x)+x)=f(-x-2)$
=$\Rightarrow$ $-2f(x)+x=-x-2$
and therefore only solution is $\fbox{f(x)=x+1}.$
and we are finally done!!  ;)

Let $P(x,y)$ denote the given assertion. Clearly $f$ is bijective. If $f(x_1)=f(x_2)$ then $P(x,x_1)$ and $P(x,x_2)$ implies $x_1=x_2.$

Take $f(x_0)=0.$ If $x_0=0$ then $P(x,x_0)$ implies $f(x)=2x$ which fails. Otherwise $P(x_0,0)$ implies $f(0)=1.$ And $P(-1,-1)$ implies $f(-1)=0.$ $P(-1,x)$ implies $f(-f(x))=-x$ and $P(x,-1)$ implies $f(f(x))=2f(x)-x$; so, by iterating, $f(Z)=Z+1$ for integers.

$P(x,-2)$ implies $f(f(x)-x)=2f(x)-2x$ and finally $P(f(x)-x,-3)$ yields $f(x)=x+1$ which indeed satisfies.


find all function f, g, h : R to R such that
f(   $ {y}^{3}$    + x) + g( y +    $ {x}^{3}$    )   = h(xy).

Let $ z_1\equal{}z_1(x,y)\equal{}y^3\plus{}x,z_2\equal{}z_2(x,y)\equal{}z_1(y,x)\equal{}x^3\plus{}y$. Then $ f(z_1)\plus{}g(z_2)\equal{}h(xy)\equal{}f(z_2)\plus{}g(z_1)$.
Let $ x\equal{}\minus{}y^3, z\equal{}y\minus{}y^9$, then $ f(0)\plus{}g(z)\equal{}h(\minus{}y^4)$. Therefore $ f(z)\plus{}g(0)\equal{}h(\minus{}y^4)\equal{}g(z)\plus{}f(0)$.
Because for any z txist y, suth that $ z\equal{}y\minus{}y^9$, we get $ f(z)\equiv g(z)\plus{}const, const\equal{}a\equal{}f(0)\minus{}g(0),f(0)\plus{}g(0)\equal{}h(0).$
$ y\equal{}0\to f(x^3)\plus{}g(x)\equal{}h(0)\to g(x^3)\plus{}g(x)\equal{}2g(0)\to g(x^9)\equal{}g(x).$
If $ f(x),g(x),h(x)$ are solution, then $ f(x)\plus{}a_1,g(x)\plus{}a_2,h(x)\plus{}a_1\plus{}a_2$ is solution too.
Therefore we can consider only case, when $ f(0)\equal{}g(0)\equal{}0\to h(0)\equal{}0$.
It give $ g(x^3)\equal{}\minus{}g(x), g(x)\equiv f(x)$. We get $ g(y\minus{}y^9)\equal{}h(\minus{}y^4)$, therefore $ g(0)|_{y\equal{}0}\equal{}h(0)\equal{}g(0)_{y\equal{}1}\equal{}h(\minus{}1)$.
It give $ g(x^3\minus{}x^{\minus{}1})\plus{}g(x\minus{}x^{\minus{}3})\equiv 0$. By $ g(x^3)\equal{}\minus{}g(x)$ we get at least for continiosly $ g\equiv f\equiv h\equiv 0$.
Therefore solutions only $ f\equiv a,g\equiv b,h\equiv a\plus{}b$.


function
number theory
relatively prime
complementary counting
prime numbers
totient function

How many numbers are relatively prime to 441?

So what I did was basically use Euler's totient function.

441 Prime factorizes: $ 7^2\cdot3^2$

and then (by Euler's product)

$ 441(1\minus{}\frac{1}{7})(1\minus{}\frac{1}{3})\equal{}441(\frac{6}{7})(\frac{2}{3})\equal{}\boxed{252}$

Is there a different way to approach this problem?

Complementary Counting+PIE

As in, there are $ \frac{441}{3}\equal{}147$ numbers divisible by 3, $ \frac{441}{7}\equal{}63$ numbers divisible by 7, and $ \frac{441}{21}\equal{}21$ numbers divisible by both, so there are $ 147\plus{}63\minus{}21\equal{}189$, and that's how many numbers are not relatively prime, so subtract this from 441 and you get $ \boxed{252}$

not really faster, but a different way, and probably one way how Euler's product is derived.

[quote="alanchou"]Complementary Counting+PIE

As in, there are $ \frac {441}{3} \equal{} 147$ numbers divisible by 3, $ \frac {441}{7} \equal{} 63$ numbers divisible by 7, and $ \frac {441}{21} \equal{} 21$ numbers divisible by both, so there are $ 147 \plus{} 63 \minus{} 21 \equal{} 189$, and that's how many numbers are not relatively prime, so subtract this from 441 and you get $ \boxed{252}$

not really faster, but a different way, and probably one way how Euler's product is derived.[/quote]

It probably is, since thats is also one of the more expanded forms of it.

