[quote="Interval"]A letter is chosen at random from a given word.  Find the probability that the letter is a vowel if the word is:

a.  ALGEBRA

b.  PROBABILITY[/quote]

[hide="a"]Theres only 3 vowels in ALGEBRA, and 7 total words. $\frac{3}{7}$.[/hide]
[hide="b"]There is 11 letters and 4 vowels. So its just $\frac{4}{11}$.[/hide]

[quote="ckck"]
[hide="b"]There is 11 total words and 4 vowels. So its just $\frac{4}{11}$.[/hide][/quote]

I think you meant 11 total letters  :lol:

Oops. I always get them mixed up. Thanks. Edited.

[quote="Interval"]A letter is chosen at random from a given word.  Find the probability that the letter is a vowel if the word is:

a.  ALGEBRA

b.  PROBABILITY[/quote]
[hide]
a. $\frac{3}{7}$ (3 vowels of 7 letters)
b. $\frac{4}{11}$ (4 vowels of 11 letters)
[/hide]

[hide="b"]
If you count "y" as a vowel, then you would get $\frac{5}{11}$ (11 letters, 5 vowels)[/hide]

[quote="tjhance"]it depends on whether or not u count "y" as a vowel. If it is a vowel, it would be $\frac{5}{11}$. Otherwise, it would be $\frac{4}{11}$[/quote]
If this question was on a test they would probably say, "...vowels (a, e, i, o, u) ..." to specify the letters that are vowels.


geometry
3D geometry
tetrahedron
octahedron

At the Word Store, each vowel sells for a different price, but all consonants are free. The word "triangle" sells for $6, "square" sells for $9, "pentagon" sells for $7, "cube" sells for $7 and "tetrahedron" sells for $8. What is the dollar cost of the word "octahedron"?

We have \[ i+a+e=6\] \[ u+a+e=9\] \[ e+a+o=7\] \[ u+e=7\] \[ e+a+e+o=8\]

Subtracting the fourth equation from the second, we get $ a=2$.  Subtracting the third equation from the fifth, we get $ e=1$.  Plugging these into the third equation, we get $ 1+2+o=7$, so $ o=4$.  The dollar cost for "octahedron" is then
\begin{align*}
o+a+e+o &= 4+2+1+4 \\
&= \boxed{11} \\
\end{align*}


function
calculus
derivative
real analysis
real analysis unsolved

i) Does there excists a function which is continuous at $0$ and discontinuous at $\mathbb{R} \setminus \{0\}$?
ii) Does there excists a function which is continuous at $0$ and discontinuous at $\mathbb{R} \setminus \{0\}$ and is twice differentiable  at 0?

For question ii)
If not... can you write up a proof for that.

ii/ is a bit a nonsense because if you want to define the second derivative at $0$, then the function must have a derivative in a small neightbourhood around $0$.

For i) the function $\ f(x)=\{\begin{array}{cc} x & x\in\mathbb{Q} \\ 0 & x\in\mathbb{R}\setminus\mathbb{Q} \end{array}$ satisfies the conditions.


I'm writing a speech about the way math is taught badly in America (which tends to be an approach focusing memorization-type learning over problem-solving), and I'm looking for any resources that might discuss this situation, explain it in detail, lend statistics or other info, etc. I'd like to get a very clear picture about what this problem is all about and get sources to back up what I say.

So if you know anything good, please post it :)

PS: I already know that RR's article on Problem Solving is probably a good place to start.

Also: sorry if this isn't the right forum, but it seems to be about "problem solving" in general.

Maybe the geometry rant on [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86087[/url]
will help.

I for one believe that if a student can prove a result not included in the syllabus they follow and make an application of the result they should be given full credit for it. Given of course that they prove it then and there.

Also in the match of rigour against intuition in proving results perhaps the basis we have on rigour is too high. Not that intuition is always reliable.


Find all c>0 s.t there is a non-constant function f:R->R satisfying the condition
[tex]f(x)=f(x^2+c)[/tex] for all [tex]x\in R[/tex]

Euhm... Take $f$ constant and any value of $c$ works.

Are you sure that is the correct question?

sorry,f is non-constant function.I have edited my post

Unless you have some other condition on $f$ (like continuity), all values of $c$ work; we can partition the nonnegative reals into infinite chains $x_n$ (possibly double-ended) with $x_n^2+c=x_{n+1}$ as well as possibly fixed points on this map. For negative $x$, $f$ is even.

If you require continuity, the condition is $c>\frac14$, so none of these chains has a limit point. In that case, any continuous function on $[0,c]$ with $f(0)=f(c)$ can be extended. If $c\le\frac14$, all of the chains have at least one limit point at a fixed point of the key transformation.


analytic geometry
algebra proposed
algebra

asume that P is set of the point in Rn with rational coordinate .for any A and B in P we can move from A to B if AB=1prove that we can move from any point to any point of P iff n \geq 5

I already solved this problem. For a solution and more details about the situation see :
http://www.mathlinks.ro/viewtopic.php?t=2635&highlight=rational+lagrange

Pierre.


[quote="Inequalities Master"]Let $ a,b,c\geq0$ s.t. $ abc \equal{} 1$. Prove that:
$ \dfrac{1}{(1 \plus{} a)^3} \plus{} \dfrac{1}{(1 \plus{} b)^3} \plus{} \dfrac{1}{(c \plus{} 1)^3} \plus{} \dfrac{5}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}\geq1$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77407


trannguyenquyND_VN and macduchuykhtn:you can post your solution?
It very nice :)  :blush:  :D

[quote="quykhtn-qa1"]Solve the system of equations:
$ y(x^2 \minus{} 6x \plus{} 13) \equal{} 20$
$ z(y^2 \minus{} 6y \plus{} 13) \equal{} 20$
$ x(z^2 \minus{} 6z \plus{} 13) \equal{} 20$
Good luck. :)[/quote]
We have :
 $ y(x^2 \minus{} 6x \plus{} 13) \equal{} y(x^{2} \minus{}6x \plus{}9 \plus{}4)\equal{}y(x\minus{}3)^{2} \plus{}4y \Rightarrow y(x\minus{}3)^{2}\equal{}20 \minus{}4y \Rightarrow y(x\minus{}3)^{2}\equal{}4(5\minus{}y)$
So we have the cases :$ (x\minus{}3)^{2}\equal{}4 \Longrightarrow x\minus{}3\equal{}\pm 2$ or $ y\equal{}4$ ...

 In analoge way can be solved other equations , this is just idea. But finally when you find $ x$,$ y$,$ z$ from equations , should find the common solutions that holds for three equations.But,however this is just an idea of solving this problem.

[b]ee"[/b]

[quote="enndb0x"][quote="quykhtn-qa1"]Solve the system of equations:
$ y(x^2 \minus{} 6x \plus{} 13) \equal{} 20$
$ z(y^2 \minus{} 6y \plus{} 13) \equal{} 20$
$ x(z^2 \minus{} 6z \plus{} 13) \equal{} 20$
Good luck. :)[/quote]
We have :
 $ y(x^2 \minus{} 6x \plus{} 13) \equal{} y(x^{2} \minus{} 6x \plus{} 9 \plus{} 4) \equal{} y(x \minus{} 3)^{2} \plus{} 4y \Rightarrow y(x \minus{} 3)^{2} \equal{} 20 \minus{} 4y \Rightarrow y(x \minus{} 3)^{2} \equal{} 4(5 \minus{} y)$
So we have the cases :$ (x \minus{} 3)^{2} \equal{} 4 \Longrightarrow x \minus{} 3 \equal{} \pm 2$ or $ y \equal{} 4$ ...

 In analoge way can be solved other equations , this is just idea. But finally when you find $ x$,$ y$,$ z$ from equations , should find the common solutions that holds for three equations.But,however this is just an idea of solving this problem.

[b]ee"[/b][/quote]

That's an integer solution, try for real numbers. And it's incorrect, as 4 isn't a prime, we can have y=2 (ok, there aren't any solutions, but you forgot the case)


function
calculus
derivative
algebra
linear equation
real analysis
real analysis unsolved

Hello,

Our term project involves alot of work with a bessel function.  I've finished the rest of the project (involving expressing the solution to a differential equation, the bessel function, as a series), but I'm still stuck on the first part.

The problem states: The bessel function is a solution to the differential equation xy"+y'+xy=0 with initial condition y(0)=1.  Implicitly differentiate the differential equation to find y'(0) and y"(0).

I've tried a few approaches to this, but i only get higher derivatives of y that i can't seem to shake.

Thanks for your help!

First, $ y'(0)\equal{}0$ straight from the equation. Next, differentiate once and evaluate at zero- you get a linear equation relating $ y$ and $ y''$. It's essentially the same as what you did to get the power series.

(Note- there are many Bessel functions. This one is $ J_0$. The equation is usually written $ x^2y''\plus{}xy'\plus{}(x^2\minus{}n^2)y\equal{}0$ for nonnegative integers $ n$, and $ J_n$ is the solution that doesn't blow up at zero, with some normalization.)

[quote="moinsegal"]
The problem states: The bessel function is a solution to the differential equation xy"+y'+xy=0 with initial condition y(0)=1.  Implicitly differentiate the differential equation to find y'(0) and y"(0).
[/quote]

$ y''(0)\equal{}\minus{}\frac{1}{2}$.


I'm thinking about taking one of the Intermediate courses and I just wanted to know how many students are in each class.

Sorry if this question was asked and answered before, but I did a search through the forum and didn't find anything :oops:

The MATHCOUNTS problem series this fall was over 70 kids.

[quote="ffdbzathf"]it varies by class.  the largest populated classes are the problem series, i think.[/quote]

I'll have to agree with first sentence but in some angle, I disagree on the second sentence. AIME Problem Series I heard is very popular. On the other hand, I don't think my AMC Problem Series class was "that" big. I remember that my Intermediate Algebra class was much bigger and that Mr.Crawford once mentioned that it was "biggest" class that he ever taught for that subject (I think I remember right).

But the thing is, it doesn't matter how big the class is. That's what I like about AoPS class even though it's pretty expensive. They'll teach you in same quality no matter how many people are in there.  :)

[quote="Silverfalcon"]That's what I like about AoPS class even though it's pretty expensive. [/quote]

AoPS classes are not expensive at all. AoPS classes are almost 1/4th the price of other online schools such as EPGY.

Also, AoPS teaches purely problem solving, classes that you wouldn't find elsewhere. Whereas EPGY teaches pretty standard curriculum, such as AP classes.

AoPS is not expensive at all.  It comes out to like 5 dollars an hour for most classes...extremely affordable.  Anyway, class size really doesnt matter all that much because it's in no way like regular school classes.  There are helpers in the room, and if you ever have a question you can always ask it and have it answered.

Olympiad Classes = about 20 kids (15 of which show up to class..  :blush: )
Intermediate Classes = about 40 kids (of which like 25 show up to class)
Intro classes = 40 kids (of which like 30 show up to class)

AIME Problem Series is pretty popular, and I think it will continue to grow.
As will Mathcounts.

In my experience at least.


Go here for the course "Programming Problems Course" @ Berkeley as preparation for IOI: http://www.cs.berkeley.edu/~hilfingr/csx98/index.html


[quote="can_hang2007"]Let $ a,b,c,d$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\plus{}d\equal{}4$. Prove that
$ a^3b\plus{}b^3c\plus{}c^3d\plus{}d^3a\plus{}23abcd \le 27.$
Equality holds if and only if $ a\equal{}b\equal{}c\equal{}d\equal{}1$ or $ a\equal{}3,b\equal{}1,c\equal{}d\equal{}0$ and its cyclic permutations. :)[/quote]
Let $a=\min\{a,b,c,d\}$, $b=a+x,$ $c=a+y$ and $d=a+z,$ where $x,$ $y$ and $z$ are non-negatives.
Hence, our inequality is equivalent to $A(x,y,z)a^2+B(x,y,z)a+C(x,y,z)\geq0$,
where  $A(x,y,z)\geq0$, $B(x,y,z)\geq0$ are obviously true and $C(x,y,z)=27(x+y+z)^4-256(x^3y+y^3z)$.
Let $f(z)=C(x,y,z).$
Hence, $f'(z)=108(x+y+z)^3-256y^3,$ which gives $z_{min}=\frac{y-3x}{3}.$
If $y\leq3x$ so $f(z)\geq f(0)=27(x+y)^4-256x^3y=27\left(3\cdot\frac{x}{3}+y\right)^4-256x^3y\geq$
$\geq27\left(4\sqrt[4]{\left(\frac{x}{3}\right)^3y}\right)^4-256x^3y=0.$
If $y\geq3x$ so $f(z)\geq f\left(\frac{y-3x}{3}\right)=256xy(y^2-x^2)\geq0.$ Done!


function
IMO Shortlist
calculus
algebra
functional equation

Find all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying the equation \[
	f(x^2+y^2+2f(xy)) = (f(x+y))^2.
\] for all $x,y \in \mathbb{R}$.

Please guys, you should learn how to $LATEX$   ;)
Let's $LATEX$ it so it can become clearer :  :lol:
Find all the functions $f: \mathbb R \rightarrow \mathbb R$ satisfying :
$f(x^2+y^2+2f(xy))=(f(x+y))^2$ for all real numbers  $x,y$

[quote="erdos"]Take $x=y=0$ and put $k=f(0)$ we get $f(2k)=k^2$
take $y=0$ we get $f(x^2+2k)=(f(x))^2=(f(-x))^2$ for all $x \in \mathbb R$
So for $x=k^2$ we get $f((-k)^2)=f(k^2)$ or $f((-k)^2)=-f(k^2)$
Suppose $f((-k)^2)=f(k^2)$, then take $x=y=k$ we get $f(2k^2+2f(k^2))=k^4$
now take $x=-y=k$ we get $f(2k^2+2f((-k)^2))=k^2$
so $k^4=k^2$ which means that $k \in \{0,-1,1\}$
we can check that $k=-1$ and $k=1$ are not possible cases.
So $f(0)=0$
Take $y=0$ in the original functional equation we get $f(x^2)=(f(x))^2$ for all $x \in \mathbb R$
and this is has been solved before, we find that $f(x)=x$ for all $x \in \mathbb R$
Have I made a mistake somewhere ? :?[/quote]

for one thing, you're writing $ f((-k)^2) = f(k^2) $ or $ f((-k)^2) = -f(k^2) $ where you really mean $ f(-k^2) = f(k^2) $ or $ f(-k^2) = -f(k^2) $.  and you're missing the case where $ f(-k^2) = -f(k^2) $

Oh yeeh , I have totally forgotten about it  :blush:
So my solution posted above is not totally correct. This sounds a hard problem. Does anybody have an idea about it ?

there's also the fact that $ f $ can be identically $ 0 $ or identically $ 1 $.  heh, this turned out to be a harder problem than i thought... (or maybe it's actually very easy and i'm just blind?)

f(x^2)=f(x)^2
It was claimed above that this equation can be solved here!
I don't think so. Not unless some sort of assumption on continuty is given.
[/quote]

I think that even the Most experienced people on this forum haven't found a solution yet. Please can anyone help ?? :?

Could you please let the people think? I really do not think it is the end of the world if we do not solve it in 4 days.

