analytic geometry
function
trigonometry
geometry
geometric transformation
homothety
induction

How'd everyone do?

On #2 I took one look at it, saw "graph theory", and decided to dumb-ass it. It happened to be the official solutions  :lol:

#4 was fail. I did everything right and then said f(x)=x,-x both work :wallbash_red: :wallbash:

#1 was simple. Can I get full credit if I gave a little more work than nesscary? What I did was I let x=m/n and did some quick casework.

I'm hoping for a 14, which will most likely get me bronze (depends on rd 2).

Can someone please confirm wheter this is correct or incorrect:

For #4, I put

f(x)=x for all x greater or equal to 0.

However, the answer says that f(x) for all x.  Since f is only defined on nonnegatives, will my answer still be counted as correct? (I didnt know how to express when defined only on certain values)

I also said that $ f(x)\equal{}x$ and $ f(x)\equal{}\minus{}x$ were the solutions to #4. In the official solution, they go from $ b^2\equal{}1$ to $ b\equal{}1$, ignoring $ b\equal{}\minus{}1$. If $ f(x)\equal{}\minus{}x$, then

$ f(f(m) \plus{} f(n)) \equal{} f(\minus{}m \minus{} n) \equal{} f(\minus{}(m\plus{}n)) \equal{} m\plus{}n$

In addition, $ f(x)\equal{}\minus{}x$ is defined for all nonnegative integers and takes nonnegative integers values whenever $ x$ is a nonpositive integer (for all of which $ f$ is defined). Is there some condition in the problem that prevents $ f(x)\equal{}\minus{}x$ from working?

f(x)=-x is only nonnegative for x=0 for a nonnegative value of x (but that point is on f(x)=x).  If x were nonpositive,then f(x)=-x is nonnegative,but x cannot be nonpositive (unless it equals 0).

Its has to be nonnegative>nonnegative

Also its cool that #4 is just a variation of Cauchys.

[quote="the future"]f(x)=-x is only nonnegative for x=0 for a nonnegative value of x (but that point is on f(x)=x).  If x were nonpositive,then f(x)=-x is nonnegative,but x cannot be nonpositive (unless it equals 0).[/quote]
Why can't x be nonpositive except at 0? The problem says that f is "defined on nonnegative integers"; it doesn't say it's undefined elsewhere. Indeed, in the official solution, the answer is "$ f(m) \equal{} m$ for all m."

Will my solution for number 3 be accepted? I had a 24-equation system and I solved it with Mathematica.

Here's how i did:

#1: perfect
#2: I dumb :censored: ed it, and got right answer and same solution as official. perfect
#3: nothing, epic fail
#4: I had overcomplicated solution that worked. I said f(1)=a and f(2)=b and bashed until I got something.
#5: i tried coordinate bash. epic fail again. (didn't send anything in)

So will I get a 5 on #4 if my solution works, but sucks? (its a very long and complex one...)

MCTP: If a problem says a function f is defined on set A, it implies that it is not defined anywhere else.

EDIT: If you want to take a look at my BS #4 solution, it's in the file.

1- 5 perfect
2- 5 perfect
3- 5 perfect
4- 4 (included f(x) = -x)
5- 2-3 used coordinates
so 19 + 3 = 22?

Meh, Silver Medal

Other than forgetting to reduce on #5 (...?) my solutions are good. I liked my solution for 4 much better than the official one; it's synthetic.

When I put f(x)=-x works for #4, will I only lose a point?

Worthawholebean, you had a typo :P

[quote]
For $ m \equal{} n \equal{} 9, f(2f(0)) \equal{} 0$.[/quote]

And here are my solutions. My solution to 3 was a bit more wordy than needed, but still gets the point across. Hoping for a 25.

I just briefly looked through the solutions, and confirmed my answers were all correct. I felt this round was the easiest so far...

Yeah, #5 could be solved with pythagorean theorem instead of cosine law, but they are pretty much equivalent.

For #3, after I came up with the system, I just said you can solve it and gave the answer (which I found using a calculator). Would that lose any points (in particular, more than 1?)

I need to go now. I will put my solution to #4 later.

This round was definitely the easiest.

All correct, I MUST get a 25 this round to get a gold...

Hmm... opinions:

1. This is just some really easy brute force...
2. This isn't that hard...
3. I managed to misread it... the solution is just brute force...
4. This is extremely well known....
5. Way too well-known... it was even posted on aops a few times. I've seen the problem on aops before.

I solved 5 using homothety, median length formula and herons. Hmm?

I'm pretty sure that #2 has a nice generalization: if the groups are of size m, and n ranges from 0 to $ \binom{m}{2}$, then the minimum value of k is 2m-1.  Can anyone confirm this?

Agree that this round was really easy; #1 especially. I solved #1 and #2 correctly, #3 I used a convoluted method sort of similar to the official solution but somehow got 50/61 instead of 50/63 (hopefully that nets me a 4 still), #4 I included f(x)=-x by accident as well, and #5 I coordinate bashed and got it right. Bleh I need 23 for bronze but I don't think I got it.

[quote="pythag011"][quote="kstan013"]I'm feeling rather proud of myself; got them all right for the first time!

I still need 23 for a silver but I just might get it.

[hide="my answers"][hide="1(5)"]1) Perfect, almost exactly like official solutions[/hide]

[hide="2(5)"]2) Showed an example, showed it couldn't be less than 8, showed how I derived the example. If not given a 5 I will probably protest.[/hide]

[hide="3(5)"]3) Used a chart based upon whether the particle will move up or down. After a slanted upward movement, the particle can move up two more or move horizontally. Made a chart showing probability it would first have 3 more downs than ups. Called the probability of there first being 3 more d's when the particle moves horizontal along the line y=1 'x', then solved for x with another probability tree, plugged it in and presto: 50/63[/hide]

[hide="4(3)"]4) Thought the question was kind of common sense, but I just showed that if it was a 1st degree polynomial it was f(x) = x, then showed that it couldn't be a degree 2 or higher, so the only function was f(x) = x. I'll probably end up getting a 3.[/hide]

[hide="5(5)"]5) Used some right triangles off of the isoceles triangles formed with the radii along with some Pythagoras. Mine made the official solutions seem too complicated. I ended up with a 4th degree function to solve for x with, but there were no x^3 or x terms so it was pretty easy to solve. [/hide]

That's 23 so we'll see if I end up with my silver medallion blingage.[/hide][/quote]

Hm... but what if its not a polynomial for 4?[/quote]

Yeah...that's probably why I'll end up getting a 3. I didn't know how to prove this since it seemed like common sense. Always the hardest to do, those easy proofs.

You could have just said that the Taylor series expansion of the function is a polynomial...  :P

Just kidding  :)

sigh every problem ive missed this year has been algebra (i.e. stuff we're supposed to learn in school, like functions and trig and algebraic manipulations). I got every problem but number 4, but I probably will get a point or 2 counted off for my other solutions, especially since my number 3 was not very organized. I get silver if I get 19.

Hm I need to learn to play around with algebra some.

Hmm this probably means I should not wait to be taught calculus in school

[quote="jvenezuela716"]Did anyone else feel that $ \frac {d}{dt}(\text{difficulty}) < 0$ this year?[/quote]

I didn't feel that, but in my opinion difficulty was a concave function....

I felt that Round 3 > Round 2 > Round 1 > Round 4 in difficulty.

[quote="jvenezuela716"]Did anyone else feel that $ \frac {d}{dt}(\text{difficulty}) < 0$ this year?[/quote]

No, I didn't really feel like the difficulty was a differentiable function.  I think the difficulty of the rounds this year would be best modelled using a Weierstrass function.

[quote="jvenezuela716"]Did anyone else feel that $ \frac {d}{dt}(\text{difficulty}) < 0$ this year?[/quote]

Since I am lacking in calculus skills, I will say what I want to say the dumb way:

I thought the difficulty was like an upside-down parabola.

EDIT: What's a differentiable function?

Well, if it's an upside-down parabola, then $ \frac{d^2}{dt^2}(\text{difficulty})<0$.

i felt that $ \frac {d}{dt}(\text{difficulty})$ was monotone decreasing, but not linear

Based mainly on the difficulty on the #5's (The rest kind of average out to the same difficulty):

Round 2>Round 3>Round 1>Round 4

You'll get no such calc expressions from me :P

I feel rather that $ \sum_{i\equal{}1}^{4} \frac{\text{difficulty}_i}{2^i} \approx \frac{\pi}{2}$.

I completed all my solutions for round 4 about 2 days after the problems were released -- typed up and everything.  I just proofread them three times or so in the month and a half following.

Based on #5s,
Round 1 > Round 2 > Round 4 in difficulty.

I didn't do the Round 3 one.

1) Perfect
2) Perfect
3) Wrote a system of 15 equations, probably all incorrect, and I was tired and it was the last night so I wrote "Solving this is not fun, and I didn’t do it." 0 points, if I'm lucky.
4) Perfect, I think?
5) Coordinate-bashed, got the right answer.

Round 3 > Round 2 > Round 1 > Round 4

[quote="someperson01"]1) Perfect
2) Perfect
3) Perfect, had a system of 2 equations :D
4) Perfect
5) Coordinates, right answer.

Definitely Round 3 > Round 2 > Round 1 > Round 4[/quote]
same system of 2 equations, I let x be the desired probability and p be the probability that the particle returns to the x-axis traveling horizontally at y = 1. Notice it is the same as the probability as the particle reaching y=6 traveling horizontally at y=5


[quote="sandu2508"][quote="math10"]Can anyone post Moldoval 2009?  :D  :D[/quote]

It was already posted: http://www.mathlinks.ro/viewtopic.php?search_id=1781420138&t=279983.[/quote]
thank you very much.


limit
logarithms
calculus
calculus computations

$ L_{n \plus{} 1} \equal{} L_n \plus{} \prod_{i \equal{} 1}^{n}(\frac {L_i}{L_i \minus{} 1})   and L_1 > 1$
1. how to prove (or disprove) that  $ lim_{n\rightarrow\infty} \frac {L_n}{nlnL_n} \rightarrow 1$
2. can you prove directly(without using prime number thereo) $ lim_{n\rightarrow\infty} \frac {L_n}{P_n}\rightarrow 1$ where  is the nth prime number.

I am assuming (1) should have stated
\[ \lim_{n\to \infty}\frac {L_n}{n\log n} \equal{} 1
\]
since, if (1) stands as is, it implies $ \lim_{n\to \infty}L_n \equal{} 1$.  If the above is actually what you meant then, given (1), (2) would be equivalent to the prime number theorem just by properties of limits.  Actually, [i]any[/i] sequence $ \{a_n\}$ satisfying $ \lim_{n\to \infty}\frac {a_n}{n\log n} \equal{} 1$ would satisfy $ \lim_{n\to \infty}\frac {a_n}{p_n} \equal{} 1$ if and only if the prime number theorem holds.  In light of this, proving (2) 'without PNT' is equivalent to proving PNT.  Proving (1) looks like it'll be a mess, but after doing some clumsy asymptotic estimates I do believe it's true.


geometry
rectangle
quadratics
trigonometry
circumcircle
cyclic quadrilateral
AMC

1. In $\triangle ABC, AB=360, BC=507, CA=780$.  $M$ is the midpoint of $CA$ and $D$ is on $CA$ such that $BD$ bisects $\angle B$.  Also, $F$ is on $BC$ so that $BD \perp DF$.  Let $BM$ and $DF$ intersect at $E$.  Find $\frac{DE}{EF}$.

2. A 10X11 rectangle contains an equilateral triangle inside of it.  What is the maximum possible area of the triangle?

I doubt it, but is the answer for #2 $221\sqrt{3}-330$? If so, I'll post the solution.

Oh, cool.

[hide="2"]The maximum area of the equilateral triangle must be given with one vertex on a vertex of the rectangle. Call the rectangle $ABCD$ and the triangle $AEF$, with $AB=11$ and $BC=10$, and $E$ on $BC$ and $F$ on $CD$. $BE=m$ and $DF=n$, so the side of the triangle has length $\sqrt{121+m^2}=\sqrt{100+n^2}=\sqrt{(10-m)^2+(11-n)^2}$. We can square the entire equation to remove the square root signs, and leave our answer in that form, as the area of the triangle is found with $s^2$. From this, we have $121+m^2=100+n^2\Rightarrow m^2=n^2-21$, and then $100-20m+m^2+121-22n+n^2=121+m^2\Rightarrow n^2-22n+100=20m$. From the first equation, $m^2=n^2-21\Rightarrow m=\sqrt{n^2-21}$, so we plug this in and square the equation, and come out with a quartic. This looks ugly, but we see that $n^2+100$ conveniently factors out, and we are left with the quadratic $n^2-44n+184$, which gives roots $x=22\pm 10\sqrt{3}$. We know that addition gives an answer far too large, so $n=22-10\sqrt{3}$. We plug this into the side length given with $n^2$, then plug that into the formula $\frac{s^2\sqrt{3}}{4}$, and we have $\boxed{221\sqrt{3}-330}$.[/hide]

[hide="Here is the solution of the first problem:"] I suppose $a>c$. Thus, $MD=AM-AD=\frac b2 -\frac{bc}{a+c}=\frac{b(a-c)}{2(a+c)}$. I note the point $G\in (BC)$ for which $BG=AB$. Then, $GC=a-c$ and $BD\perp AG$ $\Longrightarrow$ $DF\parallel AG$ $\Longrightarrow$ $\frac{FG}{FC}=\frac{DA}{DC}=\frac{BA}{BC}=\frac ca$ $\Longrightarrow$

$\frac{FG}{c}=\frac{FC}{a}=\frac{a-c}{a+c}$ $\Longrightarrow$ $GF=\frac{c(a-c)}{a+c}$ $\Longrightarrow$ $BF=BG+GF=c+\frac{c(a-c)}{a+c}$ $\Longrightarrow$ $\boxed {BF=\frac{2ac}{a+c}}$.
In a different way, $BD=\frac{2ac}{a+c}\cos \frac B2$, $BD=BF\cos \frac B2$ $\Longrightarrow$ $BF=\frac{2ac}{a+c}$.