 :D

I was wondering if there was a quick trick because if this wasn't a Target problem it would take a while (comparatively) to get the solution.

Are you talking about all the relatively prime numbers to 441 less than 441? Cuz there's an infinite amount of relatively prime number thingys without constraints.

The problem [i]is[/i] asking you to calculate a value of Euler's totient function; that's a definition, not a technique.

The technique:  $ \varphi(7^2 \cdot 3^2) \equal{} \varphi(7^2) \cdot \varphi(3^2) \equal{} (7 \cdot 6) \cdot (3 \cdot 2) \equal{} 42 \cdot 6 \equal{} 252$.  Here we have used two lemmas that allow us to calculate $ \varphi(n)$ fairly quickly.

[b]Lemma:[/b]  If $ a, b$ are relatively prime, $ \varphi(ab) \equal{} \varphi(a) \varphi(b)$.

[b]Lemma:[/b]  If $ p$ is a prime, $ \varphi(p^n) \equal{} (p\minus{}1) p^{n\minus{}1}$.


I don't know how to use the angle relation...... :?

Hmm... maybe try to work the other way. Suppose there are $PQ = AB = CD$ and $PQ$ intersects $AL$ at $R$ and $DL$ at $S$ and show $\angle ALC = 2\angle ROS$. Does it work?

[quote="warut_suk"]Let $O$ be the center of circle $w$. Two equal chords $AB$, $CD$ of $w$ intersect at $L$ such that $AL>LB$, $DL>LC$. Let $M$,$N$ be points on $AL$, $DL$ respectively such that $\angle ALC = 2\angle MON$. Prove that the chord of $w$ passing through $M$ and $N$ is congruent to $AB$ and $CD$.[/quote]

seems there is some error in the data.
AL has M, DL has N on them

AB, CD and MN have to be congruent....I think it is not possible.

[quote="probability1.01"] It looks like there should be an easier way though.[/quote]

I don't think so...

[quote="probability1.01"]Drop perpendiculars to all the chords, then show that OM and ON are bisectors of AMN and DNM using similarity or whatever.  It looks like there should be an easier way though.[/quote]

I have attached a figure for one of the possible cases( not to scale).

It gives me an impression that there is no possibility of the proof for what is asked.

Your angle MON is way too large.  Remember it is half the angle of the angle between the chords.

I don't feel like posting again for some reason, so in response to below post... concurrent $\ne$ congruent...

[quote="probability1.01"]Your angle MON is way too large.  Remember it is half the angle of the angle between the chords.[/quote]

could be.  However, M and N will be placed somewhere as I have shown.

What I want  to emphasise is....

whether it is possible to have chord through MN, concurrent to AB and CD.


Let a, b, c, d real numbers such that:
[tex]a = \sqrt{4 + \sqrt{5 + a}}[/tex] ; [tex]b = \sqrt{4 - \sqrt{5 + b}}[/tex] ; [tex]c = \sqrt{4 + \sqrt{5 - c}}[/tex] ; [tex]d = \sqrt{4 - \sqrt{5 - d}}[/tex]
find [tex]abcd[/tex]

we note that a,b,c,d are four solutions of the equation

[tex]((x^2-4)^2-5)^2=x^2[/tex]

the other four are -a, -b, -c, -d so [tex](abcd)^2=(4^2-5)^2=121[/tex] and so abcd=11


trigonometry
integration
logarithms
function
calculus
algebra
functional equation

$ \sin x \equal{} \operatorname{Im} \left( {e^{ix} } \right),\,\cos x \equal{} \operatorname{Re} \left( {e^{ix} } \right).$

Is this a definition or does exist a proof?

$ \exp(z): \equal{} \sum_{k \equal{} 0}^\infty \frac {z^k}{k!}$ is the definition for the exponential function.

$ \cos(z): \equal{} \frac {\exp(iz) \plus{} \exp( \minus{} iz)}{2}$ , $ \sin(z): \equal{} \frac {\exp(iz) \minus{} \exp( \minus{} iz)}{2i}$ are the definitions for cosine and sine.

Particularly for a real number $ x \quad \exp( \minus{} ix) \equal{} \exp(\,\overline{ix}\,) \equal{} \overline{\exp(ix)}$

So $ \cos(x) \equal{} \frac {\exp(ix) \plus{} \overline{\exp(ix)}}{2} \equal{} \text{Re}\Big(\exp(ix)\Big)$ and $ \sin(x) \equal{} \frac {\exp(ix) \minus{} \overline{\exp(ix)}}{2i} \equal{} \text{Im}\Big(\exp(ix)\Big)$ .