OK. I am really sorry.  :blush:  :blush:  :blush:

Thank you harazi, I spend 3 hours to find this solution.
 :)

Put f(0)=a

we have f(x 2 +y 2 +2f(xy))= [f(x+y)] 2

Put now X=x+y and Y=xy then the equation becomes :

         f(X 2 -2Y+2f(Y))=f(X) 2

Put X=0 Then   f(2f(Y)-2Y)=f(0) 2 =a 2

Put now g(x)= 2f(x)-2x   then f(x)= 1/2g(x)+x

Composing we have :  f(g(x))=1/2g(g(x))+g(x)=a 2 (*)

Linear equation with constant coefficient:

So g(x) must be in the forme  : g(x)=Ax+B

Reporting this to (*) and computing we mill have:

   A(A+2)=0 and B(A+2)=2a 2

1°case:   A=0  then B=a 2  and f(x)=x+1/2a 2

or [f(x)] 2 =[f(-x)] 2 so a=0

And f(x)=x.

2°case: A=-2  then a=0  and g(x)=-2x+B then f(x)=B/2 or f(0)=a=0

Then f(x) =0.

Correct me if not ;)

you're also missing the possibility $ f(x) \equiv 1$.
and there are a couple of other points in your solution i'm unclear about...

First of all X=x+y and Y=xy implies that X^2>=4Y.
So if you let X=0 then Y has to be negative. Furthermore
How can you possibly say that g(g(x)) + g(x) is linear and thus....
This again brings me to a point in one of the last posts:
f(x^2)=f(x)^2 was said to imply f(x)=x^2, where as it is completely wrong. One can construct many funmctions satisfying that equation.
The solution needs to be more detalied...

Just to be explicit:
take the equation g(g(x)) + 2g(x) =2a^2 that you derived.
x has to be negative.(I explained this above)
So you can't really use the above equation, Because you don't know how g(x) behaves when x>0.

for Ali

IF WE HAVE  g(g(x))+2g(x)=2a 2 then g(x)=Ax+B

For this take x(0)=x and x(n+1)=g(x(n)) then x(n+1)+2x(n)=a 2

take y(n)=x(n)-2/3a 2  then y(n+1)+2y(n)=0 and y(0)=x-2/3a 2

The caracteristical equation is : x+2=0 and go on :?

my sol uses limit idea :!:
 all sol are F(x)=x ,F(x)=0 and all functions of the form F(x)=1 or -1,according to x does not belong to X and it does belong, :) X is proper subset of(-infinity,-2\3)

, :| ,sorry i didnt see sol ofn bojan basic

This is an interesting f.e. problem.
I am sorry I did not notice there are three pages following this post until I submitted my message. Pls forget about this proof if you are familiar with this post.

Let $ a \equal{} f(0)$.
$ P(x,0): f(x^2 \plus{} 2a) \equal{} f(x)^2$.(*)
Suppose $ a < 0$, we let $ x \equal{} \sqrt { \minus{} 2a}$ and (*) gives $ a \equal{} f(x)^2\ge0$. Contradiction!
Hence $ a\ge0$.
Plug this into the RHS of the original formula,
$ f(x^2 \plus{} y^2 \plus{} 2f(xy)) \equal{} f(x^2 \plus{} y^2 \plus{} 2xy \plus{} 2a)$. (**)

If $ f(x) \equal{} x \plus{} a,\forall x$, it is easy to verify $ a \equal{} 0$ and $ f(x) \equal{} x$ is one solution.
From now on, we assume there exists $ b_0$ with $ c_0 \equal{} f(b_0)\ne b_0 \plus{} a$.

1) Let $ c \equal{} f(b)$ for arbitrary $ b\ne0$, then $ f(u \plus{} 2c \minus{} 2b) \equal{} f(u \plus{} 2a),\forall u\ge 2|b| \plus{} 2b$
When $ u\ge 2|b| \plus{} 2b$, we can always solve $ (x \plus{} y)^2 \equal{} u$ and $ xy \equal{} b$
Then we can plug in (**) and show the claim.
Hence, $ 2c \minus{} 2b \minus{} 2a$ is a period for $ f(x)$ when $ x\ge 2|b| \plus{} 2b \plus{} 2a$.
In particular, we see $ T_0 \equal{} |2c_0 \minus{} 2b_0 \minus{} 2a_0| > 0$ is a period when $ x\ge N_0 \equal{} 4|b_0| \plus{} 2a$.

2) $ f(x) \equal{} C$ when $ x\ge2a$, where $ C \equal{} 0$ or $ C \equal{} 1$.
Let $ b\in [N_0,N_0 \plus{} 1]$ be arbitrary with $ c \equal{} f(b)$.
Then we have $ f(b^2 \plus{} 2a) \equal{} c^2$ and $ f((b \plus{} T_0)^2 \plus{} 2a) \equal{} f(b \plus{} T_0)^2 \equal{} c^2$.
By 1), when $ x\ge 4b \plus{} 4T_0 \plus{} 2a$, both $ 2c^2 \minus{} 2b^2 \minus{} 6a$ and $ 2c^2 \minus{} 2(b \plus{} T_0)^2 \minus{} 6a$ are periods.
So $ [2c^2 \minus{} 2b^2 \minus{} 6a] \minus{} [ 2c^2 \minus{} 2(b \plus{} T_0)^2 \minus{} 6a] \equal{} 2T_0^2 \plus{} 4bT_0$ is also a period.
Now let $ b_0 > b_1$ be arbitrary on $ [N_0,N_0 \plus{} 1]$, we see for $ x \ge N_1 \equal{} 4N_0 \plus{} 4 \plus{} 4T_0 \plus{} 2a$,
$ [ 2T_0^2 \plus{} 4b_0T_0] \minus{} [ 2T_0^2 \plus{} 4b_1T_0] \equal{} 4(b_0 \minus{} b_1)T_0$ is a period.
In other words, any number in between $ 0$ and $ 4T_0$ is a period of $ f(x)$ when $ x \ge N_1$.
It is easy to see that $ f(x) \equal{} C$ for some constant $ C$ when $ x\ge N_1$.
Then use (*), we have $ C^2 \equal{} C$. So $ C \equal{} 0$ or $ C \equal{} 1$.
Again from (*), we know $ f( \minus{} x) \equal{} \pm f(x)$.
For any $ x\ge0$, we can find $ b\le \minus{} N_1$ and $ x \plus{} 2c \minus{} 2b\ge N_1$ because $ c \equal{} 0,\pm1$.
By 1), $ f(x \plus{} 2a) \equal{} f(x \plus{} 2c \minus{} 2b) \equal{} C$. This shows the claim.

3) If $ a \equal{} 0$, $ f(x) \equal{} 0$.
By 2), $ f(x) \equal{} C,\forall x\ge0$. Since $ f(0) \equal{} 0$, $ f(x) \equal{} 0$ for all $ x\ge0$. (*) implies $ f( \minus{} x) \equal{} \pm f(x) \equal{} 0$.

4) If $ a > 0$, $ f(x) \equal{} 1,\forall x\ge \minus{} \frac23;f(x) \equal{} \pm1,\forall x < \minus{} \frac23$.
By (*), $ f(2a) \equal{} a^2$. By 2), $ a^2 \equal{} C$. Since $ a > 0$, we have $ f(x) \equal{} 1$ when $ x\ge2$.
Now LHS of (*) is 1,  $ f(x) \equal{} \pm 1,\forall x$.
Now we look back the original equation.
If $ f(b) \equal{} \minus{} 1$ for some $ b \equal{} xy$, we have $ f(x^2 \plus{} y^2 \minus{} 2) \equal{} 1$.
In other words, $ f(u) \equal{} 1,\forall u\ge2|b| \minus{} 2$.
Once $ 2|b| \minus{} 2\le b$, we will get contradiction.
Hence, $ b < \minus{} \frac23$. It is not hard to see that is sufficient for the equation to hold.

Conclusion:
1) $ f(x) \equal{} x$.
2) $ f(x) \equal{} 0$.
3) $ f(x) \equal{} 1,\forall x\ge \minus{} \frac23;f(x) \equal{} \pm1,\forall x < \minus{} \frac23$.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2170259]here[/url] for official solution.

Substitute $y=0$ gives us $f(x^2+2f(0))=f(x)^2$ for all real number $x$.
This also gives $f(x)^2=f(-x)^2$ for all real number $x$.
Note that for any real number $a,b$ that $a^2\geq 4b$, there exists real numbers $x,y$ that $x+y=a$ and $xy=b$.
From $f(x^2+y^2+2f(xy)) =f(x+y)^2$ for all real numbers $x,y$, we get that
$$f(a^2-2b+2f(b))=f(a)^2=f(a^2+2f(0))$$ for all real numbers $a,b$ that $a^2\geq 4b$.
This can be simplify to
$$P(x,y): f(x-2y+2f(y))=f(x+2f(0))$$ for all real numbers $x,y$ that $x\geq \max \{ 0,4y\}$.
Also, we've
$$Q(x,y): f(x^2-y^2+2f(-xy))=f(x-y)^2$$ for all real numbers $x,y$.

If $f(y)=y+f(0)$ for all real number $y$, substitute in $f(x^2+y^2+2f(xy))=f(x+y)^2$ gives us
$$x^2+y^2+2xy+3f(0)=(x+y+f(0))^2$$ for all real numbers $x,y$.
This easily gives $f(x)=x$ for all real number $x$, which clearly is one of the solution.

Now, we’ll consider only the case when there exists real number $k$ that $f(k)\neq k+f(0)$.
Let $M=2f(k)-2k-2f(0)$. Note that $M\neq 0$.
From $P(x-2f(0),k)$, we’ve $f(x+M)=f(x)$ for all real number $x\geq 2f(0)+\max \{ 0,4k\}$.
If $M>0$, we get $f(x+|M|)=f(x)$ for all real number $x\geq 2f(0)+\max \{ 0,4k\}$.
If $M<0$, we get $f(x)=f(x-M)=f(x+|M|)$ for all real number $x-M\geq 2f(0)+\max \{ 0,4k\}\iff x\geq M+2f(0)+\max \{ 0,4k\}$.
In both cases, we have $f(x)=f(x+|M|)$ for all real number $x\geq 2f(0)+\max \{ 0,4k\}$.
We can prove by induction that $f(x)=f(x+v|M|)$ for all positive integer $v$ and real number $x\geq 2f(0)+\max \{ 0,4k\}$.

For any real numbers $c,d$ that $f(c)=f(c+d)$, we’ve the following results:
1) $f(c^2+2f(0))=f(c^2+2cd+d^2+2f(0))$
This follows from $f(c^2+2f(0))= f(c)^2=f(c+d)^2=f(c^2+2cd+d^2+2f(0))$.
2) $f(x)=f(x-2d)$ for all real number $x\geq 2f(c)-2c+\max \{ 0,4c,4c+4d\}$
This follows from $P(x,c),P(x,c+d)$, which gives us $f(x-2c+2f(c))=f(x-2c-2d+2f(c))$ for all real number $x\geq \max \{ 0,4c,4c+4d\}$.
3) $f(x)=f(x-2(2cd+d^2))$ for all real number $x\geq 2f(c)^2-2(c^2+2f(0))+\max \{ 0,4c^2+8f(0),4c^2+8f(0)+8cd+4d^2\}$.
This is true by applying the result 1) combine with 2) and note that $f(c^2+2f(0))=f(c)^2$.

Back to the problem.
In this paragraph, we'll show that $f(\ell )=f(\ell -\Delta)$ for all real numbers $\Delta \in (0,|M|)$ and $\ell \geq 2f(0)+\max \{ 0, 4k\} +|M|$.
From $f(x)=f(x+|M|)$ for all real number $x\geq 2f(0)+\max \{ 0,4k\}$, using result 3), we get that
$$f( t)=f(t-2(2x|M|+M^2))$$ for all real number $t\geq 2f(x)^2-2(x^2+2f(0)) +\max \{ 0,4x^2+8f(0), 4x^2+8f(0) +8x|M|+4M^2 \}$ where $x\geq 2f(0)+\max \{ 0,4k\}$.
There exists real number $h$ that $2(2h|M|+M^2) =\Delta $.
There also exists positive integer $n$ that $h+n\geq 2f(0)+\max \{ 0,4k\}$.
So, we’ve $$f(t)=f(t-2(2(h+n)|M|+M^2))=f(t-\Delta -4n|M|)$$ for all real number $t\geq 2f(h+n)^2-2((h+n)^2+2f(0))+ \max \{ 0,4(h+n)^2+8f(0),4(h+n)^2+8f(0)+8(h+n)|M|+4M^2\}$.
For any real number $\ell \geq 2f(0)+\max \{ 0,4k\}+|M|$, there exists positive integer $m>4n$ that
$$\ell +m|M|\geq 2f(h+n)^2-2((h+n)^2+2f(0))+ \max \{ 0,4(h+n)^2+8f(0),4(h+n)^2+8f(0)+8(h+n)|M|+4M^2\}.$$
So, $f(\ell +m|M|)=f(\ell -\Delta -4n|M|+m|M|)=f(\ell -\Delta +(m-4n)|M|)$.
Since $\ell -\Delta >\ell -|M|\geq 2f(0)+\max \{ 0,4k\}$ and $\ell \geq 2f(0)+\max \{ 0,4k\}+|M|\geq 2f(0) +\max \{ 0,4k\}$, we get
$$f(\ell +m|M|)=f(\ell)\text{ and } f(\ell -\Delta +(m-4n)|M|)=f(\ell -\Delta).$$
Hence, $f(\ell )=f(\ell -\Delta)$, which we aim at the beginning of this paragraph.

Using the result in preceding paragraph, together with $f(x)=f(x+|M|)$ for all real number $x\geq 2f(0)+\max \{ 0,4k\}$, we get that $f(x)$ must be constant for all real number $x\geq 2f(0)+\max \{ 0,4k\}$.
For convenient, we write it as $f(x)=C$ for all real number $x\geq D$. Note that we can force $D$ to be positive.
Since there exists real number $r$ that $r$ and $r^2+2f(0))$ are greater than $D$, we get $C=f(r^2+2f(0))=f(r)^2=C^2$.
Hence, $C\in \{ 0,1\}$. So we've two following cases:

1) $C=0$.
Since $f(x)=0$ for all real number $x\geq D$ and $f(x)^2=f(-x)^2$ for all real number $x$, we get that $f(x)=0$ for all real number $x$ that $|x|\geq D$.
For any real number $e$, there exists positive real number $u$ that $u(u+e)>D$ and $(u+e)^2+u^2>D$.
The former gives $f(-u(u+e))=0$ and so $f((u+e)^2+u^2+2f(-u(u+e)))=f((u+e)^2+u^2)=0$.
From $Q(u+e,u)$, we get that $0=f(e)^2\implies f(e)=0$.
Hence, $f(x)=0$ for all real number $x$, which clearly is one of the solution.