I apply the Menelaus' theorem to the transversal $\overline {BEM}$ for the triangle $CDF$ :

$\frac{BF}{BC}\cdot \frac{MC}{MD}\cdot \frac{ED}{EF}=1$ $\Longrightarrow$ $\frac{ED}{EF}=\frac{BC}{BF}\cdot \frac{MD}{MC}=\frac{a(a+c)}{2ac}\cdot \frac{a-c}{a+c}=\frac{a-c}{2c}$ $\Longrightarrow$

$\boxed {\frac{ED}{EF}=\frac{a-c}{2c}}$. Therefore, $a=507$, $c=360$ $\Longrightarrow$ $\boxed {\frac{ED}{EF}=\frac{49}{240}}$.[/hide]

Please discuss #2 here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66710

Thanks :D

[quote="Optimosis"]1. In $\triangle ABC, AB=360, BC=507, CA=780$.  $M$ is the midpoint of $CA$ and $D$ is on $CA$ such that $BD$ bisects $\angle B$.  Also, $F$ is on $BC$ so that $BD \perp DF$.  Let $BM$ and $DF$ intersect at $E$.  Find $\frac{DE}{EF}$.
[/quote]
[hide="monkey"]Draw the circumcircle of ABC. Extend BD to X on the circumcircle. By congruent arcs AX=CX. Hence XM is perpendicular to AC.  $\angle XAC = \angle DBC$

Let K be a point on BD such that AK is perpendicular to BD.
$\angle AKX =\angle AMX = 90^{\circ}$, which makes AKMX a cyclic quadrilateral hence
$\angle XAM = \angle XKE = \angle DBC$. So $MK$ is parallel to $BC$.
$\frac{BK}{BD}=\frac{CM}{CD}$.
Now $\triangle ABK \sim \triangle FBD$, so
$\frac{BK}{BD}=\frac{AB}{BF}=\frac{CM}{CD}$.
So $BF=360*\frac{CD}{CM}$
Now Menelaus says
$\frac{DM}{MC}*\frac{BC}{BF}*\frac{EF}{DE} = 1$
substitute BF in we get
$\frac{DM}{MC} * \frac{507}{360}*\frac{CM}{CD}*\frac{EF}{DE} = 1$

$\frac{DM}{CD}*\frac{507}{360}*\frac{EF}{DE} = 1$

$\frac{AD}{CD}=\frac{360}{507}$,
$CD=\frac{507AC}{360+507}$, $CM=\frac{AC}{2}$,
$CD-CM=DM=\frac{147AC}{2*867}$
$\frac{DM}{CD}=\frac{147}{507*2}$
$\frac{147}{507*2} * \frac{507}{360} = \frac{DE}{EF} = \frac{147}{720} = \frac{49}{240}$[/hide]


About how much do quants make at firms like D.E. Shaw or Jane Street Capital?

Look here:
http://news.efinancialcareers.com/NEWS_ITEM/newsItemId-4131

You can find some other good articles on quants here:
http://news.efinancialcareers.com/newsArchiveSearchhandler.cfm

Mr. Rusczyk's a good person to ask about DE Shaw matters on this forum.  If I remember correctly though, he said that on salaries can range widely but for someone of 4-5 or more years of experience mid hundred thousands is definitely not unheard of.

A gazillion dollars.

It varies.  See, financial firms randomly pay most of the compensation through what's called a bonus that the quants/traders/whatever receive at the end of year.  The salary makes up only a very small portion of the total compensation.  As far as I know, the size of the bonus one receives depends on a whole host of factors.  Different companies pay bonuses differently, but there are some general rules, so to speak.  If the company did very well that year, it will have more money to pay out, and will be most likely to give larger bonuses as compared to other years.  If one was a star quant/trader/whatever who made more money for the company as compared to other quants/traders/whatever at the same level of the hierarchy, he/she would be very likely to receive a larger bonus than his/her colleagues.  There are many more variables taken into account than those I've mentioned, so check out some of the articles Joml linked you to.  But again, in one's 4th or 5th year, one million dollars is not unheard of, but neither is  low hundred thousands.  It all just depends.

Wow, that's pretty amazing.  Are these salaries typical at many moderately sized firms, or only at very prominent firms such as D.E. Shaw and Jane Street?

The variance in quant pay a few years out is very, very large.

Bubala's comments (post-gazillion) are pretty much on-mark.  While it's not expected that you'll make a million dollars in year 5, neither is it impossible.  If you're not making at least middle six-digits by year 6-8 as a quant in a hedge fund, then something has gone very wrong for you.

(My experience may somewhat skew these remarks - I happened to have a very quick rise because I got lucky in the group I landed in when I arrived at Shaw.  That said, nearly everyone I know who has been involved in trading for 7-10 years is very, very wealthy.  FWIW, I was only there for 4 years.)

[quote="rrusczyk"][quote="themonster"]Is the pay related to how much education one has?  Like if you have a PhD in physics will you get more compensation than if you just have an undergraduate degree?[/quote]

When you start, yes.  Five years in, it won't really matter so much.

In general, I'd say that if you can get a position at a good trading firm right out of undergraduate, in the long run you'll be way ahead of where you would be if you waited until you had a PhD.[/quote]

very true, getting in as an undergrad will be tougher, but a star out of a Ivy level CS, Math, or Physics program who can really code shouldn't have a problem


Let $a_{0}$ be a positive integer. The sequence $(a_{n})$  has properties: $a_{i+1}=a_{i}/2$ if $a_{i}$ is even and $3a_{i}-1$ if $a_{i}$ is odd.
Prove that if n such that $a_{n}=a_{0}$ then $2^{n}>a_{0}$

Posted before. It comes from some Polish MO.

http://www.mathlinks.ro/Forum/viewtopic.php?t=64610
http://www.mathlinks.ro/Forum/viewtopic.php?t=107848

I think these threads should be merged or so..


linear algebra
matrix
abstract algebra
linear algebra unsolved

A is a mn matrix.Prove:$Ker{A^{n}}=Ker{A^{n+1}}$

I don't know what he meant when he said "a mn matrix" but I don't think he meant $m\times n.$

So I assume that $A$ is a square matrix. Observe that $\mathrm{ker}\, A^{k}=\mathrm{ker}\, A^{k+1}$ if and only if $A$ acts as an isomorphism on the range of $A^{k}$. Since the range of $A^{k}$ does not expand as $k$ increases, equality $\mathrm{ker}\, A^{k}=\mathrm{ker}\, A^{k+1}$ implies $\mathrm{ker}\, A^{j}=\mathrm{ker}\, A^{j+1}$ for all $j>k$.

If $A$ is an $n\times n$ matrix, the kernel of $A^{k}$ will stabilize in at most $n$ steps.


algebra
polynomial
function
induction
greatest common divisor
number theory proposed
number theory

Prove that for any given integer $n$ you can represent each integer $a$ in infinitely many ways as $a=\pm 1^{n}\pm 2^{n}\pm 3^{n}\pm ... \pm m^{n}$ when choosing the signs and $m$ the right way.

This can be generalized to any non-trivial (see below) polynomial $P(x)$ instead of $P(x) = x^n.$
More precisely, we have the following result :

If $P(x)$ is a polynomial with degree $n$, taking only integer values on the integers, and such that $\gcd (P(0), \cdots , P(n)) = 1$ then for each integer $k$ there exists an integer $m \geq 1$ such that :
\[k = \sum_{i=1}^m \epsilon_i P(i)\]
where $\epsilon_i = \pm 1$ for all $i$.

The proof is a combinatorical one, as did Erdös for the case $P(x) = x^2.$

Pierre.

Yes, our proof (I would not call it combinatorial  ;) ) also works for that generalisation.  :)

Can someone post the proof please? Thanks.

Obviously, the difference of two powers (i.e. $x^n - (x+k)^n$) is (because of $(x+k)^n$ being $x^n + \cdots$ an polynom of the Degree $n-1$. So the difference of differences is of Degree $n-2$ etc., by continuing this, you will get a constant after $n$ steps.

Let us call this constant $c$.

By choosing the Signs the right way, you may get a Sequence looking like $ \pm 1 \pm 2^n \pm \cdots + c - c + c - c $. So, if you have got one way expressing a number in that way, using this, you'll get infinitely many ways of doing this.

Because of $(a+1)^n = \cdots + 1$, we know that $a \equiv 1 \mod c \rightarrow a^n \equiv 1 \mod c$.
We also know, that in a sequence of $c$ successive integers there is always one Number with the remainder 1 divided by $c$. If we have a sequence $x^n, (x+1)^n, \cdots, (x+2c)^n$, at least ther must be two such numbers, and we know, that $x+k \equiv x + k + c \mod c$.

If we choose the signs correctly (and subtract the sums of the subsequences from $x$ to $x+c-1$ and from $x+2c - 1$, we will get a number divisible by $c$. Now if we change the signs of one of the Numbers with remainder 1, we will get a number with remainder 2.

If we do the same thing twice, we get remainder 4, 6, 8, ...

To get an odd reminder, we only have to postpone building the subsequences after $1^n$, so we may add $1^n$.

At the end, we have to add (or subtract) $c$ (as shown) often enough, to express every integer.

(Please excuse my bad english)

Here's my solution (similar to Schoppenhauer's but also proving Pierre's generalisation):

Define for some function $f$ and real $i$ the (well known¿) difference operators $\Delta_i (f) (x) := f(x+i) - f(x)$ and $\Delta ^0 (f) = f , \Delta ^ k (f) := \Delta_{2^{k-1}} ( \Delta ^{k-1} (f) )$
Someone then easily sees that you can write $ \Delta ^ k (f) (x)$ as some sum $\pm f(x) \pm f(x+1) \pm f(x+2) \pm ... \pm f(x+2^k-1)$

Then if $p$ is a polynomial and not constant $=0$, you get $\deg ( \Delta_i (p) ) = \deg (p) -1$ for all positive $i$ where additional $deg(0)=-1$ and so if $0\leq k \leq \deg (p)+1$, it follows by induction that $\deg ( \Delta ^ k (p) ) = \deg (p) - k$.

Now let be $d := \deg(p)$ where $p$ is a polynomial as mentioned in Pierres post.
Let be $C:= \Delta ^ d (p)$ and $D=\Delta ^ {d+1} (p)$
Then $\deg ( C ) = \deg (p) - d = 0$, so $C$ is constant $=c$ and not $0$. Also $D$ is constant, but $=0$.
Define now $E(x) = p(x) + p(x+1) + ... + p(x+c-1)$.

So we only have to reach every residue class $\mod c$ and it's done, because than we can add and subtract $c$ as often we want.
So let be $e:= p(1) + p(2)  + ... + p(k)$ where $k$ is the smallest positive integer such that the sum that defines $d$ is odd (such $k$ exists, otherwise you can't have greatest common divisor $1$)
Now (by Bezout's theorem) there exists integers $x_i$ with $x_1 p(1) + x_2 p(2) + ... + x_n p(n) = 1$.

Now write out the sums of the type $E(1) - E(1 + c) + E(1+2c) - ... - E( 1 + 2n \cdot c)$ in terms of $p$.
Then you can change some signs in that sums (as Schoppenhauer did) to reach the residue class of $2$. Do this again and again to get all 'multiples' of $2 \mod c$ (a little bit more exact: change the signs of the residue class of $1\leq j \leq n$ exactly $|x_j|$ often (the sign's at $mc + j$ and $m(c+1)+j$ are different))
If $c$ would be odd (which will not happen), we are done.
If $c$ is even, you can add the 'odd' residue class of $e$ at the end.


Find all integral solutions to $p(m+n)=mn$, whith p a prime.

Since $p$ is prime at least one of $m, n$ is divisible by $p$. By symmetry let $m=pk$. Then our equation rewrites as:
$pk=n(k-1)$
Since $(k, k-1)=1$ $n$ is divisible by $k$. So $n=kt$. Then we have:
$p=t(k-1)$
From this we deduce: $k=2, t=p$ or $t=1, k=p+1$.
Answer: $(m, n) \in \{(2p, 2p), (p(p+1), p+1), (p+1, p(p+1))\}$

Supose that m+n=a and m-n=b then p(m+n)=mn so pa=(a^2-b^2)/4 so a^2-4pa-b^2=0 D=16p^2+4b^2 so 4p^2+b^2=x^2 pithagorean triples.....

Another quick way to do this is factor

$(m-p)(n-p) = p^{2}$

and then all we have to do is consider the divisors of $p^{2}$. We get the same solutions as shown by TomciO.

Gah, and I was trying to do something ugly with prime factorizations. I hate being stupid.

[hide="More direct solution"]Rearrange

\begin{eqnarray*}p(m+n)=mn &\iff& pm+pn=mn\\ &\iff& mn-pn=pm\\ &\iff& n(m-p)=pm\\ &\iff& n=\frac{pm}{m-p}\\ &\iff& n=\frac{pm-p^{2}+p^{2}}{m-p}\\ &\iff& n=\frac{p(m-p)+p^{2}}{m-p}\\ &\iff& n=p+\frac{p^{2}}{m-p}\end{eqnarray*}

Since $p$ is prime, $\frac{p^{2}}{m-p}$ can be integer only if $m-p\in\{1,p,p^{2}\}\iff m\in\{p+1,2p,p^{2}+p\}$ etc.[/hide]

And this is basically the same as [b]paladin8[/b]'s solution.


A standard deck of 52 cards is shuffled well, and cards are turned over in order. What is the expected value of number of cards before the second ace is turned over?

[hide="Possibly wrong partial solution"]Call $f(n)$ the probability that you have to flip over $n$ cards before finding the second ace. Thus the expected value of number of cards is $v=\sum_{k=1}^{50}(f(k)k)$. There are $n$ ways to put the first ace somewhere in the sequence of cards and only $1$ way of putting the second ace at the end out of $\binom{52}{n+1}$ possible combinations of the first $n+1$ cards, so $f(n)=\frac{n}{\binom{52}{n+1}}$, and thus $v=\sum_{k=1}^{50}\frac{n^{2}}{\binom{52}{n+1}}$. I don't know how to do the sum out.[/hide]

I
would
use [hide="magic."] State diagrams! [/hide]


For $a,b,c$ positive real numbers peove that

$\displaystyle \frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2} \ge \frac{a+b+c}{3}$

Maybe you're wrong,maverick because:
$9a^3 \ge (4a-b)(2a^2+b^2)$
$\leftrightarrow 9a^3 \ge 8a^3+4ab^2-2a^2b-b^3$
$\leftrightarrow a^3+b^3+2a^2b \ge 4ab^2$
$\leftrightarrow (a-b)(a^2+3ab-b^2)\ge 0$
Which is false.

After a lot of computations ( suggested to me by hungkhtn)  you have to prove that

$S_a(b-c)^2+S_b(c-a)^2+S_c(a-b)^2 \geq 0$ (*)

where

$S_a=c^3a^2+b^3c^2+b^2c^3+3a^3c^2-a^3b^2-2b^2c^2a$

$S_b=a^3b^2+c^3a^2+c^2a^3+3b^3a^2-b^3c^2-2c^2a^2b$

$S_c=b^3c^2+a^3b^2+a^2b^3+3c^3b^2-c^3a^2-2a^2b^2c$

These are correct at 100% as I have checked it many many times.