For all rational numbers $ \frac {p}{q}$ you can show $ e^{\frac {p}{q}} \equal{} \exp\left(\frac {p}{q}\right)$ .

And for all other expressions $ z\notin \Bbb{Q}$ define $ e^z$ as $ \exp(z)$ .

So the equality $ e^z \equal{} \exp(z)$ holds for all $ z$ for which the expression $ \exp(z)$ is sensible.

Let $ f(\theta)=\cos\theta+i\sin\theta$.  Hence,
\[ \begin{eqnarray*}df(\theta)&=&(-sin\theta+i\cos\theta)d\theta=if(\theta)d\theta\\ \int \frac{1}{f(\theta)}df(\theta)&=&\int id\theta\\\ln f(\theta)&=&i\theta\iff f(\theta)=e^{i\theta}\end{eqnarray*},\]
as required.

My answer to the original question is "both". You can use the identities as definitions of $ \sin$ and $ \cos$, but then you'll have to explain why these are the same functions that we use to find the sidelengths in a right triangle. Or you can define all three functions independently, and then prove that they are related. I prefer the latter. In Calculus 2 we show that
$ e^x \equal{} \sum_{n \equal{} 0}^{\infty}\frac {x^n}{n!}$,  $ \sin x \equal{} \sum_{n \equal{} 0}^{\infty}( \minus{} 1)^n\frac {x^{2n \plus{} 1}}{(2n \plus{} 1)!}$, and $ \cos x \equal{} \sum_{n \equal{} 0}^{\infty}( \minus{} 1)^n\frac {x^{2n}}{(2n)!}$ for all $ x\in\mathbb R$.
The power series also converge for all $ x\in \mathbb C$. Splitting the (absolutely convergent) series for $ \exp$ into odd and even-numbered terms, we find
$ e^{ix} \equal{} \sum_{n \equal{} 0}^{\infty}\frac {i^n x^n}{n!} \equal{} \sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^k x^{2k}}{(2k)!} \plus{} i\sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^k x^{2k \plus{} 1}}{(2k \plus{} 1)!}$, etc.

As part of that last approach, I like to go through the proof that the function

$ f(z)\equal{}\sum_{n\equal{}0}^{infty}\frac{z^n}{n!}$

does satisfy the functional equation $ f(z\plus{}w)\equal{}f(z)f(w).$ This is the key algebraic property that we would expect something called an exponential function to have, which at least partly justifies us using the notation $ e^z$ for it.

And once we've shown that, then $ e^z\equal{}e^{x\plus{}iy}\equal{}e^xe^{iy}.$ In that, $ e^x$ is the real exponential that we're already familiar with, while we deal with $ e^{iy}$ as mlok has indicated.


hallo,
i think this is an easy one. solve the equation: \[x!+y!+z!=2^{v!}.\] :)

We may assume that $x\le y\le z$. Then \[x!(1+\frac{y!}{x!}+\frac{z!}{x!})=2^{v!},\] thus $x\le2$.
1)x=1. If y>1,  $1+y!+z!$ is odd and >1. So y=1 and $2+z!=2^{v!}\qquad\Rightarrow\qquad z\le3$. Finally we have
(x,y,z,v)=(1,1,2,2).
2)x=2. Then y<4 and so on. In this case we have no sols.
3 solutions altogether.

Там v!, а не vt.


[quote="mysmartmouth"]Is [url=http://talk.collegeconfidential.com/showthread.php?t=108828&page=5&pp=20]this[/url] illegal?  I certainly think it is.  According to the SAT rules they read us, it certainly looks illegal.  Should we report this to someone?

BTW, on a lighter note, did anyone take the November 5th SAT?  What are you hoping for?[/quote]

I thought the math was way too easy, verbal way too hard, and writing was allright, my predictions:

Math: 800
Verbal: 550
Writing: 680

Essay: 10/12

BTW: yeah, it is illegal, he can't discuss the question in detail due to copyright infringements (but I'm just glad that I got all the questions right by looking at it :P )

The College Confidential boards can be very useful, but they reveal that a lot of college applicants who haven't got enough brainpower to get into the top schools through their own effort will try ANYTHING--honest or dishonest--to gain an edge. I like the fact that AoPSers generally apply their own effort, honestly, to gaining entrance into college.