2) $C=1$.
Since $f(x)=1$ for all real number $x\geq D$ and $f(x)^2=f(-x)^2$ for all real number $x$, we get that $f(x)\in \{ -1,1\}$ for all real number $x$ that $|x|\geq D$.
For any real number $e$, there exists positive real number $u$ that $u(u+e)>D$ and $(u+e)^2+u^2-2>D$.
The former gives $f(-u(u+e))\in \{ -1,1\} $ and so $(u+e)^2+u^2+2f(-u(u+e)) \geq (u+e)^2+u^2-2>D$.
Hence, $f((u+e)^2+u^2+2f(-u(u+e))) =1$.
From $Q(u+e,u)$, we get that $1=f(e)^2\implies f(e)\in \{ -1,1\}$.
In other words, $f(x)\in \{ -1,1\}$ for all real number $x$.
Let $A=\{ x\in \mathbb{R}\mid f(x)=1\}$ and $B=\{ x\in \mathbb{R}\mid f(x)=-1\}$. Note that they form a partition of $\mathbb{R}$.
From $f(x^2+y^2+2f(xy)) =f(x+y)^2$ for all real numbers $x,y$, we get that
$$f(x^2-2y+2f(y))=f(x)^2=1$$ for all real numbers $x,y$ that $x^2\geq 4y$.
This gives us $f(x-2y+2f(y))=1$ for all real number $x\geq \max \{ 0,4y\}$.
Using this equation, we get that
$$\begin{cases}
 x-2-2y\in A  \ \text{for all} \ x,y\in \mathbb{R}\text{ that }x\geq \max \{ 0,4y\} ,y\in B\\
 x+2-2y \in A  \ \text{for all} \ x,y\in \mathbb{R}\text{ that }x\geq \max \{ 0,4y\} ,y\in A
\end{cases} $$
In case that $y>0$, we may (and will) substitute $x$ by $4y+z$ where $z$ is real number that $z\geq 0$.
So, we can rewrite the condition in the following four cases
\begin{align*}
1) & \ x-2y+2\in A \ \text{for all} \ x,y\in \mathbb{R}\text{ that }x\geq 0,y\leq 0,y\in A \\
2) & \ x+2y+2\in A \ \text{for all} \ x,y\in \mathbb{R}\text{ that }x\geq 0,y\geq 0,y\in A \\
3) & \ x-2y-2\in A \ \text{for all} \ x,y\in \mathbb{R}\text{ that }x\geq 0,y\leq 0,y\in B \\
4) & \ x+2y-2\in A \ \text{for all} \ x,y\in \mathbb{R}\text{ that }x\geq 0,y\geq 0,y\in B.
\end{align*}
If there exists real number $0\leq w\leq 2$ that $w\in B$, using condition 4), we get that $w<2w-2 \implies w>2$, contradiction.
So, $x\in A$ for all real number $x$ that $0\leq x\leq 2$.
Using condition 2), we can extend it to $x\in A$ for all real number $x$ that $x\geq 0$.
If there exists real number $\frac{-2}{3} \leq w\leq 0$ that $w\in B$, using condition 3), we get that $w<-2w-2\implies w<\frac{-2}{3}$, contradiction.
So, $x\in A$ for all real number $x$ that $\frac{-2}{3}\leq x\leq 0$.
Hence, now we’ve $f(x)=1$ for all $x\geq \frac{-2}{3}$.
This means there exists subset $\mathcal{S}$ of $\{ x\in \mathbb{R}\mid x<\frac{-2}{3} \}$ that
$$f(x)=\begin{cases}
-1& , \text{if} \ x\in \mathcal{S}\\
1& , \text{otherwise}
\end{cases}.$$
Not hard to verify that this function is one of the solution for all subset $\mathcal{S}$ of $\{ x\in \mathbb{R}\mid x<\frac{-2}{3} \}$.

Finally, we have three type of solutions to the problem:
1. $f(x)=0$ for all $x\in \mathbb{R}$,
2. $f(x)=x$ for all $x\in \mathbb{R}$, and
3. $f(x)=\begin{cases}
-1& , \text{if} \ x\in \mathcal{S}\\
1& , \text{otherwise}
\end{cases}$ for any subset $\mathcal{S}$ of $\{ x\in \mathbb{R}\mid x<\frac{-2}{3} \}$.

[hide = Solution]

We claim that the solutions are $f \equiv 0, f \equiv 1, f \equiv x,$ and the following monster:

\[ f(x)=\{\begin{array}{rr}1,&x\in [\frac{-2}{3}, +\infty)\\ \pm 1,&x\not\in (-\infty, \frac{-2}{3})\end{array} \]

Now, we will show that they are all such solutions. As usual, let $P(x, y)$ denote the assertion that $f(x^2 + y^2 + 2f(xy)) = f(x+y)^2$. By $P(0, y)$ we have that:

$$f(y^2 + 2f(0)) = f(y)^2. \qquad (1)$$
Observe that as a direct corollary of $(1)$, we have that:

$$f(x) \ge 0, \forall x \ge 2f(0). \qquad (2)$$

If $f$ is injective, then we have that $f((x+y)^2 + 2f(0)) =^{(1)} f(x+y)^2 = f(x^2 + y^2 + 2f(xy)),$ and so injectivity now implies that $(x+y)^2 + 2f(0) = x^2 + y^2 + 2f(xy) \Leftrightarrow f(xy) = xy + f(0).$ This clearly implies that $f(x) \equiv x + c$ for some constant $c$ ($= f(0)$), and so plugging back in gives us our solution $f \equiv x.$

Otherwise, let's suppose that $f$ is not injective. We will use the following lemmas to help us gain information about $f$.

$\mathbf{Lemma}$ $\mathbf{1.}$ If $a, b \in \mathbb{R}$ so that $f(a) = f(b),$ then $f(c) = f(d)$ whenever $c, d$ are $>\max (2 \max(\sqrt{a}, \sqrt{b}), 2f(0))$ and $c^2 - d^2 = 2a - 2b.$

$\mathbf{Proof.}$ Observe that for $p > 2 \max(|a|, |b|),$ there exist real numbers $t, u, v, w$ such that $tu = a, vw = b, t^2 + u^2 = v^2 + w^2 = p.$ Hence, we know by $P(t, u), P(v, w)$ that $f(\sqrt{p+2a})^2 = f(t+u)^2 = f(v+w)^2 = f(\sqrt{p+2b})^2.$ Now, by simply letting $p$ be big so that $\sqrt{p + 2a}, \sqrt{p+2b} > 2f(0),$ $(2)$ implies the lemma.

$\blacksquare$

With this in mind, we will show the following lemma.

$\mathbf{Lemma}$ $\mathbf{2.}$ There exists a constant $C$ for which $f$ is constant over reals $\ge C.$

$\mathbf{Proof.}$ Note that in the notation of the above lemma, as $c^2 - d^2 = 2a-2b$ and $c, d$ vary so that both are $>\max (2 \max(\sqrt{a}, \sqrt{b}), 2f(0))$ and $<\max (2 \max(\sqrt{a}, \sqrt{b}), 2f(0)) + N$ for some huge enough $N$, it's clear that $c-d$ can cover all values in the interval $[r, r+\epsilon]$ for some $r, \epsilon \in \mathbb{R}$, with $\epsilon > 0$ sufficiently small. Hence, if we use the above lemma with $c, d$ in lieu of $a, b$, since any real number can be expressed as a linear combination of real numbers in $[2r, 2(r+\epsilon)],$ we can use Lemma 1 repeatedly to imply Lemma 2.

$\blacksquare$

With these in mind, we can now prove the following useful lemma.

$\mathbf{Lemma}$ $\mathbf{3.}$ $f(x)^2$ is either the $0$ function or the $1$ function.

$\mathbf{Proof.}$ Let $d$ be a constant such that $f(x) = d$ for all $x \ge C,$ as guaranteed by Lemma 2. By letting $x >0$ be a sufficiently large positive real, we have that $d^2 = f(2x)^2 = f(2x^2+ 2f(x^2)) = f(2x^2 + 2d) = d,$ and so hence $d \in \{0, 1\}.$ For small enough negative reals, we have that $f(2r)^2 = f(2r^2 + 2f(r^2)) = f(2r^2 + d) = d,$ and so hence $f(x)^2 = d$ for negative enough $x$. Now, for any real $r$, letting $x$ be a sufficiently large positive real gives that $f(r)^2 = f(x + (r-x))^2 = f(x^2 + (r-x)^2 + 2f(x(r-x))) = f(x^2 + (r-x)^2 + 2d) = d,$ and so we're done.

$\blacksquare$

If $f(x)^2$ is the $0$ function, then we arrive at our solution $f(x) \equiv 0, \forall x \in \mathbb{R}.$ Otherwise, let's suppose that $f(x)^2 = 1$ for all $x \in \mathbb{R}.$ Then, we know that $f(x) = \pm1$ for all real $x$. This allows us to rewrite $P(x, y)$ as:

$$f(x^2 + y^2 + 2f(xy)) = 1.$$

Suppose now that $a$ is a real such that $f(a) = -1.$ Then, for any $p \ge 2|a|$, we can find $x, y \in \mathbb{R}$ such that $x^2 + y^2 = p$ and $xy = a$. Hence, $P(x, y)$ would give that $f(p - 2) = 1.$ Therefore, since $p = a+2$ would clearly give a contradiction, we know that $a+2 < 2 |a|.$ In other words, any $a \in \mathbb{R}$ satisfying $f(a) = -1$ has to satisfy $a+2 < 2|a|$ as well. However, this clearly implies that $a < \frac{-2}{3},$ and so we therefore know that $f(x) = 1, \forall x \ge \frac{-2}{3}.$ We have therefore finally arrived at our final solution, and so the problem is solved.

$\square$

[/hide]

Amazing practice problem.
The answer is $f(x) = 0$, $f(x) = x$, and $f(x) = -1$, for any subset of $(-\infty, -\frac{2}{3})$ and $f(x) = 1$ otherwise. \\
It's easy to check that these are the only solutions.  \\
We'll prove that these are the only ones. \\
Let $x + y = a$ and $xy = b$. Therefore we can rewrite the functional equation as
\[ P(a,b) : f(a^2 - 2b + 2f(b)) = f(a)^2 \ \forall a^2 \ge 4b \]
Now, define $2f(b) - 2b = g(b)$.
\[ P(a,b) : f(a^2 + g(b)) = f(a)^2 \]
It is quite immediate to notice that $f(-a)^2 = f(a)^2$ and $f$ is eventually non-negative (by fixing one value of $\ell$ and take $a \ge \max \{0, 4 \ell \}$ if $g(\ell) \ge 0$ and $a \ge \max \{ 4 \ell, \sqrt{-g(\ell)} \} $ if $g(\ell) < 0$.
-----------
$\textbf{Case 01.} \ \text{If } g \ \text{is constant}$ \\
Then $f$ is linear. We can check that $f(x) = x$ is the only solution. \\
------------
$\textbf{Case 02.} \ \text{If } g \ \text{is non constant}$ \\
Suppose not, we will prove that $f$ is eventually constant. Firstly, notice that if $g(b)$ is non-constant, there exists two distinct real numbers $\alpha$ and $\beta$ such that $g(\alpha) - g(\beta) = c > 0$. Notice that
\[ f(u)^2 = f(u^2 + g(\alpha) ) = f(u^2 + g(\beta) + c) = f((\sqrt{u^2 + c})^2 + g(\beta)) = f(\sqrt{u^2 + c})^2 \]
Denote $\sqrt{u^2 + c} = v$. Now, we have that if $v^2 - u^2 = c$, then $f(u) = f(v)$. \\
Now, we have
\[ g(v) - g(u) = (2f(v) - 2v) - (2f(u) - 2u) = 2(f(v) - f(u)) + 2(u - v) = 2(u - v) = \frac{2c}{u + v} \]
Now, we have
\[ f(s^2 + g(u)) = f(s^2 + g(v)) = f(s)^2 \]
Let $u,v \in [A,3A]$ where $A$ is large. Let $\varepsilon = g(v) - g(u) = \frac{k}{u + v} \in [\delta, 2 \delta]$. Therefore,
\[ f(s^2 + g(u) + \varepsilon) = f(s^2 + g(v)) = f(s)^2 = f(s^2 + g(u)) \]
This forces that for large $a,b > 12A$, we then have $f(a) = f(b)$ if $a^2 - b^2 \in [\delta, 2 \delta]$. Repeating this process by changing the interval, we get that $f$ is eventually constant.
\\
Now, take two large integers $a,b > M$ such that $b^2 - a^2 = g(u)$ for some integer $u$. Suppose the constant is $c$. However,
\[ c^2 = f(b)^2 = f(a)^2 = f(a^2 + g(u)) = f(b^2) = c \]
This gives us $c = 0$ or $c = 1$. \\
Take a large negative number $-P$. We know that $|f(-P)| = |f(P)| = c$ which is a constant. Therefore, $f(x) \in \{ -c, c \}$ as well for $x \le -P$.
 Hence,
\[ f(s^2 + 2f(-P) -2(-P)) = f(s^2 + 2P - c_0) = f(s)^2 \]
for all large integer $P$ and some constant $c_0$. This forces $f(s)^2 = c$ for all integer $s$.
Hence, if $c = 0$, we get $f \equiv 0$ which satisfy the equation.  \\
Now, assume that $c = 1$. \\
\[ R(x,y): f(x^2 + y^2 + 2f(xy)) = 1 \]
Notice that if $f(0) = -1$. Then $R(\sqrt{2},0)$ gives us $f(0) = 1$, a contradiction. \\
Therefore, $f(0) = 1$ which therefore gives $f(x) = 1$ for all $x \ge 2$.
\\
Now, suppose $f(t) = -1$. Suppose $-\frac{2}{3} \le t \le 2$. Then there exists $x,y \in \mathbb{R}$ such that
\[ x^2 + y^2 - 2 = xy = t \]
as $2 + t = x^2 + y^2 \ge 2|xy| = 2|t|$. \\
Therefore, this gives
\[ 1 = f(x^2 + y^2 + 2f(xy)) = f(t) \]
a contradiction. Therefore, $x^2 + y^2 - 2 = xy = t$ must gives no solution. \\
Hence, if $f(t) = -1$, then $t < -\frac{2}{3}$. \\
This gives a solution $f(x) = -1$, for any subset of $(-\infty, -\frac{2}{3})$ and $f(x) = 1$ otherwise. \\

[hide=Solution]First, for clarity, rewrite the equation as
\[f((x+y)^2+2(f(xy)-xy)) = f(x+y)^2. \tag{1}\]
Let $y=0$ in (1) gives $f(x^2+2f(0)) = f(x)^2 \ge 0$, so $f(x)\ge 0$ for all $x\ge 2f(0)$. In particular, $f(0)\ge 0$. In addition, $f(-x)^2 = f(x^2+2f(0)) = f(x)^2$. Let $y=-x$ in (1) gives
\[f(2x^2+2f(-x^2))=f(0)^2. \tag{2}\]

If $f$ is injective, then (2) implies $f(x)=x+f(0)$ for any $x\le 0$, so
\[f(x)^2 = \begin{cases} (f(0)-x)^2 & \text{if } x\ge 0\\ (f(0)+x)^2 & \text{if } x\le 0. \end{cases}\]
Take $x\ge 0$, then
\[x^2+f(0) = f(x^2+2f(0)) = f(x)^2 = x^2-2xf(0)+f(0)^2\]
for all $x\ge 0$, so $f(0)=0$. This gives  a solution $f(x) \equiv x$.

From now on, suppose that $f(a)=f(b)$ for some distinct $a,b$. We will first prove the following claim:

[b]Claim.[/b] If $f(a)=f(a+k)$ with $k>0$, then $f(c)=f(\sqrt{c^2+2k})$ for any \[c^2 \ge \max(4(a+k),4f(0)^2).\]
[i]Proof.[/i] Because $c^2 \ge 4a$, there exist $x,y$ such that $x+y=c$ and $xy=a$. Because $\sqrt{c^2+2k} > c^2 \ge 4(a+k)$, there exist $x,y$ such that $x+y=\sqrt{c^2+2k}$ and $xy=a+k$. Therefore,
\begin{align*}
f(c)^2
&= f(c^2 + 2(f(a)-a))\\
&= f(c^2 + 2k + 2(f(a+k)-(a+k)))\\
&= f(\sqrt{c^2+2k})^2.
\end{align*}
Since $c>2f(0)$, both $f(c)$ and $f(\sqrt{c^2+2k})$ must have the same sign, so $f(c)=f(\sqrt{c^2+2k})$ as desired. $\square$

Observe that $\sqrt{c^2+2k}-c > \frac{k}{2}$ whenever $c>2$ (this ensures that the ``distance'' between a non-injective pair converges to zero). We repeatedly apply the claim, each time replacing $(a,a+k)$ by $(c,\sqrt{c^2+2k})$ where we choose \[c = \max(2, 2\sqrt{|a+k|}, 2f(0))\] which is $o(a)$. This ensures that there exists an absolute constant $M$ such that for all $a,b>M$, $f(a)=f(b)$, i.e. $f$ is eventually constant.