So now good work to prove (*)

Vasc, have you a proof for this one or have you only conjectured it is true?

[quote]I have a serious doubt for the case $a \geq c \geq b$[/quote]

I have spent a lot of hours on it without any result :(

[quote]I really can prove it only for

$a \geq b \geq c$  but for $a \geq c \geq b$ it seems impossible.[/quote]

Have you any idea for this case ? ;)

No interested :?

To solve this is a great result ( my opinion  ;) )

Manlio, I have a proof, but it is a secret proof in this year.  ;)

I think you have too many secrets, Vasc.

And this is not in the aim of the forum.

i hv a strange solution.
Assume
\[
\frac{{a^3 }}{{2a^2  + b^2 }} \ge ma - nb
\]
so that \[
m - n = \frac{1}{3}
\]
then if there's real m,n satisfying these two conditions, we have the desired inequality.

Some simplifcation gives
\[
(1 - 2m)a^3  + nb^3  + 2na^2 b \ge mab^2
\]
Applying weighted am-gm on l.h.s. we have
\[
\begin{array}{l}
 (1 - 2m)a^3  + nb^3  + 2na^2 b \\
  \ge 5\sqrt[5]{{(1 - 2m)n^4 \frac{2}{{27}}}} \\
  = 5\sqrt[5]{{(1 - 2m)(m - \frac{1}{3})^4 \frac{2}{{27}}}} \\
 \end{array}
\]
Set this to be m, then \[
m = 5\sqrt[5]{{(1 - 2m)(m - \frac{1}{3})^4 \frac{2}{{27}}}}
\]
this is a degree 5 equation as the coefficient of m^5 cannot be cancelled out, so there must be a real root. (As complex roots are in pairs) so there must exist some m,n satisfying the above ineq and we are done.

[quote="manlio"]To apply AM-GM

$(1-2m)$

must be positive.

Is it positive,siuhochung?

I don't think you can affirm that. ;)

Am I wrong? :?[/quote]

well i can't affirm that...  :(
maybe it's better if someone can find the sol of the degree-five equation..

Perhaps it is sufficient to prove the 5 grade equation has a solution m between 0 and 1/2 and this is easy.

if m is smaller than 1/3 (as it's near to 0.27 said by stryker), then my method fails because we need n to be positive too.

[quote="mecrazywong"][quote="StRyKeR"]My calculator shows that there are two real solutions, one that is a tiny fraction above 0.5, and the other well below 0.5, (it is close to 0.27203), so it seems that the problem is solved. But now the question is, how do we determine the root without a calculator?[/quote]
Isn't that simple?
Setting ${f(m)=m^5-(1 - 2m)(m - \frac{1}{3})^4 \frac{2}{{27}}}$. Since $f(0)<0,f(1/2)>0$, there must be a root in [0,1/2].[/quote]
me, he said determine the root, not the range of the root.

I also tried to do like this but it gave no result:
After simplifying $\frac{a^3}{2a^2+b^2}\ge ma-nb$ we get $a^3(1-2m) +2a^2bn-b^2am+b^3n\ge 0$
This holds for all $a,b>0$.Put $a=0$ to get $n>0$,put $b=0$ to get $1/2>m$ (*),put $a=1/2,b=1$ to get $12n+1\ge  6m$ using $m-n=1/3$ we get $12(m-1/3)+1\ge 6m$ so $m\ge 1/2$ but this contradicts(*)
I think I didn't make any mistakes.

[quote="siuhochung"]
Assume
\[
\frac{{a^3 }}{{2a^2  + b^2 }} \ge ma - nb
\]
so that \[
m - n = \frac{1}{3}
\]
then if there's real m,n satisfying these two conditions, we have the desired inequality.[/quote]
This inequality is true if and only if its form is $(a-b)^2(pa+qb) \ge 0$ with $p \ge0$ and $q \ge0$ .
We get $m=5/9$ and finally $(a-b)^2(2-a) \ge 0$, which is not true.

This inequality also can be proved using substitutions $b=a+x$, $c=a+x+y$ if $a\leq b\leq c$ or $b=a+x+y$, $c=a+x$ if $a\leq c\leq b$. It reduces to something obvious.
I don’t know are you satisfied with this solution?

I'm satisfied with any solution that takes less than 30 minutes to write up. =] So does it take less than 30 minutes?

Yes, if you use computer it will take you less than one minute.
I'm not contestant any more and it is allowed for me!

Well, it seems that nobody posted a solution for this one.
If manlio's results are correct, let a = min(a,b,c), we have
$S_a \ge 0$
and
\[
\begin{array}{l}
 S_b  + S_c  \\
  = 2a^3 b^2  + c^2 a^3  + 4b^3 a^2  - 2c^2 a^2 b + 3c^3 b^2  - 2a^2 b^2 c \\
  = (c^3 b^2  + a^3 b^2  + a^3 b^2  - 2a^2 b^2 c) + (c^3 b^2  + c^3 b^2  + b^3 a^2  + c^2 a^2 b - 2c^2 a^2 b) \\
  + 3b^3 a^2  \\
  \ge (3a^2 b^2 c - 2a^2 b^2 c) + (4ab^2 c^2  - 2c^2 a^2 b) + 3b^3 a^2  \\
  \ge a^2 b^2 c + 2c^2 a^2 b + 3b^3 a^2 (Sinceb \ge a,ab^2 c^2  \ge c^2 a^2 b) \\
  > 0 \\
 \end{array}
\]
So we are done.

[quote="xtar"]Dear siuhochung

Can you show me a complete and clear solution![/quote]

see manlio's post first!
Then if $S_a$ and $S_b+S_c \ge 0$,
$\sum S_a(b-c)^2 \ge S_a(b-c)^2+(S_b+S_c)(min(b-a,c-a))^2 \ge 0$

[quote="manlio"]After a lot of computations ( suggested to me by hungkhtn)  you have to prove that

$S_a(b-c)^2+S_b(c-a)^2+S_c(a-b)^2 \geq 0$ (*)

where

$S_a=c^3a^2+b^3c^2+b^2c^3+3a^3c^2-a^3b^2-2b^2c^2a$

$S_b=a^3b^2+c^3a^2+c^2a^3+3b^3a^2-b^3c^2-2c^2a^2b$

$S_c=b^3c^2+a^3b^2+a^2b^3+3c^3b^2-c^3a^2-2a^2b^2c$

These are correct at 100% as I have checked it many many times.

So now good work to prove (*)

[/quote]

Sorry, but I don't understand

why

 $a=min(a,b,c)$   then    $ \displaystyle S_a=c^3a^2+b^3c^2+b^2c^3+3a^3c^2-a^3b^2-2b^2c^2a\ge0$

and why

$S_b+S_c\ge0$ then $\displaystyle S_b(c-a)^2+S_c(a-b)^2 \geq (S_b+S_c)(min(b-a,c-a))^2 $

Thank you very much.

Consult Kedlaya's notes and you'll see why

Bomb

While solving this very hard inequality I came to this ineq (I can't solve it, but it's surely true (checked by computer) and if you're able to prove it, then ineq4 is also proven ;)):
For positive reals $a,b,c$ prove:
$17(a+b+c)+10(\frac{b^3}{a^2}+\frac{c^3}{b^2}+\frac{a^3}{c^2})\geq 27(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c})$

[quote="ondrob"]And I found that the last inequality I wrote can be a bit more better. My computer says that
\[17.655(a+b+c)+9.345(\frac{b^3}{a^2}+\frac{c^3}{b^2}+\frac{a^3}{c^2})\geq 27(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}) \]
holds, but:
\[17.656(a+b+c)+9.344(\frac{b^3}{a^2}+\frac{c^3}{b^2}+\frac{a^3}{c^2})\geq 27(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c})\]
does not hold, so best constant is somewhere between $17.655$ and $17.656$. But I still have no proof for all these my (+my computer) conjectures.[/quote]
I find your inequality is equal to:
find the max number $k$,such that for all positive numbers $a,b,c$:
$\frac{b(a-b)^2}{a^2}+\frac{(b-c)^2}{b^2}+\frac{a(a-c)^2}{c^2} \geq k\frac{(a-b)^2}{a}+\frac{(b-c)^2}{b}+\frac{(a-c)^2}{c}$(or a similar one)
but it is useless to me,I hope it will be useful to you :)
indeed I guess you may also find it. ;)

thanx, I started new thread about ineq I wrote...

[quote="siuhochung"][quote="Vasc"]As xtar, I think that
$S_b+S_c\ge0$ doesn't imply $\displaystyle S_b(c-a)^2+S_c(a-b)^2 \geq (S_b+S_c)(min(b-a,c-a))^2 $. :o[/quote]
but i have a the minimum of a,b,c...[/quote]
Sorry so the problem is still unsolved, anyone have an idea
I think that if we assume a =min{a,b,c} so not only S(b)+S(c)>0 and S(a)>0 but also     S(c)>0
but i can't develop it....


integration
real analysis
function
calculus
real analysis unsolved

Show,that if $f(x)$ is not continuous every $x$ from $[a,b]$ then does not exsist $\int_{a}^{b}f(x)dx$

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=37648]this thread[/url] for an argument that doesn't use Lebesgue tools. We only get a dense set of points of continuity, but that's enough.

Wouldn't that be if f(x) is not piece-wise continuous?

No. A function can be Riemann integrable even if it is discontinuous on a dense set. The function $f(x)=\begin{cases}0& \text{if }x\text{ is irrational}\\ \frac1q&\text{if }x=\frac pq\text{ in lowest terms}\end{cases}$ is the standard example; it is Riemann integrable on $[0,1]$ with integral zero.

A function cannot be Riemann integrable if it is discontinuous everywhere- that's the problem of the thread.

Oy, sorry.  Completely misread the top.  Thanks.


inequalities
Cauchy Inequality
inequalities proposed

[quote="pvthuan"]1) Prove that if $a,b,c\geq0$, then
\[\frac a{\sqrt{b^{2}+\tfrac14bc+c^{2}}}+\frac b{\sqrt{c^{2}+\tfrac14ca+a^{2}}}+\frac c{\sqrt{a^{2}+\tfrac14ab+b^{2}}}\geq2. \]
2) Prove that if $a,b,c\geq0$, then
\[\frac a{\sqrt{a^{2}+\tfrac14ab+b^{2}}}+\frac b{\sqrt{b^{2}+\tfrac14bc+c^{2}}}+\frac c{\sqrt{c^{2}+\tfrac14ca+a^{2}}}\leq2. \]
[/quote]
Nice, pvthuan.
To prove part 1/, we can use Cauchy-Schawrz inequality. A old my inequality is the general for it :
\[\frac a{\sqrt{b^{2}+kbc+c^{2}}}+\frac b{\sqrt{c^{2}+kca+a^{2}}}+\frac c{\sqrt{a^{2}+kab+b^{2}}}\ge min(2;\frac{3}{\sqrt{k+2}}\]
where $k \ge-2$.
So $\frac{1}{4}$ is the best constant.
Part 2/ is too :
\[\frac a{\sqrt{a^{2}+kab+b^{2}}}+\frac b{\sqrt{b^{2}+kbc+c^{2}}}+\frac c{\sqrt{c^{2}+kca+a^{2}}}\le max(2;\frac{3}{\sqrt{k+2}}) \]

Here is a solution from the book. For the first one,
\[A\geq\frac{(a+b+c)^{2}}{a\sqrt{b^{2}+\tfrac14bc+c^{2}}+b\sqrt{c^{2}+\tfrac14ca+a^{2}}+c\sqrt{a^{2}+\tfrac14ab+b^{2}}}. \]
It is sufficient to prove that
\[a\sqrt{b^{2}+\tfrac14bc+c^{2}}+b\sqrt{c^{2}+\tfrac14ca+a^{2}}+c\sqrt{a^{2}+\tfrac14ab+b^{2}}\leq\frac12(a+b+c)^{2}. \]
Let $B$ be the left hand side of the last inequality, we write
\[B=\sqrt a\sqrt{ab^{2}+\tfrac14abc+ac^{2}}+\sqrt b\sqrt{bc^{2}+\tfrac14abc+ba^{2}}+\sqrt c\sqrt{ca^{2}+\tfrac14abc+cb^{2}}. \]
By Cauchy inequality,
\[B^{2}\leq (a+b+c)\Bigl[\frac34abc+a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)\Bigl]. \]
Finally, we need to prove that
\[\frac34abc+a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)\leq\frac14(a+b+c)^{3}, \]
which is just I. Schur inequality after simple manipulation.

Another idea is to find $r$ such that
\[\frac a{\sqrt{b^{2}+\tfrac14bc+c^{2}}}\geq\frac{2a^{r}}{a^{r}+b^{r}+c^{r}}.\]

For the second one, Cauchy also works, and we need a classical substitution in the final step.

 :P

The second one is solved using Cauchy Schwarz and $ p,q,r$ method.

[quote="pvthuan"]Here is a solution from the book. For the first one,
\[ A\geq\frac {(a \plus{} b \plus{} c)^{2}}{a\sqrt {b^{2} \plus{} \tfrac14bc \plus{} c^{2}} \plus{} b\sqrt {c^{2} \plus{} \tfrac14ca \plus{} a^{2}} \plus{} c\sqrt {a^{2} \plus{} \tfrac14ab \plus{} b^{2}}}.
\]
It is sufficient to prove that
\[ a\sqrt {b^{2} \plus{} \tfrac14bc \plus{} c^{2}} \plus{} b\sqrt {c^{2} \plus{} \tfrac14ca \plus{} a^{2}} \plus{} c\sqrt {a^{2} \plus{} \tfrac14ab \plus{} b^{2}}\leq\frac12(a \plus{} b \plus{} c)^{2}.
\]
Let $ B$ be the left hand side of the last inequality, we write
\[ B \equal{} \sqrt a\sqrt {ab^{2} \plus{} \tfrac14abc \plus{} ac^{2}} \plus{} \sqrt b\sqrt {bc^{2} \plus{} \tfrac14abc \plus{} ba^{2}} \plus{} \sqrt c\sqrt {ca^{2} \plus{} \tfrac14abc \plus{} cb^{2}}.
\]
By Cauchy inequality,
\[ B^{2}\leq (a \plus{} b \plus{} c)\Bigl[\frac34abc \plus{} a^{2}(b \plus{} c) \plus{} b^{2}(c \plus{} a) \plus{} c^{2}(a \plus{} b)\Bigl].
\]
Finally, we need to prove that
\[ \frac34abc \plus{} a^{2}(b \plus{} c) \plus{} b^{2}(c \plus{} a) \plus{} c^{2}(a \plus{} b)\leq\frac14(a \plus{} b \plus{} c)^{3},
\]
which is just I. Schur inequality after simple manipulation.

