geometry
circumcircle
trigonometry
geometry proposed

$ABC$ is an triangle. $M$ centroid of $ABC$. $AM,BM,$ and $CM$ meet the circumcicle of $ABC$ at $A',B'$ and $C'$ recpectively. $S_M$ it's area of interseqtion of the interior triangle $ABC$ and the interior of triangle $A'B'C'$.
      Prove  That    $S_M\geq (2/3)S_{ABC}$

[quote="Tiks"]$S_M$ it's area of interseqtion of the interior triangle $ABC$ and the interior of triangle $A'B'C'$[/quote]
What do you mean?

[i]Congratulations, Tiks ![/i] I like the your proposed problem. Always a difficult (for me) problem is equivalently with the interesting (for me) problem. Generally, a problem with a median is difficulty. First time I remark some interesting identities w.r.t. the medians of the tiangle $ABC$:

$\blacksquare\ 1^{\crc}.\ 4a^2m^2_a=16S^2+(b^2-c^2)^2$ (wellknown relation);
$\blacksquare\ 2.1^{\circ}.\ 16b^2m^2_a=16S^2+(3b^2+c^2-a^2)^2;$
$\blacksquare\ 2.2^{\circ}.\ 16c^2m^2_a=16S^2+(3c^2+b^2-a^2)^2$ (not at all wellknown !).
$\blacksquare\ 3^{\circ}.\ x=m(\widehat {CAG}),\ y=m(\widehat {BAG})\Longrightarrow \frac{c}{\sin x}=\frac{b}{\sin y}=\frac{2m_a}{\sin A}.$

I shall use these identities for the proving your inequality. Soon I shall come back, [i]Tiks [/i]!

Thank you Virgil! :) .
I wont say that I also have same interesting results(similiar this problem)and I am going to say ther in other topics :) (I think you have seen my problem in the topic whos name
  is $geo ineq.1$).

I did not see the your problem "[b]Geo. 1[/b]". I solved it enough quickly. See the my reply for this problem.

Dear Virgil this ineq. is true :)
In my solution I proved it. :D

[quote="Virgil Nicula"]$\cdots \Longleftrightarrow [AA_1A_2]=\frac{a^4}{(2a^2+b^2)(2a^2+c^2)}\cdot S$ a.s.o. Therefore,
$\sum [AA_1A_2]\le \frac 13 S\Longleftrightarrow \sum \frac{a^4}{(2a^2+b^2)(2a^2+c^2)}\le \frac 13$ a.s.o.
I think that isn't the case to enter in the detail !.[/quote]
Virgil if you havn't seen,this ineq. was be solved this
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=395465#p395465]www.mathlinks.ro/Forum/viewtopic.php?p=395465#p395465[/url]


I need some help with this problem:

Let G be an oriented graph with vertices of "out-degree"=2("out-degree" is the number of edges going out from a vertice), and without cycles of length smaller than m. Prove that G have at least 2m vertices.

Here's a suggestion:

Fix some vertex $v$ in the graph, and divide all the other vertices into classes based on the length of the shortest path starting from $v$ and ending at that vertex.   How small can each class (or at least the classes at small distance) be?


Ok so this is a problem that me and my friend cannot agree on

you have a 3 reeled slot machine

each reel is a number from 0-9

if two are the same number you get 3 times what you bet
if all three are the same but NOT 0 you get 15x what you bet
If all three are 0 you get 50x what you bet.

What are the statistical gains/losses in this system?

[hide="Solution"]$ \bullet$ There are $ 10\cdot9\cdot8 \equal{} 720$ possible outcomes that are $ 3$ distinct numbers (in this case, you win nothing).

$ \bullet$ There are $ 10\cdot9\cdot3 \equal{} 270$ outcomes of that contain 2 same numbers (in this case, you win 3x your bet).

$ \bullet$ There are $ 9$ cases in which $ (n,n,n)$, where $ n\ne0$, is the outcome (in this case, you win 15x your bet).

$ \bullet$ There is exactly $ 1$ case in which $ (0,0,0)$ is the outcome (in this case, you win 50x your bet).

For simplicity, let the bet be $ \$1$ per game.

The gross expected winnings after $ 1000$ games are $ \$50\cdot1 \plus{} \$9\cdot15 \plus{} \$270\cdot3 \equal{} 50 \plus{} 135 \plus{} 810 \equal{} \$995$. However, since you pay $ \$1$ for each game, your net expected outcome will be a loss of $ \$5$ after $ 1000$ games. Thus, you are expected to lose $ \$0.005$ each time.[/hide]

Arent there 10*9*2 for 2 numbers that are the same?

edit: wait... never mind, you're right. I suck at counting and probability. :blush:

im not sure about this, but dont you expres the statistical gains/losses as a percent? in this case, you lose .5% per turn

That's basically what was stated, only without percentages.

For a $ \$$1 bet, you are expected to lose $ \$$0.005, or 1% of the bet.

[quote="mathemonster"]Wouldnt it be $ 10 * 9 * 2$ number of choices for two that are the same b/c after the first 2 there are 2 numbers that will make it :|[/quote]
Let the result be $ (a,b,c)$.

There are $ 10$ possibilities for $ a$. Once $ a$ is determined, $ b$ will be the same number. The last number can then be anything other than $ a$, which is $ 9$ possibilities. Now, multiply by $ 3$ to account for permutations: $ (c,a,b)$ and $ (a,c,b)$. Note that $ (a,b,c)$ and $ (b,a,c)$ are the same thing.

Thus, there are a total of $ 10\cdot9\cdot3 \equal{} 270$.

i learned how to do these types of questions in school few weeks ago. i still don't get it. probability is my weak spot. y is it 10*9*3??? y 3??? someone tell me plz.

Life is therapy itself... -Dean Koontz

[quote="iluvjiyongx3"]y is it 10*9*3??? y 3??? someone tell me plz.[/quote]
By simply multiplying $ 10\cdot9$, you don't count the permutations of the numbers. $ (a,b,c)$ is what you count, but there are also $ (a,c,b)$ and $ (c,a,b)$. Thus, you have to multiply by $ 3$.

However, $ (a,b,c)$ is the same as $ (b,a,c)$ since $ a \equal{} b$, and they should not be counted as different outcomes.

[b]iluvjiyongx3[/b] — Look at it this way: when you have two numbers in a row, there are three ways to arrange them: $ (a,a, b)$, $ (a,b,a)$, and $ (b,a,a)$. We have ten numbers, so it would be logical to assume that there are $ 3\cdot10$ ways of getting two numbers that are the same. However, we have to account for the fact that $ b$ can be any of the $ 9$ other numbers ($ b$ can't be $ a$, though, or we would have $ 3$ numbers that are the same), so we multiply by $ 9$, to get $ 3\cdot9\cdot10 \equal{} 270$.

Yeah, [b]i_like_pie[/b], there is a 0.5% expected loss.


Hello,i think the limit is
$ \lim_{n \to \infty}\left({\frac{(2n)!}{n^nn!}}\right)^{\frac{1}{n}}\equal{}\frac{4}{e}$.
Sonnhard.

Yes that's it

give me please a hint to your solution, I want to know if it's the same as mine.

Thanks :wink:

Use Stirling's approximation:

$ n! \approx \sqrt{2 \pi \n} \frac{n^n}{e^n}$


geometry
parallelogram
geometric transformation
rotation
invariant
group theory

Hi,

You have a regular hexagon of side length $n$.
You want to fit in it the maximum number $f(n)$ of parallelograms which all have the same form that is $4$ equilateral triangles of side $1$ in a row (up, down, up, down). You can use rotations and symmetries, but the tiles must have side parallel to the sides of the hexagon.

What is the exact value of $f(n)$?

I've found that $f(2n) = 6n^2$ (you can cover entirely the regular hexagon) and I think that $f(2n+1) = 6n(n+1)$ which means that you can cover the whole hexagon except 6 little equilateral triangles. Clearly, this cover is possible, but I can't prove that it is impossible to cover the whole hexagon except 2 little triangles. If done, it would prove my conjecture because of consideration modulo 4.
There must be some kind of invariant modulo 3  :?

Myrtille.

Ok, I think I've got it. It is not really an invariant modulo 3.

The sides of the hexagon determine 3 distinct directions, say $a,b,c$.
Now any parallelogram which is put into the hexagon has its longest side with direction among these three ones.
Let's call $x,y,z$ the number of parallelograms for which this direction is $a,b,c$ respectively.

First, as noted above, we may divide the hexagon of side length $n$ into $6n^2$ equilateral triangles of sides 1.
Each parallelogram covers exactly 4 of these triangles. Let $m$ be the number of parallelogram (we will call it a tile) used. Then the number of triangles covered by these tiles is divisible by 4. In the case of odd $n$, we have $6n^2 = 2$ mod[$4$], so that the maximum number of covered equilateral triangles is $6n^2-2$ or, if not possible (as we will prove it below : this is your conjecture), $6n^2-6$ (which is indeed possible and give the result you mentioned).

Now, assume that we may cover exactly $6n^2 - 2$ equilateral triangles.
According to the direction $a$, color the rows alternatively black and white. Then a $a$-tile cover either 4 or 0 black triangles whereas any other tile covers exactly 2 black triangles. Since the number of black triangles is $3n^2$ it follows that the number of covered black triangles is $4t + 2(y+z)$ (where $t$ denotes the number of $a$-tiles which cover 4 black triangles). In another hand since there are exactly two triangles (not necessarly black) which are not covered, we deduce that $4t+2(y+z) = 3n^2$ or $3n^2-1$ or $3n^2-2$.
By parity consideration (recall that $n$ is supposed to be odd), we deduce that the only possible case is $4t+2(y+z) = 3n^2-1 = 2$ mod[4]. It follows that $y+z$ is odd.

Similarly, we deduce that both $x+z$ and $x+y$ are odd.
Therefore, $x,y,z$ have pairwise distinct parities, a contradiction.

Thus, to cover $6n^2-2$ triangles is impossible and we are done.

Pierre.

I think it can be solved by using boundary words and some simple group theorey, but I'm not that experienced in this field (actually, I'm not experienced at all :)). Anyway, here's what I thought:

A long time ago I downloaded and read an article regarding boundary words. It's a group theoretical approach to problems such as this one.

Assume each horizontal segment is $a$ when we go through it from lweft to right, and $a^{-1}$ when we go through it in the opposite direction, each segment maing an angle of $60^{\circ}$ with $(Ox$ is $b$ when we go through it from left-down to right-up and $b^{-1}$ otherwise, and every segment making an angle of  $120^{\circ}$ is $c$ when we go through it from right-down to left-up and $c^{-1}$ otherwise.

If you look at the possible shapes of a parallelogram, you'll see that the constraints we impose on $a,b,c$ are that $a^2,b^2,c^2$ commute with everything. In other words, the group we're working with is the one with presentation $<a,b,c|ab^2=b^2a,ac^2=c^2a,bc^2=c^2b,ba^2=a^2b,ca^2=a^2c,cb^2=b^2c>$. These are the conditions for the boundary word along the boundary of a parallelogram to be $1$ (the identity element in the group). However, the boundary of our hexagon is $a^{2n+1}b^{2n+1}c^{2n+1}a^{-(2n+1)}b^{-(2n+1)}c^{-(2n+1)}$. If we take our constraints into account, we see that the element mentioned above is $x=abca^{-1}b^{-1}c^{-1}$.

It’s not hard to see that for a region consisting of two equilateral triangles, the boundary word cannot be $x$, so the uncovered area can’t consist of two equilateral triangles.

I’m missing lots of details and it might not even be a correct approach. I’ve never applied this method before. Could this work?

Do you have the reference of the paper you mentioned?

Pierre.

I think the first time I read something on the topic was in the following paper:


given triangle ABC and point P and Q lie on BC prove that $ r(ABP) \equal{} r(AQC) \leftrightarrow r(ABQ) \equal{} r(APC)$
r(XYZ) means to radius of the inner circle of triangle XYZ

[hide]p/s it would be nice if your solution use trigonometry[/hide]


I have no clue to find the value of 2^sqrt2, if I use my cal, 2.6655... comes out.
How do you get this number?
I went as far as log x= sqrt2 with a base 2...
I don't know any In and calculus, please explain in easy way as possible. :)

You already "got" that number: it is $ 2^\sqrt{2}$.
For anything more, you need to define what you want.

I think he wants to know how to find this answer 2^sqrt2 without a calculator or massive guess and check.

Answer for what¿ He asked only for the value of a number. I gave him the number.

I believe he wants to know how to manually find the decimal approximation.

Every real number is the limit of some sequence of rational numbers. If we take the sequence $ a_n$ such that $ \lim_{n\rightarrow\infty} a_n\equal{}\sqrt2$, consider the limit of the sequence $ b_n\equal{}2^{a_n}$; it will be $ 2^{\sqrt2}$.


Is anyone making a team and requires a team member?

Or better, is anyone looking to join a team?

Must be in Toronto.

I am currently Gr. 10 and I am living in Toronto. I  am not  great, but I qualified for USAMO. I would love to join your team. It is on April 17, isn't it? If you need one more team member email me at
[color=blue][u]alanwalkalone@hotmail.com[/u][/color]

April 17 is the regular Mathbattle.

When is the Junior Mathbattle?

No idea, but Prof. Ivrii told me that it should be late April or early May.

I assume your name is Alan Guo, are you sure it would't be trouble to drive over?

If not, your very welcome.

[quote="Singular"]I wouldnt mind having the person 'startogalaxy' on my team (East Asian Gangstars United)

my mail is alpha(underscore)wice(at)hotmail(dot)com[/quote]

According to the rule, you are not allowed to add any person now. Plus, even if you do, we will beat you guys this time anyways!
c ya on sunday!

I thought this is for junior math battle , isn't it? The regular one will be hold this weekend and the junior is somewhere in may, according to the website [url=http://www.math.toronto.edu/oz/turgor/news.php]www.math.toronto.edu/oz/turgor/news.php[/url] .Is it too late to join the junior team?If I am welcome, 'hitseme' would  do tell me your team name and contact me when the date of the battle is finally settled. The ride shouldn't be a problem. And you are absolutely right; to clear my identity, I am Alan Guo.

I wouldn't mind join your East Asian Gangstar United team either, but as "11th dimension" mentioned above, is it too late to join according to the rule. Beside, I don't think I am good enough either. But if you are short of a team member, I would love to come.

Actually, there is no such rule.  I remember in junior math battle, I joined Jarvis2 at the last second.

Besides, our roster currently has 5 members, and there is a maximum of 6 people on our team.

If you want to join (startogalaxy) please email me at that address.  I will confirm with Olga right now.

BTW, if you made USAMO, you are better than me!  (heheh)

EDIT:

"person "Alan Guo" wants to join our team, is this okay?

O'K. It would be 6 then. "

So olga confirms it.

if you want to join (startogalaxy) either email me and show up, or just show up to the battle, if you know where it is.


Hi everyone,

Could someone show me how to do this problem, please?

The weight of adult males in the U.S. follows a normal distribution with a mean of 171 pounds and a standard deviation of 30 pounds.

Suppose a random sample of 40 men was selected and the mean weight of the 40 men was calculated. What is the probability that the sample mean is more than 186 pounds?

I think I need to use the equation z=(x-x bar)/standard deviation. Is that correct?
n=40

Thank you very much

suppose that $ X_i$ are iid random variables, $ X_i \equal{} N(m, \sigma^2)$. Then what is the law of $ \frac{\sum_1^n X_i}{n}$. I you know how to do that, then you can solve your question.