Suppose that $f(x)=C$ eventually. Then choosing large $x,y$ yields $C^2=C$, so $C=0$ or 1.

[b]Case 1:[/b] $C=0$, then (2) implies $f(0)=0$, so $f(x^2)=f(x)^2$. So $f(x)=0$ for all $x>\sqrt[2^n]{M}$ holds for any positive integer $n$. This, combined with $f(-x) = \pm f(x)$, implies that $f(x)\equiv 0$, which is a valid solution.

[b]Case 2:[/b] $C=1$, then similar to above we have $f(x)^2=1$ for all $x$. Plugging this back to (1) gives
\[f(x^2+2f(xy)+y^2) = 1 \tag{3}\]
If $f(a)=-1$, then (3) implies that for all $x\ge 2|a|-2$, $f(x)=1$. Therefore $a < 2|a|-2$ so either $a>2$ or $a< - \frac{2}{3}$. But $f(0)=1$ implies $f(x)>0$ for all $x>2$, so $a$ must be smaller than $-\frac{2}{3}$.

On the other hand, for any set $\mathcal{A} \subseteq (-\infty, -\frac{2}{3})$, the function
\[f(x) = \begin{cases} -1 & \text{if } x\in \mathcal{A}\\ 1 & \text{otherwise}\end{cases}\]
is a valid solution.[/hide]

The solutions are $f(x)=x$, $f(x)=0$, and for any set $A\subseteq \left(-\infty, -\tfrac 23\right)$
$$f(x) = \begin{cases}
-1 & x\in A \\
1 & x\notin A;
\end{cases}$$
the first two are easy to verify. To verify the last solution, it's helpful to reparametrize the functional equation as
$$f(a^2+2f(b)-2b) = f(a)^2 \text{ whenever } a^2\geq 4b \qquad (*)$$
[list]
[*] If $b<0$, then it's easy to check that $2f(b)-2b \geq -\tfrac 23$ for any $b$, which makes the equation visibly true;
[*] if $b\geq 0$, then $a^2+2f(b)-2b \geq 2b+2 >0$, which again makes the equation visibly true.
[/list]
This completes the verification. Now we prove that these are all solutions.
-----
[b][color=red]Claim 1:[/color][/b] If $f$ is injective, then $f$ must be an identity function.

[i]Proof:[/i] For any $r,s$, take $x$ large enough so that
$$f(x^2+2f(r)-2r) = f(x)^2 = f(x^2+2f(s)-2s),$$
or $f(r)-r=f(s)-s$ after injectivity. This gives $f(x)=x+c$ for some constant $c$; it's easy to check that $c=0$. $\blacksquare$
-----
[b][color=red]Claim 2:[/color][/b] Assume that $f(r)=f(s)$ such that $r> s$, then there exists $K$ which $f$ is constant on $[K,\infty)$.

[i]Proof:[/i] First of all, we note that $f(x)\geq 0$ for all sufficiently large $x>M_0$. This can be easily seen by fixing $b$ and taking $a$ sufficiently large. Now, consider
\begin{align*}
f(\sqrt{x+2r})^2 &= f(x+2r + 2f(r)-2r) \\
&= f(x+2f(r)) \\
&= f(x+2f(s)) \\
&= f(\sqrt{x+2s})^2,
\end{align*}
for all $x$ large enough (say $>100|r|$) Thus, $f(\sqrt{x+2r}) = f(\sqrt{x+2s})$ for all sufficiently large $x$, say $x>M_1$. We also take $\varepsilon \in (0, \sqrt{M_1+2r} - \sqrt{M_1+2s})$.

Now, fix $u,v$ such that $\tfrac{\varepsilon}{2} < u^2-v^2 < \varepsilon$, and $u,v$ are larger than $M$ (which will be told later). Then, we can find $B$ so that $\sqrt{B+2r} - \sqrt{B+2s} = u^2-v^2 \quad (\spadesuit)$. This choice of $B$ must be bounded above by a constant depending only on $r,s,f$, since we have a fixed lower bound of $u^2-v^2$. So, let's suppose that it's bounded above by $M_2$.

By repeating the argument above, we find that
$$f\left(\sqrt{A+\sqrt{B+2r}}\right) = f\left(\sqrt{A+\sqrt{B+2s}}\right)$$
for any $A > \max\{100\sqrt{B+2r},M_0\}$. Thus, let $M_3 = \max\{100\sqrt{M_2+2r},M_0\}$. The above equations hold for $A>M_3$.

Hence, if $M > \sqrt{M_3+\sqrt{M_2+2r}}$, we can find a suitable $A$ which $A+\sqrt{B+2r} = u^2$. From $(\spadesuit)$, we get that $A+\sqrt{B+2s}=v^2$. Therefore $f(u)=f(v)$ for any $u,v>M$ and $u^2-v^2 \in (0.5 \varepsilon, \varepsilon)$. By applying this fact repeatedly, it's easy to see that $f$ is constant for a large enough value. $\blacksquare$
-----
Thus, we see that $f$ is a constant $c$ over $[K,\infty)$. By fixing $b$ and taking $a$ very large in the equation, we deduce that $c^2=c$ or $c\in\{0,1\}$. Since $f(a)^2 = f(-a)^2$, we get that $f$ must be $\pm c$ in $(-\infty, -K]$ too.

Now, we take $b$ very negative, so that $2f(b)-2b > K$. This gives $f(a)^2 = c$ or $f(a)=\pm c$ for any $a$. In particular, if $c=0$, then we have forced $f$ to be a zero function.

The only case left is $f(x)=\pm 1$ for all $x$. The equation is then reduced to
$$f(a^2+2f(b)-2b) = 1 \text{ whenever } a^2\geq 4b.$$
It remains to show that if $f(t)=-1$ for some $t$, then $t\leq -\tfrac 23$. We note that the equation gives
$$f(a^2 - 2 - 2t)=1 \text{ whenever } a^2\geq 4t.$$
[list]
[*] If $t\geq 0$, then the above translates to $f(k)=1$ for any $k\geq 2t-2$. Thus, $t\leq 2t-2$ or $t\geq 2$. However, since $f(a^2+2f(0))=1$ for any nonnegative $a$, we get that $f(x)=1$ for any $x\geq 2$, contradiction.
[*] If $t<0$, then the above translates to $f(k)=1$ for any $k\geq -2t-2$. Thus, $t<-2t-2$ or $t< -\tfrac 23$ as desired.
[/list]

We claim the solutions are $f(x)\equiv 0,x,f_S(x)$, where $S\subseteq(-\infty,-2/3)$ is arbitrary, and $f_S(x)$ is defined to be $1$ outside of $S$ and $-1$ on $S$.

Let $P(x,y)$ be the given FE. Note that \[P(x,0)\implies f(x^2+2f(0)) = f(x)^2,\] so we can rewrite our FE as \[Q(x,y): f(x^2+y^2+2f(xy)) = f(x^2+y^2+2xy+2f(0)).\]

We have the following key lemma.

[b]Lemma:[/b]  If $f(a)=f(b)$, then \[f(z+2a)=f(z+2b)\] for all $z\ge 2f(0)+2\cdot\max\{|a|,|b|\}$.

[i]Proof:[/i]  Fix $z\ge 2f(0)+2\cdot\max\{|a|,|b|\}$. By the inequality condition on $z$, we may pick reals $x$ and $y$ such that $xy=a$ and $x^2+y^2 = z-2f(0)$, and similarly we may pick reals $x'$ and $y'$ such that $x'y'=b$ and $x'^2+y'^2=z-2f(0)$. Comparing $Q(x,y)$ detail $Q(x',y')$, we see that $f(z+2a)=f(z+2b)$, as desired. $\blacksquare$

We also have the following easy lemma.

[b]Lemma:[/b]  There exists a positive constant $A$ such that $f(x)\ge 0$ for all $x\ge A$.

[i]Proof:[/i]  Note that \[P(x,1/x)\implies f(x^2+1/x^2+2f(1))=f(x+1/x)^2\ge 0,\] so $A=\max\{2+2f(1),0\}$ suffices. $\blacksquare$

We can use these two lemmas to get the following killer claim.

[b]Claim:[/b]  If $f(a)=f(b)$ for some $a\ne b$, then there exists constants $B$ and $k$ such that $x\ge B\implies f(x)=k$.

[i]Proof:[/i]  Using the lemma, we may shift up the values of $a$ and $b$ arbitrarily, so assume that they are greater than $A$. Now, fix some $T\ge 2\cdot\max\{a,b\}$.

Let $x$ and $y$ be given by $x+y=\sqrt{T+2a}$, $x-y=\sqrt{T-2a}$, and let $x'$ and $y'$ be given by $x'+y'=\sqrt{T+2b}$, $x'-y'=\sqrt{T-2b}$. We see that $xy=a$, $x'y'=b$, and $x^2+y^2=x'^2+y'^2=T$, so comparing $P(x,y)$ and $P(x',y')$ tells us that \[f(\sqrt{T+2a}) = f(\sqrt{T+2b})\] for $T\ge 2a$ (assuming that $a>b$ WLOG). Focusing our attention to $T\in[2a,3a]$, we see that by the lemma, there exists a constant $B$ such that for $z\ge B$, \[f(z+2(\sqrt{T+2a}-\sqrt{T+2b})) = f(z)\] for all $T\in[2a,3a]$. In this interval, the expression $\sqrt{T+2a}-\sqrt{T+2b}$ can take all the values in some interval $[\alpha,\beta]$ with $0<\alpha<\beta$, so we have the following statement.

There is a positive constant $B$ and an interval $[\alpha,\beta]$ with $0<\alpha<\beta$ such that \[f(z+\gamma)=f(z)\] for all $z\ge B$ and $\gamma\in[\alpha,\beta]$. Let $0<\delta<\beta-\alpha$. Then, \[f(z)=f(z+\beta)=f((z+\delta) + (\beta-\delta))=f(z+\delta)\] for all $z\ge B$, since $\beta-\delta\in[\alpha,\beta]$. So $f(z)=f(z+\delta)$ for all $z\ge B$ and $\delta\in(0,\beta-\alpha)$, so an easy induction implies that $f$ is constant beyond $B$, as desired. $\blacksquare$

If $f$ is injective, then $Q(x,y)$ implies that $f(xy)=xy+f(0)$ for all real $x,y$, so $f(x)=x+f(0)$. Plugging this into the FE shows that only $f(x)=x$ works, so in this case, we the solution $\boxed{f(x)\equiv x}$. Now, suppose that $f$ is not injective. By the above claim, we have that $f(x)=k$ for all $x\ge B>0$.

From $P(x,0)$, we see that $f(-x)\in\{f(x),-f(x)\}$ for all $x$. Fixing any real $z$, we have \[P(z+x,-x)\implies f(x^2+(z+x)^2+2f(-x(z+x))) = f(z)^2\] for all real $x$. By making $x$ sufficiently large, we see that $f(-x(z+x))\in\{k,-k\}$ and $x^2+(z+x)^2-2|k|\ge B$, so in fact $f(z)^2=k$ for all real $z$. This implies that $k\in\{0,1\}$. If $k=0$, then this implies that $\boxed{f(x)\equiv 0}$. Now assume $k=1$, so this implies $f(x)\in\{1,-1\}$ for all real $x$.

Our FE now becomes \[f(x^2+t^2/x^2+2f(t)) = 1.\] If $f(t)=-1$ for some real $t$ (potentially zero), then we see that \[f(x^2+t^2/x^2-2) = 1\] for all real $x\ne 0$, so $t$ must be outside of the image of the function \[x\mapsto x^2+t^2/x^2-2,\] where $x$ varies over all nonzero reals. The image of this function is $[2|t|-2,\infty)$, so $t<2|t|-2$, or $t\in(-\infty,-2/3)\cup(2,\infty)$. This means that $f(x)=1$ for all $x\in[-2/3,2]$.

Suppose we know that $f(x)=1$ for all $x\in[0,C]$ where $C>0$. Then, \[f(x^2+y^2+2)=1\] for all nonnegative real $x$ and $y$ such that $xy\le C$. Thus, we have $f(x)=1$ for all $x\in[0,2C+2]$. We know the statement is true for $C=2$, and since iterating the map $C\mapsto 2C+2$ sends the value to infinity, we see that $f(x)=1$ for all $x\ge -2/3$. This gives $\boxed{f(x)\equiv f_S(x)}$ for any set $S\subseteq(-\infty,-2/3)$. This covers all the cases, so we're done.
>>>

[b]Can someone please check the below solution[/b] (it seems very different from the above ones, so maybe I messed up somewhere)

[rule]

The solutions are $f(x) \equiv x$ ; $f(x) \equiv 0$ and
$$f(x) = \begin{cases} -1 \qquad & \text{if } x \in T \\ 1 \qquad & \text{otherwise}     \end{cases}$$
for any set $T \subseteq \left( - \infty , \tfrac{-2}{3} \right)$. The first two functions trivially work. For the third function: RHS is always $1$. Now just consider two cases according to the value of $f(xy)$ and use the upper bound on elements of $T$.

Now we prove the above are the only solutions. Let $P(x,y)$ be the given assertion. $P(x,0)$ gives
$$f(x)^2 = f(x^2 + 2f(0)) \qquad{(1)}$$
Let $2f(0) = 2c$. Then the original equation becomes
$$f(x^2 + y^2 + 2f(xy)) = f(x^2 + y^2 + 2xy + 2c) \qquad{(2)}$$
Lets first assume $f$ is injective. Then $(2)$ directly gives us $f(x) \equiv x+c$, and it is not hard to note that only $c = 0$ works ; i.e. $f(x) \equiv x$ is the only solution here.

Now we consider the crucial case when $f$ is not injective. Let $f(a) = f(b)$ with $a \ne b$.

[b][color=#f00]Claim: [/color][/b] There are constants $M,r > 0$ such that $f(x+r) = f(x) ~ \forall ~ x \ge M$

[b][color=#00f]Proof: [/color][/b] It is not hard to see that $(2)$ implies
$$f(x + 2f(y)) = f(x + y + 2c) ~ \forall ~ x,y \in \mathbb R \text{ with } x \ge 4|y|$$
Now taking $y=a,b$ in the above implies our claim. $\square$

[b][color=#f00]Claim: [/color][/b] For some $d \in \{0,1\}$ we have $f(x) = d ~ \forall ~ x \ge M$.

[b][color=#00f]Proof: [/color][/b] So replacing $x$ by $x+r$ in $(1)$ gives
$$f((x+r)^2 + 2c) = f(x^2 + 2c) ~ \forall ~ x \ge M \qquad{(3)}$$
Now any fix any $p > q \ge M$.  Note that a $x \ge M$ can be chosen such that
$$(x+r)^2 + 2c,x^2 + 2c \ge M ~~ \text{and} ~~ \frac{\Big(((x+r)^2 + 2c) - (x^2 + 2c) \Big) - (p-q)}{r} \in \mathbb Z$$
Then that along with $(3)$ and our previous claim would give that $f(p) = f(q)$. Hence there is a constant $d$ such that $f(x) = d ~ \forall ~ x \ge M$. Taking $x > 0$ sufficiently large in $(1)$ yields $d^2 = d$, i.e. $d \in \{0,1\}$. $\square$

[b][color=#f00]Claim: [/color][/b] $f(x) \in \{d,-d\} ~ \forall ~ x \in \mathbb R$.