Another idea is to find $ r$ such that
\[ \frac a{\sqrt {b^{2} \plus{} \tfrac14bc \plus{} c^{2}}}\geq\frac {2a^{r}}{a^{r} \plus{} b^{r} \plus{} c^{r}}.
\]
For the second one, Cauchy also works, and we need a classical substitution in the final step.

 :P[/quote]

I have a short solution  :D


function calculus derivative algebra domain limit

problem 1) Let  f be a function that have derivative in R.If  f : R --------> R  and  f(x+y)=f(x)+f(y)+2xy then find  f  functions.

problem 2) Let f be a function that have domain in R.If  f is continuous at x=0 and for every x element of   real numbers  f satisfies   2f(2x)=f(x)+x .Find  f functions.

[quote="mmrproblem"]problem $1)$
Let  $f$ be a function that have derivative in $\mathbb{R}$.
If  $f : \mathbb{R} \to \mathbb{R}$  and  $f(x+y)=f(x)+f(y)+2xy$ then find  $f$  functions.

[/quote]

[hide]At first, notice that $f(x)=x^2$ satisfies the conditions, because $(x+y)^2=x^2+y^2+2xy$

set $x=0\Rightarrow$

$f(0+y)=f(0)+f(y)+0 \Rightarrow f(0)=0$

$f(x+y)=f(x)+f(y)+2xy \Rightarrow$

$f(x+y)-f(x) = f(y)+2xy \Rightarrow$

$\frac{f(x+y)-f(x)}{y} = \frac{f(y)}{y}+2x \Rightarrow$

$\lim_{y\to 0}\frac{f(x+y)-f(x)}{y} = \lim_{y\to 0}\frac{f(y)}{y}+2x \Rightarrow$

$f'(x) = f'(0)+2x$

$f'(0)=c \neq 0$ , constant

$f'(x) = 2x+c \Rightarrow f(x) = x^2+cx$

$f(x+y) = (x+y)^2+c(x+y) = (x^2+cx)+(y^2+cy)+2xy = f(x)+f(y)$, so it's ok [/hide]

[quote="scorpius119"][hide="2"]First find a particular solution, which can be reached through differentiation; since $2f(2x)=f(x)+x$, $4f^\prime (2x)=f^\prime (x)+1$ and $2^{n+1}f^{(n)}(2x)=f^{(n)}(x)$ for $n>1$. Using $x=0$, find that a solution would have $f(0)=0$, $f^\prime (0)=\frac 13$, and $f^{(n)} (0)=0$ for $n>1$. So a particular solution is $f(x)=\frac 13 x$, which checks into the original equation.

Write $f(x)=\frac 13 x+R(x)$. Then $2f(2x)=\frac 43 x+2R(2x)$ and $f(x)+x=\frac 43 x+R(x)$. So $2R(2x)=R(x)$, which means that $2^kR(2^kx)=R(x)$ for $k\in \mathbb{N}$. Also $R$ is continuous at $x=0$ because $f$ is.

Let $y=2^kx$, then $2^kR(y)=R(2^{-k}y)$, or $R(y)=2^{-k}R(2^{-k}y)$. As $k\rightarrow \infty$, $2^{-k}\rightarrow 0$ and because $2^{-k}y\rightarrow 0$ and $R$ is continuous at $x=0$, $R(2^{-k}y)\rightarrow R(0)$. But plugging 0 into $2R(2x)=R(x)$, $R(0)=0$ so $R(2^{-k}y)\rightarrow 0$.

Since $R(y)$ can be expressed as a product of two reals that each get arbitrarily close to 0, $R(y)$ must be 0. This makes the only solution $\boxed{f(x)=\frac 13 x}$.[/hide][/quote]
But we don't know if $f$ is defferentiable.

[hide="Here is my approach"]First of all let $x=0 \Rightarrow 2f(0)=f(0)$

So $f(0)=0$

Let $x_0\in \mathbb{R^*}$. We will find $f(x_0)$

We define the sequences $\{a_n\},\{x_n\}$ such that $a_n = \frac{1}{2^n}$ and $x_n = x_0 \cdot a_n$

The sequence $\{x_n\}$ tends to zero, hence the secuence $\{f(x_n)\}$ tends to $f(0)=0$
(because $f$ is continuous at $0$)

$2f(2x)=f(x)+x \Rightarrow$

$2f(2x_n)=f(x_n)+x_n \Rightarrow$

$2f(x_{n-1})=f(x_n)+x_n \Rightarrow$

$2f(x_0 \cdot a_{n-1}) = f(x_0 \cdot a_n)+ x_0 \cdot a_n \Rightarrow$

$2a_n \cdot f(x_0 \cdot a_{n-1}) = a_n \cdot f(x_0 \cdot a_n)+ x_0 \cdot a^2_n$

We write all equations from $n=1$ to $n$

$\begin{array}{cc} 2a_1 \cdot f(x_0 \cdot a_0) = a_1 \cdot f(x_0 \cdot a_1)+ x_0 \cdot a^2_1 \\ 2a_2 \cdot f(x_0 \cdot a_1) = a_2 \cdot f(x_0 \cdot a_2)+ x_0 \cdot a^2_2 \\ \cdots \\ 2a_n \cdot f(x_0 \cdot a_{n-1}) = a_n \cdot f(x_0 \cdot a_n)+ x_0 \cdot a^2_n \\ \end{array} \}$

We add all these equations. Notice that $2a_2 \cdot f(x_0 \cdot a_1) = a_1 \cdot f(x_0 \cdot a_1)$ etc.
We get:

$2a_1f(x_0 a_0) = a_nf(x_0 a_n) + x_0 \sum_{k=1}^{n}{a^2_k} \ \ (1)$

When $n \to \infty$, then:

$a_n \to 0$ and $f(x_0 a_n) \to 0 \Rightarrow a_nf(x_0 a_n) \to 0$

and $\sum_{k=1}^{\infty}{a^2_k} = \sum_{k=0}^{\infty}{a^2_k}-a^2_0 = \frac{4}{3}-1=\frac{1}{3}$

$(1)$ gives

$2a_1f(x_0 a_0) = 0+\frac{1}{3} \cdot x_0 \Rightarrow$

$2 \cdot \frac{1}{2} \cdot f(x_0)= \frac{1}{3} \cdot x_0 \Rightarrow$

$f(x_0)= \frac{1}{3} \cdot x_0$

[hide="Note"]
$\sum_{k=0}^{\infty}{a^2_k} = \sum_{k=0}^{\infty}{(\frac{1}{2^k})^2} = \sum_{k=0}^{\infty}{\frac{1}{2^k} \cdot \frac{1}{2^k}} = \sum_{k=0}^{\infty}{\frac{1}{4^k}} = \sum_{k=0}^{\infty}{(\frac{1}{4})^k} = \frac{1}{1-\frac{1}{4}}=\frac{4}{3}$[/hide][/hide]

I did not require $f$ to be differentiable in my general solution. I just assumed that there was a differentiable function that worked (and there was) and worked with that particular solution.

[quote="pontios"]Ok! Hopefully it was differentiable. And you got a correct result.

But how can we be sure that there is not a function $f$ satisfying the given properties and $f$  is not differentiable ?

I mean, by assuming that $\exists f'$ we find that $f(x) = \frac{x}{3}$,ok!

But maybe there is another function $g$, which is not differentiable but $2g(2x)=g(x)+x, \ \forall x \in \mathbb{R}$, and $\lim_{x \to 0}{g(x)}=g(0)$[/quote]

Let us write $g(x) = \frac{x}{3} + h(x)$.  Then we seek a function $h(x)$ such that $2h(2x) = h(x)$.

The problem condition seems to imply that $g(x)$, and by extension $h(x)$, is defined for all $x \in \mathbb{R}$ (or else it is undefined at an infinite number of points, another possible solution.)  Now, barring that solution, $h(x)$ must be defined for all $x$.

So, suppose some $h(k) = m$.  We can write a recurrence relation $h(2x) = \frac{h(x)}{2}$ that allows us to conclude $h(2^n k) = \frac{m}{2^n}$ for all $n \in \mathbb{Z}$ (positive, zero, and negative).

Now, $\lim_{n \to - \infty} h(2^n k) = \infty$ if $m \neq 0$, a contradiction (because $g(x)$ is supposed to be continuous at $0$), so this implies $m = 0$ for any $k$.  Q.E.D.

t0rajir0u, it's ok!  :)

It is proven that, only the condition $\lim_{x \to 0}f(x)=f(0)$ is enough to get the conclusion that $f(x) = \frac{x}{3}$


probability
vector
trigonometry
linear algebra
matrix
Functional Analysis
advanced fields

What does the fact that "two random variables are non-correlated" means, intuitively speaking ?

Some ideas:

To say that $X$ and $Y$ are positively correlated is to say that, on average, the larger $X$ is, the larger $Y$ will tend to be. Negative correlation is the opposite effect: the larger $X$ is, the smaller $Y$ tends to be. Then, for $X$ and $Y$ to be uncorrelated is the borderline case between these two tendencies.

If $X$ and $Y$ are independent, then $X$ and $Y$ are uncorrelated. The converse is false, which is to say that independence is a much stronger property than vanishing correlation.

An example to make that point: Let $(X,Y)$ be chosen so that the points $(1,0),(0,1),(-1,0),$ and $(0,-1)$ each occur with probability $\frac14.$ Then $X$ and $Y$ are uncorrelated, but $X$ and $Y$ are certainly not independent.

The covariance is, in effect, an inner product. (In fact, if I'm willing to use some fancy language, it is an inner product: on a probability space $\Omega,$ let $H$ be the Hilbert space $L^2(\Omega)/\mathbb{R}$ of square integrable random variables modulo constants. Then $\text{Cov}(X,Y)$ is the inner product of the equivalence classes of $X$ and $Y$ in that space.)

With vectors, we're used to writing $v\cdot v=\|v\|^2,$ or, in what amounts to the same thing, defining $\|v\|=\sqrt{v\cdot v}.$ Following the inner product idea, then what is $\text{Cov}(X,X)?$ It is $\text{Var}(X),$ which makes the analogue of the length of a vector the standard deviation: $\sigma_X=\sqrt{\text{Cov}(X,X)}.$

We know how do define the angle between vectors: $\cos\theta=\frac{v\cdot w}{\|v\|\|w\|}$ What's the analogue of this for random variables? The correlation coefficient:

$\rho=\frac{\text{Cov}(X,Y)}{\sigma_X\sigma_Y}.$

Taking this back to your original question: uncorrelated is analogous to orthogonal.

There's more I could say from a matrix point of view.


integration
real analysis
real analysis unsolved

find the solution of the system

$ x'\equal{}\lambda x\plus{}y$
$ y'\equal{}\lambda y$

with the initial condition $ x(0)\equal{}2$ and $ y(0)\equal{}3$

Write the system in the form: $ x' \equal{} \left[ \begin{array}{cc} \lambda & 1 \\
0 & \lambda \end{array} \right] x$.

The solution is $ x \equal{} e^{\left[ \begin{array}{cc} \lambda & 1 \\
0 & \lambda \end{array} \right] t} x_0 \equal{} e^{At}x_0$. So let's Laplace-transform:
$ \mathcal{L}\{x\} \equal{} \mathcal{L}\{e^{At}x_0\} \equal{} (sI \minus{} A)^{ \minus{} 1}x_0 \equal{} \frac {\left[ \begin{array}{cc} s \minus{} \lambda & 1 \\
0 & s \minus{} \lambda \end{array} \right]}{(s \minus{} \lambda)^2} \left[ \begin{array}{cc} 2 \\
3 \end{array} \right]$
$ \equal{} \frac {\left[ \begin{array}{cc} 2(s \minus{} \lambda) \plus{} 3 \\
3(s \minus{} \lambda) \end{array} \right]}{(s \minus{} \lambda)^2}$. Remembering $ \mathcal{L}^{ \minus{} 1}[X(s)] \equal{} \frac {1}{2\pi \jmath} \int_{\gamma \minus{} \jmath \infty}^{\gamma \plus{} \jmath \infty}e^{st}X(s)ds$, the solution is $ x(t) \equal{} \left[ \begin{array}{cc} e^{\lambda t}(2 \plus{} 3t) \\
3e^{\lambda t} \end{array} \right]$


[quote="GameBot"]What is the smallest positive integer with exactly 14 positive divisors?[/quote]

The number is in the form
$ p_1^13$
OR
$ p_1 \cdot p_2^6$.
The smallest in the first case is $ 2^13$, which is clearly larger then $ 3 \cdot 2^6\equal{}\boxed{192}$.

Note that $ 14\equal{}2\cdot 7$ so there is only one way to partition one 2 and one 7: $ (2)\cdot (7)\equal{}(1\plus{}1)(6\plus{}1)$.  Using the formula for $ d(n)$, the number of positive divisors of $ n$, we see that $ \alpha_{1}\equal{}1$ and $ \alpha_{2}\equal{}6$ (or the other way around).  So now just plug in the smallest possible primes and see which number is smallest: $ 2^{1}\cdot 3^{6}\equal{}1458$ or $ 2^{6}\cdot 3^{1}\equal{}192$.  Clearly 192 is the smallest such element then.


Here is a very difficult generalization of USAMTS R3 Q2:

Prove that min(n) | (kn+1)((k+4)n+1) is a perfect square (n>0) is always =k+2, for ALL nonnegative integer k. To start things off, the analysis is already difficult, because it is not always true that (kn+1,(k+4)n+1))=1 unless k=0 (mod 4).

I ran a mathematica program to test this conjecture and it holds for k=0 to 10000, so it is highly likely that the statement is valid, though it seems quite difficult to prove

I don't think it was that tough after all.  Here is my hint in spoiler:[hide]First show that if n < k + 2, then (kn + 1)((k + 4)n + 1) is strictly between ((k + 2)n - 1)2 and ((k + 2)n + 1)2.  Thus the only possibility for a square is ((k + 2)n)2.  But it's easy to rule out that case.[/hide]

Ravi - neat.

I have comments in USAMTS thread.. other possible values of n are:
(a+1)(a+2)(a+3)

and b(b^2-2)(b^4-3b^2+1) (where b=a+2)
But other values do exist! (depending on the value of a)


One mole of an ideal monoatomic gas is confined in a cylinder fitted with a frictionless piston. When 63 J of heat are supplied and the gas is allowed to expand at constant pressure, the temperature rises by 3.0K. What heat input would be required to raise the temperature of the same mass of gas through the same interval at constant volume?