Thank you very much

We haven't done that yet. Could you also do it with a z score?

Thank you very much


Let $ a,b,c\geq 0$, s.t. $ ab\plus{}bc\plus{}ca\equal{}3$.Prove that:

$ \frac{a(b^2\plus{}c^2)}{a^2\plus{}bc}\plus{}\frac{b(c^2\plus{}a^2)}{b^2\plus{}ca}\plus{}\frac{c(a^2\plus{}b^2)}{c^2\plus{}ab}\geq 3$.

The Vasc's hint: prove that $ \sum_{cyc}\frac {ab(a^2 \plus{} c^2)(b^2 \plus{} c^2)}{(a^2 \plus{} bc)(b^2 \plus{} ac)}\geq3.$ :wink:

[quote="Inequalities Master"]Let $ a,b,c\geq 0$, s.t. $ ab \plus{} bc \plus{} ca \equal{} 3$.Prove that:

$ \frac {a(b^2 \plus{} c^2)}{a^2 \plus{} bc} \plus{} \frac {b(c^2 \plus{} a^2)}{b^2 \plus{} ca} \plus{} \frac {c(a^2 \plus{} b^2)}{c^2 \plus{} ab}\geq 3$.[/quote]

Problem #2 in http://reflections.awesomemath.org/2008_1/unexpected_ineq.pdf

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3208806#p3208806

The following inequality is also true
[img]http://s5.sinaimg.cn/middle/0018zOAxgy70CockMxS94&690[/img]


algebra
polynomial
vector
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $A$ be a commutative unital ring, which is a finitely generated algebra over a field $k$ (that is, there are elements $x_{1},\ldots,x_{s}$ of $A$ such that every element of $A$ is a polynomial in the $x_{i}$ with coefficients in $k$).

Prove that $A$ is an Artinian ring iff it is a finite dimensional vector space over $k$.

$A$ fin gen $k$-algebra ==> Noetherian ==> (Artinian <==> $\dim{A}= 0$ <==> $A$ fin gen as $k$-module (beacause of Noether normalisation and dim(k-Alg) = tr.deg(A/k)) <==> fin dim $k$ vector space)


Let ABC be a given triangle and let A'BC, AB'C, ABC' be equilateral triangles erected outwards on the sides of triangle ABC (A'B'C' – first Fermat triangle). Let [tex]{\cal C}[/tex] be the circumcircle of A'B'C' and let A", B", C" be the respective intersections of  [tex]{\cal C}[/tex] with the lines AA', BB', CC'.

Prove that:  AA" + BB" + CC" = AA' = BB' = CC' .

In [url=http://groups.yahoo.com/group/Hyacinthos/message/8067?expand=1]Hyacinthos message #8067[/url], I proved AA" + BB" + CC" = AA'. Of course, AA" + BB" + CC" = BB' and AA" + BB" + CC" = CC' can be proven analogously (moreover, AA' = BB' = CC' is a well-known and very easy fact).

  Darij

Very nice. Thank you for the answer !!!!


calculus
integration
real analysis
real analysis unsolved

I think it's easy but I don't see how find it:
\[\int_0^{+\infty}\frac{ln(t)}{t^2+1}dt\]

Definatly not easy

The answer is actually 0

$\int_0^1\frac{\ln(t)}{t^2+1}\,dx=-C$, where C is the Catlan number
But
$\int_1^\infty\frac{\ln(t)}{t^2+1}\,dx=C$ so the entire integral is 0

Unforunatly I cant prove either :(

let $x=\frac{1}{t}$
then it is easily find
$\int_0^1\frac {ln(t)}{t^2+1}dt=- \int_1^{+ \infty}\frac {ln(x)}{x^2+1}dx$
the answer is 0.


inequalities
group theory
abstract algebra
superior algebra
superior algebra unsolved

Determine all integers k with the property that there exists only one isomorphism type of group of order k using Schur-Zassenhaus  Theorem: a finite group G has cyclic Sylow subgroups $ \iff G \cong \mathbb{Z}_{m} \rtimes \mathbb{Z}_{n}$, where $ gcd(m, n) \equal{} 1$.

With this theorem it is very simple. If there is exactly one isomorphism type, then it must be the cyclic one.

If $ k$ is such a number, then $ k$ is squarefree. If $ p^2 | k$, then $ \mathbb{Z}_{k/p^2} \times \mathbb{Z}_p^2$ and $ \mathbb{Z}_k$ would be two non-isomorphic groups of order $ k$. So $ p^2$ can't be a divisor of $ k$.

Using the theorem every group of order $ k$ is a semidirect product $ \mathbb{Z}_n\ltimes\mathbb{Z}_m$. If every such semidirect product is in fact a direct, this means, that there can be no homomorphism $ \mathbb{Z}_m\to\mathbb{Z}_n^\ast$ which is equivalent to $ gcd(m,\varphi(n))\equal{}1$.

Now it is easy to see, that this can only hold for all factorizations $ k\equal{}nm$, if $ gcd(k,\varphi(k))\equal{}1$.

On the other hand: If $ gcd(k,\varphi(k))\equal{}1$, then $ k$ must be squarefree. Hence all sylowgroups are cyclic, so every group of order $ k$ is a semidirect product $ \mathbb{Z}_n\ltimes\mathbb{Z}_m$. Because $ m$ is relative prim to $ \varphi(n)$, the only semidirect product is the direct one. So every group of order $ k$ is cyclic.


Expand by use of Minors

3 4 2
6 4 3
0 4 2

Solve the equation
0 0 -1
3 x 0 =-x squared
2 0 3

I assume that these are both determinant problems?

$\left|\begin{array}{ccc}3&4&2\\6&4&3\\0&4&2\end{array}\right|= \det\begin{bmatrix}3&4&2\\6&4&3\\0&4&2\end{bmatrix}$

and

Solve for $x: \,\left|\begin{array}{ccc}0&0&-1\\3&x&0\\2&0&3\end{array}\right|= -x^2$


Does this relate with the computational complexity theory in any way?

i'm confused about what the big deal is...

[quote="keeyounglee"]a friend of mine showed me this:

$P = NP$

$P-NP = 0$

$P(1-N) = 0$

$P=0; N=1$

what do you guys think about this?[/quote]

A friend of yours either has an interesting sense of mathematical humor or doesn't understand the P vs. NP problem. That "proof" isn't related to the actual computational complexity problem at all.

[quote="mathnerd314"]A friend of yours either has an interesting sense of mathematical humor or doesn't understand the P vs. NP problem. That "proof" isn't related to the actual computational complexity problem at all.[/quote]

Is there anyone on this site who [i]doesn't[/i] have an interesting sense of mathematical humor?

[quote="keeyounglee"]perhaps there should be something in there about not deviating from the topic and posting text that has no correlation to the topic at hand.[/quote]Hi keeyounglee. There is a forum especially designed for these types of discussions. The current forum is a problem solving forum and so this current thread has deviated topic from what the the current forum is designed for.

This is the correct forum for this kind of discussions: [url=http://www.mathlinks.ro/Forum/index.php?f=139]GAMES AND FUN FACTORY[/url]


[quote="xxxxx"]0.428571... repeating, can be expressed as a simplified fraction in the form a/b. FInd b-a[/quote]
[hide]$\frac{3}{7}$ is the fraction so the answer is 4[/hide]

[quote="Ubemaya"]do you have that memorized?[/quote]
no. Here's how my mind works.
[hide]Hmmm THis looks a lot like $\frac{1}{7}$ I should know what this is but I don't feel like thinking. Maybe I can multiply $\frac{1}{7}$ by 10 and subtract 1 or something and find it. Hey that works! Well, we have our answer now.[/hide]

[quote="Ubemaya"]do you have that memorized?[/quote]
it's nice to have decimal expansions of a/7 memorized.
Dunno how often it shows up in problems(never seen one other than this) but it's cool
$1/7=.142857...$
$2/7=.285714...$
$3/7=.428571...$
$4/7=.571428...$
$5/7=.714285...$
$6/7=.857142...$
Just the digits shifted over....

The point was to illustrate how the fractional equivalent can be found.  I used that specific value because it had already been used in this topic.  The specific fraction doesn't matter and was just used to illustrate the process.

[quote="bpms"][quote="roadnottaken"]A more rigorous way to find the fractional equivalent of a repeating decimal is to multiply the original value by $10^{n}$, where $n$ is the number of digits of repetition, subtract the original number, put $99\ldots99$ in the denominator, where there are $n$ $9$'s, and reduce.  For example,

$\begin{eqnarray*}x &=& .142857\ldots\\ \\ 1000000x &=& 142857.142857\ldots\\ \\ 1000000x-x &=& 142857.142857\ldots-.142857\ldots\\ \\ 999999x &=& 142857\\ \\ x &=& \frac{142857}{999999}\\ \\ &=& \frac{1}{7}$[/quote]
Well how many of us can make that last reduction without a calculator? I couldn't unless I knew the answer *cough* *cough*[/quote]

Lol, learn to factor and divide.

$9|1+4+2+8+5+7=27$, so $9|142857$.
$\frac{142857}{999999}=\frac{9\cdot 1 5873}{111111 \cdot 9}= \frac{1 5873}{111 \cdot 1001}$.
Then, try dividing out 111 from 15873. It works, so we have $\frac{143}{1001}$. Then, notice that $1001 = 143 \cdot 7$ by guess-and-check or w/e you want to call it, so the final reduced fraction is $\boxed{ \boxed{ \boxed{ \frac{1}{7}}}}$

you can also be totally 1337 and use the euclidean algorithm...

$(999999,142857)=(0,142857)$ subtracting $142857$ a few times, then the common factor is $142857$...so you divide... and get 1/7

[quote="moogra"][quote="xxxxx"]0.428571... repeating, can be expressed as a simplified fraction in the form a/b. FInd b-a[/quote]
[hide]
thats 3/7 (memorized for math contests). I like roadnottaken's solution though, but I think he coud have just skipped to x=142857/999999 because all 9's are repeating[/hide][/quote]
I wanted to demonstrate why the process works, so that people seeing this will be able to apply it to other repeating decimals.  Also, note that this process still works when there are non-repeating digits between the decimal point and the repitend.

i didnt have this exact EDIT : fraction (i cant type) memorized, but i saw it was the x/7 digits, and just less than 1/2, so i approxamated 3/7.  i guess thats not very mathematical, but it worked in this case.

[quote="roadnottaken"]A more rigorous way to find the fractional equivalent of a repeating decimal is to multiply the original value by $10^{n}$, where $n$ is the number of digits of repetition, subtract the original number, put $99\ldots99$ in the denominator, where there are $n$ $9$'s, and reduce.  For example,

$\begin{eqnarray*}x &=& .142857\ldots\\ \\ 1000000x &=& 142857.142857\ldots\\ \\ 1000000x-x &=& 142857.142857\ldots-.142857\ldots\\ \\ 999999x &=& 142857\\ \\ x &=& \frac{142857}{999999}\\ \\ &=& \frac{1}{7}$[/quote]
Ahem. Something to look at also [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120350[\url] if you're interested.

[quote="bigboy82892"]Ahem. Something to look at also [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120350[\url] if you're interested.[/quote]
No, I did not look at that topic before posting this.  I also learned the method several years ago.

[quote="Walk Around The River"]i saw it was the x/7 digits, and just less than 1/2, so i approxamated 3/7.[/quote]

This is actually a very good way to think for contest math.  You can save a lot of time on calculations if you come up with stuff like this  :)

well, if you know it's x/7, then 14x is approximately equal to the first two digits of the expansion. (approximately takes care of carries)

Or, if you notice that it's x/7, you can tell by the first digit which one it is.
[edit]If the first digit is ... [/edit]
1=1/7
2=2/7
4=3/7
5=4/7
7=5/7
8=6/7


AMC
USA(J)MO
USAMO
algebra
polynomial
inequalities
complex numbers

Thanks to Youngwhan Lim for some insights on this problem:

My first USAMO problem I ever solved (with just a bit of help  :) ), and refined my proof after some discussion today..... it's super-duper easy, but still..... :) .

I would like to know what I would get on it, and any comments would be appreciated:

Find all solutions, real or complex to the system
$ x + y + z = 3$
$ x^2 + y^2 + z^2 = 3$
$ x^3 + y^3 + z^3 = 3$

My solution:

$ x^2 + y^2 + z^2 = (x + y + z)^2 - 2( xy + yz + zx )$

This implies:

$3 = 9 - 2( xy + yz + zx )$ , so $\fbox{ xy + yz + zx = 3}$

Now, $ x^3 + y^3 + z^3 = ( x + y + z )(( x + y + z)^2 - 3 ( xy + yz + zx)) + 3(xyz)$

This implies:

$3 = 3 ( 9 - 3 (3)) + 3xyz$ , so so $\fbox{ xyz = 1}$

We have:

$x + y + z = 3$
$xy + yz + zx =3$
$xyz = 1$

So the distinct ordered pair $(x, y, z)$ is a permutation of the roots of the polynomial $ a^3 - 3a^2 + 3a - 1 $

We can factor this as $(a - 1)^3$ , suggesting $ x = y = z =1$

There is only one permutation of this, so the only solution to the above ordered pair is $(x, y, z) = (1, 1, 1)$

You might want to mention something like "by Viete's Formulas" before jumping into the polynomial part, but other than that it looks good. Nice proof, btw.

On a side note, is it possible to use the equality condition on Chebyshev's to prove it really quickly?

The only part I don't understand is the cubic at the end. Otherwise, very nice solution. :)

[quote="paladin8"]On a side note, is it possible to use the equality condition on Chebyshev's to prove it really quickly?[/quote]

Not easily.  Thinking big picture, we would like to induce some ordering on $x,y,z$ that allows us to order $x^2,y^2,z^2$ and $x^3,y^3,z^3$.  But we can't necessarily simultaneously order $x,y,z$ and $x^2,y^2,z^2$ if negative numbers are allowed, and we can't order complex numbers, period.

Nice solution!
And I think in the last step, we can use AM-GM:
$x+y+z = 3 \sqrt{xyz}$, so we must have $x=y=z$.

If $x,y,z>0$ then $x=y=z=1$.It follows immediately from Chebyshev's inequality. ;)

err i did this problem using a different method (this is my first USAMO proof too) so i wanted to know if this was good enuff for a proof

u can set $x,y,z$ as the roots to the polynomial $P(x) = a_3x^3 + a_2x^2 + a_1x + a_0 = 0$

also,
$x+y+z = 3 = s_1$
$x^2+y^2+z^2 = 3 = s_2$
$x^3+y^3+z^3 = 3 = s_3$

using newton's sums, you can get three equations:
$a_3s_1 + a_2 = 0 \Rightarrow 3a_3 + a_2 = 0$
$a_3s_2 + a_2s_1 + 2a_1 = 0 \Rightarrow 3a_3 + 3a_2 + 2a_1 = 0$
$a_3s_3 + a_2s_2 + a_1s_1 + 3a_0 = 0 \Rightarrow 3a_3 + 3a_2 + 3a_1 + 3a_0 = 0$

looking at the last equation, you see that:
$a_3 + a_2 + a_1 + a_0 = 0$, which means that the sum of the coefficients of $P(x)$ is 0, implying that one of the roots is equal to $1$

so, we can set $x= 1$ without affecting the answer, and you get:
$y+z = 2, y^2+z^2 = 2, y^3 + z^3 = 2$

now we can do the same thing, using a new polynomial, $f(x) = a_2x^2 + a_1x + a_0 = 0$ with roots $y,z$
thus getting:
$2a_2 + a_1 = 0$
$2a_2 + 2a_1 + 2a_0 = 0$

looking at this last equation, you see that the sum of the coefficients of $f(x)$ again equals 0, which means that $f(x)$ has a root that is $1$

so now we can set $y=1$, and from that get $z = 1$, so the only solution is for $x,y,z$ is $(1,1,1)$

[quote="1234567890"]Thanks for the replies, guys! So would I get a 7?