Math: 800
CR: 680
Writing Skills: 750

Essay: 11/12

Isn't CR, critical reading, which is the same as verbal? ;)  ;)

[quote="236factorial"]I don't see how this is illegal. It was discussed according to the times, after the test.[/quote]

Yes, but I just looked at the thread regarding the PSAT after October 12, however, a lot of people took the same PSAT on October 15, so it is VERY easy to cheat that way, which I think a lot of people have done :mad:  :mad:

[quote="bleumoose"]For those of you who never pay attention during the obligatory announcements on test day:
Yes, it is illegal. Multiple choice questions are not allowed to be "carried" from the testing room in any form- including your memory. (of course, this does not mean you are not allowed to remember the questions, just sharing them). The essay questions, however, are allowed to be discussed: AFTER YOU HAVE RECIEVED YOUR SCORES FROM COLLEGE BOARD'S WEBSITE.
So there you are. Straight from the proverbial horse's mouth.[/quote]

Yes indeed, in fact, during the SAT, you have to rewright and sign that statement that you're not going to share any info about ANY of the problems, even if no two SATs are teh same on different dates. Although I don't see why not though, I guess Collegeboard would rather have people buy the tests from their practice books.

It's because of time differences, partly anyway.

[quote="Iversonfan2005"][quote="Sly Si"]Isn't the PSAT given on the alternate day a different version?[/quote]

Apparently not the October 12 and October 15 ones...this really makes me mad if someone makes a perfect on Math by cheating :mad:  :mad:[/quote]

There were different tests.

Form W and Form S.

Blah, there will be a point in life, where they won't be able to cheat... and they'll find themselves very lonely in a dark dark hole. Feel pity, not anger.

I don't think that was cheating.  It's more like, trying to find out you SAT score as fast as you can by discussing the questions and stuff. Not saying that it's illegal, but I'm pretty sure they aren't trying to cheat.

but that could still allow other people who are taking the test later to cheat... the SAT exam recycles MC questions from test to test apparently- same for AP exams

[quote="bleumoose"]For those of you who never pay attention during the obligatory announcements on test day:
[/quote]

Well, I've [i]never[/i] taken the SAT.


function
calculus
integration
algebra unsolved
algebra

Prove that there is a function $f$ from the set of all natural numbers to itself such that for any natural number $n$, $f(f(n)) = n^{2}$.

$f(f(n))=n^2$ (from the given equation)
$f(f(f(n)))=f(n^2)$ (from my previous equation)
$f(f(f(n)))=f(n)^2$ (from the given equation)
So, $f(n^2)=f(n)^2\Rightarrow f(n)=\sqrt{f(n^2)}$

[quote="draperp"]$f(n)=\sqrt{f(n^2)}$[/quote]
Why is this useful? How do you know there's a $f: \mathbb{N}\to\mathbb{N}$ with $f(n)=\sqrt{f(n^2)}$ for every $n$?

I derived that equation from the equations shown.  The equation [u]must[/u] hold for all $n$ if such a function exists.  One use of the equation is to show that when the function is to applied to a perfect square, the result is a perfect square (otherwise $f$ would not be $\mathbb{N}\to\mathbb{N}$.
It is not surprising that you found an infinite number of functions, Hawk Tiger, since Shobber merely asked us to prove that one exists.
Let every natural number that cannot be written $k^{2^n}$ for natural $k$ and $n$ be a "M-number."
Then, M-number $a$ and M-number $b$ can be uniquely paired as follows:
$f(a^{2^n})=b^{2^n}$ and $f(b^{2^n})=a^{2^{n+1}}$ (for nonnegative integral $n$.)

These unique pairs are linked to no others, and it is very easy to see that the solution satisfies.

[quote="draperp"]I derived that equation from the equations shown.  The equation [u]must[/u] hold for all $n$ if such a function exists.  One use of the equation is to show that when the function is to applied to a perfect square, the result is a perfect square (otherwise $f$ would not be $\mathbb{N}\to\mathbb{N}$.
It is not surprising that you found an infinite number of functions, Hawk Tiger, since Shobber merely asked us to prove that one exists.  I found that for any natural $a$ and $b$,
$f(a^{2^n})=b^{2^n}$ and $f(b^{2^n})=a^{2^{n+1}}$ works.[/quote]I don't know what you mean clearly,but we can make infint pairs of integer(a,b),where a,b is not square,and every natuaral number in one and only one pair. then as you said,define the function f as $f(a^{2^n})=b^{2^n}$ and $f(b^{2^n})=a^{2^{n+1}}$for all $n>0$
this one is a fuction satisfying the condition.as the method to make the pairs is infinit,so the fuction is infinit.in fact ,all the fuction can be gained  using this method.
bye

Yes, you are right.  I should have said non-squares; in fact, what I said was wrong since (I don't know what I was thinking).  Oh, and I just made-up "M-number."  I guess I could have just defined a set or something like that...but it's wrong anyway.
I did make another mistake: $f(1)=1$ is defined by necessity.
(This is one application of $f(n)=\sqrt{f(n^2)}$.)
This seems quite unusual to use "infinite pairing," but I am certain that it is mathematically valid.