A. 87.9 J
B. 64.3 J
C. 54.7 J
D. 38.1 J

Hmmm, now I have a question.  So apparently you get (D) by noting that $ 63 \mbox{ J} \equal{} nC_p dt$, so that $ n \equal{} 1$ and $ dt \equal{} 3 \implies C_v \equal{} C_p \minus{} R \equal{} 21 \minus{} R$, whence $ dQ \equal{} nC_v dt \equal{} 3(21\minus{}R) \equal{} 38.1$.

I'm wondering what's wrong with the following reasoning: the internal energy of the gas is just $ \frac{3}{2}nRT \equal{} \frac{3}{2}RT$ in this case, and at constant volume, $ dW \equal{} 0$, so $ dQ \equal{} dU \equal{} \frac{3}{2}R dt \equal{} \frac{3}{2}R(3) \equal{} 37.4$.

The theoretical value of Cv is $ \frac {fR}{2}$, where f is the degree of freedom. This may not always agree with the observed value.


Prove that for [tex]a, b, c > 0[/tex] and [tex]a+b+c=1[/tex],

[tex]
\\
\displaystyle\frac{1}{a+bc} + \displaystyle\frac{1}{b+ca} + \displaystyle\frac{1}{c+ab} \geq \displaystyle\frac{27}{4}
\\
[/tex]

I'll post my sol'n after everyone's worked on it for a bit.

[hide]1=(a+b+c):^2:  = a:^2: +b:^2: +c:^2: +2(ab+bc+ca):ge: 3(ab +bc+ca)

ab+bc+ca :le: 1/3

c+ab+a+bc+b+ca :le: 4/3

Lemma: if x+y+z :le: n, 1/x+1/y+1/z :ge: 9/n

1/(c+ab) + 1/(a+bc) + 1/(b+ca) :ge: 27/4

Proof of lemma:

let x + y + z = n

by AM-GM xyz :le: n:^3:/27

also (xy + yz + xz) :le: n:^2:/3

1/x + 1/y + 1/z = 1/(xyz) * (xy+yz+xz)  :le: 27/n:^3: * n:^2:/3 = 9/n

[/hide]

edited: sorry, started writing before you said you'd post it

By AM-HM

cycsum(a+bc)/3  :ge: 3/LHS

So LHS  :ge: 9/(a+b+c+ab+bc+ca)  :ge: 9/(1+1/3)  :ge: 27/4 as desired.

....

I'm sure other people have posted something similar.

Since a+b+c = 1, we change this inequality to

[tex]

\\
\displaystyle\frac{1}{1-b-c+bc} + \displaystyle\frac{1}{1-a-c+ca} + \displaystyle\frac{1}{1-a-b+ab} \geq \displaystyle\frac{27}{4}
\\
[/tex]

Factoring, we get:

[tex]
\\
\displaystyle\frac{1}{(1-b)(1-c)} + \displaystyle\frac{1}{(1-a)(1-c)} + \displaystyle\frac{1}{(1-a)(1-b)} \geq \displaystyle\frac{27}{4}
\\
[/tex]

Substitution gives us:

[tex]
\\
\displaystyle\frac{1}{xy} + \displaystyle\frac{1}{yz} + \displaystyle\frac{1}{xz} \geq \displaystyle\frac{27}{4}
\\
[/tex]

[tex]
\\
\displaystyle\ x + y + z \geq \displaystyle\frac{27xyz}{4}
\\
[/tex]

Since x + y + z = 2,

[tex]
\\
\displaystyle\ x+y+z \geq \displaystyle\frac{27xyz}{(x+y+z)^2}
\\
[/tex]

And this simplifies to the AM-GM:

[tex]
\\
\displaystyle\ (x+y+z)^3 \geq \displaystyle\ 27xyz
\\
[/tex]

[tex]
\\
\displaystyle\frac{(x+y+z)}{3} \geq \displaystyle\ (xyz)^{1/3}
\\
[/tex]

By AM-HM,

[tex]
\displaystyle\frac{\frac{1}{a+bc} + \frac{1}{b+ac} + \frac{1}{c + ab}}{3} \geq \frac{3}{a+b+c+ab+ac+bc}
[/tex]
[tex]
\displaystyle\frac{1}{a+bc} + \frac{1}{b+ac} + \frac{1}{c + ab} \geq \frac{27}{3(1+ab+ac+bc)}
[/tex]

Thus 3(1+ab+ac+bc) must be less than 4.

[tex]
\displaystyle1+ab+ac+bc \leq \frac{4}{3}
[/tex]
[tex]
\displaystyle ab+ac+bc \leq \frac{1}{3}
[/tex]

We can prove the inequality above by AM-GM:
[tex]
\displaystyle\frac{a+b+c}{3} \geq \sqrt[3]{abc}
[/tex]
[tex]
\displaystyle\frac{1}{9} \geq \sqrt[3]{a^2 b^2 c^2}
[/tex]

But we know that

[tex]
\displaystyle\frac{ab+bc+ac}{3} \geq \sqrt[3]{a^2 b^2 c^2}
[/tex]

Thus ab+bc+ac must be approximately 1/3.

[tex]
\displaystyle\frac{1}{a+bc} + \frac{1}{b+ac} + \frac{1}{c+bc} \geq \frac{27}{3(1 + 1/3)}
[/tex]
[tex]
\displaystyle = \frac{27}{4}[/tex]

[quote="nr1337"]Thus ab+bc+ac must be approximately 1/3.
[/quote]

???

Since a + b + c = 1, then (a+b+c)^2 = 1, which means that a^2 + b^2 + c^2 + 2(ab + bc + ca) = 1. Since ab + bc + ca :le: a^2 + b^2 + c^2 by rearrangement, we have that the maximum of ab + bc + ca = 1/3, which is exactly what we need.

oh ok, i was just confused about the "approximately" part...

Actually, nrleet didn't prove that part correctly -- essentially, he said a > b and c > b, and then used that to draw a relationship between a and c.  Masamune's method does work, however.


probability
geometry
3D geometry
analytic geometry
induction
calculus
integration

Real numbers X_1, X_2,... are chosen randomly and independently from the interval (0,1). Let N be the smallest integer such that
X_1+...+X_N > 1

Find the average or expected value of N.

[hide="Answer"]$ e$[/hide]
[hide="Solution"]Consider the probability that $ N > n$. Visualize all possible values of $ x_{1}, x_{2},\ldots x_{n}$ in an $ n$-dimensional unit cube. The desired ($ n$-dimensional) volume (that is, the probability that $ N > n$) is all points in the unit cube with the sum of their coordinates less than one. Using induction and basic integral calculus, you can determine that the volume is $ 1/n!$.

Now the probability that $ N$ is some fixed number is $ P(N) \equal{} 1/(N\minus{}1)!\minus{}1/N!$. We want the expected value, so now we compute $ \sum_{i\equal{}1}^{\infty}{N\cdot P(N)}$. First, note $ P(1) \equal{} 0$, so eliminate that term.

$ \sum_{i\equal{}2}^{\infty}{N\cdot\left(\frac{1}{(N\minus{}1)!}\minus{}\frac{1}{N!}\right)}$

$ \sum_{i\equal{}2}^{\infty}{\frac{N}{(N\minus{}1)!}\minus{}\frac{1}{(N\minus{}1)!}}$

$ \sum_{i\equal{}2}^{\infty}{\frac{N\minus{}1}{(N\minus{}1)!}}$

$ \sum_{i\equal{}2}^{\infty}{\frac{1}{(N\minus{}2)!}}$

$ \sum_{i\equal{}0}^{\infty}{\frac{1}{N!}}$

You can recognize that the last thing is $ \boxed{e}$.[/hide]


This place seems sort of dead. Maybe it wasnt a great idea after all.  :(

I was the only person keeping it alive anyway, so I kind of gave up on that venture. By all means, post something. I'm still checking it regularly.

And I may throw something in later. Not competition stuff, as that seems to be failing miserably.

[quote="tokenadult"]I read this forum for every new post.[/quote]
It's just that it's a bit hard to find here, Arts and Sciences is a bit of a dark corner, many people probably don't know about it.

Go on! Continuez! Ga door!  :D


The point with coordinates $ (5,\minus{}11)$ is the midpoint of the segment with one endpoint at $ (9,1)$. Find the sum of the coordinates of the other endpoint.

By the fact that the average of two coordinates is their midpoint, it is clear that this is $ (5\minus{}(9\minus{}5),\minus{}11\minus{}(1\minus{}(\minus{}11))\equal{}(1,\minus{}23)\implies1\minus{}23\equal{}\boxed{\minus{}22}$.


conics
ellipse
geometry
calculus
calculus computations

is this differential equation volume-preserving?

$ x''\minus{}q(t)x\equal{}0$

If $ q$ is constant, then the phase space curves are ellipses whose eccentricity depends on $ q$ (in this case the areas are preserved). Now make $ q$ piecewise constant: as the eccentricity changes, so does the area.

but $ q(t)$ depends on $ t$ so wouldn't the volume be non-preserving?


Hi friends,
I am stuck with a problem which I am facing in video.

There are 2 functions:

w(n) = 1 , when n= f*Q + W + l ,
               (where, f is a integer from 0 to INF.
                         W is a constant integer.
                         Q is a constant integer(which is greater than W+L)
                         l can vary from 0 to L.)
       = 0, elsewhere

AND,

i(n) = 1 , when i = R+k*P,
               (where k is a integer from 0 to INF.
                         P is a constant integer
                         R can take any value b/w 1 to P-1.)

I want to find out for which values of 'f' and k, w(n) = i(n).
Variables are --- f, k and l.

So I write,

f*Q + W < R + k*P < f*Q+W+L        ------ (1)

equation 1 reduces to,

1 - (L/P) <= {(f*Q+W-R)/P},     ----------(2)
where {x} is the fractional part of x. (e.g. {3.44} = 0.44)

L is always less than P. W is always less than Q.

Can somebody tell me what will be the maximum 'f' (< P ofcourse) which can satisfy equation (2) .... given that R can vary from 1 to P-1 , and W is constant.

If, L=0, I can see that maximum 'f' will be P-1 which will satisfy the equation.
But as I increase L, maximum 'f' decreases. I want to find this upper bound on 'f'.

For example,

when Q= 450450
P=6758
W=468
L=0,
max bound on 'f' is P-1 = 6757 for some R.

But, as I make L=183, max 'f' that satisfies the equation comes down to 55.

Can somebody help me out??

Thanks,
Abhishek.

Please do not double-post ( http://www.mathlinks.ro/Forum/viewtopic.php?t=98836 ).
Could you type it in $\LaTeX$ please, it's very hard to read and understand (for a tutorial: see http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 ).


Hi guys could somebody give me a procedure to find the equation of the conic that passes through the points
$ (0,0), (0,5), (1,5), (\minus{}1,\minus{}3), (\minus{}3,\minus{}2)$  it is a hyperbola i guess

I think this belongs in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=149]High School Basics[/url] forum.

i didnt realize that this was mid-school..I'll post it there , thanks powerofpi


inequalities
algebra
polynomial
inequalities unsolved

Prove that for $ n$ distinct positive integers $ a_1,a_2,\ldots,a_n$,the following inequality holds:
\[ {a_1}^3\plus{}{a_2}^3\plus{}...{a_n}^3\ge (a_1\plus{}a_2\plus{}\ldots\plus{}a_n)^2\]

Simple proof can be based on induction. To prove the step from $ n\minus{}1$ to $ n$ it is enough to show that $ a^2_n\ge a_n \plus{}2(\sum\limits_{i\equal{}1}^{n\minus{}1} a_i)$. Now just use obvious inequalities $ a_{n\minus{}i}\le a_n \minus{}i$  and $ a_n\ge n$ :wink:

[quote="skywalkerJ.L."][quote="MeKnowsNothing"]Am I right that because of Newton's sums, one can reduce this to three variable inequality?[/quote]

Newton's sum?
What is the content of this theorem?

By the way,thanks for leshik, i finally prove it because of your hint.[/quote]
My idea was to consider to polynomial $ (x \minus{} a_1)\cdot\ldots\cdot (x \minus{} a_n)$. Then sum of $ a_i$'s and $ a_i^3$'s can be represented as a rational expressions with three unknowns.


Here's a poll, what language do you primarily use (like what do you think in)?

I am an ABC (American Born Chinese), so naturally I use English most often, but my Chinese isn't too bad.  :)

i talk in english and hindi, and depends what im thinking about for which lang i use, but i ususally dream in hindi.  wierd, isnt it.

Horrible classification of languages in your poll, they shouldn't be classified by geography, but of the roots. Finnish in your poll would be included with Spanish and they are very different.

[size=150][b]ROMANIAN[/b][/size]  :wink:

I am [i]Romanian[/i] so i mostly use [i]Romanian[/i].However i have taught myself a lil bit of English too  :) .

Your categorisation is a bit weird, for instance : Portuguese originated in Western Europe but it has way more speakers in Brazil than in Europe!
Maybe you should split it into Slavic (Russian, Polish,...), Germanic (German, Dutch, English, Swedish,..) and Romance languages (Italian, French, Spanish, Romanian,..)

Well I speak Dutch, with a Flemish accent of course.  It's a Germanic language very similar to German (but still different enough to require years of study  :( ) and it is spoken in the Netherlands, the northern region Flanders in Belgium (the majority of Belgians speaks Dutch) and Surinam.

i was born speaking romanian so i guess its my primary language, but my english is much better.

I was born in canada, so egyptian is my primary language! [color=white]Not realy, its english![/color]

[quote="funcia"]ENGLiSH FOR ALL! :!:  :!:  :!:[/quote]
 :| What do you mean?

i am stating i think that all of us(on this thread not on all of AoPS) use English mos of the time on AoPS(not in real life-I wouldn't know)

[quote="rem"]It's [b]Russian[/b] for me.
because this forum is nearly all in English language, there are more ppl with primamry language being Englsih. However there is still a very wide diversity on this forum. :)[/quote]
I wish I could speak Russian.  I actually tried this summer, but I didn't get much further than the Cyrillic alphabet (and I'm still unsure about that weird "hard sign" :( )  Right now, even when including Latin, I'm only familiar with Germanic and Romance, not Slavic. :(

No Southeastern Asian?  :huh:

Edited: Forget the Asian :P

[quote="BoesFX"]No southeastern?  :huh:[/quote]  What exactly do you mean?

:o

It's "Afrikaans"!  Please change it if possible. :!:

...and it's "your primary language" not "you're"

"your" is a possessive pronoun, while "you're'"="you are"

[quote="rem"]@fredbel: by "hard sign" do you mean "ъ"?