Paladin: I keep hearing Viete's, but I'm not sure what it is. So I really can't mention it.[/quote]

I'm not sure of the exact score, but I'd say 7 (no flaws really).

Viete's (sp?) Formulas: http://mathworld.wolfram.com/VietasFormulas.html

It's just the formulas that deal with the coefficients of a polynomial in relation to its roots.

[quote="tetrahedr0n"]AM-GM with negative numbers?[/quote]

They're positive...

And I think it's Yongwhan Lim. Using it as roots of polynomials was some serious bit of insight. But I still like he AM-GM one better. Btw, what's chevaler's?

Edit: well, to see that they're positive, look at xyz, which is positive. That means either they're positive or two are negative. wlog, let it by y and z negative, with x positive. x+y+z is positive, so x>|y| and x>|z|. The sum xy+yz+xz is positive. Terms xy and xz are negative and yz is the only positive one. However, we know both |xy| and |xz| are greater than yz by x>|y| and x>|z|. This makes the sum negative.

It would be a 7. Good job. Like my proof on 1986 USAMO #3(or 4), u didnt mention Viete's formula, but the graders would know what you mean. But, you never know, so it is always good to be as detailed as possible.

The whole discussion about AM-GM or other inequalities is kind of nullified by the presence of potential negative or complex roots. You have to do some work (as everyone has) before you can get a positive restriction on things to use those ineqs. I mean, otherwise the problem would be worthless. Consider

If $x, y, z > 0$, find all solutions of the same system:

Well then Power-Meaning any pair of the equations would force equality.


geometry calculus integration probability

how would you do these countdown questions quickly:

the sum of the areas of two circles with integral radii is 202pi. what is the radius of the larger circle?

at an intersection, 30% of the cars turn left, 40% turn right, and 30% go straight. what is the probability that the next car turns lseft and the folowing one turns right? express your answer as a common fraction.

[hide]2. (3/10)*(2/5)=3/25...I'm not sure about #1 though.  I think you would find two square numbers that add up to 202, but I'm not sure which ones those are[/hide]

1) 81+121=202.
the larger is 11.

basically, I looked for the squares that ended in 1.  seems to have worked

For the second one you multiply 30% by 40% getting 12%. I think.

[quote="eryaman"]2. (3/10)*(2/5)=3/25...I'm not sure about #1 though.  I think you would find two square numbers that add up to 202, but I'm not sure which ones those are[/quote]
eryaman already said that, brian...


First there is a circle with center O and radius 2. Points A and B lie on the circle such that m<AOB = 90. A semicircle with diameter AB is constructed (the semicircle lies outside the circle.)

Find the difference in area between the two marked regions in the figure.

The triangle's area would be 2 because it's a 45, 45, 90 triangle. The semi circle's area would be pi because you do (2^1/2)^2 x pi and then divide it by 2 because it's a semi circle so you get the area of pi. Then you find out the area of the sector of the bigger circle which would be 1/4 of the whole circle so it would be pi. So then the area of the the region just outside of the triangle would be pi-2. So then you do the area of the semi circle-the area of the region right outside the triangle thus getting pi-(pi-2) and then subtract that with the area of the triangle so it would be [b](pi-(pi-2))-2which is equal to 0[/b].  :showoff:  :bruce:

[quote="Slizzel"]We note $R$-the big circle radius and $r$-the semicircle radius

The area of sector $AOB$ is $\frac{\pi R^2}{4}$ and $Area_{\Delta ABC}=\frac{R^2}{2}$. If $\Delta ABC$ is a right triangle we can find the semicircles radius $r=\frac{AB}{2}=\frac{R\sqrt{2}}{2}$. Now the area of the semicircle is $\pi r^2=\frac{\pi R^2}{4}$.

$Area_{sector}-A_{\Delta ABC}=\frac{R^2(\pi-2)}{4}$ so the area of the marked pice of semicircle is $\frac{\pi R^2}{4}-\frac{R^2(\pi-2)}{4}=\frac{R^2}{2}$.

The diference between the marked region is $\frac{R^2}{2}-\frac{R^2}{2}=0$[/quote]

 :huh:  :huuh:  :wacko: You're way seems kinda complicated to me, I think I might have done it an easier way....but it's still cool.

[quote="Slizzel"]It's not confusing. You don't even need $R$ ;)[/quote]

Oh well, as long as we both got the right answer  :gleam:

Another way to do it:
Look at the area between the two regions. Call it X. If you add X to the each of the two regions, the problem remains unchanged. Then you need to find the difference between a semicircle and a quartercircle.

And where do you get all these smilies??

When you write a post , in the left side is a box with all kind of smiles. This is my favorite  :icecream:

:wallbash: stupid me, i didn't see the "view more" option.
Thanks!

This was kind of a fun problem.  :D
[hide]The area of triangle AOB is $2\times2\div2=2$.  Since AOB is a 45-45-90 triangle, AB is $2\sqrt{2}$, so the radius of the semicircle is $\sqrt{2}$, and its area is $\frac{(\sqrt{2})^{2}\pi}{2}=\pi$.
Now we need to find the area of circular segment AB.  This is equal to area of arc AB-area of triangle AOB, which simplifies to $\frac{2^{2}\pi}{4}-2=\pi-2$.  So the area of the shaded area of the semicircle is $\pi-(\pi-2)=2$.
Because the area of the triangle is $2$, the difference is $0$.[/hide]


$\triangle ABC,\ M,N\in (BC),\ MB=MC,\ \widehat{BAN}\equiv \widehat{CAN}\Longrightarrow$

$\overline{\underline{| m(\widehat{BAC})=90^{\circ} +m(\widehat{MAN})\Longleftrightarrow \frac bc =1-2\cdot \cos A| }}$

[b]Remark.[/b] ${b=2c\Longrightarrow \overline {\underline | A=120^{\circ}\Longleftrightarrow m(\angle MAN)=30^{\circ}| }}.$

You can prove the following $tg^2{\frac{A}{2}} = tg{\alpha}. cotg{\frac{B-C}{2}}$, where $\alpha$ is the angle between the median and the bisector.


[b]For positive x,y,z such that xyz=1, show that min max {x+y,x+z,y+z}=2[/b]

im a little confused on these types of problems. what's the general strategy/approach?

$ x \plus{} y\leq max\{x \plus{} y, x \plus{} z,\ y \plus{} z\},\ y \plus{} z\leq max\{x \plus{} y, x \plus{} z,\ y \plus{} z\},$
$ \ z \plus{} x\leq max\{x \plus{} y, x \plus{} z,\ y \plus{} z\}$, adding these inequalities,
$ 3max\{x \plus{} y, x \plus{} z,\ y \plus{} z\}\geq 2(x \plus{} y \plus{} z)\geq 2\cdot 3\sqrt [3]{xyz} \equal{} 6$, yielding $ max\{x \plus{} y, x \plus{} z,\ y \plus{} z\}\geq 2.$

[quote="mihail911"]but it says to find the min max of the set[/quote]That's the min of all the max's

[quote="Temperal"]Another problem from AoPS 2, problem 214 from Mandelbrot #1. Mihail, I'd recommend you'd stop posting problems from the book that the website itself publishes, please. it seems kind of redundant to me. I might be wrong, and you should post them, though...

The min max is simply consider the max of all sets satisfying the given conditions, and then find the min of those max elements. Generally, you'll want to put the min max on the greater than side of AMGM and vice versa for max min.[/quote]
sorry.  :blush:
sometimes, i just find a problem that i havne't seen before on the website and/or have a question about so i choose to post it.

[hide="Discussion"] The general strategy I like to employ on problems symmetric in the variables, especially on problems involving the $ \text{min}$ or $ \text{max}$ function, is to assume WLOG that $ x \ge y \ge z$.  Then clearly $ \text{max}(x \plus{} y, x \plus{} z, y \plus{} z) \equal{} x \plus{} y$ and it remains to show that $ x \plus{} y \ge 2$.  But since

$ x \ge z$

and

$ y \ge z$

it follows that

$ (x \plus{} y) \plus{} \frac {x \plus{} y}{2} \ge x \plus{} y \plus{} z$

hence that

$ x \plus{} y \ge \frac {2}{3} \left(x \plus{} y \plus{} z \right) \ge 2 \cdot \sqrt [3]{xyz} \equal{} 2$. [/hide]

[quote="Elio (n)"]What does it mean $ y,x$[/quote]
Possible  :spam:  or perhaps be more clearer what you are asking about... $ x,y$ are positive reals here...  :ninja:

[quote="mihail911"]
sorry.  :blush:
sometimes, i just find a problem that i havne't seen before on the website and/or have a question about so i choose to post it.[/quote]

Have you the solutions manual? That would be helpful.

What if he doesn't understand the solution?

[quote="Elio (n)"]In {x+$ y,x$+$ z,y$+z}  what are $ y,x$ and $ z,y$[/quote]
Hi Elio (n)
$ y,x$ is not a quantity

Look at the whole thing, it is a [b][[set]][/b] (I hope you know what a set is...) where its [b]elements[/b] are separated by commas, like {1,2,3,4,5,6}

For this problem, we have 3 elements, namely,
x+y
x+z
y+z
3 elements we can put them in a set ---->$ \{x\plus{}y\quad\quad,\quad\quad x\plus{}z\quad\quad,\quad\quad y\plus{}z\}$, with each element separated by commas.

[quote="Temperal"]The AoPS official solutions are likely to be much clearer than anything on here.[/quote]
that's not always true. the only time ill post a problem from a book like aops v2 that i already have the solutions manual for is if i don't understand the problem. plus, i don't see anything wrong against posting a problem that hasn't been posted before.  :wink:

oops i did not looked it with atention , it samed me like a sum x+x,y  :blush:

Thankyou 10000thuser for the explication ironic


[b]Let [/b]$ A \in M_{n}{(\mathbb{R})}$ [b]Such that [/b]: $ Rank(A) + Rank(I_n + A^2) \leq n$
[b]Prove inequality  :[/b]
        $ \det(A + xI_n)\det(A + yI_n) \geq \det(A + \sqrt {xy}I_n)^2$ ,$ \forall x,y \in \mathbb{R^{ + }}$
$ I_n$ is identity matrix

We can assume $ A$ in Jordan normal form, i.e. ${ A=J_{t_1}(z_1)\oplus J_{t_1}(\bar{z_1})\ldots\oplus J_{t_m}(z_m)\oplus J_{t_m}(\bar{z_m})\oplus J_{s_1}(r_1})\ldots\oplus J_{s_k}(r_k)$, where $ J_{s}(x)$ is a Jordan block of size $ s$ corresponding to the eigenvalue $ x$ and $ z_1,\ldots,z_m\in\mathbb{H},r_1,\ldots,r_k\in\mathbb{R}$. If we denote  $ l=\{j=1,\ldots,k|r_j=0\}$ and $ l'=\{j=1,\ldots,m|z_j=i\}$, we have the inequality $ n\geq Rank(A)+Rank(I+A^2)=n-l+n-2l'$, i.e. $ n\leq l+2l'\leq \sum_{j=1}^ks_j+2\sum_{j=1}^mt_j=n$. This implies $ A=iI_m\oplus -iI_m\oplus 0_{n-2m}$ and for arbitrary $ w\in\mathbb{R}$ we calculate $ \det (A+wI)=(w^2+1)^mw^{n-2m}$. Now for $ x,y\geq 0$ we compute:

$ \begin{array}{rl}\det(A+xI)\det(A+yI)-\det(A+\sqrt{xy}I)^2= & [(x^2+1)(y^2+1)]^m(xy)^{n-2m}-(xy+1)^{2m}(xy)^{n-2m}\\
 = & (xy+1)^{2m}(xy)^{n-2m}\left\{\left[1+\frac{(x-y)^2}{(xy+1)^2}\right]^m-1\right\}\geq 0\end{array}$


Is it true that $\sqrt{2\sqrt{3\sqrt{4...\sqrt{n}}}}<3$ for all $n\in N$?

[quote="Rushil"]yup, although I don't remember the RHS of the eqn. AS far as I can recall , i know a proof based on some form of induction[/quote]

Which form, then?  :)

let the term be $x$,
so $x^2=2\sqrt{3\sqrt4...}$ - 1
     $x^4=3\sqrt{4\sqrt5...}$ - 2
dividing.... 1 and 2, let $\sqrt{4.....}$ be $n$
$x^2=\frac{3\sqrt{n}}{2\sqrt{3n}}$
So, Solving, we get,
$x^2=\frac{\sqrt{3}}{2}$
so, $x<3$
correct?

This problems's general case has been solved in Arthur englel's classic"Problem solving strategies". He makes use of something called reverse induction on $m$ which is the general term which replaces $2$.

Thus the general result is
$ \sqrt {m\sqrt{m+1 \sqrt {m+2 ....} } } < m +\ 1 $

Just use reverse mathematical induction on $m$ starting from $ m =\ N $ and moving till $ m =\ 2$

Use that $ \sqrt{N} < N + 1 $

why, isnt my solution right? it seems pretty right :huh:

shouldn't $X^4 $have 12 as first term ( 4 *3)

oh no! My solution is all wrong! :mad:  Ill have to use yours then :oops:

I too was wondering why a great like Arthur engel would intoduce ' reverse induction' and other tactics to solve this problem if there existed your solution.

Have you understood my solution??

not quite... what is reverse induction?.... what is induction? :blush:

induction refers to the Principle of Mathematical Induction. I was hoping you'd be knowing that. No offence meant, but which class are you in??? :P  :o

It basically states that if a result is true for integers n =1,2,3,.....n (by assumption) and we prove it to be true for n+1 , then the result is , logically, true for all integers. Don't pay heed to what I just wrote. It might coz you trouble when you try to understand it. Just search on the web or ask your teacher about Mathematical Induction (PMI). It's probably the most useful method in maths for positive integers. You better understand the concept fully.It will be extremely useful ( i used it in the '2n turns' problem implicitly)

The only place where I have found anyone use 'reverse induction' is in this particular problem in engel's book. If you have access to it or don't , i recommend you to buy it if you are preparing for the olympiad. It (and Paul Zeitz's Art and Craft of Problem solving) are the best books in the market. If you do all the problems in these books properly without looking at the soln, you don't need to do anything else for the olympiads.

SO if you want to learn about 'reverse induction' , look up engel's book

I am going to go in 10th standard :blush:
what class are you in?

I am in class 12 (our schools start in different months - mine in april - yours in september)

Anyway, it's great you are doing olympiad stuff now.I too started such things in class 10. I assure you, yo'll really enjoy it all. I suggest, you search on the net for olympiad resources in form of books etc. If you do everything properly, you will be able to perform outstandingly.The UK has great resources. Get all the books in all topics by excellent authors. You can serch for them on the net.