The sum of the first $n$ positive even integers is 90.  What is the sum of the first $n$ positive odd integers?

[hide]the formula to find the sum of the nth even integer if
n 2 +n
soooo....  n 2 +n=90
so you can see with out using quadratics that n=9
the formula to find the sum of the nth odd integer is n 2
so put 9 for n and 9 2 =81
thats the answer[/hide]


probability
search
geometry
3D geometry
AMC
AMC 10
algebra

While I was searching for a certain probability topic, I came across a text-based game thread here in G&FF. Unfortunately, it hasn't been posted on since 2005. So now I'm starting a new one. It follows pretty much the same rules as the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=105878715&t=177263]original[/url].

Welcome to AoPS Adventure II!

You are standing in the doorway of a tall stone building. It looks grand, like a palace.
>explore building
>exit
>check stats

>check stats

(What else would you do?)

Energy: 100
Agility: Level 1
Speed: Level 1
Strength: Level 1
Brains: Level 1
Math(obviously): Level 1

>exit stats

You are standing in the doorway of a tall stone building. It looks grand, like a palace.
>explore building
>exit
>check stats

>exit

You are outside in a random field with flowers.

>pick the flowers and frolick happily
>lie down and look at the clouds
>go back to the building

>pick the flowers and frolic happily

you pick a carnivorous flower, it bites your finger.

> scream nonstop until something happens
> tear it (and your finger) off
>pry off and give to worst enemy

@sailingdog: You do not have enemies yet.

> scream nonstop until something happens

A gardener runs up and pours maple syrup on your finger, at which the flower lets go.

Gardener: "You crazy kid! You aren't supposed to pick the flowers!"

*you are in a conversation with the Gardener*

> "Sorry, I didn't know."
> Cry
> "Get lost, sir."
> "Sorry, I'll bury it."
> run away(exit conversation)

>scream nonstop until something happens.

a pretty girl comes up to you

what do you do/say?

>"OMFG THIS FLOWER BIT MY FINGER!!!!"
>"uh.. *flustered* hi.. umm... don't worry about my finger. it's.. umm... fine.. hehe.."
>don't talk to her at all

already posted something, but I'll go along with yours.

>"OMFG THIS FLOWER BIT MY FINGER!!!!"

Girl: Sing it a song.
You: WHAT?!
Girl: Just do it.

You sing "The Star Spangled Banner", and the flower lets go.

*You are in conversation with a pretty girl*

> "Thanks."
> "Get lost."
> "My name is User0001. What is yours?"
> Exit conversation

>Thanks.
you continue on you way. when you come to a small village and meet an elderly gentleman who invites you into his home, you

> punch him in the stomach and continue walking into the village
> say hello and go in for a cup of tea
> ignore him and walk past

> say hello and go in for a cup of tea.

You chat up a typhoon with the gentleman. Level up brains.

Energy: 100
Agility: Level 1
Speed: Level 1
Strength: Level 1
Brains: Level 2
Math(obviously): Level 1

You learn about a horrid tax collector that is practically stealing from people. He comes in right after the gentleman finishes talking.

Tax collector: Taxes.
Gentleman: Can't they wait?
Tax collector: No, they can't.
Gentleman: Well they will have to.
Tax collector: Do you want to be locked up?
Gentleman: Just give me two more days, and I will come to your place and give you your taxes.
Tax collector: No, NOW.
Gentleman: I don't have it.
Tax collector: Then the guards will be here in a minute.

The tax collector leaves. You say "I can fight them.", but the gentleman counters: "You haven't seen these guys. They are on steroids. You should practice on people in the bar.

The gentleman leaves the house, and so do you.

The gentleman walks down the street into an alley.

> follow him
> walk away
> go to bar

>follow him

Man asks you why you are following him. You say

>I have absolutely no clue. (exit conversation)
>I thought you would lead me somewhere safe from the tax collector and his guards.
>Isn't this the way to the bar?

>I thought you would lead me somewhere safe from the tax collector and his guards.

Gentleman: You are clever for one so young. All right, you can follow me. If you ask around, you might find a way to the bar.

The gentleman walks into a building.

You are standing in an alley. An old woman is washing her cats in an unorthodox way: scrubbing them on the side of a barrel.

>follow gentleman
>walk away
>ask old woman about way to bar.

>walk away

You stop following the gentleman and go back the way you came. Suddenly, a small rabbit runs past you and jumps into what appears to be a solid wall. There is a man sitting next to the spot where the rabbit jumped through.