[b][color=#00f]Proof: [/color][/b] Vary $x,y$ in $P(x,y)$ such that both of $x,y$ are negative and $xy = M$. We obtain there is a constant $M_0 < 0$ such that
$$f(x)^2 = d ~ \forall ~ x \le M_0 \qquad{(4)}$$
Now consider any real number $u$. Take $n$ to be a sufficiently large positive number. Consider $P(n+u,-n)$. We obtain $f(u)^2 = d$, which proves our claim. $\square$

So now our $P(x,y)$ becomes
$$Q(x,y) ~:~ f(x^2 + y^2 + 2f(xy)) = 1 $$
If $d=0$, then we directly have $f(x) \equiv 0$.

Now assume $d=1$. Let $S := \{x : f(x) = 1\}$ and $T := \{x : f(x) = -1 \}$.

[color=#f00][b]Claim: [/b][/color] $\left[ \tfrac{-2}{3},\infty \right] \subseteq S$, and consequently $T \subseteq \left( - \infty , \tfrac{-2}{3} \right)$.

[b][color=#00f]Proof: [/color][/b] Consider any $t \in T$. Fixing $xy = t$ in $Q(x,y)$ and varying $x,y$ we obtain all real numbers $\ge 2|t| - 2$ must be in set $S$ . Thus $t < 2|t| - 2$. This forces $t \notin \left[\frac{-2}{3},2 \right] $. Thus $\left[\frac{-2}{3},2 \right] \subseteq S$.  Then $Q(x,0)$ gives $[2,\infty] \subseteq S$. This proves our claim. $\square$

This completes the proof of the problem. $\blacksquare$

The answer is $f(x)\equiv x, f(x)\equiv 0$ and $f(x)=1\forall x\ge -\frac 23$ and $f(x)\in \{-1,1\}$ for all $x<-\frac 23$.

Let $P(x,y)$ denote the assertion in the question.

Case 1: $f$ is injective.

In this case, let $s=x+y$. Then $s^2-2xy+2f(xy)$ is constant if $s$ stays the same, so $f$ is linear for the interval $(-\infty, \frac s2)$. Picking $s$ large enough implies $f(x)\equiv x+C$. A simple substitution suggests $C=0$ so $f(x)\equiv x$.

Case 2: $f$ is not injective.

Suppose $f(a)=f(b)$. WLOG $a<b$. Pick $x,y,u,v$ such that $x^2+y^2=u^2+v^2=K$ for some $K>2\max\{|a|,|b|\}$, $xy=a$ and $uv=b$.

Now, $P(x,y)$ and $P(u,v)$ gives $f(u+v)^2 = f(K+2f(b))=f(K+2f(a))=f(x+y)^2$. We have $u+v=\sqrt{K+2b}$ and $x+y=\sqrt{K+2a}$. It follows that $|f(\sqrt{K+2a})| = |f(\sqrt{K+2b})|$ for all $K>\max\{2|a|, 2|b|\}$.

Claim: There exists a constant $C$ such that $f(x)$ is constant for $x\ge C$.

First, I prove that $f(x)\ge 0$ for a constant $x$ independent of $a,b$: if $t>2+2f(1)$ then there exists $x^2+y^2=t-2f(1), xy=1$ and thus $f(t)=f(x^2+y^2+2f(xy))=f(x+y)^2\ge 0$.

Picking $K$ such that $\sqrt{K+2a} > 2+2f(1)$ we obtain that $f(\sqrt{K+2a})=f(\sqrt{K+2b})$, or $f(\sqrt{K}) = f(\sqrt{K+nd})$ where $d=b-a, n\in \mathbb{N}$.

Suppose $\sqrt{K+d}-\sqrt{K}=c$ and $\sqrt{K+\epsilon+d}-\sqrt{K+\epsilon}=c-\delta$. Then, for all $x>2(K+\epsilon+d)$, we have $f(\sqrt{x}) = f(\sqrt{x+c}) = f(\sqrt{x+c-\delta})$ for some arbitrarily small $\delta$ that is not affected by other external variables other than $\epsilon$. By continuity, any sufficiently small $\delta$ is constructible, so we can carve a "path" from $f(m)$ to $f(n)$ where $n-m=\delta_1+\delta_2+\cdots+\delta_t$ and have $f(m)=f(m+\delta_1)=f(m+\delta_1+\delta_2)=\cdots=f(m+\delta_1+\cdots+\delta_t)$.

Say $f(x)=c$ for all $x>C$. Then $P(C+5, C+5)$ implies $c=c^2$ so $c\in \{0,1\}$.

Also, $P(x,y)$ and $P(-x,-y)$ suggests $f(x+y)^2 = f(-x-y)^2$ so it follows that $f(x)=\pm c$ for all $x<-C$.

Case 1: $c=0$. Now, $P(x,-x-d)$ for $x$ large suggests $f(x^2+(x+d)^2+2f(-x(x+d))=f(d)^2$. Taking $x(x+d)>C$ gives $|f(-x(x+d))|=c$ so LHS is c, so $f(x)\equiv 0$.

Case 2: $c=1$. Again we can prove that $f(x)\in \{1,-1\}$ for all $x$. Thus, $f(x^2+y^2+2f(xy))=1$.

Suppose $f(x)$ is not identically 1. Then if $f(t)=-1$, then we have 3 cases.

If $f(0)=-1$ then $P(0,y)$ suggests $f(y^2-2)=1$. Substituting $y=\sqrt{2}$ gives a contradiction.

If $f(t)=-1$ for some $t>0$ then for any $xy=t$ we have $x^2+y^2\ge 2t$. Since the range of $x^2+y^2-2$ is $\mathbb{R}_{\ge 2t-2}$ it follows $t<2t-2$ i.e. $t>2$. However, this implies that if $x^2+y^2=t-2$ then $f(xy)=-1$, and we can pick $0<xy<2$ to get a contradiction. Thus, $f(t)=1$ for all $t\ge 0$.

If $f(-t)=-1$ for some $t>0$ then take $x<0, y>0$ and then $x^2+y^2-2 \ne t$. Since the range of $x^2+y^2$ is $\mathbb{R}_{\ge 2t-2}$ it follows that $2t-2 > -t$ i.e. $t>\frac 23$. Thus, $f(x)=1$ for all $x\ge -\frac 23$. Observe that $f(x)=1\forall x\ge -\frac 23$ and $f(x)\in \{1,-1\}$ for all $x<-\frac 23$ is a solution because if $x^2+y^2+2f(xy)<0$ then $xy\le -\frac 23$ and thus $x^2+y^2 \ge \frac 43$.

Critical.

[hide=Solution]
Let $a=xy$ and $b=x+y.$ Observe that $b^2\geq 4a,$ define $g(a)=2f(a)-2a.$ So the original equation is $$f(b^2+g(a))=(f(b))^2 ~~\forall a,b \in \mathbb{R}.~~~~~(*)$$ If not $b^2\geq 4a$, then $x,y$ are complex.
$a\to 0\implies f(b^2+g(0))=(f(b))^2~~~~~(**)$
$\implies f(x)\geq 0 \text{ if } x\geq g(0).~~~~~(\phi)$
It follows that $g(0)\geq 0.$ Let $g(0)=c,$ where $c$ is some non negative constant.

Case 1: $g$ is constant.
Then $g(0)=g(a)=c \implies f(a)=a+c/2.$ But $c=0 \text{ in } (**) \implies \boxed{f(x)=x},$ which fits.

Case 2: $g$ is not constant.

Claim: $f$ is constant in some interval.
[hide=Proof] Suppose $\exists s,t \in \mathbb{R^+}:g(e)-g(f)=n ~~\forall n\in [s,2s]$ with $e,f\leq t.$
Take $y>x\geq 2\sqrt{t}: y^2-x^2\in [s,2s] \implies y^2-x^2=g(e)-g(f)\implies x^2+g(e)=y^2+g(f)\implies (f(x))^2=(f(y))^2.$ Note that by increasing $t: t\geq (c/2)^2$ we get $f(x)=f(y).$ So the function $f(\sqrt{x})\in [4t,+\infty)$ is periodic, so it is constant in this interval. It follows that $f\in [2\sqrt{t},+\infty)$ is constant. $\blacksquare$[/hide]

Claim: If $x\in H$ then $f(x)=-1, f(x)=1$ otherwise. ($H$ is any subset of $(-\infty,-2/3)$)
[hide=Proof]So $\exists N,Q \in \mathbb{R}:f(x)=Q ~~\forall x\geq N>0.$ Taking very large $b$ in $(**)$ we get $Q=0$ or $Q=1.$ It also follows that $f(-b)=\pm f(b) ~~\forall b.$ So $|f(x)|=Q \leq 1 ~~\forall x\in (-\infty,-N] \implies$ $$g(a)= -2a-2 ~~\forall a\in (-\infty,-N]~~~~~(\$),$$ it shows that $g$ can take extremely large values in this interval. So $\forall b \in \mathbb{R} \exists a\in [N,0): b^2+g(a)\geq N\implies f(b^2+g(a))=Q.$ By $(*), |f(b)|=Q ~~\forall b\in \mathbb{R}.$

Subcase 1:
If $Q=0,$ then $\boxed{f(x)=0},$ which obviously fits.
Subcase 2:
If $Q=1,$ then $c=2f(0)=2.$ Easy to get contradiction with $f(a)=-1.$ So $a-g(a)<0$ or $0\leq a-g(a)<4a.$ But $(\$)$ yields $a<-2/3$ or $a>2.$ Easy to get contradiction with $a>2.$ So if $H$ is any subset of $(-\infty,-2/3)$ it follows that $\boxed{f(x)=-1~~\forall x\in H}.$ It is not hard to see that $\boxed{f(x)=1 ~~\forall x\notin H}.$ Both of these work. $\blacksquare$ [/hide]

Claim: Exists such $s$ and $t.$
[hide=Proof]Take $u,v:g(u)-g(v)=n>0$ and take $i\geq \max\{2\sqrt{|u|},2\sqrt{|v|},c\}=M,$ also take $j=\sqrt{i^2+n}.$ It follows that $i^2+g(u)=j^2+g(v)$ for $i^2\geq 4u, j^2\geq 4j$ and $j>i\geq c.$ From $(*)$ and $(\phi)$ we have $f(i)=f(j)\implies g(i)-g(j)=2(j-i)=\frac{2(j^2-i^2)}{j+i}=\frac{2(g(a)-g(b))}{j+i} =\frac{2n}{i+\sqrt{i^2+n}}=q(i).$ Let $q(M)=2s$ and for $L>M, q(L)=s.$ So $\forall z\in [s,2s], \exists i\in [M,L]: q(i)=i \implies z=g(i)-g(j).$ We may choose $t=L+\sqrt{n}$ so that $i,j\leq t.$ Thus the proof is complete. $\blacksquare$[/hide]
[/hide]

Solved with [b]lrjr24[/b]

Note that this is equivalent to \[ f(s^2+2(f(p)-p)) = f(s)^2 \] for $s^2 \ge 4p$. Denote this assertion with $P(s, p)$.
By $P(s, p)$ and $P(-s, p)$ for $p < 0$ it follows that $|f(s)| = |f(-s)|$.
If $f$ is injective, then $f(x) - x$ is constant so $f(x) = x + c$. Checking gives that only $c = 0$ works.
Else, suppose that there are two distinct $a, b$ such $f(a) - a > f(b) - b$.
Then, $f$ is periodic with period $t = (f(a) - a) - (f(b) - b)$ on some interval $I = (c, \infty)$.
Since sufficiently large $f$ is nonnegative, $f$ must be nonnegative on $I$.

[b][color=pink]Claim:[/color][/b] $f$ is eventually constant.
[i]Proof.[/i] Note that for $\sqrt{x} \in I$ \[ f\left(\sqrt{x}\right)^2 = f(x + 2(f(p) - p)) = f(x + 2(f(p) - p) + t) = f\left(\sqrt{x + t}\right)^2. \]
Then, by $P(x, a)$ for $x \ge 2a$ it follows that $f\left(\sqrt{x + 2a}\right) = f(x + 2f(a))$
As such, for $y \ge 2\sqrt{x+t}$, \[ f\left(\sqrt{y + 2\sqrt{x}}\right) = f\left(\sqrt{y + 2\sqrt{x + t}}\right) \]
Then, take sufficiently large $a > b$. It follows that for $r^2, s^2 > 4\sqrt{a + t}$ and \[ 2\sqrt{b + t} - 2\sqrt{b} < r^2 - s^2 < 2\sqrt{a + t} - 2\sqrt{a}, \] $f(r) = f(s)$.
Fix a $r$. By repeating the above we have that $f(r) = f(s)$ if \[ s^2 \in \left[r^2 + n \cdot \left(2\sqrt{b + t} - 2\sqrt{b}\right), r^2 + n  \cdot \left(2\sqrt{a + t} - 2\sqrt{a}\right)\right] \] for some $n$.
From taking a sufficiently large $n$ it follows that $f$ is constant on some interval with length larger than $t$ in $I$. By periodicity it follows that $f$ is constant on $I$. $\blacksquare$
Let this constant on sufficiently large inputs be $d$.
Since $f$ is constant for sufficiently large, taking $P(s, p)$ for large enough $s$ means that $d^2 = d$ so either $d = 0$ or $d = 1$.
Thus, by $P(s, p)$ for sufficiently negative $p$ it follows that \[ f(s^2 + k) = f(s)^2 \] for arbitrarily large $k > 0$.
As such, $f(x)^2 = d$ for all $x$.
If $d = 0$, then $f(x) = 0$.
From here on, suppose $d = 1$.

[b][color=pink]Claim:[/color][/b] $f$ is $1$ on $\left(-\frac23, \infty\right)$.
[i]Proof.[/i] By $P(s, 0)$ it follows that \[ f(s^2 + 2f(0)) = 1 \]
Since $f(0) = \pm 1$, it follows that $f$ is constant on $[2, \infty)$.
For the sake of contradiction suppose that for some $x \in \left[0, 2\right)$ that $f(x) = -1$.
Then, by $P\left(s, x\right)$ it follows that \[ f(s^2 - 2 - 2x) = 1 \] for $s^2 \ge 4x$. Since $2x - 2 \le x$ this leads to a contradiction.
Now, suppose that for some $x \in \left(-\frac23, 0\right)$ that $f(x) = -1$. Then, by $P\left(s, x\right)$ \[ f(s^2 - 2 - 2x) = 1 \] However, $-2 - 2x \le x$ holds, contradiction. $\blacksquare$
As such, either $f(x) = x$, $f(x) = 0$, or \[ f(x) = \begin{cases} 1 & x \ge -\frac23 \\ \pm 1 & x < -\frac23 \end{cases} \] pointwise.


inequalities
trigonometry
logarithms
calculus
calculus computations

$1.\blacktriangleright$ Ascertain $x>-2$ for which $2^{x}+6^{x}+3^{x}=2x^{2}+6x+3\ .$
$2.\blacktriangleright$ Prove that $x\in(0,\frac{\pi}{2})\Longrightarrow x+\cos x-1<\ln (x+1)<x\ .$
$3.\blacktriangleright$ Solve the system $\{\begin{array}{c}e^{x}+e^{y}=2\\\ xy=x+y\end{array}$

exactly what are u asking for in number 1?