Best of luck!!
I can help you whenever you want.

[quote="Rushil"]I too was wondering why a great like Arthur engel would intoduce ' reverse induction' and other tactics to solve this problem if there existed your solution. [/quote]

He sometimes gives harder proofs than necessary, especially if they're really creative. His two proofs for IMO 86 6 are longer than the shortest and neatest proof.

[quote="riddler"]Thank you for your help :D  Ive got friends who are preparing for IIT aswell like EMINEM and SHADYSAYSURSPAMMED. In England they start very very late, that's why I keep up with the Indian Standards :D
Thanks for your offer.[/quote]

Are you preparing for IIT? Come and discuus in my IIT forum in the college section. I never though people prepared for IIT from the UK. Who's gonna come here when you have Oxbridge???

I want to know more. Please post in the particular forum I have mentioned above

[quote="Rushil"] Oxbridge??? [/quote]
It is Oxford and Cambridge. :D  :P

I really dont know what to discuss in the IIT forum?

didnt know about ur forum section,rushil.will come there.
riddler's pretty stingy n he does not tell me books,n in indore,u limited books,so i could need some help from u :)

ps-10 in india is totally muggin and huge n borin

yes all of them are preparing for IIT.
And Shady you better shut up! look at yourself never tell me anything!
[quote="RUSHIL"]I was asking about your remarh that your friends(in UK i suppose) are preparing for IIT[/quote]
they are in India. :D  and they all are preparing for IIT, I guess IIT is the only thing anykid would want to do nowadays in India.

Yes that is true. engineering is the craze in India and IITs are the place to be.

It seems a bit off trach about your being in UK and having friends in India who are also on this forum. Anyways, instead of discussing such stuff in wake of a mathematics question, post about yourself in the India forum in National communitites section.Also plz read my IIt forum and post your views and queries.

Let's talk there from now.If you have (all Indians) Yahoo or Hotmail , let's talk on messengers from now

do you have messenger? I have it, you can download it from the net if you dont. I know you're e-mail but I dont think you know mine I will PM it too you.

Please use the private messaging feature of the forum to coordinate getting together on instant messenger. ;)


modular arithmetic
number theory unsolved
number theory

For any nonnegative integer $ n$, let $ f(n)$ be the greatest integer such that $ 2^{f(n)} | n \plus{} 1$. A pair $ (n, p)$ of nonnegative integers is called nice if $ 2^{f(n)} > p$. Find all triples $ (n, p, q)$ of nonnegative integers such that the pairs $ (n, p)$, $ (p, q)$ and $ (n \plus{} p \plus{} q, n)$ are all nice.

[hide="kind of a messy solution, but..."]There is the obvious solution $ (0,0,0)$, so we assume that $ n>0$.  Let $ 2^{k\plus{}1}>n\ge2^k$ where $ k\ge0$.  Then, since $ (n\plus{}p\plus{}q,n)$ is nice, we must have $ n\plus{}p\plus{}q\equiv \minus{}1\pmod{2^{k\plus{}1}}$.  So, let $ 2^i\parallel{}n\plus{}1$.  We have 3 cases:  $ i\equal{}2k\plus{}1$, $ i\equal{}0$, and $ 2k\plus{}1>i>0$:

$ \text{Case 1 } i\equal{}2k\plus{}1$

Since $ n<2^{k\plus{}1}$, we have that $ n\equal{}2^{k\plus{}1}\minus{}1$.  Then since $ 2^{f(n)}\equal{}2^{k\plus{}1}>p$ and $ 2^{k\plus{}1}\le2^{f(p)}>q$, we hav $ n\plus{}p\plus{}q<3\cdot2^{k\plus{}1}\minus{}1$.  Hence, since $ n\plus{}p\plus{}q\equiv \minus{}1\pmod{2^{k\plus{}1}}$, we have that $ n\plus{}p\plus{}q\equal{}2^{k\plus{}1}\minus{}1$ or $ 2^{k\plus{}2}\minus{}1$.  In the former case, we have $ p\plus{}q\equal{}0\Rightarrow (p,q)\equal{}(0,0)$.  In the latter case, we have $ p\plus{}q\equal{}2^{k\plus{}1}$.  If $ p$ is even, then $ q\equal{}0$ and so $ p\equal{}(2^{k\plus{}1},0)$, but $ 2^{f(n)}\not>2^{k\plus{}1}$.  If $ p\equal{}2^{k\plus{}1}\minus{}1$, then $ q\equal{}1$.  Otherwise, we have $ p\equal{}2^{k\plus{}1}\minus{}(2^tj\plus{}1),q\equal{}2^tj\plus{}1$ where $ j$ is odd.  But, we have $ 2^{f(p)}\equal{}2^t>q\equal{}2^tj\plus{}1$, a contradiction.  So we have solution sets $ (n,p,q)\equal{}(2^{k\plus{}1}\minus{}1,0,0),(2^{k\plus{}1}\minus{}1,2^k\minus{}1,1)$.

$ \text{Case 2 } i\equal{}0$

If $ i\equal{}0$ then $ n$ is even.  So, $ 2^f(n)\equal{}1>p$.  Thus, we have $ p\equal{}0$.  So $ 2^{f(p)}\equal{}1>q$.  So, $ q\equal{}0$.  But then $ n\plus{}p\plus{}q$ is even.  So we have no solution sets.

$ \text{Case 3 } 2k\plus{}1>i>0$

Moreover, let $ n\equal{}2^{k\plus{}1}\minus{}(2v\plus{}1)2^i\minus{}1$.  Now, we have $ 2^{f(n)}\equal{}2^i> p$ and so we also have $ 2^{f(p)}\ge2^i> q$.  Thus, we have $ n\plus{}p\plus{}q< 2^{k\plus{}1}\minus{}(2v\plus{}1)2^i\minus{}1\plus{}2^i\plus{}2^i\equal{}2^{k\plus{}1}\minus{}(2v\minus{}1)2^i\minus{}1$.  Since $ 2^{k\plus{}1}\minus{}1\le n\plus{}p\plus{}q$, we must have $ 2^{k\plus{}1}\minus{}1< 2^{k\plus{}1}\minus{}(2v\minus{}1)2^i\minus{}1\Rightarrow (2v\minus{}1)2^i<0\Rightarrow v\equal{}0$.  So, we have $ n\equal{}2^{k\plus{}1}\minus{}2^i\minus{}1$.  Then, since $ n\plus{}p\plus{}q\equiv\minus{}1\pmod{2^{k\plus{}1}}$ and since $ 2^{f(n)}\equal{}2^i>p\ge q$, we must have $ p\plus{}q\equal{}2^i$.  If $ p$ is even, then $ q\equal{}0$ and so $ p\equal{}2^i$, but $ 2^{f(n)}\equal{}2^i\not>p$.  If $ p\equal{}2^i\minus{}1$, then $ q\equal{}1$.  Otherwise, we have $ p\equal{}2^i\minus{}(2^tj\plus{}1),q\equal{}2^tj\plus{}1$ where $ j$ is odd.  But, we have $ 2^{f(p)}\equal{}2^t>q\equal{}2^tj\plus{}1$, a contradiction.  Thus, we have the solution sets $ (n,p,q)\equal{}(2^{k\plus{}1}\minus{}2^i\minus{}1,2^i\minus{}1,1)$.

So all solutions are

$ (n,p,q)\equal{}(2^{k\plus{}1}\minus{}1,0,0),(2^{k\plus{}1}\minus{}1,2^{k\plus{}1}\minus{}1,1),(2^{k\plus{}1}\minus{}2^i\minus{}1,2^i\minus{}1,1)$.

[/hide]

Note on the solution above: You cannot have $ p \equal{} 0$ because then $ f(p) \equal{} 0$, so we cannot have $ f(p) > q$. (I just typed this; I have not yet read the entire above solution.) EDIT - no, he is right.  :blush:

[hide]
$ 2^{f(n)}|n \plus{} 1$, $ 2^{f(n)} > p$

$ 2^{f(p)}|p \plus{} 1$, $ 2^{f(p)} > q$

$ 2^{f(n \plus{} p \plus{} q)}|n \plus{} p \plus{} q \plus{} 1$, $ 2^{f(n \plus{} p \plus{} q)} > n$

So

$ 2^{f(n \plus{} p \plus{} q)} > n$
$ n \plus{} 1\ge 2^{f(n)}$
$ 2^{f(n \plus{} p \plus{} q)}\ge 2^{f(n)}$ (1)

$ 2^{f(n)} > p$
$ p \plus{} 1\ge 2^{f(p)}$
$ 2^{f(n)}\ge 2^{f(p)}$ (2)

$ 2^{f(p)} > q$ (3)

So again $ 2^{f(n \plus{} p \plus{} q)}|n \plus{} p \plus{} q \plus{} 1 \equal{} (n \plus{} 1) \plus{} (p \plus{} 1) \plus{} (q \minus{} 1) \equal{} 2^{f(n)}a \plus{} 2^{f(p)}b \plus{} (q \minus{} 1)$. (a,b are odd integers.)

Because $ f(n \plus{} p \plus{} q)\ge f(n)\ge f(p)$ (by (1) and (2)), we must have $ 2^{f(p)}|q \minus{} 1$. Given that $ 2^{f(p)} > q$, this can only happen if:

a) $ q \equal{} 0$, and $ 2^{f(p)} \equal{} 1\implies f(p) \equal{} 0$.

$ 2^{f(n \plus{} p)}|2^{f(n)}a \plus{} p$

We know that $ 2^{f(n)} > p$, but since we need $ 2^{n}|p$, we must have $ p \equal{} 0$.

$ 2^{f(n)}|2^{f(n)}a \equal{} n \plus{} 1$
$ 2^{f(n)} > n$

so $ n \equal{} 2^{f(n)} \minus{} 1$.

b) $ q \equal{} 1$. So now

$ 2^{f(n \plus{} p \plus{} q)}|2^{f(n)}a \plus{} 2^{f(p)}b$.

But again, as $ f(n \plus{} q \plus{} q)\ge f(n)\ge f(p)$, we must have $ f(n) \equal{} f(p)$. If we examine how equation (2) was derived, we must have equality so $ 2^{f(n)} \equal{} p \plus{} 1 \equal{} 2^{f(p)}$. Let $ n \equal{} 2^{f(p)}a \minus{} 1$

$ 2^{f(n \plus{} p \plus{} q)}|(n \plus{} 1) \plus{} 2^{f(p)} \equal{} 2^{f(n)}a \plus{} 2^{f(p)} \equal{} (a \plus{} 1)(2^{f(p)})$
$ 2^{f(n \plus{} p \plus{} q)} > n \equal{} 2^{f(p)}a \minus{} 1$

Suppose $ a \plus{} 1 \equal{} 2^{c}d$, with $ d$ odd. Then $ f(n \plus{} p \plus{} q) \equal{} f(p) \plus{} c$, so

$ 2^{f(n \plus{} p \plus{} q)} \equal{} 2^{f(p) \plus{} c} > 2^{f(p) \plus{} c}d \minus{} 1$

which is true only if $ d \equal{} 1$. Therefore we can say that $ a \plus{} 1$ is a power of $ 2$.

So now we know we have to possible solution sets, and I will show that they satisfy the conditions.

$ n \equal{} 2^{f(n)} \minus{} 1$
$ p \equal{} 0$
$ q \equal{} 0$

for some integer $ f(n)$

$ (n,p)$ is nice because $ 2^{f(n)} > 0$.

$ (p,q)$ is nice because $ 2^{f(p)} \equal{} 1 > 0$.

$ (n \plus{} p \plus{} q,n)$ is nice because $ 2^{f(n)} > 2^{f(n)} \minus{} 1 \equal{} n$.

$ q \equal{} 1$
$ p \equal{} 2^{f(p)} \minus{} 1$
$ n \equal{} 2^{f(p)}(2^c \minus{} 1) \minus{} 1$

for some $ f(p) \ge 1$ and $ c\ge 1$.

$ (n,p)$ is nice because $ 2^{f(p)} > \equal{} 2^{f(p)} \minus{} 1p$

$ (p,q)$ is nice because $ 2^{f(p)} > 1 \equal{} q$

$ (n \plus{} p \plus{} q,n)$ is nice because $ f(n \plus{} p \plus{} q) \equal{} 2^{f(p) \plus{} c} > 2^{f(p)}(2^c \minus{} 1) \minus{} 1$.

so these solutions work.[/hide]

I don't think so.  $ 2^{f(0)} \equal{} 2^0 \equal{} 1$ so if $ (0,q)$ is nice, we can have $ q \equal{} 0$.

Oops, my bad. (For a second, I thought it was $ f(p)>q$, not $ 2^{f(p)}>q$.)


Hallo!
1.
Wenn ihr beispielsweise Aufgaben von der IMO löst, habt ihr dann standardisierte Lösungsverfahren oder denkt ihr euch jedes mal individuell eine Lösung aus?
2.
Wie wichtig ist Training für den Erfolg beim Lösen von Aufgaben? Wie viel besser ist man, wenn man sich täglich ~2h mit Mathematik beschäftigt?

Hallo!
[quote]1.
Wenn ihr beispielsweise Aufgaben von der IMO löst, habt ihr dann standardisierte Lösungsverfahren oder denkt ihr euch jedes mal individuell eine Lösung aus? [/quote]
das kann ich f"ur (sorry, kein die ü auf dieser tastatur) die geometrie bejahen... ich habe in den letzten jahren gemerkt, wie man an geometrieaufgaben herangehen kann um sie dann in den meisten f"allen (dito@ "a) auch zui l"osen (dito@ "o). damit meine ich, da"s man sich systematisch "uberlegt, ob man den beweis am besten vorne, hinten oder in der mitte anf"angt, und nicht einfach drauf losr"at.
dies hilft aber auch nicht in allen f"allen, denn manchmal ist es entscheidend, zus"atzlich einige s"atze und standardsituationen zu kennen (damit meine ich vor allem die ganzen s"atze "uber sehnen, tangenten, sekanten am kreis, ceva, menelaos... nicht so wichtig erscheinen mir sinus- und cosinussatz, aber auch inversion (wird meiner meinung nach "uberbewertet))

in anderen gebieten, vor allem der "au"serst mysteri"osen zahlentheorie, sind mir solche l"osungsverfahren allerdings unbekannt geblieben.
[quote]2.
Wie wichtig ist Training für den Erfolg beim Lösen von Aufgaben? Wie viel besser ist man, wenn man sich täglich ~2h mit Mathematik beschäftigt? [/quote]
training hilft bestimmt weiter... man muss halt nur die zeit dazu haben[/quote]

[quote="ZetaX"][quote="Werwulff"]Hallo!
[quote]1.
Wenn ihr beispielsweise Aufgaben von der IMO löst, habt ihr dann standardisierte Lösungsverfahren oder denkt ihr euch jedes mal individuell eine Lösung aus? [/quote]
das kann ich f"ur (sorry, kein die ü auf dieser tastatur) die geometrie bejahen
[...]
in anderen gebieten, vor allem der "au"serst mysteri"osen zahlentheorie, sind mir solche l"osungsverfahren allerdings unbekannt geblieben.[/quote]

Ja genau, interessant ist also was man standardisiert durch systematische Verfahren lösen kann !¿
Echt ne tolle Einstellung...[/quote]

geometrie ist sehr sch"on, weil es sehr vielf"altig sein kann!  da"s man so leicht auch noch systematisch l"osungen finden kann, macht es noch sch"oner. au"serdem bleiben die aufgaben immer noch recht trickreich, selbst wenn man derart systematisch vorgeht...
ich finde es halt "ode, wenn man wie in der zahlentheorie nur planlos rumraten kann, und mit etwas gl"uck kommt halt das richtige raus... gerade wo die aufgaben an sich schon uninteressant genug sind.

ps: doch! (@zetanoob) :!:

Ich sags kurz und knapp: You have NO idea!

doch!