>examine wall
>walk away
>talk to man

> examine wall

upon close examination, you find you are able to walk through the wall. you do. when you get to the other side, you find you are in bunnyland. a bunny walks up to you and asks if you would like to become a bunny.

> become bunny
> shoot him
> say no thank you and continue to walk into bunnyland
> say no and leave through the wall

> become bunny.

Bunnies are fast and agile, but they lack brains. +1 Speed, +1 Agility, -1 Brains.
Energy: 100
Agility: Level 2
Speed: Level 2
Strength: Level 1
Brains: Level 1
Math(obviously): Level 1

You meet a mother bunny who asks you to find her lost child.
> help bunny
> walk away
> randomly scream at the top of your lungs

>Help bunny

There are woods, a river, and mountains in Bunnyland.

>check forest
>check near river
>check [i]in[/i] river
>check mountains
>give up on helping

>eat a carrot.

You turn back into a bunny.  While searching around you, you find a bunny Wii.

You:
>play the bunny Wii
>wish you were a human so you could play a regular Wii
>continue looking around you

>continue looking around.

You're blind.

>play the bunny Wii
>wish you were a human so you can play a real Wii
>wish erabel a happy birthday.

>wish erabel a happy birthday

You can see!

>play the bunny Wii
>wish you were a human so you can play a real Wii
>continue looking around

>continue looking around

You find Erabel.
You

>wish him a happy birthday
>eat his cake
>invite him to play wii with you
>eat a carrot
play wii and ignore him

>wish him a happy birthday

You're blind again. Oh well.

>wish him another happy birthday
>wish him two more happy birthdays
>scream "I'M BLIND!!!!!!!!!!"
>learn Braille
>eat thirty carrots. They're good for your eyes

>eat thirty carrots

Energy: 120
Agility: Level 32
Speed: Level 33
Strength: Level 37
Brains: Level 32
Math(obviously): Level 36

You:
>throw up the thirty carrots
>run around randomly
>scream and run into a wall

>scream and run into a wall.

The wall falls on top of you.
You

>keep screaming
>eat more carrots
>sing happy birthday at the top of your lungs
>try to push the wall off of yourself

>sing happy birthday at the top of your lungs.

You suffocate.

>continue
>restart
>quit
>switch game carts

>continue

You almost suffocate.

>apply the Heimlich maneuver to nothing
>scream
>clear your throat loudly
>scream more
>eat another carrot
>take the sixth vitamin

>take the sixth vitamin

Suddenly, you hear a great booming voice.
[b][color=darkblue][size=150]YOU HAVE OFFICIALLY JOINED THE LEAGUE OF THIEVES. YOU MAY NOW STEAL ANYTHING IN THE WORLD. THIS WILL, HOWEVER, DRAIN YOUR ENERGY.[/size][/color][/b]

You

>shout "who is this" at the top of your lungs
>steal the voice
>steal a mathbook
>steal a PSP
>steal a crown
>steal a carrot
>sleep

>steal the voice

You do. -10 energy
Energy: 110
Agility: Level 32
Speed: Level 33
Strength: Level 37
Brains: Level 32
Math(obviously): Level 36

You say in the voice "who's there?" and the same voice replies, "I am!"

You:
>freak out
>say, "Who is I" in the fancy voice
>say, "Who is I" in your normal voice
>sleep
>steal something random
>commit suicide

>commit suicide...

game over..

>restart?
>play your wii
>quit
> :rotfl:
>restart
>die

>quit.

Lock this forum, please. It's getting kinda stupid.

[quote="erabel"]Lock this forum, please. It's getting kinda stupid.[/quote]

Whoa, the whole forum?  That's a bit harsh, isn't it?

[quote="lingomaniac88"]Whoa, the whole forum?  That's a bit harsh, isn't it?[/quote]

Haha yeah lemme just delete every single topic in history from G&FF  :roll: .


advanced fields
advanced fields theorems

Anyone in possess of a document to learn the basics of coding theory? It seems like a very interesting branch to me I may consider taking next year :)

Try looking [url=http://dmoz.org/Science/Math/Applications/Communication_Theory/Coding_Theory/]at the ODP[/url].  You should also look at [url=http://www.dpmms.cam.ac.uk/~twk/]Tom Koerner's home page[/url] for his notes on Coding and Cryptography.