The first is an equation on $(-2, \infty )\ .$

$2.\blacktriangleright$ Prove that $x\in\left(0,\frac{\pi}{2}\right)\Longrightarrow \ln (x+1)>x+\cos x-1\ .$

$\left\{\begin{array}{c}\underline{f(x)=\ln (x+1)-x-\cos x+1}\\\\ f'(x)=\frac{1}{x+1}-1+\sin x\\\\ f''(x)=-\frac{1}{(x+1)^{2}}+\cos x\\\\ f'''(x)=\frac{2}{(x+1)^{3}}-\sin x\\\\ f''''(x)=-\frac{6}{(x+1)^{4}}-\cos x<0 \end{array}\right|$ $\Longrightarrow$  $\left\{\begin{array}{cccccc}x&0&\ldots&\ldots&\ldots&\frac{\pi}{2}\\\\ f'''(x)&+&\searrow&0&\searrow&-\\\\ f''&0&\nearrow&+&\searrow&-\\\\ f'&0&\nearrow&+&\searrow&+\\\\ f&0&\nearrow&+&\nearrow&+\end{array}\right|$ $\Longrightarrow\left(\forall\right)\ x\in \left(0,\frac{\pi}{2}\right)\ ,\ f(x)>0$, i.e. $\ln (x+1)>x+\cos x-1\ .$

hello, your system
$ e^x\plus{}e^y\equal{}2$
$ xy\equal{}x\plus{}y$
has only the solution $ x\equal{}y\equal{}0$
Sonnhard.

hello, your equation
$ 2^x\plus{}6^x\plus{}3^x\equal{}2x^2\plus{}6x\plus{}3$
has only the solutions $ x\equal{}0$ or $ x\equal{}1$
Sonnhard.


calculus
integration
logarithms
calculus computations

hello, do you want to calculate
$ \int ln\left(\frac{x}{a\plus{}2x}\right) dx$?
Sonnhard.

hello,here integration by parts will help.
$ \int 1\cdot \ln\left(\frac {x}{a \plus{} 2x}\right)dx \equal{} x\ln\left(\frac {x}{a \plus{} 2x}\right) \minus{} a\int\frac {dx}{a \plus{} 2x}$.
Sonnhard.

so i get this?
integral(ln(x/(a+2x)))dx=xln(x/(a+2x))-integral(a/(a+2x))
thank you for your help.

hello, yes on gets
$ \int \ln\left(\frac{x}{a\plus{}2x}\right)dx\equal{}x\ln\left(\frac{x}{a\plus{}2x}\right)\minus{}\frac{a}{2}\ln|a\plus{}2x|\plus{}C$.
Sonnhard.


calculus integration trigonometry function limit logarithms inequalities

Evaluate: \[ \int\limits_0^{\frac\pi2}\sin(\omega\tan x)\,dx\]

Is $ \omega$ some constant?  The integral involves some hyperbolic trig functions and imaginary components.

[b]JRav[/b]
Thank you for reply. Yes, now I've came to similar conclusion...
May be it represents in some rational form if $ \omega\equal{}1$? For comparison, similar integral with $ \cos$ instead of $ \sin$ is evaluable exactly for all $ \omega$.

$ \int_{0}^{\infty} \frac {\sin (\omega t)}{1 + t^2} \, dt = \frac {e^{ - \omega}\text{Ei}(\omega) - e^{\omega}\text{Ei}( - \omega)}{2}$, where $ \text{Ei}$ is the [url=http://en.wikipedia.org/wiki/Exponential_integral]exponential integral[/url].

[hide="My solution"]Let $ I(\omega) = \int_{0}^{\infty} \frac {\sin (\omega t)}{1 + t^2} \, dt$.

By Leibniz's rule for integral, one can show that $ I$ satisfies the differential equation $ I'' - I + \frac {1}{\omega} = 0$.

Then the general solution for $ y'' - y + \frac {1}{x} = 0 \ \cdots \ (\ast)$ on $ (0, \infty)$ is given by $ y = Ae^{x} + Be^{ - x} + I(x)$. This shows that the only bounded solution of $ (\ast)$ satisfying $ \lim_{x \to 0^{ + }} y = 0$ is $ y = I(x)$.

We claim that $ F(x) = \frac {e^{ - x}\text{Ei}(x) - e^{x}\text{Ei}( - x)}{2}$ is the very solution for $ (\ast)$.

Using the relation $ \text{Ei}(x) = PV\int_{ - \infty}^{x} \frac {e^{t}}{t} \, dt$, one can easily show that $ y = F(x)$ does satisfy $ (\ast)$.

It remains to show that (1) $ \lim_{x \to 0^{ + }} F(x) = 0$ and (2) $ F$ is bounded on $ (0, \infty)$.

For $ x > 0$, we have $ F(x) = \sinh x \int_{x}^{\infty} \frac {e^{ - t}}{t} \, dt + e^{ - x} \int_{0}^{x} \frac {\sinh t}{t} \, dt$. Then

(1) For $ 0 < x < 1$,

$ \begin{eqnarray*} |F(x)| & \leq & \sinh x \int_{x}^{1} \frac {1}{t} \, dt + \sinh x \int_{1}^{\infty} e^{ - t} \, dt + e^{ - x} \int_{0}^{x} \frac {\sinh 1}{1} \, dt \\
& = & - \sinh x \ln x + e^{ - 1} \sinh x + x e^{ - x}\frac {\sinh 1}{1} \end{eqnarray*}.$

Applying squeeze theorem to this inequality, we obtain $ \lim_{x\to 0^{+}} F(x) = 0$ and $ F$ is bounded on $ (0, 1)$.

(2) For $ x \geq 1$,

$ \begin{eqnarray*} |F(x)| & \leq & \sinh x \int_{x}^{\infty} e^{ - t} \, dt + e^{ - x} \int_{0}^{x} \frac {\sinh x}{x} \, dt \\
& = & 2 e^{ - x} \sinh x \end{eqnarray*},$

which obviously implies that $ F$ is bounded on $ [1, \infty)$. Thus $ F$ is bounded on $ ( 0, \infty)$.

Then $ I(x) = F(x)$ for all $ x > 0$. Since both $ I$ and $ F$ is odd, this equality extends to all real $ x$ by defining $ F(0) = 0$.[/hide]


geometry
3D geometry
tetrahedron
geometry unsolved

Let $ABCD$ be a tetrahedron satisfying
i)$\widehat{ACD}+\widehat{BCD}=180^{0}$, and
ii)$\widehat{BAC}+\widehat{CAD}+\widehat{DAB}=\widehat{ABC}+\widehat{CBD}+\widehat{DBA}=180^{0}$.
Find value of $[ABC]+[BCD]+[CDA]+[DAB]$ if we know $AC+CB=k$ and $\widehat{ACB}=\alpha$.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.VMO1992Problem1

by [i]Vo Duc Dien[/i]

This is the hardest problem I've solved thus far. For more solutions please visit

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.MathematicalOlympiads


There is a wide river and a narrow old bridge across it. In order to cross the bridge during the
night, one has to carry a flashlight, and at most two people can cross the bridge together (using the same flashlight.) The bridge is so unsafe that noone can even stay on it without a flashlight. If two people cross together, they move at the speed of the slower person.

One night during a war, four soldiers end up on one side of the bridge while running from the other army. The soldiers need to cross the bridge in at most 17 minutes in order to escape. However, they have only one flashlight. Due to some injuries it is going to take each soldier 1, 2, 5 and 10 minutes, respectively, to cross the bridge individually. It is easy to see that the soldiers can cross the bridge in 19 minutes. Would they be able to cross it in 17 minutes and thus escape?

(Source = Some math PDF file in the internet)

[quote="isabella2296"][quote="xcenario"] Due to some injuries it is going to take each soldier 1, 2, 5 and 10 minutes, respectively, to cross the bridge individually. It is easy to see that the soldiers can cross the bridge in 19 minutes.
[/quote]

$ 1 \plus{} 2 \plus{} 5 \plus{} 10 \equal{} 18$, not $ 19$...[/quote]

I just wrote it exactly as it was in the PDF... and if you think about it, theres only 1 flashlight, so my guess one would be having to come back and pick up another guy to continue crossing.

Btw, here's the problem source: [url]http://mathcircle.berkeley.edu/original/Rookie-Veteran1.pdf[/url] (1-3 Problem 2)

[quote="isabella2296"][quote="xcenario"] Due to some injuries it is going to take each soldier 1, 2, 5 and 10 minutes, respectively, to cross the bridge individually. It is easy to see that the soldiers can cross the bridge in 19 minutes.
[/quote]

$ 1 \plus{} 2 \plus{} 5 \plus{} 10 \equal{} 18$, not $ 19$...[/quote]
how did u get that

yes they have time the 1 minute brings the ten throws the flashlight and the others cross

mathblitz, its not a trick question

the soluitn is

person 1 and 2 cross, 1 brings back flashlight, 5 and 10 cross, 2 brings back flasslight, 1 and 2 cross again.
=2+1+10+2+2=17 mins

[quote="isabella2296"]How did who get what?[/quote]
nevermind .

[quote="isabella2296"][quote="xcenario"] Due to some injuries it is going to take each soldier 1, 2, 5 and 10 minutes, respectively, to cross the bridge individually. It is easy to see that the soldiers can cross the bridge in 19 minutes.
[/quote]

$ 1 \plus{} 2 \plus{} 5 \plus{} 10 \equal{} 18$, not $ 19$...[/quote]

19 minutes is NOT referring to 1+2+5+10...

It'll probably take them more than 2 minutes to solve it--

Please refrain from reviving old threads, especially with spam. Thank you.  :wink:


geometry
geometric transformation
reflection
incenter
search
trapezoid
trigonometry

Consider triangle $ABC$. Let $D$, $E$, and $F$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively. The incircle touches sides of the triangle at points $X$, $Y$, and $Z$; $X$ lies on $BC$, $Y$ on $AC$ and $Z$ on $AB$. Lines $XY$ and tangent to incircle from $F$ meet at $K$. Points $L$ and $M$ are defined similarly. Prove that $K, L, M$ are collinear.

[b][size=109]LEMMA.[/size][/b] – A triangle $ \bigtriangleup ABC$ is given and let $ X,$ $ Y,$ $ Z,$ the tangents points of it’s incircle $ (I),$ to the sidelines $ BC,$ $ AC,$ $ AB,$ respectively.
We define three arbitrary points $ D,$ $ E,$ $ F,$ on $ BC,$ $ AC,$ $ AB$ respectively, such that $ R\equiv AD\cap BE\cap CF$ and we denote as $ N,$ $ P,$ $ Q,$ the second tangent points to $ (I),$ from $ D,$ $ E,$ $ F,$ respectively.
Prove that the segment lines $ AN,\ BP,\ CQ,$ are concurrent at one point.

$ \bullet$ In our configuration now, we denote the points $ K\equiv XY\cap FQ,$ $ L\equiv XZ\cap EP$ and $ M\equiv YZ\cap DN.$

Because of the segment line $ XY,$ as the polar of the vertex $ C$ of $ \bigtriangleup ABC,$ with respect to the incircle $ (I),$ passes through the point $ K,$ we conclude that the polar of $ K,$ wrt $ (I),$ passes through $ C.$

So, because of the point $ Q,$ is the tangent point to $ (I),$ from $ K,$ we conclude that the segment line $ CQ,$ is the polar of $ K,$ wrt the incircle $ (I)$ and similarly the segment lines $ BP,$ $ AN,$ are the polars of $ L,$ $ M$ respectively, with respect to $ (I).$

Because of now, the segment lines $ AN,$ $ BP,$ $ CQ,$ based on the above Lemma, are concurrent at one point, so be it $ S,$ we conclude that the orthogonal projections of $ K,$ $ L,$ $ M,$ on the segment line $ IS,$ where $ I,$ is the incenter of $ \bigtriangleup ABC,$ are coincided with the point $ S',$ as the harmonic conjugate of $ S,$ with respect to the points $ T,$ $ T',$ as the intersection points of $ (I),$ from the segment line $ IS.$

See at http://www.mathlinks.ro/Forum/viewtopic.php?t=139564

Hence the points $ K,$ $ L,$ $ M,$ are collinear and the proof is completed.

Kostas Vittas.

PS. The above Lemma is well known (at least to me) and probably it has already been posted in this forum. I don’t remember it’s solution and so, I will search my archive for the relative reference.

[quote="Huyền Vũ"]The tangent of incircle from D,E,F touch incircle at N,P,Q
We see C is the polar of XY wrt incircle so C is on the polar line of K wrt (O)
KQ is the tangent of incircle so Q is on the polar line of K wrt (O)
Thus K is the polar of CQ wrt (O) (1)
Similarly L,M is the polar of AN,BP wrt (O) (2)
Otherwise the lines AN,BP,CQ is the reflection of AX,BY,CZ to AD,BE,CF
But AX,BY,CZ and AD,BE,CF are concurent
so AN,BP,CQ are concurent (3)
From (1) (2) (3) we have K,L,M are collinear q.e.d[/quote]
Oh sorry  :blush: ! Because there are some things wrong in my solution :blush:  I'll correct it later

Finally, we don’t need any Lemma, to prove the concurrency of the segment lines $AN,$ $BP,$ $CQ,$ when $D,$ $E,$ $F,$ are the midpoints of $BC,$ $AC,$ $AB$ respectively, because of they are concurrent at the [b][size=109]Nagel’s point[/size][/b] of the given triangle $\bigtriangleup ABC.$

It is easy to show that the point $X',$ as the intersection point of $BC,$ from the segment line $AN,$ is the reflexion of $X,$ with respect to the midpoint $D,$ of $BC$ $($ well known problem, which states that the line $AX',$ passes through the antidiametric (= diametric opposite) point of $X,$ with respect to the incircle (I) $).$

That is the points $X,$ $X',$ are isotomic points wrt $BC$ and similarly the points $Y'\equiv BP\cap AC,$ $Z'\equiv CQ\cap AB,$ are the isotomic points of $Y,$ $Z$ respectively, wrt $AC,$ $AB.$

So, it is easy to show the concurrency of the segment lines $AN,$ $BP,$ $CQ,$ as the consequence of the concurrency of the segment lines $AX,$ $BY,$ $CZ,$ at the $($ [b][size=109]Gergone’s point[/size][/b] of $\bigtriangleup ABC$ $).$

Kostas Vittas.

PS. I will post here next time, a difficult but interesting proof of the Lemma I mentioned (necessary for the solution of generalized [b][size=109]''cvix’s[/size][/b] problem), which I found in my archive.