[code]The message is too small. Please make the message longer before submitting.[/code]
$\Rightarrow$ :spam:

Jedenfalls nicht von Zahlentee!
a) gibt es eine gewisse Systematik
b) gibt es keine in gescheiten Problemen
c) schreib mal wem der eine Vermutung gelöst hat eine Gratulation zum Lotteriegewinn ;)

PS: learn Short-Post-Theory

Gut, danke! Also soll ich das so verstehen, dass es zwar bei bestimmten Fragestellungen eine gewisse "Herangehensweise" gibt, diese jedoch nur einen kleinen Teil der Lösung ausmacht?
Und noch eine Frage: Wie lange beschäftigt ihr euch täglich mit Mathematik (im Durchschnitt)?

[quote="ZetaX"]c) schreib mal wem der eine Vermutung gelöst hat eine Gratulation zum Lotteriegewinn ;)[/quote]
das habe ich bisher jedesmal gemacht (sofern mir die emailadresse bekannt war)

[quote="Blender"]Gut, danke! Also soll ich das so verstehen, dass es zwar bei bestimmten Fragestellungen eine gewisse "Herangehensweise" gibt, diese jedoch nur einen kleinen Teil der Lösung ausmacht?[/quote]
so in etwa. manchmal macht es nur einen kleinen teil aus, manchmal einen gr"o"seren.
[quote="Blender"]
Und noch eine Frage: Wie lange beschäftigt ihr euch täglich mit Mathematik (im Durchschnitt)?[/quote]
mindestens 6 stunden (aber auch erst, seitdem ich studiere)

[quote="Blender"]und vorher? (Seit wann studierst du)?[/quote]
vorher bin ich nur sehr wenig zu madde gekommen (weniger als  2h pro tag), da die kollegstufe doch recht anstrengend war (jedenfalls wenn man schule nicht ganz schleifen l"asst (@kanonische n00bs!  :bye: )).
(ich bin jetzt im 2. semester) :surf:

Ich fühl mich mal unter den kanonischen angesprochen  :bruce:
 Aber bevor ich noch mehr hier  :spam: -e:

zu 1.: ich zumindest nicht (Mathematik lebt vom Unerwarteten! / alles was straightforward ist ist langweilig

zu 2.: ich trainiere eigentlich nicht so direkt; ich mache einfach Aufgaben die mir gerade sympatisch erscheinen (seltenst Geo, meist Zahlentee).

[quote="ZetaX"]zu 2.: ich trainiere eigentlich nicht so direkt; ich mache einfach Aufgaben die mir gerade sympatisch erscheinen (seltenst Geo, meist Zahlentee).[/quote]

Der Trainingseffekt davon sollte aber auch relativ groß sein, wenn dir viele Aufgaben sympatisch erscheinen... wie viel Zeit ist das dann täglich, die du damit ca. verbringst?

So viel erscheint mir nicht sympatisch (hier auf ML z.B. sind recht viele hässliche "Durchrechenaufgaben"; auch sind einige oft trivial, was wohl auch kein Training bringt). Bezüglich Zeit habe ich kA da ich das immer zwischendurch mache.

Hallo Blender,

zur Frage nach standardisierten Vorgehensweisen:
Schau doch einmal auf
[url]http://www.probleme-und-strategien.de[/url] vorbei.
(Die Seite ist nichtkommerziell.)
Dort sind Ideen zum systematischen Lösen mathematischer Probleme zusammengestellt, die sich in der Literatur meines Wissens so nicht finden.
Eine besondere Rolle spielt dabei die Kombination von Mindmaps mit mathematischen Lösungswerkzeugen:
Man benutzt eine "Problem-Map", in der das eigentliche Problem untersucht wird, und eine oder mehrere "Werkzeug-Maps" mit Lösungswerkzeugen.

Viele Grüße
Thomas


[hide="Wonder if this can be shortened"]Suppose $ f(a) \equal{} f(b) \equal{} c$. Then $ f(xc) \minus{} f(x \plus{} c) \equal{} af(x) \minus{} a \minus{} f(x) \equal{} bf(x) \minus{} b \minus{} f(x)$. So as long as there exists an $ x$ such that $ f(x) \neq 1$, we get $ f$ injective. $ f(x) \equal{} 1$ for all $ x$ does not satisfy the original equation, so we can assume the existence of such an $ x$.

Put in $ x \equal{} 0$. We get $ f(f(y)) \equal{} y(1 \minus{} f(0)) \plus{} f(0)$. Suppose $ f(0) \equal{} 1$. Then put $ y \equal{} 0$ into that equation to get $ f(1) \equal{} 1$, contradicting $ f$ injective. Therefore $ f(0) \neq 1$ and $ f$ is bijective.

This means there exists a real $ c$ such that $ f(c) \equal{} 0$. Put in $ y \equal{} c$ to get $ f(0) \plus{} c \equal{} cf(x)$. The only way for this to be true if $ f$ is injective is for $ c \equal{} 0$. So $ f(0) \equal{} 0$. Now $ x \equal{} 0$ gives $ f(f(y)) \equal{} y$ for all reals $ y$.

Replace $ y$ in the given equation with $ f(y)$ to get the neater looking
$ f(xy) \plus{} f(x) \plus{} f(y) \equal{} f(x \plus{} y) \plus{} f(x)f(y)$.

Now for some substituting fun.
--- Put $ x \equal{} y \equal{} 2$ in to get $ f(2)^2 \minus{} 2f(2) \equal{} 0$. So $ f(2) \equal{} 2$ or $ f(2) \equal{} 0$. The latter contradicts $ f$ injective ($ f(2) \neq f(0)$), so $ f(2) \equal{} 2$.
--- Put $ x \equal{} y \equal{} 1$ in to get $ f(1)^2 \minus{} 3f(1) \plus{} 2 \equal{} 0$. So $ f(1) \equal{} 1$ or $ f(1) \equal{} 2$. The latter also contradicts $ f$ injective ($ f(1) \neq f(2)$), so $ f(1) \equal{} 1$.
--- Put $ x \equal{} 1$ in to get $ f(x) \plus{} 1 \equal{} f(x \plus{} 1)$. This also means $ f(x) \equal{} f(x \minus{} 1) \plus{} 1$. We also conclude $ f(n) \equal{} n$ for all integers $ n$. In particular $ f( \minus{} 1) \equal{} \minus{} 1$.
--- Put in $ x \equal{} \minus{} 1$ to get $ f( \minus{} x) \plus{} f(x) \equal{} f(x \minus{} 1) \plus{} 1 \minus{} f(x)$. The RHS is 0 by the previous line. So $ f( \minus{} x) \equal{} \minus{} f(x)$.
--- Put in $ y \equal{} \minus{} x$ to get $ f( \minus{} x^2) \plus{} f(x) \plus{} f( \minus{} x) \equal{} f(x)f( \minus{} x)$, or $ f(x^2) \equal{} f(x)^2$. There's actually only one thing important about this equation: $ f(x) > 0 \Leftrightarrow x > 0$, and because $ f$ is odd, $ f(x) < 0 \Leftrightarrow x < 0$.

We now show $ f(nx) \equal{} nf(x)$ for all positive integers $ n$; we do this by induction. Base case $ n \equal{} 1$ is trivial. If $ f(nx) \equal{} nf(x)$, then $ y \equal{} nx$ gives $ f(nx^2) \plus{} f(x) \plus{} f(nx) \equal{} f((n \plus{} 1)x) \plus{} f(x)f(nx)$, or $ f(x) \plus{} nf(x) \equal{} f((n \plus{} 1)x)$, or $ (n \plus{} 1)f(x) \equal{} f((n \plus{} 1)x)$ as wanted. Now by rewriting the identity as $ f(x) \equal{} nf(x/n)$ we see that identity actually holds for all rationals as well. Additionally, this gives $ f(x) \equal{} x$ for rationals since $ f(x) \equal{} xf(1) \equal{} x$.

Now suppose $ a$ is a rational, $ b$ is any real, and $ a < b$. Put in $ x \equal{} a, y \equal{} b \minus{} a$, and note $ y > 0$. We get $ f(a(b \minus{} a)) \plus{} f(a) \plus{} f(b \minus{} a) \equal{} f(b) \plus{} f(a)f(b \minus{} a)$, or $ a \plus{} f(b \minus{} a) \equal{} f(b)$. Recall from $ f(x^2) \equal{} f(x)^2$ that $ f(x) > 0 \Leftrightarrow x > 0$. As $ b \minus{} a > 0$, $ f(b \minus{} a) > 0$ and $ a \equal{} f(a) < f(b)$.

Finally if $ a$ is a rational, $ b$ is any real, and $ a > b$, then $ x \equal{} a, y \equal{} a \minus{} b$ gives similar stuff that leads to $ f(a) > f(b)$. By transitivity of $ >$, this shows that $ f$ is always increasing. As $ f$ is fixed on the rationals and the rationals are dense, this gives $ f(x) \equal{} x$ for all reals.[/hide]

I think my solution is shorter than you :
we first prove $ f(1) \equal{} 1$ and $ f$ is injective and $ f(xy) \plus{} f(x) \plus{} f(y) \equal{} f(x \plus{} y) \plus{} f(x)f(y)$ (as the way MellowMelon shows)
now if $ x \equal{} 1$ then $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$ . now $ x \plus{} 1 \longrightarrow x$ which gives :
$ f(xy \plus{} y) \plus{} f(x \plus{} 1) \plus{} f(y) \equal{} f(x \plus{} y \plus{} 1) \plus{} f(x \plus{} 1)f(y)$ now $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$ and $ f(x \plus{} y \plus{} 1) \equal{} f(x \plus{} y) \plus{} 1$ which gives :
$ f(xy \plus{} y) \plus{} f(x) \plus{} f(y) \equal{} f(x \plus{} y) \plus{} f(x)f(y) \plus{} f(y)$
$ \longrightarrow f(xy \plus{} y) \minus{} f(y) \equal{} f(x \plus{} y) \plus{} f(x)f(y) \minus{} f(x) \minus{} f(y)$ but the $ RH$ is $ f(xy)$ so we have :
$ f(xy \plus{} y) \equal{} f(y) \plus{} f(xy)$ but we can put $ xy \equal{} a$ and $ y \equal{} b$  for all reals $ a,b$ and then :
$ f(a \plus{} b) \equal{} f(a) \plus{} f(b)$ and then by $ f(xy) \plus{} f(x) \plus{} f(y) \equal{} f(x \plus{} y) \plus{} f(x)f(y)$ u can get :
$ 1)f(a \plus{} b) \equal{} f(a) \plus{} f(b)$
$ 2)f(ab) \equal{} f(a)f(b)$
so $ f(x) \equal{} x$ for all $ x$ or $ f(x) \equal{} 0$ but $ f(x) \equal{} 0$ is not an answer so we have $ f(x) \equal{} x$

What are these links? The problem is different from these links.

[quote="BaBaK Ghalebi"][url]http://www.mathlinks.ro/viewtopic.php?t=187785[/url]
[url]http://www.mathlinks.ro/viewtopic.php?search_id=1361591936&t=178957[/url]
 :huh:[/quote]

Yeah! but the problem can be reduced to that functional equation, so that's probably what he meant.

[quote="MellowMelon"] Put in $ x \equal{} 0$. We get $ f(f(y)) \equal{} y(1 \minus{} f(0)) \plus{} f(0)$[/quote]

I don't think it is true(maybe a typo) but the solution is correct (it is not affected ).For x=0 you get $ f(0) \plus{} y \plus{} f(0) \equal{} f(f(y)) \plus{} yf(0)$ or $ f(f(y)) \equal{} y(1 \minus{} f(0)) \plus{} 2f(0)$    :wink:

Here is my solution,check it please.Thanks a lot.May be there are some similar ideas in the above solutions.
[quote]$ f(xf(y)) \plus{} y \plus{} f(x) \equal{} f(x \plus{} f(y)) \plus{} yf(x)$            (*)[/quote]
Setting $ x \equal{} 0$ gives $ f(f(x)) \equal{} (1 \minus{} a)x \plus{} 2a$,where $ a \equal{} f(0)$.
If $ a \equal{} 1$:
Then $ f(f(x)) \equal{} 2 \forall x \in R$ and $ f(0) \equal{} 1$.
Since $ f(0) \equal{} 1,$,we have $ f(f(0)) \equal{} f(1)$=>$ f(1) \equal{} 2$=>$ f(f(1)) \equal{} f(2)$=>$ f(2) \equal{} 2$
So $ f(1) \equal{} f(2) \equal{} 2$.
Set $ x \equal{} 1$ into (*),we get $ f(f(y) \plus{} 1) \equal{} 4 \minus{} y$,so if $ f$ is injective,contradiction.(because we have $ f(1) \equal{} f(2)$)
So $ a$ is not equal to 1.
$ f(f(x)) \equal{} (1 \minus{} a)x \plus{} 2a$ and $ f$ is injective.
Since RHS of the above condition=$ R$,there exists $ x_0$ such that $ f(x_0) \equal{} 0$
Set $ y \equal{} x_0$ into (*),we get $ x_0f(x) \equal{} x_0 \plus{} a$.
If $ x_0$ is not equal to $ 0$,then $ f(x)$ is a constant.No solution in this case.
Hence $ x_0 \equal{} 0$,this gives us $ f(0) \equal{} 0$ and $ f(f(x)) \equal{} x$.
Replacing $ x$ by $ f(x)$ into (*) and using the condition $ f(f(x)) \equal{} x$,we get:
$ f(f(x)f(y)) \plus{} x \plus{} y \equal{} f(f(x) \plus{} f(y)) \plus{} xy$
Setting $ x \equal{} f(x),y \equal{} f(y)$ into the above condition,we get:
$ f(xy) \plus{} f(x) \plus{} f(y) \equal{} f(x \plus{} y) \plus{} f(x)f(y)$,which is trivial.See [hide="here"]http://www.mathlinks.ro/viewtopic.php?search_id=1361591936&t=178957[/hide]  to continue.