Ruussian is hard because the endings of nearly all words change, and they have many different forms, unlike french, or english, or german.[/quote]
Yes I meant that sign.  Very hard (for me at least) is that that hard sign looks like your symbol for "b" and your symbol "v"! Confused! :(  :wink:
But trust me , French doesn't make much sense, considering its endings.

[quote="rem"]No, French is ok I think. The hard thing about the Fench language is that it has very many exceptions, making it hard to memorize it. Like for every rule, there is always an excpetion.[/quote]
I don't know how much you know about French, but I don't speak any Russian so I definitely can't compare.

But... in French :

the "eu" in "j'ai eu" does not sound like the "eu" in "ils veulent", and "je veux" sounds different as well

the "ai" in "parlerai" does not sound like the "ai" in "aide" but does sound like the "ais" in "parlerais"

"aider", "aidez" and "aidé","aidée", "aidés" and "aidées" all sound the same

the "s" in "des" in "des livres" is not pronounced but it is pronounced in "des amis" but there it becomes a "z"

there is a difference in pronunciation between "haine" et "homme", but the spelling does not indicate that

final consonants in "ils parlent", "un grand chien", "vous parlez", "il parlerait" are all unnecessary...

....

Are you telling me that Russian is even harder. :(  :!:
(In my opinion the French spelling is a spelling that didn't keep up.  It has so many ways to write the same sound, but on the other hand the spelling doesn't differ when it really really needs to....)

[quote="Derek"]East Asian languages like Cantonese and Mandarin[/quote]

Yeah but those you don't have to deal with lots and lots of verb conjugations (unlike English)


geometry proposed
geometry
U sviti matematyky

Let $ O$ be the midpoint of the side $ AB$ of triangle $ ABC.$ Points $ M$ and $ K$ are chosen at sides $ AC$ and $ BC$ respectively such that $ \angle MOK\equal{}90^{\circ}.$ Find angle $ ACB,$ if $ AM^2\plus{}BK^2\equal{}CM^2\plus{}CK^2.$

[hide="Solution"]Let point D such that O is the midpoint of MD, then KM=KD, AM=BD.
$ CM^2 \plus{} CK^2 \minus{} KM^2 \equal{} BK^2 \plus{} BD^2 \minus{} KD^2\Rightarrow \angle C \equal{} \angle KBD \equal{} \angle A \plus{} \angle B\Rightarrow\angle C \equal{} 90^o$.[/hide]


analytic geometry
combinatorics unsolved
combinatorics

Let $ n$ be a positive integer and let $ T$ denote the collection of points $ (x_1, x_2, . . . , x_n) \in R^n$ for which there exists a permutation $ \sigma$ of $ 1, 2, . . . , n$ such that $ x_{\sigma( i )} \minus{} x_{\sigma( i \plus{} 1 )} \ge 1$ for each $ i \equal{} 1, . . . , n\minus{}1$. Prove that there is a real number $ d$ satisfying the following condition: For every $ (a_1, . . . , a_n) \in R^n$ there exist points $ (b_1, . . . , b_n)$ and $ (c_1, . . . , c_n)$ in $ T$ such that, for each $ i \equal{} 1, . . . , n$, $ a_i \equal{} (b_i \plus{} c_i)/2$, $ |a_i \minus{} b_i| \le d$, and $ |a_i \minus{} c_i| \le d$.

Here's something I think works.

Rewording, $ T$ consists of the sequences that, when sorted in increasing order, increase by at least $ 1$ at every step.  In particular, this implies that the difference between the largest and smallest coordinates of an element in $ T$ must be at least $ n \minus{} 1$.  This means the $ d$ must be allowed to depend on $ n$ (no point of $ T$ is within distance $ n/2$ of $ (0, 0, \dots, 0)$.

Let $ x \equal{} (1, 5, 9, \dots, 4n \minus{} 1)$.  For an integer $ t$ let $ b_t \equal{} \frac {1}{2}(a \plus{} tx)$ and $ c_t \equal{} \frac {1}{2} (a \minus{} tx)$ (addition and subtraction being done pointwise).   It suffices to show that there is some $ t \leq d$ such that both $ b_t$ and $ c_t$ are in $ T$.   Call a pair $ (i,j)$ is a [b]b-witness[/b] (respec. [b]c-witness[/b]) for $ t$ if the $ i^{th}$ and $ j^{th}$ coordinates of $ b_t$ (respec. $ c_t$) differ by at most $ 1$ in absolute value.

By construction, each pair can be a $ b \minus{}$witness for at most one integer $ t$, and the same holds for $ c$.  Therefore there is some integer $ 0 \leq t \leq 2\binom{n}{2}$ which does not have a witness.  But this implies that both $ b_t$ and $ c_t$ are in $ T$.


A rectangle with a diagonal length of $x$ is twice as long as it is wide. What is the area of the rectangle?

A. $\frac{1}{4} x^2$
B. $\frac{2}{5} x^2$
C. $\frac{1}{2} x^2$
D. $x^2$
E. $\frac{3}{2} x^2$

I edited it for you... did I do it right?

Put your mouse cursor over the latex things to see the codes I used, you have to put dollar signs around latex codes to activate them. Look on the left toolbar for more info.

Yea it is good but I took out the last two lines so people won't keep answering.  I am trying to learn latex so I can use it.

I have a question solverfalcon.  Why is this a question in a contest for grade 12s? Don't you think they should be more challenging?  Thanks in advance.

Masoud Zargar


The Table of Contents is [url=http://www.artofproblemsolving.com/Books/Precalculus/toc.pdf]here[/url].  Does that answer your question?

[quote="benzi455"]What exactly does it mean that the Precalculus text "covers vectors and matrices"?[/quote]

Vectors and matrices are included and taught about in the book...

@rrusczyk: Cool!  I'm impressed.

@AwesomeToad: Yes, I got that much.


algebra
polynomial
search
algebra unsolved

Find all polynomials $P(x)\in \mathbb Z[x]$ such that for any $n\in \mathbb N$, the equation $P(x)=2^{n}$ has an integer root.

[quote="cancer"]Find all polynomials $P(x)\in Z[x]$ such that for any $n\in N$, the equation $P(x)=2^{n}$ has an integer root.[/quote]
. http://www.mathlinks.ro/Forum/viewtopic.php?search_id=834109250&t=26076


geometry
geometric transformation
reflection
trigonometry
ratio
perpendicular bisector
trig identities

Here's an interesting problem from an old Euclid contest that I dug up:

For a given triangle ABC, angle C = angle A + 60 degrees. If BC = 1, AC = r and AB =$r^{2}$, where $r>1$, prove that $r< \sqrt{2}$

[hide="hmm... i get r less than the square root of 2"]
Reflect $B$ over the perpendicular bisector of $\overline{AC}$ and let its image be $D$. Since the image of $\overline{BC}$ is $\overline{AD}$ and the image of $\overline{AB}$ is $\overline{CD}$, we have $AD=BC=r$ and $CD=AB=r^{2}$. Ptolemy's theorem on quadrilateral $ACBD$ (since $\triangle DAB\cong \triangle DCB$, $\angle DAB=\angle DCB$ so it's cyclic) gives $1\cdot BD+r\cdot r=r^{2}\cdot r^{2}$, or $BD=r^{4}-r^{2}$.

Also, since $\triangle ACD\cong \triangle CAB$, $\angle ACD=\angle CAB$ and $\angle BCD=60^\circ$. Now the Law of Cosines on $\triangle BCD$ gives
\[(r^{4}-r^{2})^{2}=r^{4}+r^{2}-r^{3}\]
Since $r\neq 0$, this results in
\[0=r^{6}-2r^{4}+r-1=r^{4}(r^{2}-2)+r-1\]
Suppose $r\geq \sqrt{2}$. Then $r^{4}(r^{2}-2)+r-1\geq r-1>0$, a contradiction. Therefore $r<\sqrt{2}$.
[/hide]

[quote="Glen Mann"]foreword: i suck at proofs, but am practicing
[hide]
By the sine law we have
$\frac{\sin A}{1}= \frac{\sin A+60}{r^{2}}$
$r^{2}\sin A = \sin(A+60) = \sin A\cos60+\cos A\sin60 = \frac{\sin A+\sqrt{3}\cos A}{2}$
$2r^{2}\sin A-\sin A = \sqrt{3}\cos A$
$(2r^{2}-1)\sin A = \sqrt{3}\cos A$
$\frac{\sin A}{\cos A}= \frac{\sqrt{3}}{2r^{2}-1}$
$\tan A = \frac{\sqrt{3}}{2r^{2}-1}$

Using $\sqrt{2}$ as our value for r we obtain

$\tan A = \frac{\sqrt{3}}{3}$
Which leaves us with our angle value for A to be $30^{\circ}$

Since A = 30, B = 90, and thus C = 60, so we are left with a 30-60-90 triangle.

$r^{2}$ must be our base and also the longer side, so it is AB
r is side A C
CB is our opposite angle and it is also 1

with $r = \sqrt{2}, r^{2}= 2$

The ratios of the sides of this 33-60-90 triangle are $1: \sqrt{3}: 2$

if AB = 2, then AC must equal $\sqrt{3}$, but AB = $r^{2}$ and AC = $r$.
This implies that $2 = \sqrt{3}$ which is not the case and as such r must be greater than or less than $\sqrt{2}$

Checking the greater than case for $r > \sqrt{2}$ i pick $r = 2$
as such
$\tan A = \frac{\sqrt{3}}{7}$ which gives us a rough estimate of angle A as $13.897^{\circ}$, meaning B = $73.897^{\circ}$ and C = $92.206^{\circ}$

Checking via sinelaw we find that
$\frac{\sin 73.897}{\sin 92.206}= 2$
Which is not the case, and as such r cannot be greater than $\sqrt{2}$
[/hide][/quote]
That doesn't quite work. You need to show that it fails for [b]all[/b] $r\geq \sqrt{2}$. Showing that a specific value fails isn't enough because there's only one value of $r$ that works. (Besides, $r\geq \frac{1+\sqrt{5}}{2}$ is absurd even without $C=A+60^\circ$.)

[quote="Glen Mann"][quote="scorpius119"]That doesn't quite work. You need to show that it fails for [b]all[/b] $r\geq \sqrt{2}$. Showing that a specific value fails isn't enough because there's only one value of $r$ that works. (Besides, $r\geq \frac{1+\sqrt{5}}{2}$ is absurd even without $C=A+60^\circ$.)[/quote]
where did I say $r\geq \frac{1+\sqrt{5}}{2}$?

how would I go about making my proof proper? am I atleast in the right direction?  :oops:[/quote]

-when you looked at $r=2$, which was greater than $\frac{1+\sqrt{5}}{2}$, to show that 2 failed

-you're almost there. From $\tan A$, you get $\cos A$, so you could use the Law of Cosines to get a giant equation for $r$.


geometry
analytic geometry
geometry unsolved

Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD; AB=3CD; and the area of the quadrilateral is 4. If a circle can be drawn touching all the sides of the quadrilateral, find its radius.

Let CD=a, AB=3a, AD=2b. AD has to be twice the radius of the circle because the center of the circle is equidistant from CD and AB, so we wish to find b. From the area, we have that $ ab \equal{} 1$. Now, set up a coordinate system with the origin at A. The center of the inscribed circle is located at (b,b), and it's distance from BC must be b. BC is the line through $ (a,2b)$ and $ (3a,0)$, from which we find its equation is $ y \minus{} 3a \equal{} \minus{} \frac {b}{a}x$, or $ \minus{}bx\plus{}ay\minus{}3a^2\equal{}0$. So using the formula for the distance from a point to a line, we have
$ \frac { \minus{} b^2 \plus{} ab \minus{} 3a^2}{\sqrt {a^2 \plus{} b^2}} \equal{} b$
$ (b^2 \plus{} \frac {3}{b^2} \minus{} 1)^2 \equal{} b^4 \plus{} 1$
$ \frac {3}{b^4} \plus{} 6 \equal{} 2b^2 \plus{} \frac {6}{b^2}$
The answer is the solution to this, my calculator gives 1.3392911......

Call M, N and P the contact points of AB, BC, CD with the incircle, I the center of this circle, let r be its radius and PC=x.
From the area formula we get r(r+x)=1  ( 1 ), then see that  BM=2r+3x . See that BIC is a rightangled triangle and get
r.r=x(2r+3x)  ( 2 ) (from NI.NI = BN.CN). But (2) is equivalent to r.r+r.x=3x.(r+x) or, with (1), 3x(r+x) = 1 ( 3 ). From (1) and (3) we get r=3x which, replaced to (1) gives 12x.x = 1, or r=sqrt(3)/2 .

Best regards,
sunken rock


cf249
mathelete
chenshi
Wickedestjr
ernie
aleph0
evilhamster

Each player has 3 grenades in their arsenal. There is no concept of turns. No consecutive turns. When you go, you have five options. Do only one of them.

1. Choose another player, and subtract points. The number of points subtracted is the post number.

2. Same as above, but add points instead.

3. Throw a grenade. Convert the post number to binary, then add up the digits. The grenade will go off that many posts later. If it does not go off in 4 days, the grenade is defused.

4. Bump the grenade to the player below you on the list. If you are at the bottom, it will go to the top.

5. Pick up a defused grenade. It goes into your arsenal.

Everybody starts off with 0 points, and they can have a negative number of points. After 256 posts, the game ends. When the grenade goes off, whoever it is by falls for the rest of the game. If a person brings another person's points back to 0, then that person earns a mushroom. If a person doubles a positive number of points, that person earns a star. These are the sixth and seventh options. A mushroom gives yourself points. When a star is used on post x, the person is invincible till post 2x. An invincible person automatically bumps a grenade without posting.

We shouldn't start unless we make sure that everybody knows that they are playing.

I don't even think that ernie has logged in since the end of the first game.

And mathlete, stop changing rules in the middle of a game.

cf249
mathelete
chenshi
Wickedestjr
ernie
aleph0 2 grenades
evilhamster Grenade 5

The "grenade gambit"

cf249
mathelete
chenshi 2 grenades
Wickedestjr
ernie Grenade 6
aleph0 2 grenades
evilhamster Grenade 5

cf249 5
mathelete
chenshi 2 grenades
Wickedestjr
ernie Grenade 6
aleph0 2 grenades
evilhamster PWNT

cf249 5
mathelete 6
chenshi 2 grenades
Wickedestjr
ernie PWNT
aleph0 2 grenades
evilhamster PWNT

cf249 5
mathelete 6
chenshi 2 grenades
Wickedestjr 7
ernie PWNT
aleph0 2 grenades
evilhamster PWNT

cf249 5
mathelete 6
chenshi 1 grenades
Wickedestjr 7
ernie PWNT
aleph0 2 grenades GRENADE 8
evilhamster PWNT

cf249 PWNT
mathelete 6
chenshi 1 grenades
Wickedestjr 7
ernie PWNT
aleph0 2 grenades
evilhamster PWNT

actually, aleph0 is PWNT as chenhsi post on a number where $ 2^x$ equals the number and thus goes off on the same post.