[quote="vittasko"] ....
It is easy to show that the point $X',$ as the intersection point of $BC,$ from the segment line $AN,$ is the reflexion of $X,$ with respect to the midpoint $D,$ of $BC$ $($ well known problem, which states that the line $AX',$ passes through the antidiametric (= diametric opposite) point of $X,$ with respect to the incircle (I) $).$
....[/quote]

[b][size=109]PROPOSITION.[/size][/b] – A triangle $\bigtriangleup ABC$ is given and let $X,$ $Y,$ $Z,$ the touch points of it’s incircle $(I),$ to the sidelines $BC,$ $AC,$ $AB$ respectively. If $X'',$ the antidiametric point of $X,$ with respect to $(I),$ prove that the segment line $AX''$ intersects $BC,$ at the touch point of the A-excircle of $\bigtriangleup ABC.$

[b][size=109]PROOF.[/size][/b] – We denote as $Y',$ $Z',$ the intersection point of $AC,$ $AB$ respectively, from the tangent line of $(I)$ at point $X''.$

Also, we denote the point $X'\equiv BC\cap AX''$ and it is enough to prove that $DX = DX',$ where $D,$ is the midpoint of $BC.$

Because of the trapezium $BCY'Z',$ is cicrcumscribed to $(I),$ by [b][size=109]Newton’s theorem,[/size][/b] we conclude that the segment line $XX'',$ passes through the point $V,$ as the intersection point of it’s diagonals $BY',$ $CZ'.$

So, from the similar triangles $\bigtriangleup BXV,$ $\bigtriangleup Y'X''V$ and $\bigtriangleup CXV,$ $\bigtriangleup Z'X''V,$ we have that $\frac{XB}{X''Y'}= \frac{VX}{VX''}= \frac{XC}{X''Z'}$ $\Longrightarrow$ $\frac{XB}{XC}= \frac{X''Y'}{X''Z'}$ $,(1)$

From $Y'Z'\parallel BC$ $\Longrightarrow$ $\frac{X''Y'}{X''Z'}= \frac{X'C}{X'B}$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $\frac{XB}{XC}= \frac{X'C}{X'B}$ $\Longrightarrow$ $XB = X'C$ $\Longrightarrow$ $DX = DX'$ and the proof is completed.

Kostas Vittas.

PS. Relatively to [b][size=109]''cvix’s[/size][/b] problem, if $N$ is the intersection point of $(I),$ from the segment line $AX''X',$ it is easy to show that the segment line $DN$ tangents to the incircle $(I),$ at the point $N$ $($ $DN = DX = DX',$ from the triangle $\bigtriangleup XNX',$ right angled at $N$ $).$

[quote="vittasko"][b][size=109]LEMMA.[/size][/b] – A triangle $\bigtriangleup ABC$ is given and let $X,$ $Y,$ $Z,$ the tangents points of it’s incircle $(I),$ to the sidelines $BC,$ $AC,$ $AB,$ respectively.
We define three arbitrary points $D,$ $E,$ $F,$ on $BC,$ $AC,$ $AB$ respectively, such that $R\equiv AD\cap BE\cap CF$ and we denote as $N,$ $P,$ $Q,$ the second tangent points to $(I),$ from $D,$ $E,$ $F,$ respectively.
Prove that the segment lines $AN,$ $BP,$ $CQ,$ are concurrent at one point.[/quote]

[b][size=109]PROOF  OF  THE  LEMMA.[/size][/b] – We denote as $Z',$ the intersection     point of $AB,$ from the segment line $CQ.$

Also, we denote the points $W\equiv AQ'\cap CF,$ $U\equiv AC\cap FQ,$ $Q'\equiv BC\cap FQ$ and $K'\equiv QZ\cap XY.$

In the configuration of circumscribed quadrilateral $AFQ'C,$ by [b][size=109]Newton’s theorem,[/size][/b] we have that the cords $XZ,$ $QY,$ pass through the point $W$ and so, the triangles $\bigtriangleup CXY,$ $\bigtriangleup FZQ$ are perspective.

Hence, by [b][size=109]Desarques’s theorem,[/size][/b] we conclude that the points $U\equiv CY\cap FQ,$ $B\equiv CX\cap FZ$ and $K'\equiv XY\cap ZQ,$ are collinear.

That is the intersection point $K',$ of $QZ,$ $XY,$ lies on the line connecting the  points $U,$ $B,$ as the intersedtion points of the opposite sidelines of the quadrilateral $AFQ'C.$

By the same way we can prove that the point $U'\equiv XQ\cap YZ,$ also lies on the segment line $UB$ and so, the triangles $\bigtriangleup AYZ,$ and $\bigtriangleup Q'XQ$ are perspective.

Hence, the segment lines $AQ',$ $YX,$ $ZQ,$ are concurrent at the point $K',$ as the intersection point of $ZQ,$ $YX.$

$\bullet$ We consider the pencil $K'.Q'BFQ$ intersected from the segment lines $AB,$ $QU$ and so we have that $(A,B,F,Z) = (Q',U,F,Q) = (C.Q'UFQ)$ $,(1)$

But because of the pencil $C.Q'UFQ,$ is intersected from the sideline line $AB$ of $\bigtriangleup ABC,$ we have that $(Q',U,F,Q) = (B,A,F,Z') = (A,B,Z',F)$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $(A,B,F,Z) = (A,B,Z',F)$ $,(3)$

From $(3)$ $\Longrightarrow$ $\frac{FA}{FB}: \frac{ZA}{ZB}= \frac{Z'A}{Z'B}: \frac{FA}{FB}$ $\Longrightarrow$ $\frac{Z'A}{Z'B}= \frac{(FA)^{2}}{(FB)^{2}}\cdot \frac{ZB}{ZA}$ $,(4)$

From $(4),$ we see that when the point $F,$ is coincided with the midpoint of the sidesegment $AB,$ then the points $Z,$ $Z',$ are isotomic conjugates with respect to $AB.$

By the same way, we can prove that $\frac{X'B}{X'C}= \frac{(DB)^{2}}{(DC)^{2}}\cdot \frac{XC}{XB}$ $,(5)$ and $\frac{Y'C}{Y'A}= \frac{(EC)^{2}}{(EA)^{2}}\cdot \frac{YA}{YC}$ $,(6)$

From $(4),$ $(5),$ $(6)$ $\Longrightarrow$ $\frac{Z'A}{Z'B}\cdot \frac{X'B}{X'C}\cdot \frac{Y'C}{Y'A}= 1$ $,(7)$ $($ $X'\equiv BC\cap AN,$ $Y'\equiv AC\cap BP$ $).$

From $(7),$ by [b][size=109]Ceva’s theorem,[/size][/b] we conclude that the segment lines $AN,$ $BP,$ $CQ,$ are concurrent at one point,so be it $S$ and the proof is completed.

Kostas Vittas.

[b][size=109]TWO EXTENTED RESULTS OF THE GENERALIZED PROBLEM.[/size][/b]

[b]1)[/b] - To prove that the points $K'\equiv XY\cap QZ,$ $L'\equiv XZ\cap PY$ and $M'\equiv YZ\cap XN,$ are collinear.

[b]2)[/b] - To prove that the lines $KLM,$ $K'L'M',$ are parallel each other, when the points $D,$ $E,$ $F,$ are the midpoints of $BC,$ $AC,$ $AB,$ respectively.

Kostas Vittas.

The idea of the generalized problem is the famous Steinbart's theorem and some pole and polar knowledge.

[b]Steinbart's theorem[/b] If $N, P, Q$ are three arbitrary points on the incircle, then the two following statements are equivalent.
  (a) AN, BP, CQ are concurrent
  (b) XN, YP, ZQ are concurrent or the intersections of $XN, YP, ZQ$ with the corresponding sides of $XYZ$ are collinear.

For the proof, I think you should check Darij's webpage.

To make this problem easier to see, I will assume that $D, E, F$ are collinear instead of concurrent. The concurrent case is similar. Since $D, E, F$ are collinear, their polars with respect to $(I)$ which are $XN, YP, ZQ$ are concurrent (at the pole of the line $DEF$). Hence by Steinbart's theorem, we obtain $AN, BP, CQ$ are concurrent.

Many thanks [b][size=109]''treegoner'',[/size][/b] for your remarks and advices.

''Nothing is new under the geometry sun'', as Francois Rideau said and indeed the Darij’s web page, is an interesting part of this sun.

By the way of this problem, I learned about the [b][size=109]Steinbart’s theorem[/size][/b] ( many nice problems appear in Hellenic textbooks, published the last 50 years, but unfortunately, without references ).

Thank you again, Kostas Vittas.

Let triangle made of those tangents to incircle at N,P,Q   be called
  triangle W.

  Show that  W is perspective with DEF at P1.

  Then show that line connecting I and P1 is perpendicular to line KLM.

  M.T.

To vittasko: I think you were too formal  :lol: . I didn't give any advice.

To armpist: For the sake of generalization, I will state the following theorem from which your problem will follow.

[b]Theorem[/b] Let $ABC$ and $A'B'C'$ are two perspective triangles such that they are excribed by a common circle $(I)$. Then the axis of homology will be the polar of the perspector wrt $(I)$ (and thus is perpendicular to the line connecting $I$ and the perspector.

[b]Solution[/b] Let's call $P$ the perspector. Denote by $D, E, F$ the points of tangency of the sides of $ABC$ with the circle $(I)$. Define $D', E', F'$ analogously.

We show that $DD', EE', FF'$ are concurrent at a point $S$ which is the pole of the axis of homology wrt $(I)$..

[hide="Proof"]This is indeed not hard if you know pole and polar. Clearly, $BC$ is the polar of $D$ and $B'C'$ is the polar of $D'$. Hence, $DD'$ is the polar of the intersection of $BC$ and $B'C'$. Since the intersections of the corresponding sides of $ABC$ and $A'B'C'$ form the axis of homology, it follows easily that $DD', EE', FF'$ must be concurrent at $S$ which is the pole of the axis of homology wrt $(I)$. [/hide]

We now show that $S$ is in fact $M$.

[hide="Proof"] We use the degenerated Pascal's theorem for 4 points. Consider 4 points $E, F, E', F'$, then $A, S, A'$ are collinear. Hence $S$ is $M$. [/hide]

I restated the lemma mentioned by dear [b]Kostas Vittas[/b] in #7 as follow:
[quote]In a triangle $ ABC$, let $ P$ a point in the interior of the triangle $ ABC$ and consider its cevians $ AA_1$, $ BB_1$, $ CC_1$. Denote by $ X$, $ Y$, $ Z$ the tangents from $ A_1$, $ B_1$, $ C_1$ to the incircle $ \omega (I)$ of triangle $ ABC$ such that $ X\notin BC$, $ Y\notin CA$, $ Z\notin AB$. Prove that the lines $ AX$, $ BY$, $ CZ$ are concurrent.[/quote]

[b]Solution.[/b] We first introduce a nice lemma.

[i]Lemma.[/i] Let $ ABC$ be a triangle and let $ D$, $ E$, $ F$ be the points of tangency of the incircle $ \omega (I)$ of triangle $ ABC$ with the sides $ BC$, $ CA$, $ AB$. Let $ A_1$ be a point lies on the segment $ BC$. Denote by $ X$ the tangent from $ A_1$ to $ \omega (I)$, $ A_2$ the second intersection of $ AX$ with $ \omega (I)$, and $ A_3$ the intersection of $ AA_1$ with $ EF$, then we have $ D$, $ A_2$ and $ A_3$ are collinear.
[i]
Proof.[/i] Denote by $ K$, $ L$ the intersections of $ AA_1$ with the circle $ \omega (I)$ ($ L$ nearer $ A$ than $ K$). We know the division $ (ALA_3K)$ is harmonic $ (1)$.

Denote $ A_3' \equal{} A_2D\cap EF$. We have the quadrilateral $ DKXL$ is harmonic, so the pencil $ A_2(DKXL)$ is harmonic. By intersecting it with the line $ AA_1$, we have the division $ (ALA_3'K)$ is harmonic $ (2)$.

From $ (1)$ and $ (2)$ we have $ A_3\equiv A_3'$. The conclusion follows.

$ \bullet$ Back to the proposed problem.

Denote by $ D$, $ E$, $ F$ be the tangency points of $ \omega (I)$ with the sides $ BC$, $ CA$, $ AB$, respectively. Let $ A_2$, $ B_2$, $ C_2$ be the second intersections of $ AX$, $ BY$, $ CZ$ with $ \omega (I)$ and $ A_3$ the intersection of $ AA_1$ with $ EF$, $ B_3$ the intersection of $ BB_1$ with $ FD$, $ C_3$ the intersection of $ CC_1$ with $ DE$.

Since the lines $ AX$, $ EX$, $ FX$ are concurrent, by the trigonometric form of Ceva's theorem, we have $ \frac {\sin\angle XAE}{\sin\angle XAF}\cdot\frac {\sin\angle XEF}{\sin\angle XEA}\cdot\frac {\sin\angle XFA}{\sin\angle XFE} \equal{} 1$, which implies $ \frac {\sin\angle XAE}{\sin\angle XAF} \equal{} \frac {\sin\angle XEA}{\sin\angle XEF}\cdot\frac {\sin\angle XFE}{\sin\angle XFA} \equal{} \left(\frac {\sin\angle XFE}{\sin\angle XEF}\right)^2$. Therefore $ \frac {\sin\angle XAE}{\sin\angle XAF} \equal{} \left(\frac {XE}{XF}\right)^2 \equal{} \left(\frac {A_2E}{A_2F}\right)^2$ $ (1)$. By the above lemma, $ D$, $ A_2$ and $ A_3$ are collinear, therefore $ \left(\frac {A_2E}{A_2F}\right)^2 \equal{} \left(\frac {A_3E}{A_3F}\cdot\frac {FD}{DE}\right)^2 \equal{} \left(\frac {\sin\angle A_1AC}{\sin\angle A_1AB}\cdot\frac {FD}{DE}\right)^2$ $ (2)$.

From $ (1)$ and $ (2)$, we have $ \frac {\sin\angle XAE}{\sin\angle XAF} \equal{} \left(\frac {\sin\angle A_1AC}{\sin\angle A_1AB}\cdot\frac {FD}{DE}\right)^2$. Similarly, $ \frac {\sin\angle YBF}{\sin\angle YBD} \equal{} \left(\frac {\sin\angle B_1BA}{\sin\angle B_1BC}\cdot\frac {DE}{EF}\right)^2$, and $ \frac {\sin\angle ZCD}{\sin\angle ZCE} \equal{} \left(\frac {\sin\angle C_1CB}{\sin\angle C_1CA}\cdot\frac {EF}{FD}\right)^2$. Multiplying these three equations,
\[ \frac {\sin\angle XAE}{\sin\angle XAF}\cdot\frac {\sin\angle YBF}{\sin\angle YBD}\cdot \frac {\sin\angle ZCD}{\sin\angle ZCE} \equal{} \left(\frac {\sin\angle A_1AC}{\sin\angle A_1AB}\cdot \frac {\sin\angle B_1BA}{\sin\angle B_1BC}\cdot\frac {\sin\angle C_1CB}{\sin\angle C_1CA}\right)^2
\]
Note that $ AA_1$, $ BB_1$ and $ CC_1$ are concurrent, therefore $ \frac {\sin\angle XAC}{\sin\angle XAB}\cdot\frac {\sin\angle YBA}{\sin\angle YBC}\cdot \frac {\sin\angle ZCB}{\sin\angle ZCA} \equal{} 1$. It is equivalent to $ AX$, $ BY$, $ CF$ are concurrent, and we are done.


I'm pretty bad at geometry, and I was wondering that if I can master all of the geometry in AOPS 1, if I would still need to get Intro to Geometry. I took the post test and could do all of them pretty easily except for 7 (it took me awhile) and 10 (I didn't even try.)

AoPS Vol. 1 doesn't go into Geometry in nearly as much detail as Introduction to Geometry. Keep in mind that AoPS Vol. 1/2 are mainly for contest preparation and are best used as review. Do Introduction to Geometry and Challenge Problems (maybe later) then you'll be in great shape.


[hide] Firstly, draw the lines (equalities) and see where they intersect.  We see that the 3 intersection points are (0,4), (5,9), and (5,4).  When you draw it out, you will notice that the intersection of the inequalities is simply the area of triangle formed by the intersections made.  The area of the triangle thus formed is $\frac{25}{2}$, which is our answer. [/hide]

I asked for the longest side of the triangle.