[i]A very good solution , MellowMelon , congratulation ! :lol: [/i]

  [i] The functional equation

$ f(x \plus{} y) \plus{} f(x)f(y) \equal{} f(x) \plus{} f(y) \plus{} f(xy)$ is a result that should be remebered[/i]

 [i][u] Using the lemma [/u]: if $ f: R \ \to \ R ; f(xy) \equal{} f(x)f(y) ; f(x \plus{} y) \equal{} f(x) \plus{} f(y) \ \forall \ x ; y \ \in \ R$

 then $ f(x) \equal{} 0 \forall x \ \in \ R$ or $ f(x) \equal{} x \forall x \ \in \ R$

  We can solve an other nice problem that had appeared in M&Y Magazine :

   Find all function $ f: R \ \to \ R$ such that :

   $ f(x \minus{} y) \plus{} f(xy) \equal{} f(x) \minus{} f(y) \plus{} f(x)f(y) \ \forall \ x ; y \ \in \ R$[/i]

[i]Try with that problem:wink: [/i]

I liked this problem very much!
Find all functions $f\colon\mathbb{R}\rightarrow\mathbb{R}$ such that
$f(xf(y))+y+f(x) = f(x+f(y))+yf(x)$
Let $P(x,y)$ be the assertion
[hide="Proving that the function is injective"]If $f(y)=f(y'),y\neq y'$
Then $P(x,y')\implies f(xf(y'))+y+f(x) = f(x+f(y'))+y'f(x)\implies f$ is injective as $f$ is non-constant, $\therefore \exists x_0$ such that $f(x_0)\neq 1$[/hide]
[hide="Proving that the function is bijective"]$P\left(\frac{f(y)}{f(y)-1},y\right)\implies f\left(\frac{f(y)^2}{f(y)-1}\right)+y+f\left(\frac{f(y)}{f(y)-1}\right)$
$=f\left(\frac{f(y)^2}{f(y)-1}\right)+yf\left(\frac{f(y)}{f(y)-1}\right)$
$\implies f\left(\frac{f(y)}{f(y)-1}\right)=\frac{y}{y-1}\forall y\neq 1 \text{ and } \forall y\text{ for which }f(y)\neq 1\cdots \cdots (1)$
$P(0,y)\implies f(0)+y+f(0)=f(f(y))+yf(0)\implies f(f(y))=f(0)(2-y)+y$, take $y=2$ to get $f(f(2))=2$
$P(x,f(2))\implies f(2x)+f(2)+f(x)=f(x+2)+f(2)f(x)$, take $x=2$ to get $f(4)+2f(2)=f(4)+f(2)^2\implies f(2)(f(2)-2)=0$
Take $x=0$ to get $2f(0)+f(2)=f(2)+f(2)f(0)\implies f(0)(f(2)-2)=0$
If $f(2)=0\implies f(2)-2\neq 0\implies f(0)=0$, not possible as $f$ is injective, therefore $f(2)=2$
If $\exists y_0$ such that $f(y_0)=1$
$P(x,y_0)\implies f(x)+y_0+f(x)=f(x+1)+y_0f(x)\implies f(x+1)$
$=f(x)(2-y_0)+y_0$
Take $x=y_0$ to get $f(y_0+1)=2$, but as $f(2)=2$, due to injectivity, $f(y_0)+1=2\implies f(y_0)=1$
So, if $\exists y_0$ such that $f(y_0)=1$, then $y_0=1$
So, now we can say that we can say that there exists a pre image in the domain for all elements in the co domain except $1$.
So, $\exists a$ such that $f(a)=0$
$P(x,a)\implies f(0)+a+f(x)=f(x)+af(x)\implies af(x)=a+f(0)$, take $x=2$ to get $a=f(0)$
$P(a,y)\implies f(af(y))+y=f(a+f(y))$, take $y=a$ to get $f(0)+a=f(a)=0\implies a=-f(0)$
Therefore, $a=f(0)=-f(0)\implies f(0)=0$
$P(0,y)\implies f(f(y))=y$,take $y=1$ to get $f(f(1))=1$, so $\exists t$ such that $f(t)=1$, and therefore, $t=1$
Thus $f(1)=1$
$f(f(y))=y$ implies $f$ is bijective.[/hide]
[hide="Proving that the function is an identity function if x is a rational"]$P(x,1)\implies f(x+1)=f(x)+1$, so it is easy to prove that $f(n)=n\forall n\in\mathbb{Z}$ with an easy induction.
So, $f(x+n)=f(x)+n\forall x\in\mathbb{R},\forall n\in\mathbb{Z}$,
$P(x,n)$, where $n$ is an integer, implies $f(xn)+n+f(x)=f(x+n)+nf(x)\implies f(nx)=nf(x)\forall x\in\mathbb{R},n\in\mathbb{Z}$
Let $x=\frac{p}{q}$, where $p\in\mathbb{Q}$ and $q\in\mathbb{N}$ be a rational,
then take $x=\frac{p}{q}$ and $n=q$ in the equation $f(nx)=nf(x)\forall x\in\mathbb{R},n\in\mathbb{Z}$
To get $f(r)=r\forall r\in\mathbb{Q}$[/hide]
[hide="The function maps non negative reals to non negative reals"]$n=-1$ in the equation $f(nx)=nf(x)\implies f(x)=-f(-x)$
$P(f(y),y)\implies f(-f(y)^2)+y+f(-f(y))=f(0)+f(-f(y))-y^2$
$\implies f(f(y))^2=y^2$, so $f(x)\ge 0\forall x\in\mathbb{R}\ge 0$[/hide]
[hide="Proving that the function is monotone increasing"]
We have proved that $f(x)\ge 0\forall x\in\mathbb{R}\ge 0$
So, if $a>b$ then $f(a-b)\ge 0$
$P(a,f(b))\implies f(ab)+f(a)+f(b)=f(a+b)+f(a)f(b)\implies f(ab)-f(a)f(b)=f(a+b)-f(a)-f(b)$
$P(a,-f(b))\implies f(ab)-f(a)f(b)=f(a-b)-f(b)+f(a)$
Therefore, $f(a+b)-f(a)-f(b)=f(a-b)-f(b)+f(a)$
$\implies f(a+b)=f(a-b)+2f(a)$, but as $f(a-b)>0$,
$f(a+b)<2f(a)\implies f(a)>\frac{f(a+b)}{2}$
$f(nx)=nf(x)\implies f(2x)=2f(x)\implies \frac{f(x)}{2}=f\left(\frac{x}{2}\right)$
So, $\frac{f(a+b)}{2}=f\left(\frac{a+b}{2}\right)$
$ f(a)>f\left(\frac{a+b}{2}\right)$
As $a>b\implies a>\frac{a+b}{2}$, so the function is monotone increasing.[/hide]
[hide="Using the density of Q in R"]We have $f(f(y))=y$
Let $f(x)>x$ for some real $x$, of course $x$ isn't rational as we have already proved that $f(r)=r$ if $r$ is rational.
By the density of $\mathbb{Q}$ in $\mathbb{R}$, $\exists r\in\mathbb{Q}$
such that $f(x)>r>x$
As the function is monotone increasing,
$f(f(x))>f(r)>f(x)\implies x>r>f(x)$, a contradiction
Similarly, we may prove that $f(x)<x$ is not possible.[/hide]
So, $f(x)=x\forall x\in\mathbb{R}$ is the only solution.

[quote="Omid Hatami"]Find all functions $ f: \mathbb R\longrightarrow \mathbb R$ such that for each $ x,y\in\mathbb R$:
\[ f(xf(y)) \plus{} y \plus{} f(x) \equal{} f(x \plus{} f(y)) \plus{} yf(x)\][/quote]
We will prove that $f(0)=0$.
Let $f(0)=a$ and $f(1)=b$.
In $(1)$ replacing $x=y=0$ :
\[f(a)=2a\]
In $(1)$ replacing $x=0,y=1$ :
\[f(b)=a+1\]
In $(1)$ replacing $x=1,y=0$ :
\[f(a+1)=f(a)+b\]
But $a+1=f(b)$ and $f(a)=2a$, we have :
\[f(f(b))=2a+b\;\;\;\;\;(*)\]
In $(1)$ replacing $x=0,y=b$ :
\[f(f(b))=2a+b-ab\;\;\;(**)\]
From $(*)(**)$ we deduce :
\[2a+b-ab=2a+b\].
Implies $a=0$ or $b=0$.
If $b=0$, we have $a=f(0)=f(b)=a+1\;\;\;\;(!)$. Hence $a=0$ or $f(0)=0$.
Replacing $x=0$ in $(1)$ :
\[f(0)+y+f(0)=f(f(y))+yf(0)\]
Implies $f(f(y))=y$. In $(1)$ replacing $y$ by $f(y)$ and remembering $f(f(y))=y$ :
\[f(xy)+f(y)+f(x)=f(x+y)+f(x)f(y)\]
This is a familiar problem.
See http://julielltv.wordpress.com/2014/08/15/functional-equation-58/

Very happy I solved this.
[hide="solution"]
Let $P(x,y)$ be the FE.
Let $f(0)=c$. $P(0,0)\implies f(c)=2c$.
$P(c, x)\implies f(cf(x))+x+f(c) = f(c+f(x))+xf(c)\implies f(cf(x))+x+2c = f(c+f(x))+2cx$
$P(f(x), 0)\implies f(cf(x))+f(f(x))=f(f(x)+c)$
Then $P(c, x)-P(f(x), 0)\implies x+2c-f(f(x))=2xc\implies f(f(x))=2c+(1-2c)x$
But $P(0, x)\implies 2c+x=f(f(x))+cx\implies f(f(x))=2c+(1-c)x$
So $P(0,x)-[P(c, x)-P(f(x), 0)]\implies c=f(0)=0$
Then $P(0, x)\implies f(f(x))=x$ so $P(x, f(y))\implies f(xy)+f(x)+f(y)=f(x+y)+f(x)f(y)\quad (1)$.
Also, $f(f(x))=x$ means $f$ is bijective.
------------------------------
Let $Q(x,y)$ be the FE stated at $(1)$.
$Q(2, 2)\implies f(2)^2=2f(2)\implies f(2) \in \{0, 2\}$
$Q(x, 1)\implies f(x+1)+f(x)f(1)=2f(x)+f(1)\implies f(x+1)=f(x)(2-f(1))+f(1)\quad (2)$
$Q(1, 1)\implies f(2)+f(1)^2=3f(1)$.
If $f(2)=0$ then $f(1)\in \{0, 3\}\implies f(1)=3$ since $f$ is bijective.
Then by $(2)$, $f(x+1)=-f(x)+3$ so $f(3)=3$ contradiction with $f$ bijective.
Thus, $f(2)=2$. Then $f(1)\in \{1, 2\}\implies f(1)=1$ since $f$ is bijective. Thus, by $(2)$, $f(x+1)=f(x)+1$, so $f(n)=n\,\forall n\in\mathbb{Z}$.
$Q(x, -1)\implies f(-x)+f(x)-1=f(x-1)-f(x)\implies f(-x)=-f(x)$.
$Q(x, y)+Q(-x, -y)\implies f(xy)=f(x)f(y)\implies f(x+y)=f(x)+f(y)$. But this is just Cauchy, and since $f(x^2)=f(x)^2 \ge 0$ so $f$ is bounded in a non-trivial interval, our only solution is $f(x)=x$ or $f(x)=0$, But since $f(1)=1\ne 0$, our only solution is $\boxed{f(x)=x}$. Plugging back in shows it works, so we are done. $\Box$
[/hide]

Oops this solution was longer than it should have been

[hide=Solution]
Let $P(x, y)$ denote the given assertion.

$P(0, y)$ gives $y(1-f(0))+2f(0)=f(f(y))$. Because the LHS of this equation spans all reals, $f$ is surjective. From this surjectivity, there exists a constant $a$ such that $f(a)=0$. Then $P(a, a)$ implies $a=0$, so $f(0)=0$. Plugging $f(0)=0$ into $P(0, y)$ yields $f(f(y))=y$, so $f$ is bijective.

$P(f(2), 2)$ gives \[f(f(2)^2)+4=f(2f(2))+4\implies f(2)=0 \ \text{or} \ 2.\] However, because $f(f(2))=2$, we can't have $f(2)=0$ so $f(2)=2$.

$P(1, f(1))$ yields \[3f(1)=2+f(1)^2\implies f(1)=1 \ \text{or} \ 2.\] However, $f(1)\neq 2$ because $f(f(1))=1$, so $f(1)=1$.

$P(x, 1)$ gives $f(x)+1=f(x+1)$; subtracting the equation \[P(x, y-1): f(xf(y)-x)+y+f(x)-1=f(x+f(y)-1)+yf(x)-f(x)\] from \[P(x, y): f(xf(y))+y+f(x)=f(x+f(y))+yf(x)\] gives \[f(xf(y))-f(xf(y)-x)=f(x).\] Replacing $y$ with $f(y+1)$ in the above yields $f(xy+x)=f(xy)+f(x)$, so $f$ is additive. Then \[P(x, f(y)): f(xy)+f(y)+f(x)=f(x+y)+f(x)f(y)\implies f(xy)=f(x)f(y),\] so $f$ is multiplicative as well. From here, it's well-known that $\boxed{f(x)=x}$ is the only solution and we can easily check that it satisfies the original FE.
[/hide]

[quote=Omid Hatami]Find all functions $ f: \mathbb R\longrightarrow \mathbb R$ such that for each $ x,y\in\mathbb R$:
\[ f(xf(y)) \plus{} y \plus{} f(x) \equal{} f(x \plus{} f(y)) \plus{} yf(x)\][/quote]

[hide=solution] Let $P(x,y)$ denote the given FE. I claim that only the identity function satisfies $P$. To see it indeed works is easy. We shall see why no other function works in a moment. Take a function $f$ that satisfies $P$ and consider the following

[b]Lemma 1.[/b] $f(0)=0$.

If $f(0)=1$ then $P(x, 0) \implies f(x+1)=2f(x)$ for all $x$; so $f(1)=2$ and $f(2)=4$. However, $P(0,y) \implies f(f(y))=2$ for all $y$. But $y=1$ violates the previous condition.

Now, $P(x, 0) \implies f(xf(0))+f(x)=f(x+f(0))$, so plugging $x=\tfrac{f(0)}{f(0)-1}$ we conclude that the equation $f(t)=0$ has a solution $x_0 \in \mathbb{R}$. Plug $y=x_0$ in $P$ to conclude that $x_0f(x)=f(0)+x_0$ for all $x \in \mathbb{R}$. As no constant function satisfies $P$; we get $x_0=0 \implies f(0)=0$ as claimed. $\blacksquare$

Now, the second crucial step.

[b]Lemma 2.[/b] $f(1)=1$.

Note that $P(0,y)$ together with $f(0)=0$ yields $f(f(y))=y$ for all $y \in \mathbb{R}$. Thus, there exists $t_0 \in \mathbb{R}$ such that $f(t_0)=1$. Unless $t_0 \ne 1$ we have $$P\left(t_0, \tfrac{t_0}{t_0-1}\right) \implies \frac{t_0}{t_0-1}=\frac{f(t_0)}{f(t_0)-1},$$ contradiction! Hence our claim must be true. $\blacksquare$

Consider $$P(x, f(y)) \implies f(xy)+f(x)+f(y)=f(x+y)+f(x)f(y)$$ for all $x, y \in \mathbb{R}$. Thus, $f(x+1)=f(x)+1$ for all $x \in \mathbb{R}$. Consequently, $$P(x, f(y+1)) \implies f(xy+x)=f(x)f(y)+f(x+y)-f(y)=f(x)+f(xy),$$ hence $f$ is additive. Apply additivity to $P(x, f(y))$ to get that $f$ is multiplicative; hence it must be the identity function. $\blacksquare$

[/hide]

This problem is trivial!
We claim that $f \equiv x \ \forall x \in \mathbb{R}$ is the only solution.Denote by $P(x,y)$ the given assertion.Then,
$$\rightarrow P(0,y) \implies 2\cdot f(0) + y = f(f(y)) + y\cdot f(0).$$
Now, if $f(0)=1$, then we easily get that $f$ is a constant function, which is not true. Thus, we conclude that $f(0)\neq 1$ and consequently, $f$ is a bijection.
Now, by surjectivity of $f$, we know that there exists a real number $\alpha$ such that $f(\alpha)=0$. Then, using $P(x,\alpha)$, we get that $\alpha = 0$. Thus, $f(0)=0$. Hence, $f(f(y))=y \forall y \in \mathbb{R}$.
Then, $$P(x,f(y)) \implies f(xy)+f(x)+f(y)=f(x+y)+f(x)\cdot f(y)$$
Now, plug in $x=y=2$, we get that $f(2)=2$.Then, since $f(f(1)+1)=f(2)$, it follows that $f(1)=1$.Thus, $P(x,1)$ gives $$f(x+1)=f(x)+1$$
Replace $x \rightarrow x+1$ in $P(x,y)$ and get:
$f(xy+y)+f(x)+1+f(y)=f(x+y)+1+f(x)\cdot f(y) + f(y) \iff f(xy+y) = f(xy)+f(y)$
Now, this gives $f$ is both additive and multiplicative and hence the conclusion follows.