THREE GRENADES ARE TOO MANY

This is unfair.

I'm never on after 8 pm EST.

Yeah, aleph, you are dead

Basically, you started your turn, then died.

cf249 5
mathelete 6
chenshi 11: 1 grenades
Wickedestjr 7
ernie PWNT
aleph0 PWNT
evilhamster PWNT

Wait but I bumped the grenade and then it exploded. It isn't fair to have automatic KO.

Why do you think I waited to use my grenade? If I couldn't do that, I want to take my greande back.

Aleph is still in, because you can bump the grenade at the last minute. so cf249 is dead.

WHAT???

Current scores:

cf249 5
mathelete 6 GRENADE 19
chenshi FORFEIT
Wickedestjr 7
ernie PWNT
aleph0 PWNT
evilhamster PWNT

After I throw the greanade, I quit. Permanently. And don't add me on to your future games unless you address all issues before game, and you give us notice at least 24 hours beforehand.

cf249 PWNT
mathelete 6
chenshi FORFEIT
Wickedestjr 7, GRENADE 19
ernie PWNT
aleph0
evilhamster PWNT

Bump grenade

cf249 PWNT
mathelete 6 GRENADE 17
chenshi FORFEIT
Wickedestjr 7, GRENADE 19
ernie PWNT
aleph0
evilhamster PWNT

[quote="aleph0"]cf249 PWNT
mathelete 6
chenshi FORFEIT
Wickedestjr 7, GRENADE 19
ernie PWNT
aleph0
evilhamster PWNT[/quote]
You cannot do that. You cannot make a grenade go off at the same post.

cf249 PWNT
mathelete: 6
chenshi FORFEIT
Wickedestjr PWNT
ernie PWNT
aleph0: -20
evilhamster PWNT

No more grenades, because it is one-on-one. -20 from you

[quote="chenhsi"][quote="mathelete"]No more grenades, because it is one-on-one.[/quote]

It does not say that in the rules.[/quote]
But we can stall and defuse the grenades. Since no third person can get in the way, grenades are out. Convert them to points. So:

mathelete: 9
aleph0: -17


function
calculus
integration
search
algebra unsolved
algebra

Find all real values of $\alpha$, for which there exists one and only one function $f: \mathbb{R} \mapsto \mathbb{R}$ and satisfying the equation \[ f(x^2 + y + f(y)) = (f(x))^2 + \alpha \cdot y \] for all $x, y \in \mathbb{R}$.

Umm.. for $\alpha=0$ we can have $f(x)=0$ or $f(x)=1$, so $\alpha=0$ isn't good.

[quote="Namdung"]Grobber + Harazi = Great pair[/quote]

:D Not really.. Harazi did all the work, remember? :D

that's the best role, grobber !
harazi, don't be sad; your sol was quite boring anyway (without offense of course)

Well, my dear Arthur, I expect a truly genial and non-boring solution from you!

[quote="harazi"]Oh, boy, this is a hell of a problem! I didn't think at the case a=0 but I think it is not so difficult as the opposite one. So, let us suppose that a is not 0. Take $ y=-\frac{f^2(x)}{a}$ and you will find that $ f( x^2-\frac{f^2(x)}{a}+f(-\frac{f^2(x)}{a}))=0$ (*). Now, suppose that $ f(u)=0$. Then take $ x=0, y=u$ and you will find that $ u=-\frac{f^2(0)}{a}$. Thus, from this observation and (*) we find that $ f(-\frac{f^2(x)}{a})=\frac{f^2(x)}{a}-x^2-\frac{f^2(0)}{a}$ (**). Then take $ -\frac{f^2(y)}{a}$ instead of y and use (**). You will find that $ f(x^2-y^2-\frac{f^2(0)}{a})=f^2(x)-f^2(y)$ (***). Thus, $ f(x^2-\frac{f^2(0)}{a})=f^2(x)-f^2(0)$ and so using this and (***) we find that $ f(x^2-y^2-\frac{f^2(0)}{a})=f(x^2-\frac{f^2(0)}{a})-f(y^2-\frac{f^2(0)}{a}) $ (****). Take $ g(x)=f(x-\frac{f^2(0)}{a}$ and you will find that $ g(x^2-y^2)=g(x^2)-g(y^2) $ (*****). From here it easily follows that g is odd and after that that g is additive. Thus, $ f(x^2-\frac{f^2(0)}{a})+f(y^2-\frac{f^2(0)}{a})=f(x+y-\frac{f^2(0)}{a}) $ and then $ f(x)+f(y)=f(x+y+\frac{f^2(0)}{a}) $ (******). Put here y=0 and you will find that $ f(x+\frac{f^2(0)}{a})=f(x)+f(0)$ and then come back in (******) to find that $ f(x)+f(y)=f(x+y)+f(0) $. Take y=0 in the initial relation to find that $ f(x^2+f(0))=f^2(x)$. Thus, $ f^2(x)=f^2(-x)$. But $ f(x)+f(-x)+2f(0)$. So, if f(0) is not 0, then the two relations above say that f is constant, false (a is not 0). Thus, f(0)=0 and so f is additive. Then, from $ f(x^2)=f^2(x)$ making x=x+y we find that $ f(xy)=f(x)f(y)$. It's classical to deduce that f is either 0 (false) or f(x)=x for any x. In this case a=2. So, if I'm not wrong for any a different from 0 and 2 we cannot find such functions. For a=2 there is a unique function and for a=0 God knows.[/quote]
[quote="Namdung"]Grobber + Harazi = Great pair

One question remains: Is there non-constant function from R to R satisfying

              f(x 2 + y + f(y)) = f(x) 2 ?[/quote]
the answer is no non-constant function satisfy this equation
$f(x^2 +y +f(y)) = (f(x))^2 \forall x, y \in R   (1)$

$\rightarrow f(x^2 + f(0)) = (f(x))^2 \forall x \in R    (2)$

and $f(y+f(y))=(f(0))^2 \forall y \in R    (3)$

so that we have $x^2 +y+f(y) +f(x^2 +y+f(y)) = (f(x))^2 + x^2 +y+f(y) \forall x, y\in R$

$\rightarrow f(x^2 +y+f(y) +f(x^2 +y+f(y))) = f((f(x))^2 + x^2 +y+f(y)) \forall x, y \in R$

using (1) andd (3) we have:
$(f(0))^2 = (f(\sqrt{(f(x))^2 + x^2})^2 \forall x \in R$

Let x=0 then $(f(0))^2 = (f(\sqrt{f(0)^2}))^2.$

From (2) $\rightarrow (f(-x))^2=(f(x))^2 \forall x $

so $(f(f(0)))^2 = (f(\sqrt{f(0)^2}))^2 = f(0)^2$

$\rightarrow f(0)^2 = f(f(0))^2$

from (3) $\rightarrow f(f0))  = f(0)^2$

so $f(0)^4 = f(0)^2$

Thus $f(0) \in \{0,1,-1\}$

We will prove that $f(0) \geq 0$

suppose $f(0)< 0$
then there exist $x_1= \sqrt{-f(0)}$
from (2) let x= $x_1$ then
$f((\sqrt{-f(0)})^2 + f(0)) = f((\sqrt{-f(0)}))^2 \geq 0$

so $f(0) \geq 0 (>< suppose f(0) < 0)$

we have f(0) $\geq 0 \rightarrow f(0) \in \{0,1\}$
I) if f(0)=0
Let A = {x| f(x) =0}
from (1) we have" if x, y $\in$ A then $x^2 + y \in A$"
because f(0)  = 0 and (2)$\rightarrow f(x^2)=f(x)^2 \forall x$
$\rightarrow$ if $ x\geq 0$ and $x \in A$ then $\sqrt{x}, x^2, -x \in A$

$\rightarrow$ if $ x\geq 0, x, y \in A$ then $\sqrt{x} \in A$ so $ (\sqrt{x})^2 + y \in A$

so if x, y $\in$ A x\geq 0 then x+ y$\in$A

 from (3)we have f(y + f(y))= 0, so y + f(y) $\in$ A $\forall y \in R$.(4)
so -(y+f(y)) also $\in$ A

$\rightarrow \forall x \in R$ there exists number a $\in A$ satisfying x>a.

$f(x)=f((\sqrt{x-a})^2+ a +f(a))$   )(because f(a)=0)
$\rightarrow f(x) = (f(\sqrt{x-a}))^2 \geq$ so f(x) $\geq$ 0 $\forall x \in R$

But $(f(x))^2 =f(x^2)=(f(-x))^2$ so f(x) = f(-x) $\forall$ x.

(1) let y$\leq$ 0 and x=$\sqrt{-y + f(-y)}$ then
${f((\sqrt{-y+ f(-y)})^2 + y + f(y)) = f(\sqrt{-y + f(-y)}}^2$
$\rightarrow f(f(-y)+f(y))=f(-y + f(-y)) = 0$ $\forall y \leq 0$
so f(2f(y))= 0 $\forall y \leq 0$
because f(x) = f(-x) then f(2f(y)) = 0 $\forall y \in R$
so we have 2f(y) $\in A \forall y \in R$

in general, A satisfy:
i. 0 $\in$ A
ii. if x, y $\in$ A, x>0 then x+ y$\in$ A
iii. x in A then -x in A
iv. y+f(y) $\in$ A $\forall y \in R$
v. 2f(y) $\in$ A $\forall y \in R$

so $\forall y \geq 0$ we have $-2f(y/2) \in A$ and $\frac{y}{2} + f(\frac{y}{2}) \in A$

using ii. we have $2(\frac{y}{2} + f(\frac{y}{2})) \in A$ or $y + 2f(\frac{y}{2}) \in A$
using ii. and $y + 2f(\frac{y}{2}), -2f(\frac{y}{2}) \in A$ so $y + 2f(\frac{y}{2})+ (-2f(\frac{y}{2})) \in A$
$\rightarrow y \in A \forall y \geq 0$
because iii. we have y $\in A$ $\forall y \in R$

$ \rightarrow f(y) = 0$ $\forall y \in R$

retrying we have function f(y) = 0 satisfied original equation.

II) if f(0)=1
....
I hope you do this by yourselves. It 's done like I) but not the same.

----------------------------------------------------

I'm right. Do you think that??

Well, this problem is much more difficult than the problem given in the TST. Congratulations, the solution is very beautiful and hard.

Here is my approach:
Step 1: f(x) $ \geq $ 0 for all x.
Proof: Suppose f(y)= - $ t^2 $ with positive t.
We have f( $ t ^ 2 + y + f (y ) $ ) = f(y) = - $ t^2 $ < 0, false.
From step 1 ,since $ f^ 2 ( x ) = f^2 ( -x ) $ we have f(x)=f(-x).
Step 2: For all a $\geq $ f(y) - y we have f(a)=f(a+2y)
Proof: For all x,y :
$ f ( x^ 2 + y + f ( y ) ) = f ( x^ 2 - y + f ( y ) ) $ , so step 2 follows.
Step 3: For all b > a > f(1)-1, f(a)=f(b)
Proof: Since b >a > f(1)-1, from step 2 we have f(a+2n)=f(a) and f(b+2n)=f(b) for all natural n. Choose n sufficently large so that a+2n > f( $\frac {b-a} {2} ) - \frac {b-a} {2} $ , with this n we have f(a+2n)=f(a+2n+b-a)=f(b+2n), so
f(a)=f(b). Note that in this step we can replace 1 by an abritary positive real r and we still have the result.
Step 4: f(x)=0 for all x or f(x)=1 for all x.
Proof: From step 3 and our initial equation, clearly $f ^ 2 ( a )$ = f(a) for all
a>f(1)-1, so f(a)=1 or 0. But from step 3, it follows that f(a)=0 for all a>f(1)-1
OR f(a)=1 for all a>f(1)-1.
Choose sufficently large a such that f(a)-a <0.
Now we repeat the reasoning of step 3 to get f(x)=f(y) for all x,y>f(a)-a
In particular, f is constant for all non-negative reals. But since f is an even function, f must be constant and we obtain the desired result.

There is still a question:"Given a positive integer k, find all functions from R to R for which there exists a constant $\alpha$ sothat:
$f(x^k + y +f(y)) = (f(x))^k + \alpha y$  $\forall x, y \in R$

if k is even we easily solve this pr. (try!!!)
Can anyone solve with odd integer k?

Let a>1.Find all f:R->R satisfying the condition:
$f(x^2-ay+f(y))=f(x)^2$
It's is interesting,try to solve it

I'm probably being stupid but why does $f(f(y))+f(y)-2y=0$ imply $f(y)=y$?

I meant in the context of his solution because that equation obviously has more solutions.

If $f(f(y))+f(y)-2y=0\forall y<0$ and $f(y)<0\forall y<0$ then easy see that $f(y)=y\forall y<0$.

I still don't get it  :blush: .
Could you explain the steps in detail?

Forall $y_{0}<0$, fix it. Put $u_{0}=y_{0},u_{n+1}=f(u_{n})(n\geq 0)$, then $u_{n+2}+u_{n+1}-2u_{n}=0\forall n\geq 0(*)$ and $u_{n}<0\forall n\geq 0(**)$.

From $(*)$ we have there are $a,b\in\mathbb{R}$ such that $u_{n}=a+b(-2)^{n}\forall n\geq 0$, from this and $(**)$ we have $a=y_{0}$ and $b=0$ therefore $f(y_{0})=y_{0}$.  :wink:

[quote="behemont"]To N.T.TUAN: Why $f(-b)=0$ ?[/quote]
Sorry, but where you saw it? Please quote here!  :maybe:

[quote="N.T.TUAN"]Find all real values of $\alpha$, for which there exists one and only one function $f: \mathbb{R}\mapsto \mathbb{R}$ and satisfying the equation
\[f(x^{2}+y+f(y)) = (f(x))^{2}+\alpha \cdot y \]
for all $x, y \in \mathbb{R}$(*).

This is my solution.
Answer $\alpha =2$.
A) If $\alpha =0$ We have at least two functions satisfying, therefore $\alpha =0$ not satisfy.

B) If $\alpha \not = 0$. Suppose $f$ is function satisfy, because $\{f^{2}(x)+\alpha y|x,y\in\mathbb{R}\}=\mathbb{R}$(note: $\alpha\not =0$) we have $f$ is surjection (1).