A) Prove there are infinitely many triples $ (x,y,z)$ of positive numbers such that:

       $ (3\minus{}\frac{1}{z}\minus{}y)((3\minus{}\frac{1}{x}\minus{}z)(3\minus{}\frac{1}{y}\minus{}x)\leq(\frac{1}{x}\plus{}z\minus{}1)(\frac{1}{y}\plus{}x\minus{}1)(\frac{1}{z}\plus{}y\minus{}1)$

B) find the the general form of all (/most, if a I'm wrong) the triples

 :)

[hide="A"]Consider $ x \equal{} y \equal{} z$:

$ (3 \minus{} x \minus{} x^{ \minus{} 1})^3 \leq (x \plus{} x^{ \minus{} 1} \minus{} 1)^3 \iff 3 \minus{} x \minus{} x^{ \minus{} 1} \leq x \plus{} x^{ \minus{} 1} \minus{} 1 \iff x \plus{} x^{ \minus{} 1} \geq 2$, which is true for all $ x \in \mathbb{R}^ \plus{}$ by AM-GM.

Since this subset of the solution set is infinitely large, the solution set itself must also be infinitely large.

This also implies infinitely many integer solutions.[/hide]

[hide="Stuff about B"]
There is no "general form," there are way too many solutions.

We have that $ \minus{} (x \plus{} 1/y \minus{} 3)(y \plus{} 1/z \minus{} 3)(z \plus{} 1/x \minus{} 3) \leq (x \plus{} 1/y \minus{} 1)(y \plus{} 1/z \minus{} 1)(z \plus{} 1/x \minus{} 1)$

First of all, if all 6 terms are positive, then we win. That is, if $ x \plus{} 1/y \geq 3$, etc. This is easily satisfied if all three variables are $ \geq \frac {3 \plus{} \sqrt {5}}{2}$ or between $ 0$ and $ \geq \frac {3 \minus{} \sqrt {5}}{2}$, but there are other solutions as well.

We also auto-win if two of the terms on the right side and left side are negative. Note that these must be corresponding terms.
Wlog, we require that $ x \plus{} 1/y > 3$, but $ y \plus{} 1/z < 1$ and $ z \plus{} 1/x < 1$. However, this requires that $ z < 1$ and $ 1/z < 1$, which violates that $ z$ is positive, so this is not possible.

Then there is the case in which the three terms on the left are negative, and none or two of the terms on the right are negative, and the right side still is greater than the left side.[/hide]

mmm ok I understand it very well...

this is the storie:

what I derived was this:
For
          $ x\equal{}\frac{p}{q}$, $ y\equal{}\frac{q}{r}$ and $ z\equal{}\frac{r}{p}$

where $ p$, $ q$ and $ r$ are sides of a triangle, we have that the inequality holds.

What I wanted to know is if this are the only triples or there are more, but considering what you sayed about $ x\equal{}y\equal{}z$, I'm considering that the triples are  $ (ax, ay, az)$ with $ a$ positive and $ x, y, z$ as I say... :maybe:

Can someone find a counterexample or prove it, I will be also trying but help is always welcome.

 :)


On the website it says that the early registration deadline is February 14th, but if you print out the Application test it says Februrary 9th. Which one is it?

TYiA.

The 14th; it was originally the 9th but the date was extended.

Yes
And if you see its in [color=red]Red[/color] in AwesomeMath Site, and its written in this forum also. You should be happy   :roll:


geometry
incenter
ratio
Ross Mathematics Program
angle bisector
projective geometry
geometry proposed

Let $ (I) $ be the incircle of the triangle $ ABC $ . $ (I) $ tangents $ BC $ at $ D $ . $ H $ is the feet of the perpendicular from A to BC . The excenter of the triangle $ ABC $ tangents BC at $ P $ . $ AI $ intersects $ BC $ at $ K $ . Prove that :
 1) $ (HKDP) = -1 $
 2) $ IP $ bisects $ AH $

The two are actually equivalent. You can prove the second one by Menelaus, for example (the computations are easy to carry on), and then the first one follows.

The problem is really easy. I slightly rewrite it:

[color=blue][b]Problem.[/b] Let ABC be a triangle, and H the foot of its altitude to the side BC. The incircle of triangle ABC touches its side BC at a point D, and the A-excircle touches the side BC at a point P. Finally, the angle bisector of the angle CAB intersects the side BC of triangle ABC at a point K.

[b](1)[/b] Prove that the points D and P divide the segment HK harmonically.
[b](2)[/b] Prove that the line IP bisects the segment AH.[/color]

[i]Solution.[/i] [b](1)[/b] Let I be the center of the incircle of triangle ABC, and let $I_a$ be the center of the A-excircle. Then, since the points D and P are the points where these circles touch the side BC, we have $ID\perp BC$ and $I_aP\perp BC$. Also, since the line AH is the altitude of triangle ABC to the side BC, we have $AH\perp BC$. Thus, $ID\parallel I_aP\parallel AH$.