[hide=Solution][quote=Iran TST 2008 D1 P1]Find all functions $ f: \mathbb R\longrightarrow \mathbb R$ such that for each $ x,y\in\mathbb R$:
\[ f(xf(y)) \plus{} y \plus{} f(x) \equal{} f(x \plus{} f(y)) \plus{} yf(x)\][/quote]
[b][color=#000]Solution:[/color][/b] Clearly constant functions don't work. Hence assume $f$ is non-constant. Let $P(x,y)$ denote the given statement. If $f(0)$ $=$ $1$, then $P( 0,x)$ $\implies$ $f$ is constant, a contradiction! Hence, $f(0)$ $\neq$ $1$
\begin{align*} P \left( \frac{f(0)}{f(0)-1},0 \right) \implies f \left( \frac{f(0)}{f(0)-1} \right)=0 \end{align*}
Hence, there exists a $c$ $\in$ $\mathbb{R}$ such that $f(c)$ $=$ $0$. From $P(c,c)$ $\implies$ $c=0$. Hence, $f$ is Injective at $0$.
\begin{align*} P (0,y) \implies f \big( f(y) \big) =y \end{align*}
Thus, $f$ is Involutive and hence Bijective. Now,
\begin{align*} P(1,1) \implies f \big( f(1) \big) +f(1) +1= f \big( f(1) +1 \big) +f(1) \implies f(2)=f(1) +1 \end{align*}
\begin{align*} P \big( 1, f(1) \big) \implies 3f(1)=f(2)+f(1)^2 \end{align*}
From here, we get $f(1)=1$.
\begin{align*} P \big( x,1 \big) \implies 2f(x) +1= f(x+1)+f(x) \implies f(x+1)=f(x)+1 \end{align*}
\begin{align*} P \big( x, f(y+1) \big) ~ \& ~  P \big( x, f(y) \big) \implies f(xy)+f(x)=f(xy+x) \end{align*}
Thus, $f$ is additive. And from,
\begin{align*} P \big(x, f(y) \big) \implies f(xy) +f(x) +f(y) =f(x+y) +f(x)f(y) \implies f(xy)=f(x)f(y) \end{align*}
Thus, $f$ is multiplicative $\implies$ $f$ is linear. Plugging $f=ax+b$ in $P(x,y)$ $\implies$ $\boxed{f(x)=x ~ ~ \forall x \in \mathbb{R}}$ $\qquad \blacksquare$
[/hide]

[quote=Omid Hatami]Find all functions $ f: \mathbb R\longrightarrow \mathbb R$ such that for each $ x,y\in\mathbb R$:
\[ f(xf(y)) \plus{} y \plus{} f(x) \equal{} f(x \plus{} f(y)) \plus{} yf(x)\][/quote]

Taking p(1,f(1)\f(1)-1),for $f(1)$ not equal to $1$,we see that $1=0$ ,which is a contradiction,so now we see that $f(1)=1$.

$p(0,1)$
          $f(0)+1=f(0)=f(f(1))+f(0)$
          .
          .
           $f(0)=0$., and from $p(0,x)$ we get that $f(f(x))=x$-involutive ,also this gives us f-injective + surjective .

$p(x,1)$ gives us $f(x+1)=f(x)+1=f(x)+f(1)$.

$p(x,f(y))$,gives us  $f(xy)+f(y)+f(x)=f(x+y)+f(x)f(y)$
$p(x,f(y+1))$,gives us $f(xy+x)+f(y+1)+f(x)=f(x+y+1)+f(y+1)f(x)$
                                   $f(xy+x)+f(y)+1+f(x)=f(x+y)+1+f(y)f(x)+f(x)$ ----- this follow using $p(x,1)$
                                  $f(xy+x)+f(y)=f(x+y)+f(x)f(y)$------Now use $p(x,f(y))$ to finish this part
                                  $f(xy+x)+f(y)=f(xy)+f(y)+f(x)$
                                  $f(xy+x)=f(xy)+f(x),--> f(m+n)=f(m)+f(n)$

Now geting back to $p(x,f(y))$ we have $f(xy)+f(y)+f(x)=f(x+y)+f(x)f(y)$
                                                                  $f(xy)+f(x)+f(y)=f(x)+f(y)+f(x)f(y)$
                                                                  $f(xy)=f(x)f(y)---->$  multiplicative

And now from multiplicatve + additive  ,we get that the desired result to be f(x)=x  $\blacksquare$

[quote=Omid Hatami]Find all functions $ f: \mathbb R\longrightarrow \mathbb R$ such that for each $ x,y\in\mathbb R$:
\[ f(xf(y)) \plus{} y \plus{} f(x) \equal{} f(x \plus{} f(y)) \plus{} yf(x)\][/quote]
Let $P(x,y)$ denote the above assertion. $$P(x,0): f(xf(0))+f(x)=f(x+f(0))$$ $$P(0,x): f(f(x))=x(1-f(0))+2f(0)$$ now if $f(0)=1$ then using $P(x,0)$ we have $f(n)=2^n \forall n \in \mathbb{N}$ which is clearly impossible by substituting in the parent equation. Thus, $f(0) \ne 1$ and hence $f$ is bijective by $P(0,x) \implies$ there exists $c$ such that $f(c)=0$ $$P(x,c): f(0)+c=cf(x)$$ clearly $f$ is not a constant function, so $f(0)=0$, $P(0,x): f(f(x))=x$ $$P(x,f(y)): f(xy)+f(x)+f(y)=f(x+y)+f(x)f(y)$$ call this new assertion $Q(x,y)$ $$Q(x,1): f(x+1)=(2-f(1))f(x)+f(1)$$ if $f(1) \ne 1$ then by an easy induction $f(n)=\frac {f(1)(2-f(1))^n}{1-f(1)}-\frac {f(1)}{1-f(1)}$ which can easily be dispelled. Thus $f(1)=1$ and by an easy induction using $Q(x,1)$ we have $f(n)=n \forall n \in \mathbb{Z}$ also $f(x+n)=f(x)+n$ and so by using $Q(x,n)$ we have $f(nx)=nf(x) \forall n \in \mathbb{Z} \implies f(-x)=-f(x)$ $$Q(x,-y)+Q(x,y): 2f(x)=f(2x)=f(x+y)+f(x-y) \implies f(a+b)=f(a)+f(b) \quad \forall a,b \in \mathbb{R}$$ also by $P(a,b): f(ab)=f(a)f(b)$. Thus $f$ is both multiplicative and additive, so
$f(x)=x \quad \forall x \in \mathbb{R}$ $\blacksquare$

We easily have that $f$ is nonconstant.
$\textbf{1. }f(1)=1$
If $f(1)\ne1$ then $P\left(\frac{f(1)}{f(1)-1},1\right)\Rightarrow1=0$, contradiction. $\blacksquare$

$\textbf{Case 1: }f(0)=1$
$P(0,x)\Rightarrow f(f(x))=2$
$P(2,f(x))\Rightarrow f(2)+f(x)=f(2)f(x)$, so $f(2)=1$ since $f$ is nonconstant, but this implies that $f(2)=0$, contradiction. $\blacksquare$

$\textbf{Case 2: }f(0)\ne1$
$P\left(\frac{f(0)}{f(0)-1},0\right)\Rightarrow f\left(\frac{f(0)}{f(0)-1}\right)=0$, so let $k=\frac{f(0)}{f(0)-1}$.
$P(x,k)\Rightarrow f(0)+k=kf(x)$, we must have $k=0$, since $f$ is nonconstant.
So $f(0)=0$, hence $P(0,x)\Rightarrow f(f(x))=x$. $\blacksquare$

$\textbf{2. }f(x+y)=f(x)+f(y)$
$P(x,f(y))\Rightarrow Q(x,y):f(xy)+f(x)+f(y)=f(x+y)+f(x)f(y)$
$Q(x,1)\Rightarrow f(x)+1=f(x+1)$
$Q\left(x,\frac yx+1\right)-Q\left(x,\frac yx\right)\Rightarrow f(x+y)=f(x)+f(y)\forall x\ne0$, but since it obviously holds for $x=0$, we're done here. $\blacksquare$

Substituting this back in to $Q(x,y)$, we easily get $f(xy)=f(x)f(y)$, and the only multiplicative and additive function is the identity function, $\boxed{f(x)=x}$, which fits. $\square$

Let $P(x,y)$ denote $f(xf(y)) + y + f(x) = f(x + f(y)) + yf(x)$. If $f(0)=1$, then $P(0,0)$ gives $f(1)=2$ and $P(0,1)$ and $P(1,0)$ gives contradiction. Now $P(\tfrac{f(0)}{f(0)-1},0)$ gives $f(x_0)=0$ for some $x_0$.

$P(x,x_0)$ gives $f(0)=0$, since constant $f$ fails. So $f(f(x))=x$ from $P(x,0)$. $P(2,f(2))$ gives $f(2)=2$ and then $P(1,f(1))$ gives $f(1)=1$. Now $P(x,1)$ gives $f(x+1)=f(x)+1$. Now comparing $P(x,f(y+1))$ and $P(x,f(y))$ gives $f$ is additive. Now $P(x,f(y))$ gives $f$ is multiplicative. So $f\equiv \text{Id}$, which fits.


geometry
circumcircle
trigonometry
perpendicular bisector
geometry unsolved

In a triangle ABC  point L is taken on side BC and points M,N on the extensions of AB  and  AC  over B and C , respectively such that ALC=2AMC and  ALB=2ANB. Let O be the circumcenter of triangle AMN . Show that OL is perpendicular  to BC.

Hint:      Cherche le papillon !

  M.T.

Let circumcircles (P), (Q) of the $\triangle AMC,\ \triangle ANB$ meet BC again at U, V. Denote $\angle ALB = 2 \phi.$ Then $\angle ALU \equiv \angle ALB = 2 \phi$ and $\angle AUL \equiv \angle AUC = \angle AMC = \frac{\angle ALC}{2}= 90^\circ-\phi,$ which means that the $\triangle ALU$ is isosceles with LU = LA. Similarly, the $\triangle ALV$ is also isosceles with LV = LA. This means that L, the midpoint of UV, is the circumcenter of the right $\triangle AUV.$ Let the perpendicular bisectors of AM, AN (through P, Q) meet the perpendicular bisector of UV (through L) at O, O'. The $\triangle POL \sim \triangle ABU$ are similar, having perpendicular sides. For the same reason, the $\triangle QO'L \sim \triangle ACV$ are also similar.

Let AV, AU meet the circles (P), (Q) again at X, Y. Since $\angle XAV \equiv \angle UAV \equiv \angle UAY = 90^\circ$ is right, P, Q are midpoints of XU, YV. Then $\angle VAC \equiv \angle XAC = \angle XUC \equiv \angle PUL$ and similarly, $\angle UAB \equiv \angle YAB = \angle YVB \equiv \angle QVL.$ Since $PL \parallel AV$ (both perpendicular to AU) and $QL \parallel AU$ (both perpendicular to AV), it follows that $\angle PLU = \angle AVU \equiv \angle AVC$ and $\angle QLV = \angle AUV \equiv \angle AUB.$ Consequently, the $\triangle VQL \sim \triangle ABU$ and the $\triangle UPL \sim \triangle ACV$ are also similar. As a result, the $\triangle POL \sim \triangle VQL$ and the $\triangle QO'L \sim \triangle UPL$ are similar, so that

$OL = PL \cdot \frac{QL}{VL}= QL \cdot \frac{PL}{UL}= O'L$

which means that $O \equiv O'$ are identical. Thus the perpendicular bisectors of AM, AN meet at O, the circumcenter of the $\triangle AMN,$ on the perpendicular bisector of UV.
______________________________

Alternately, denote R = LU = LV = LA circumradius of the right $\triangle AUV$ and $p = PL,\ q = QL.$ Let P', Q' be feet of normals from P, Q to BC. Then $p' = P'L = p \cos \phi,$ $q' = Q'L = q \cos (90^\circ-\phi) = q \sin \phi,$

$BU = 2R-2(R-q') = 2q',\ \ CV = 2R-2(R-p') = 2p',$

$AU = 2R \sin \phi,\ \ AV = 2R \sin (90^\circ-\phi) = 2R \cos \phi.$

$OL = PL \cdot \frac{BU}{AU}= \frac{pq}{R}= QL \cdot \frac{CV}{AV}= O'L$

$BN,CM$ meet $(O)$ at $E,F$
$\angle ENA+\angle FMA=1/2(\angle ALB+\angle ALC)=90$
So $EF$ is the diameter of circle $(O)$
Applying Pascal theorem for $AMFENA$ we have $C = AN \cap MF$, $B = AM \cap EN$, $K = AA \cap EF$ are collinear ($AA$ is the tangent of cicle $(O)$ at $A$)
Otherwise $\angle AOK=\angle ALK(=2\angle ANE)$
so $A,O,L,K$ are cyclic
This means $\angle OLK=\angle OAK=90$ $q.e.d$


geometry perimeter inequalities trigonometry circumcircle quadratics calculus

Consider a fixed circle. Prove that the maximum perimetr from the inscribed triangles has equilateral triangle.Without use of Jensen.

[quote="riddler"]what is Jensen?[/quote]

I guess he meant Jensen Inequality  :P

Yes, of course I meant Jensen's inequality.