[quote="ckck"]Consider the intersection of theh solutions to the following inequalities:$y\le x+4, y\geq4$ and $x\le5$. What is the length of the longest side of the triangular region formed by this intersection?[/quote]
[hide]
The vertices of the triangle are $(5,9),(0,4),(5,4)$.
the longest length is between $(0,4)$ and $(5,9)$. the length is $\sqrt{41}$

edit-i miscalculated. the actual distance between $(0,4)$ and $(5,9)$ is $5\sqrt{2}$
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:blush: Oh, sorry about that, I thought you said area.  By the way, ProtestanT, you miscaculated.  Using the distance formula, we find that the distance from 0,4 to 5,9 is $\sqrt{(5-0)^{2}+(9-4)^{2}}=\sqrt{50}=5\sqrt{2}.$


Sa se determine functile bijective $f:[0,1]\to[0,1]$ care verifica relatia :$\sqrt{f^{ - 1} (x)f(x)} = x,\forall x \in [0,1].$

I'll write my solution in English so that the author of the problem will be able to understand it.
For a random $x\in(0,1)$ the sequences are $(f^{[n]}(x))_{n}$ and $(f^{[-n]}(x))_{n}$ geometrical progressions.
Let q be the ratio of the first geom. progression. This implies $f^{[n]}(x)=xq^{n-1}$, but since $f^{[n]}(x)\in(0,1)$ q must be in (0,1]. From $f(x)=qx$ we also get $f^{-1}(x)=\frac {1}{q}x$. Consequently $\frac {1}{q}$ is the ratio of the second progression considered; this is tantamount to $\frac{1}{q}\in(0,1)$; so q=1.
For x=0 we have $f^{-1}(0)f(0)=0$ so either $f(0)=0$ or $f^{-1}(0)=0$; but one equality implies the other.
The case x=1 is trivial.
Because f(x)=x satisfies the equation we conclude that this is its solution.

[quote="Valentin Vornicu"][quote="vlad"]I'll write my solution in English so that the author of the problem will be able to understand it.
[/quote]Cred că înţelege şi în Română ;)[/quote]
Moubinool Omarjee este de origine romana???

Ma indoiesc profund. Cel putin, din cate stiu eu nu stie romana. Dar bine, limba romana din matematica nu este foarte bogata.

Am zis că înţelege şi în Română demonstraţia, nu că ştie să vorbească în Română :)


parameterization
real analysis
probability
induction
function
algebra
functional equation

$ \tau$ is a random variable with integer value.$ \{\zeta_n\}_{n \equal{} 1}^\infty$ (i.i.d) are independent with $ \tau$,
$ \zeta_n \equal{} 1$,w.p.$ p$,$ \zeta_n \equal{} 0$,w.p.$ 1 \minus{} p$.
Let $ X \equal{} \sum_{i \equal{} 0}^\tau \zeta_n,Y \equal{} \tau\minus{} X$.If $ X,Y$ are independent,show that $ \tau$ is Poisson.

I know it can be done by using generating function.But how?

I would try something like
$ \phi(\lambda)\equal{}E[e^{\lambda\tau}]\equal{}E[e^{\lambda(X\plus{}Y)}]\equal{}E[e^{\lambda(X)}]E[e^{\lambda(Y)}]$(u:sing independence between X and Y

Then maybe calculations get in the end

$ \phi(\lambda)\equal{}e^{\lambda.(e^{cte }\minus{}1)}$

where cte is the parameter of the Poisson's law

no warranty on this, it's just an idea

[quote="suweijie"]Let $ X \equal{} \sum_{i \equal{} 0}^\tau \zeta_n,Y \equal{} \tau \minus{} X$.If $ X,Y$ are independent,show that $ \tau$ is Poisson.[/quote]
I wonder how would $ X$ and $ Y$ be independent?

If $ X\equal{}1$ w.p.$ p$ and $ \minus{}1$ w.p. $ 1\minus{}p$ then $ X$ and $ Y\equal{}X^2$ are obviously independent because  $ Y\equal{}1$ w.p. $ 1$. Its not the case in general, mostly we would always have $ X$ and $ Y$, where $ Y\equal{}f(X)$, are dependent
Generally, does it make sense to ask viz. when would $ X$ and $ Y$, where $ Y\equal{}f(X)$, be independent? For what kind of $ f$ etc?
Classical definition is  $ X$ and $ Y$ are independent if the corresponding sigma algebras $ \sigma(X)$ and $ \sigma(Y)$ are independent.

Can anyone clear my doubt both intuitively as well as by the classical definition?

Thanks.

limsup: OK, so let's take the definition: $ X,Y$ are independent if $ \sigma(X)$ and $ \sigma(Y)$ are independent. This means that for every event $ A,B$ then: $ P(X \in A, Y \in B) \equal{} P(X \in A)P(Y \in B)$. Moreover, it is straightforward to see that if $ X,Y$ are integer valued it suffices to check that for every $ i,j \in \mathbb{Z}$ then $ P(X\equal{}i, Y\equal{}j)\equal{} P(X\equal{}i)P(Y\equal{}j)$. Let's do it:

lets call $ \tau_i\equal{}P[\tau\equal{}i]$. Then $ P(X\equal{}n)\equal{}\sum_{k\equal{}n}^{\infty} P(\tau\equal{}k) P[X\equal{}n| \tau\equal{}n]$, which reads:
\[ P(X\equal{}n)\equal{}\sum_{k\equal{}n}^{\infty} \tau_k \binom{k}{n} p^n (1\minus{}p)^{k\minus{}n}\]
and the same computation gives
\[ P(Y\equal{}m)\equal{}\sum_{k\equal{}m}^{\infty} \tau_k \binom{k}{m} (1\minus{}p)^m p^{k\minus{}m}\]
and
\[ P(X\equal{}n,Y\equal{}m)\equal{} \tau_{m\plus{}n} \binom{m\plus{}n}{n} p^n (1\minus{}p)^{m} \]

From here you easily check that if $ \tau$ is a Poisson distribution, then $ X$ and $ Y$ are indeed independent. I'm curious about the converse ?

I guess it should read $ \tau$ is a RV with nonnegative integer values, otherwise it is not clear how the sum is to be interpreted. I'll show that $ \tau$ has to be a Poisson RV:

Let $ p(k) = P(\tau = k)$. Given $ \tau=k$, the conditional joint distribution of $ X,Y$ is multinomial (sequence of $ k$ independent trials with two possible outcomes with probability $ p$ and $ 1-p$ respectively). Using independence of $ X,Y$ we get
\[ P(X=n)P(Y=m)  =  p(n+m) \binom{n+m}{n}p^n(1-p)^m \qquad (*)
\]
We may assume that $ p(0) < 1$. (Otherwise $ \tau=0$ is constant and satisfies the given property.)  So $ p(k) >0$ for some $ k>0$. (*) implies $ P(X=l), P(Y=l)>0$ for all $ l \le k$. So (*) implies $ p(l) >0$ for all $ l \le 2k$, and so on. By induction $ P(X=n), P(Y=m),p(k) >0$ for all $ n,m,k \ge 0$. Using (*) for $ n=0$ and $ n=1$ we get
\[ \frac{P(Y=m)}{P(Y=m-1)} = \frac{p(m)}{p(m-1)}(1-p) \]
and
\[ \frac{P(Y=m)}{P(Y=m-1)} = \frac{p(m+1)}{p(m)} \frac{m+1}{m}(1-p),
\]
which gives
\[ \frac{p(m+1)}{p(m)} = \frac{p(m)}{p(m-1)}\frac{m}{m+1}.
\]
By induction
\[{ \frac{p(m)}{p(m-1)}= \frac{p(1)}{p(0)}\frac{1}{m}}.
\]
Setting $ \lambda = \frac{p(1)}{p(0)}$, again by induction $ p(m) = \lambda^m \frac{1}{m!} p(0)$. So by normalization $ \tau$ is a Poisson RV.

If you want to use generating functions, the following is a sketch of a proof (probably with mistakes and with a lot of steps left out): Calculate the joint GF of  $ X,Y$ by conditioning on $ \tau$ and you get something like $ M_{X,Y}(t_1,t_2) = M_{\tau}((1-p)t_1 + p(t_2))$ and if $ X,Y$ are independent, this factors into a function of $ t_1$ and a function of $ t_2$. If you use this carefully it follows that $ M_{\tau}$ satisfies a variant of the Cauchy functional equation, and since it satisfies some minimal regularity conditions, it basically has to be an exponential function $ M_{\tau}(t) = ce^{\lambda t}$. Use normalization to conclude that $ \tau$ is a Poisson RV. (I hope that this is at least roughly correct. Anyway it should give an idea how to use generating functions here.)


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combinatorics unsolved
combinatorics

Let $ \sigma \equal{} (a_{1}, a_{2}, a_{3}, \dots , a_{n})$ be a permutation of $ (1, 2, 3, \dots , n)$ . A pair $ (a_{i}, a_{j})$ is said to correspond to an inversion of $ \sigma$, if $ i < j$ but $ a_{i} > a_{j}$ .
(Example: In the permutation $ (2, 4, 5, 3, 1)$ , there are $ 6$ inversions corresponding to the pairs $ (2, 1), (4, 3), (4, 1), (5, 3), (5, 1), (3, 1)$ .)
How many permutations of $ (1, 2, 3, . . . , n)$, $ (n \geq3)$, have exactly two inversions.


BE and CE are the altitudes of triangle ABC.  AB > AC.  Show that AB + CE > AC + BD.

The same question was posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=180066]here[/url] :?

I quote a proof of this exercise, without  use of the triangle area:

  If $AM = AC\mathop {}\limits_{} ,\mathop {}\limits^{} MZ \bot AC\mathop {}\limits^{} ,\mathop {}\limits^{} MK \bot BD$   , we have:

$ AB-AC = BM > BK = BD-KD = BD-MZ = BD-EC\mathop{}\limits_{} $


vector
linear algebra
matrix
inequalities
algebra
system of equations
linear algebra unsolved

I wish this is not posted before. This problem came to my mind an hour ago, I have one solution for it! :)
Suppose $V\subset\mathbb R^{n}$ is a vector space of dimension $k$ on $\mathbb R$. Prove that $|V\cap\{0,1\}^{n}|\leq2^{k}$.

Could you generalize the problem?

Let $v_{1},\ldots,v_{t},\ t\le k$ be a basis for the $\mathbb R$ - vector space spanned by $V\cap Q$ where, $Q=\{0,1\}^{n}$. Since the system formed by $v_{1},\ldots,v_{t}$ and any other vector from $V\cap Q$ is linearly dependent over $\mathbb R$, it is linearly dependent over $\mathbb F_{2}$ (the field with two elements) as well. This means that the $\mathbb F_{2}$ - vector space spanned by $V\cap Q$, which, as a set, can be identified with $V\cap Q$ itself, is at most $k$ - dimensional, and the conclusion follows (a $k$ - dimensional vector space over $\mathbb F_{2}$ has $2^{k}$ elements).

[quote="grobber"]Since the system formed by $v_{1},\ldots,v_{t}$ and any other vector from $V\cap Q$ is linearly dependent over $\mathbb R$, it is linearly dependent over $\mathbb F_{2}$ (the field with two elements) as well. [/quote]

may i know why?

This is my solution:
We prove the theorem by contradiction. Assume we can choose $2^{k}+1$ vectors from $V\cap\{0,1\}^{n}$. Consider these $2^{k}+1$ vectors in $\mathbb Z_{2}$. Because any vector space of dimension at most $k$ over $\mathbb Z_{2}$ has less than $2^{k}+1$ elements, so we can choose $k+1$ of these vectors dubbed $v_{1},v_{2},\dots,v_{k+1}\in V\cap\{0,1\}^{n}$ that are linearly independent over $\mathbb Z_{2}$. Put $v_{1},v_{2},\dots,v_{k+1}$ as rows of a $(k+1)\times n$ matrix $T$. Obviously $\mbox{Rank}(T)=k+1$, thus we can choose a $(k+1)\times(k+1)$ submatrix $P$ of $T$ that is invertible (over $\mathbb Z_{2}$). Therefore $\det(P)\neq0$ (over $\mathbb Z_{2}$). So $\det(P)$ is an odd integer over $\mathbb R$. Thus matrix $P$ is invertible over $\mathbb R$ and $\mbox{Rank}(P)=k+1$, but we know that $k+1=\mbox{Rank}(P)\leq\mbox{Rank}(V)\leq k+1$(over $\mathbb R$). And the last inequality holds beacause $T$ is a $(k+1)\times n$ matrix. So $k+1$ rows of matrix $T$ are linearly independent over $\mathbb R$, and this is contradictin because all of them are elements of $V$ which has dimension $k$.
[i]
To Soarer :[/i]
That's because $v_{1},v_{2},\dots,v_{t}\in\{0,1\}^{n}$ and if they are linearly dependent there exists $\alpha_{1},\dots,\alpha_{n}\in\mathbb Q$ such that $\alpha_{1}v_{1}+\dots+\alpha_{n}v_{n}=0$ and multiplying by asuitable number all coefficients will be integers and at least on of them will be odd. Then you could do a same relation for $\mathbb Z_{2}$.

[quote="Omid Hatami"]if they are linearly dependent there exists $\alpha_{1},\dots,\alpha_{n}\in\mathbb Q$ such that $\alpha_{1}v_{1}+\dots+\alpha_{n}v_{n}=0$ [/quote]
Sorry for stupid questions again and again, but why?

[i]To Soarer : [/i]
Because a system of equations rational coefficients has answer in $\mathbb R$ iff it has answer in $\mathbb Q$.

[i]To grobber:[/i]
Also an easy generalization. If $p$ is a prime number then $|V\cap\{0,1,\dots,p-1\}^{n}|\leq p^{n}$

Another Proof!!

Let ${v_{1},v_{2},...,v_{m}}$ be any ensemble of (0,1)-vecors in a $k$-dimentional space.It is easy to see that for any 1$\leq i,j,k$ $\leq m$ we have( ${v_{i}-v_{j}}$).(${v_{k}-v_{j}}$) $\geq 0$.So this ensemble is a set of points in a $k$-dimentional space with all of it's angles less than 90.
There is a famouse problem which says:
Any ensemble of points in a$k$-dimentional space with more than $2^{k}$ members,has an angle greater 90.


Hello. Could you give me some links to online introduction to complex numbers, thanks in advance


I have to ask -

if a land having length 2m and breadth 2m its area becomes 2*2 =4m
 square
but when a land having length 0.5m and breadth 0.5m its area becomes
 0.5*0.5=0.25m square
then how it can be possible that land with length 0.5 and breadth 0.5
 decreases to 0.25m square in reality .  please clear its answer.

A square of dimensions $ 2 \times 2$ can be divided into four squares of dimensions $ 1 \times 1$ by two cuts that each divide the square in half.  Each square of dimensions $ 1 \times 1$ can be divided into four squares of dimensions $ 0.5 \times 0.5$ by two more cuts (each).

So our original square, of area $ 4$, has been divided into $ 16$ squares, each of which has area $ \frac {4}{16} \equal{} 0.25$.

The general principle is, roughly, this:  when you multiply the side lengths of something by $ k$, you multiply its area by $ k^2$.

Edit:  is your problem that you don't understand why the number $ 0.25$ is less than the number $ 0.5$?  That's because meters and square meters are two completely different units of measurement.  The magnitudes of quantities measured by each set of units have nothing to do with each other.