The line AK, being the angle bisector of the angle CAB, passes through the incenter I and the A-excenter $I_a$ of triangle ABC. The incenter I of triangle ABC also lies on the angle bisector of the angle ABC, which is, of course, simultaneously the angle bisector of the angle ABK; thus, since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we obtain, by applying this to the triangle ABK, that $\frac{AI}{IK}=\frac{AB}{BK}$. Similarly, the A-excenter $I_a$ of triangle ABC lies on the external angle bisector of the angle ABC, which is, of course, simultaneously the external angle bisector of the angle ABK; thus, since the external angle bisector of an angle of a triangle divides the opposite side externally in the ratio of the adjacent sides, we obtain, by applying this to the triangle ABK, that $\frac{AI_a}{I_aK}=-\frac{AB}{BK}$. Hence, $\frac{AI_a}{I_aK}=-\frac{AI}{IK}$. But since $ID\parallel I_aP\parallel AH$, Thales yields $\frac{AI_a}{I_aK}=\frac{HP}{PK}$ and $\frac{AI}{IK}=\frac{HD}{DK}$, so that this becomes $\frac{HP}{PK}=-\frac{HD}{DK}$. Thus, the points D and P divide the segment HK harmonically. Part [b](1)[/b] of the problem is solved.

[b](2)[/b] Since ID || AH, the lines ID and AH intersect at an infinite point. Denote this infinite point by $H_{\infty}$. Also, let $H_1$ be the point of intersection of the lines IP and AH. Then, since the points H, K, D and P lie on one line, and the points H, A, $H_{\infty}$ and $H_1$ are the projections of these points onto the line AH from the point I, by the invariance of the cross-ratio under central projection we see that $\frac{HH_1}{H_1A}:\frac{HH_{\infty}}{H_{\infty}A}=\frac{HP}{PK}:\frac{HD}{DK}$. But since $\frac{HP}{PK}=-\frac{HD}{DK}$, we have $\frac{HP}{PK}:\frac{HD}{DK}=-1$, and thus $\frac{HH_1}{H_1A}:\frac{HH_{\infty}}{H_{\infty}A}=-1$, so that $\frac{HH_1}{H_1A}=-\frac{HH_{\infty}}{H_{\infty}A}$. In other words, the point $H_1$ is the harmonic conjugate of the point $H_{\infty}$ with respect to the segment AH. But the point $H_{\infty}$ is an infinite point, and the harmonic conjugate of an infinite point with respect to a segment is the midpoint of the segment. Thus, the point $H_1$ is the midpoint of the segment AH. In other words, the line IP bisects the segment AH, and this solves part [b](2)[/b] of the problem.

By the way, an alternative solution of part [b](2)[/b] of the problem, without using cross-ratios, can be found using the results from http://www.mathlinks.ro/Forum/viewtopic.php?t=5811 . I am leaving the details to the reader, who can also look up the alternative solution in Ross Honsberger, [i]Episodes in Nineteenth and Twentieth Century Euclidean Geometry[/i], MAA 1995, chapter 3, §3.

  Darij


[quote="tony123"]How to draw  triangle using compass and straightedge

the parameter=12 and the measure of 40,60,80[/quote]

If you can construct such a triangle then you will be famous. :rotfl:

If you can construct an angle of $ 40^{\circ}$ then you will be able to costruct a 9-gon using only compass and straightedge, which was proved to be impossible.

[url]http://en.wikipedia.org/wiki/Constructible_polygon[/url]


let $f(x) = arctan\[(\sqrt{{x^2}+1}-x)$ for all réals
Prove that   $f(-x)=\frac{\pi}{2}-f(x)$    for all réals
$\mathrm{Fermat -Euler}$

[color=red]the correction of bruno's solution [/color]
 Let $Y = \frac{\pi}{2} - f(x)$ we have...
$\tan Y = \tan\left[ \frac{\pi}{2} - f(x) \right]$
$\tan Y = \cot\left[ f(x) \right]$
$\tan Y = \cot\left[\arctan(\sqrt{x^2 + 1} - x)\right]$
$\tan Y = \frac{1}{\sqrt{x^2 + 1} - x}$
$\tan Y = \frac{\sqrt{x^2 + 1} + x}{(\sqrt{x^2 + 1})^2 - x^2}$
$\tan Y = \sqrt{x^2 + 1} + x$
$Y = \arctan(\sqrt{x^2 + 1} +x)$
$Y = \arctan(\sqrt{(-x)^2 + 1} - (-x))$
$Y = f(-x)$
$\frac{\pi}{2} - f(x) = f(-x)$

By the way, $\LaTeX$ doesn't work in titles. Sorry.