According to the solution of the [url=http://www.mathlinks.ro/Forum/topic-45771.html]Geometry problem[/url], given the angle $\angle A = \alpha$ and the side a = BC of a triangle $\triangle ABC$, the triangle with the maximum perimeter is isosceles with AB = AC. Its semiperimeter s is then given by

$s = \frac{ha}{2r} = \frac{a}{2}(\frac{1}{\sin \frac \alpha 2} + 1)$

where h is the altitude from the vertex a and r is the inradius. Giving the angle $\alpha$ and the side a is equivalent to giving the angle $\alpha$ and the circumradius R, because $R = \frac{a}{2 \sin \alpha}$. Hence the semiperimeter of this isosceles triangle is

$s = R \sin \alpha (1 + \frac{1}{\sin \frac \alpha 2}) = 2R \sin \frac \alpha 2 \cos \frac \alpha 2 \frac{1 + \sin \frac \alpha 2}{\sin \frac \alpha 2} = R (\sin \alpha + 2 \cos \frac \alpha 2)$

The semiperimeter is maximum iff

$\frac{\text ds}{\text d\alpha} = R(\cos \alpha - \sin \frac \alpha 2) = 0$

$\frac{\text d^2s}{\text d\alpha^2} = -R(\sin \alpha + \frac 1 2 \cos \frac \alpha 2) < 0$

Solving the quadratic equation $\frac{\text ds}{\text d\alpha} = 0$ for $\sin \frac \alpha 2$,

$\cos \alpha - \sin \frac \alpha 2 = 1 - 2 \sin^2 \frac \alpha 2 - \sin \frac \alpha 2 = 0$

$\sin \frac \alpha 2 = \frac{1 \pm \sqrt{1 + 8}}{-4} = -\frac{1 \pm 3}{4} = -1$ [color=white].[/color] or [color=white].[/color] $\frac 1 2$

The root -1 is not acceptable, because it would imply $\frac \alpha 2 = 270^\circ,\ \alpha = 540^\circ$. Hence, $\sin \frac \alpha 2 = \frac 1 2,\ \frac \alpha 2 = 30^\circ$ or $150^\circ,\ \alpha = 60^\circ$ or $300^\circ$. The only root acceptable for a triangle angle is $\alpha = 60^\circ$. It follows that the remaining 2 angles of this isosceles triangle are $\beta = \gamma = \frac{180^\circ - \alpha}{2} = 60^\circ$, i.e., this triangle is equilateral. Now we just check the 2nd derivative value at $\alpha = 60^\circ$:

$-R(\sin 60^\circ + \frac 1 2 \cos \frac{60^\circ}{2}) = -R(\frac{\sqrt 3}{2} + \frac 1 2 \cdot \frac{\sqrt 3}{2}) = -\frac{3R \sqrt 3}{4}  < 0$

Hence, the semiperimeter s at the angle $\alpha = 60^\circ$ is maximum.

Isn't the calculus solution effectively similar to Jensen's inequality????


AMC
USA(J)MO
USAMO
analytic geometry
geometry
similar triangles

Good luck to everyone tomorrow! Get some Georgian representation at MOP. :)

Yeah Miles, Santhosh, Boru, Tianqi, Howard, Harrison, Billy, Ben, Oliver, and Eddy (every Georgian who made USAMO except for me)!  Best of luck on USAMO.  As for me, I won't be needing luck, as the best I will be able to do will probably be 1/42 (that's my goal for tomorrow. :starwars: )

Seriously, this isn't a self-confidence issue.  I will be slain tomorrow, so...yeah. :lol:

Oh come on Sitan, shoot for at least a 2! :-)

You're more likely to get a nonzero score if each day you shoot for at least one solved problem. :wink:  But practice hard for next year, so you can make red MOP.

Good luck, everybody!

So how did everyone do?

I solved #1 and #4, but I think my proof for #4 was kind of shaky (as to prove there doesn't exist any other $ n$ satisfying the condition). On Day I I got 2.5 spare hours after writing up #1, only to be wasted trying #2 futilely...On Day II I practically didn't have any time left, since I spent more than two hours organizing, writing, and modifying the proof until the last minute...

According to the discussion in AMC forum, many people think this was an easier test. If so, my chance of making Red MOP is even more minute, considering my poor performance (or, more honestly, ability) and bad timed-proof-writing skill...

OMG!  Ha, Mr. Fulton!  I actually didn't fail.  Forced to do math for 4.5 hours for two days in a locked room, I actually managed to write a proof for 5 out of the 6 problems, 3 of the five actually making sense.  I think I got numbers 4 and 5, and though I just got proved wrong on one of my lemmas for #3 on this forum several seconds ago, I'm confident that I will get at least 1 point for a fairly legitimate proof.

[quote="timwu"]So how did everyone do?

I solved #1 and #4, but I think my proof for #4 was kind of shaky (as to prove there doesn't exist any other $ n$ satisfying the condition). [/quote]

Same here :(

LOL I tried to do #2 through brute force by setting a coordinate system at A
and on the second day I kinda fell asleep for half an hour or so... so I didn't even try #6:D
Uber Phail

Tim what was your attempt at showing that the other n did not exist?

[quote="Boy Soprano II"]I got problems 1, 4, and 5, and I was a pair of similar triangles away from number 2.  Day one was pretty strange for me because I saw a solution using multiplicative generators and Dirichlet's theorem that took about twenty minutes to find and write up, and then I spent about four hours slogging through problem 2.  Day two was normal, though.

Anyhow, in years past, a problem and a little over has been enough for red MOP.  Good luck!

Hm, your room was locked?  My lock was on the inside, which meant that had I been desparate for more time, I could have tried locking myself in, but it wouldn't have worked for very long.[/quote]

Well, my room was locked on the inside, too, but you get what my exaggeration means. :P

Really?  But I thought red MOP was for freshmen only.  Or do you mean if I had taken this next year?

EDIT: Oh, you were talking to Tim... :oops:

haha same as tim. got 1 and 4, but didn't prove there weren't any other n other than 2^x+2^y. might have picked up a few points just for throwing stuff on paper for the others

Wait, I have a question.

For #4, instead of $ 2^x\plus{}2^y$, I put numbers expressed in the form $ 4\cdot2^k$ and $ 2^k\cdot(2^m\plus{}1)$.  Is that the same thing?

Please tell me it is, cuz if it's not, I'm going to turn emo. :ninja:

Yeah it's the same thing, just prettier if you have a single term. And you don't really even need the 4*2^k if you say that m can equal 0.

Yay.  Yeah, it does look better with one term, but I'm satisfied with a lower score as long as it doesn't mean that I got a 0/42. :D Thanx for the confirmation, though.

:D Yeah number 4 should be (2^k+1)2^m.....number 2 we can use a coordinate and it's pretty easy....number 6 I used 3 hours thinking about it at the test and still couldn't get a single do-able idea..... :(


La prueba de la Olimpiada del Cono Sur se está llevando a cabo del 14 al 20 de abril, estamos a la espera de resultados del equipo peruano.

Las pruebas las pueden ver en

[url]http://www.selectivos-peru.blogspot.com/[/url]

$ Tipe$

Entre ayer y hoy estan volviendo las delegaciones a sus respectivos paises. Posteo aqui los resultados por equipos (en orden de puntajes) con los premios.

1. Peru: 3 Oros, 1 Plata
2. Argentina (132): 2 Platas, 2 Bronces
3. Brasil (120): 2 Platas, 2 Bronces
4. Paraguay: 2 Bronces, 1 Mencion.
5. Bolivia: 1 Bronce, 1 Mencion.
6. Uruguay: 1 Bronce
7. Ecuador: 1 Mencion

Total de Concursantes: 27 (Uruguay llevo solo 3 participantes)
* Chile no participo de la Olimpiada.

Los puntajes precisos no recuerdo bien, pero seguro que dentro de poco alguien va a subir la tabla y los examenes para poder discutirlo. Cabe destacar que el año que viene la olimpiada se realizara en Brasil.

Felicitaciones a todos! Un saludo como siempre,

Ariel S.

Felicitaciones al equipo peruano por arrasar con los oros!!!!, y tambien al equipo brasileño y argentino por sus resultados!!!!!

Una lastima que Chile no participo, pero bueno, ojala que puedan participar en la IMO o en la Ibero.

Saludos, y ojala dejen en alto a Sudamerica en Bremen

[quote="edgar011"]aca van los puntajes de cada uno
PER 4	Julián Mejía Cordero	Perú	2	10	10	22	10	10	9	29	51	Medalla de Oro
PER 3	José García Sulca	Perú	9	10	9	28	10	9	0	19	47	Medalla de Oro
PER 2	Iván Muñoz Castillo	Perú	10	10	10	30	7	7	0	14	44	Medalla de Oro
ARG 1	Alfredo Umfurer	Argentina	5	9	3	17	10	10	0	20	37	Medalla de Plata
BRA 3	Matheus Barros de Paula	Brasil	4	10	3	17	10	10	0	20	37	Medalla de Plata
PER 1	Gianfranco Umeres Peralta	Perú	8	10	4	22	10	5	0	15	37	Medalla de Plata
BRA 1	Deborah Alves	Brasil	10	10	5	25	10	0	0	10	35	Medalla de Plata
ARG 4	Nicolás Cogorno	Argentina	10	8	0	18	10	5	0	15	33	Medalla de Plata
ARG 2	Sebastián Prillo	Argentina	4	10	3	17	10	5	0	15	32	Medalla de Bronce
BRA 4	Matheus Secco Torres da Silva	Brasil	10	3	10	23	3	5	0	8	31	Medalla de Bronce
ARG 3	Ariel Zylber	Argentina	8	7	0	15	10	5	0	15	30	Medalla de Bronce
PAR 2	Ariel Schvartzman Cohenca	Paraguay	6	9	0	15	9	5	1	15	30	Medalla de Bronce
PAR 1	Marcos Martínez Sugastti	Paraguay	4	9	0	13	10	5	0	15	28	Medalla de Bronce
URU 4	Javier Peraza	Uruguay	1	8	0	9	9	0	0	9	18	Medalla de Bronce
BOL 3	Diego Salazar	Bolivia	3	10	0	13	1	3	0	4	17	Medalla de Bronce
BRA 2	Gabriel Lopes	Brasil	1	7	0	8	5	4	0	9	17	Medalla de Bronce
ECU 1	Christian Abad Coronel	Ecuador	2	5	0	7	5	1	0	6	13	Mención de Honor
BOL 2	Anahí Viviana Castillo	Bolivia	1	2	0	3	8	0	0	8	11	Mención de Honor
PAR 3	Edgar Elizeche	Paraguay	1	4	0	5	6	0	0	6	11	Mención de Honor
URU 1	Alejo García	Uruguay	3	2	0	5	3	3	0	6	11
ECU 2	Nicolás Gómez Castro	Ecuador	3	3	0	6	2	0	0	2	8
BOL 4	Ibelice Ramírez	Bolivia	1	2	0	3	2	0	0	2	5
ECU 4	Sylviamaría Tarré	Ecuador	1	1	0	2	3	0	0	3	5
BOL 1	Rodrigo Uruchi	Bolivia	0	1	0	1	3	0	0	3	4
URU 2	Javier Rondán	Uruguay	1	1	0	2	0	0	0	0	2
ECU 3	Abraham Ordóñez	Ecuador	1	0	0	1	0	0	0	0	1
PAR 4	Freddy Ferreira	Paraguay	1	0	0	1	0	0	0	0	1

Perdon por el desorden
es el primer numero es el puntaje en el problema 1 ,el segundo en el 2 el tercero en el 3, el cuarto es la sumatoria del dia 1, el quinto es el del 4, el sexto el del 5, el septimo el del 6, el octavo es el total en el dia 2, y el ultimo es el total[/quote]

Grax por subir los puntajes!!!!!!


Rep123max,

You found a parametric form for x and y.  Are there limits on your parameter, a, that limit whether x and y can be positive?

You're almost there.  But only x and y need be positive while a can take on nonpositive value(s).

Oh, sorry, you said 8 pairs and that is correct.  I read the part where you said a could be 1, 2,...,7, but you forgot to list 0.  I think you just accidentally left it out of the list.  8 is the correct answer.  Nice work.

Can you explain how you used mod to set up the two parametric equations? That is the only part I dont understand..

i just did it with a pattern... maximum value for multiple of 11 is 550 and decreases by 77 from there... 8 is the answer.

how does the mod method work?  anyone have  a good source for learning mod?  i just dont understand it and its so important :/

[quote="MCrawford"]How many ordered pairs of positive integers (x, y) satisfy 7x + 11y = 592?[/quote]

Once you find one solution, you can easily find the rest. Notice that $(x,y)=(6,50)$ works. Now we have

\[x=6+11k, y=50-7k\]

It can easily be found that $0 \leq k \leq 7$ because $k=8$ would make $y<0$. So there are 8 solutions.


integration
advanced fields
advanced fields unsolved

Let $a>0$ and Let $(X,M,\mu)$ be a measure space such that $a<\mu(X)<\infty$. Let $f$ be a positive measurable function. Show that \[\inf\left\{\int_{E}f\,d\mu\mid E\in M, \mu(E)\geq a\right\}>0\]

The expression is clearly $\geq 0$. If it were $0$, what would you be able to say about the measurable sets $E_{n}=\{x; \ f(x) \geq \frac{1}{2^{n}}\}$, where $n \in \mathbb{N}$ ?

From here you must be able to find a contradiction, hence proving that your expression is $>0$.

Define $S(\tau) = \{x \in X | f(x) < \tau\}$ and $g(\tau) = \mu(S(\tau))$; let $S_{n}, \alpha_{n}$ be $S(\frac{1}{n}), g(\frac{1}{n})$. Note that all the $S_{n}$ are measurable as they are the preimages of open intervals under a measurable function. Then $\bigcap_{n=1}^{\infty}S_{n}= \{x \in X | \forall n, f(x) < \frac{1}{n}\}= \{x \in X | f(x) = 0\}= \emptyset$, so  as the $\alpha_{n}$ are finite, decreasing, and bounded, we see that they have a limit, which must be $0$. Hence $\exists m$ such that $\alpha_{m}< \frac{a}{2}$, in which case $\forall E$ with $\mu(E) \ge a$, we have that $\int_{E}f d\mu = \int_{E \bigcap S_{m}}f d\mu+\int_{E \setminus S_{m}}f d\mu \ge \int_{E \setminus S_{m}}f d\mu \ge \int_{E \setminus S_{m}}\frac{1}{m}d\mu$ $=$ $\frac{1}{m}\mu(E \setminus S_{m})$ $\ge$ $\frac{1}{m}\cdot \frac{a}{2}$, and thus the infimum must be positive.

Thanks Mystic.
Let me try to use julien hints and also Mystic idea......

Let $E_{n}$ be the sets defined by Julien. Let $E\in M$ such that $\mu(E)\geq a$.
Since $E_{n}\cap E\nearrow E$ then $\mu(E_{n}\cap E)\to \mu(E)<\infty$ in particular there is $n$ such that $|\mu(E_{n}\cap E)-\mu(E)|<a/2$. This implies that $\mu(E_{n}\cap E)>\mu(E)-\frac{a}{2}\geq \frac{a}{2}$.

Now
$\int_{E}f\,d\mu\geq \int_{E_{n}\cap E}f\,d\mu\geq \frac{1}{2^{n}}\mu(E_{n}\cap E)>\frac{a}{2^{n+1}}$. Thus the infimum is positive.


Find $ x \in R$ of this equation:$ x^3\minus{}4x^2\minus{}5x\plus{}6\equal{}\sqrt[3]{7x^2\plus{}9x\minus{}4}$

hello, $ x\equal{}5$ or $ x\equal{}\frac{1}{2}\sqrt{5}\minus{}\frac{1}{2}$.
Sonnhard.

Actually, there are three solutions: $ x \equal{} \left\{5,\frac { \minus{} 1\pm\sqrt5}2\right\}$. I haven't found a particularly nice solution yet, though...

[quote="doabli"]Does Anyone have hints for this problem?[/quote]

Hi doabli. This is hint for you:

Let $ u \equal{} x \plus{} 1$ and $ v \equal{} \sqrt [3]{7x^2 \plus{} 9x \plus{} 4}$, we can rewrite the equation as
$ u^3 \plus{} u \equal{} v^3 \plus{} v$. To find real solution, we have $ u \equal{} v$, which is
$ (x \minus{} 5)(x^2 \plus{} x \minus{} 1) \equal{} 0$. The rest is trivial.

xxp 2000 already answered this question, right?


See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244710&start=20]here[/url].  Same thing is apparently happening in the classroom.