By (1) exist $b$ such that $f(b)=0$, from (*) we have $f(x^{2}+b)=f^{2}(x)+\alpha b$ and $f(x^{2}-b)=f^{2}(x)-\alpha b$ , therefore $2\alpha b=f(x^{2}+b)-f(x^{2}-b)$, choose $x=0$ here we have $b=0$(note:$f(-b)=0$). So $f(x)=0$ iff $x=0$ (2).

In (*) choose $y=0$ we have $f(x^{2})=f^{2}(x)$ (3).

In (3) choose $x=1$ and by (2) we have $f(1)=1$ (4).

In (*) choose $y=1$ we have $f(x^{2}+2)=f(x^{2})+\alpha$ (5). In (5) choose $x=0$ then $\alpha=f(2)$. Therefore $\alpha^{2}=f^{2}(2)=f(4)=f(2)+\alpha=2\alpha$, so $\alpha =2$.

C)Now, for $\alpha =2$ we will prove exist unique function satisfy (*). Suppose $f$ is a function satisfy (*), then $f$ has (1),(2),(3),(4),(5).

In (*) choose $x=0$ we have $f(x+f(x))=2x$ (6).

We have $f(x)=-f(-x)$(7), in fact, if $b=f(a)=f(-a),a>0$ then $b>0$. In (*) choose $y=a$ and $y=-a$ we have $f(x^{2}+a+b)=f^{2}(x)+2a$(8) and $f(x^{2}-a+b)=f^{2}(x)-2a$ (9). In (8) choose $x=0$ we have $f(a+b)=2a$, in (9) choose $x=\sqrt{a+b}$ we have $f(2b)=0$ therefore $b=0$. Contradiction!

For any $y<0$ ,in (6) choose $x=\sqrt{-y}$ we have $f(f(y))+f(y)-2y=0$,  therefore $f(y)=y$ for all $y<0$. By (7) we have $f(x)=x$ for all $x\in\mathbb{R}$.
... :D[/quote]

Up there..  :wink:

[quote="radio"][quote="harazi"] Thus, f(0)=0 and so f is additive. Then, from $ f(x^{2})=f^{2}(x)$ making x=x+y we find that $ f(xy)=f(x)f(y)$. It's classical to deduce that f is either 0 (false) or f(x)=x for any x.[/quote]

How can we deduce it?[/quote]
If $ f(x+y)=f(x)+f(y)\forall x,y\in\mathbb{R}$(1) and $ f(xy)=f(x)f(y)\forall x,y\in\mathbb{R}$(2) then $ f=0$ or $ f=x$. In fact, By (1) $ f(r)=ar\forall r\in\mathbb Q$, by (2) and (1) : $ f$ is monotone , so $ f(x)=ax\forall x\in R$, use (2) we have $ a=0$ or $ 1$.

Oh!This old thred was brang back to life....
  Note that :this problem was proposed based on a problem from Sant Peterbough 2001 but asking for integral set.Look at here:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1578574219&t=28981

We find all such functions. Let $P(x,y)$ be the assertion $f(x^2+y+f(y))=f(x)^2+\alpha y$. Clearly $f(x)=\pm f(-x)$. Consider $\alpha =0$. If $f(x)<0$, then $P\left (\sqrt{-f(x)},x\right )$ is contradiction. So $f(x)\geqslant 0$, whence $f$ is even. Since $f(x+y+f(y))=f(x-y+f(y))$ for all $x\geqslant 0$, it easily follows $f\equiv 0$ or $f\equiv 1$.

For $\alpha \neq 0$, $f$ is surjective. Let $f(k)=0$, then also $f(-k)=0$. And $P(0,k)$ and $P(0,-k)$ shows $k=0$. So $f(0)=0$. $P(x,0)$ gives $f(x^2)=f(x)^2$ and $P(0,x)$ gives $f(x+f(x))=2x$. $P(x,-f(x)^2/\alpha )$ implies $f(-f(x)^2/\alpha)=f(x)^2/\alpha-x^2$, so $P(x,-f(y)^2/\alpha)$ implies $f(x^2-y^2)=f(x^2)+f(-y^2)$. In particular, $f$ is odd and so additive. Since $f:\mathbb R^+\to \mathbb R^+$, $f$ is linear. And hence (checking), $f\equiv \text{Id}$, which works.

[quote=ZETA_in_olympiad] Let $P(x,y)$ be the assertion $f(x^2+y+f(y))=f(x)^2+\alpha y$. Clearly $f$ is even or odd..[/quote]
No. $f(x)^2=f(-x)^2$ is not enough to conclude that $f(x)$ is even or odd. Maybe $f(-x)=f(x)$ for some $x$ and $f(-x)=-f(x)$ for some others.

[quote=pco]
No. $f(x)^2=f(-x)^2$ is not enough to conclude that $f(x)$ is even or odd. Maybe $f(-x)=f(x)$ for some $x$ and $f(-x)=-f(x)$ for some others.[/quote]

Yes, I meant this by what I said :)

First note that if $\alpha=0$ then $f(x)=0$ and $f(x)=1$ work so $\alpha \ne 0$.
Let $P(x,y)$ the assertion of the given F.E.
Clearly $f$ surjective so by $P(0,c)$ where $f(c)=0$
$$c=-\frac{f(0)^2}{\alpha} \implies c \; \text{is unique}$$
$P \left(x,-\frac{f(x)^2}{\alpha} \right)$
$$f \left(-\frac{f(x)^2}{\alpha} \right)=\frac{f(x)^2}{\alpha}-x^2-\frac{f(0)^2}{\alpha} \overset{x=c}{\implies} \frac{f(0)^4}{\alpha^2}=-\frac{f(0)^2}{\alpha} \implies f(0)=0 \; \text{or} \; \alpha=-f(0)^2$$
Note that from here we also get that if $f(a)^2=f(b)^2$ then $a^2=b^2$
[b]Case 1:[/b] $\alpha=-f(0)^2$
From here $f(1)=0$ so by $P(1,0)$
$$f(1+f(0))=0 \implies f(0)=0 \implies \alpha=0 \; \text{contradiction!!}$$
[b]Case 2:[/b] $f(0)=0$
$P(x,0)$
$$f(x)^2=f(x^2) \implies f(x)>0 \; \forall x>0$$
$P\left(0,-\frac{f(x)^2}{\alpha} \right)$
$$f(-x^2)=-f(x)^2=-f(x^2) \implies f \; \text{odd}$$
$P\left(\sqrt{x^2+y^2},-\frac{f(y)^2}{\alpha} \right)$
$$f(x^2+y^2)=f(x^2)+f(y^2) \implies f(u+v)=f(u)+f(v) \; \forall u,v  \in \mathbb R^+ \implies f(u)=k \cdot u \; \forall u \in \mathbb R^+$$
Its easy to check that $k=1$ and since $f$ is odd we get that $f(x)=x \; \forall x \in \mathbb R$ as the unique solution so this implies that $\alpha=2$ yelds only one solution thus we are done :D


the title is sort of self-explanatory.. please give me hints to this proof and tell me how to approach it..thank!  :)

[b]Prove that it is impossible for three consecutive sqres to sum to another perfect square[/b]

enjoy!

Answer

Let the three consecutive perfect squares be $ m^{2}, (m+1)^{2}, (m+2)^{2}$

We know that $ m$, $ m+1$, $ m+2$ must be congruent to $ 0$, $ 1$, $ 2$ $ (mod\,3)$ in some order

Hence their squares must be congruent to $ 0$, $ 1$, $ 1$ $ (mod\,3)$ in some order

Hence $ m^{2}+(m+1)^{2}+(m+2)^{2}\equiv 2\, (mod\,3)$

We have actually shown above that any perfect square is congruent to either $ 0$ or $ 1$ $ (mod\,3)$.

Hence it is impossible for the sum of three consecutive squares to sum to another perfect square.

oh..i think i get it..because the square of any integer is never 2 mod 3? right?

thanks sludgethrower, for the help!

just for future reference, how did you choose to do the problem in mod 3? when do you know to use which mod?

mihail911

I knew to choose $ mod\,3$ as we have [b]3[/b] consecutive integers. If we had chosen $ mod\,4$ the integers could be congruent to 0, 1, 2 or 1, 2, 3 or 2, 3, 0 or 3, 0, 1 in some order and each of these cases would have to be treated seperately. (Even more cases would occur for $ mod\,k$ if $ k > 4$). By choosing $ mod\,3$ there is only one case.

thanks...slowly, im learning how to solve these number theory problems.. :lol:


[quote="czczczczczcz1991"][size=150][color=darkred]WE have x,y,z>0 and x+y+z=1
   proof that:   1/(x^2+y)+1/(y^2+z)+1/(z^2+x) >13/2[/color][/size][/quote]

I think it is hard.

[hide="hint"]
Problem. $ a,b,c \ge 0,a\plus{}b\plus{}c\equal{}1$ then
\[ \sum \frac{1}{a\plus{}b^2} \ge \frac{13}{2}
\]
Hint.
Using the Cauchy Schwarz
\[ \sum \frac{1}{a\plus{}b^2} \ge \frac{(\sum (2a\plus{}3b\plus{}7c))^2}{\sum (2a\plus{}3b\plus{}7c)^2(a\plus{}b^2)}
\]
:)
[/hide]

[quote="shfdfzhjj"][quote="can_hang2007"][hide="hint"]
Problem. $ a,b,c \ge 0,a \plus{} b \plus{} c \equal{} 1$ then
\[ \sum \frac {1}{a \plus{} b^2} \ge \frac {13}{2}
\]
Hint.
Using the Cauchy Schwarz
\[ \sum \frac {1}{a \plus{} b^2} \ge \frac {(\sum (2a \plus{} 3b \plus{} 7c))^2}{\sum (2a \plus{} 3b \plus{} 7c)^2(a \plus{} b^2)}
\]
:)
[/hide][/quote]
Thank you for your great solution！ :)
But I still have some questions，the step of Cauchy is too amazing，are there some secrets？ :maybe:
Thank you very much！[/quote]
Well, I got some private techniques on using Cauchy Schwarz and Holder inequality. It's help me solve many hard problems, for example the Jack Garfunkel's inequality.

For you, morheli! :lol:
$ \sum_{cyc}\frac {1}{a \plus{} b^2} \equal{} \sum_{cyc}\frac {(2a \plus{} 3b \plus{} 7c)^2}{(2a \plus{} 3b \plus{} 7c)^2(a^2 \plus{} b^2 \plus{} ab \plus{} ac)}\geq$
$ \geq\frac {144(a \plus{} b \plus{} c)^2}{\sum(2a \plus{} 3b \plus{} 7c)^2(a^2 \plus{} b^2 \plus{} ab \plus{} ac)}.$
Id est, it remains to prove that $ 288(a \plus{} b \plus{} c)^4\geq13\sum_{cyc}(2a \plus{} 3b \plus{} 7c)^2(a^2 \plus{} b^2 \plus{} ab \plus{} ac).$
But $ 288(a \plus{} b \plus{} c)^4\geq13\sum_{cyc}(2a \plus{} 3b \plus{} 7c)^2(a^2 \plus{} b^2 \plus{} ab \plus{} ac)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(119a^4 \minus{} 187a^3b \plus{} 411a^3c \minus{} 235a^2b^2 \plus{} 180a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow119\sum_{cyc}(a^4 \plus{} a^3c \minus{} 2a^3b) \plus{} \sum_{cyc}(292a^3c \minus{} 235a^2b^2 \plus{} 51a^3b \plus{} 180a^2bc)\geq0,$
which true because $ 235^2 \minus{} 4\cdot292\cdot51 < 0.$ :wink:

thank you very much! arqady,you are so nice! :)

[quote="toanIneq"]can_hang2007: Can you share your private techniques to solve some hard inequalities :D[/quote]
I agree with toanIneq. I too face some problems during using $ Cauchy$ and $ H\ddot{o}lder$. Please share the secret with us!

[quote="superacidduck"]Someone said it can equal to $ 13/2$
Is $ 13/2$ best?[/quote]
I don't think that. From 10 million generated random cases my feeling is that the minimum is about 6.517577 and this holds for f(x,1-x,0), where x~0.3549184773 or cyclic permutation of (x,1-x,0).

[quote="czczczczczcz1991"][color=darkred]Let $ x,y,z > 0$ and $ x \plus{} y \plus{} z \equal{} 1,$ prove that

$ \frac {1}{x^2 \plus{} y} \plus{} \frac {1}{y^2 \plus{} z} \plus{} \frac {1}{z^2 \plus{} x} > \frac {13}{2}.$[/color][/quote]$ \sum_{cyc}{\frac {1}{x^2 \plus{} y(x \plus{} y \plus{} z)} \minus{} \frac {13}{2}}\equiv\frac {F(x,y,z)}{2(x^2 \plus{} yx \plus{} y^2 \plus{} yz)(y^2 \plus{} zx \plus{} zy \plus{} z^2)(z^2 \plus{} x^2 \plus{} xy \plus{} xz)}.$

$ F(x,y,z) \equal{} F(x,x \plus{} s,x \plus{} t) \equal{} 32x^6 \plus{} 64x^5(s \plus{} t)$

$ \plus{} 10x^4\left(9s^2 \plus{} 7st \plus{} 9t^2\right) \plus{} x^3\left(86s^3 \plus{} 141s^2t \minus{} 39st^2 \plus{} 86t^3\right)$

$ \plus{} x^2\left(53s^4 \plus{} 113s^3t \plus{} 23s^2t^2 \minus{} 67st^3 \plus{} 53t^4\right)$

$ \plus{} \frac {x}{32}\left[t(2s \minus{} t)^2(372s^2 \plus{} 540st \plus{} 343t^2) \plus{} 544s^5 \plus{} 16s^4t \plus{} 201t^5)\right]$

$ \plus{} \frac {s \plus{} t}{18750}\left[(9s \minus{} 5t)^2(397s^3 \plus{} 1830s^2t \plus{} 1650st^2 \plus{} 1500t^3) \plus{} 5343s^5 \plus{} 2375s^3t^2\right]$

$ > 0,$ which is clearly true for $ x \equal{} \min\{x,y,z\}.$[quote="can_hang2007"]Problem. $ a,b,c \ge 0,a \plus{} b \plus{} c \equal{} 1$ then
\[ \sum \frac {1}{a \plus{} b^2} \ge \frac {13}{2}
\]
[hide="Hint."]Using the Cauchy Schwarz
\[ \sum \frac {1}{a \plus{} b^2} \ge \frac {(\sum (2a \plus{} 3b \plus{} 7c))^2}{\sum (2a \plus{} 3b \plus{} 7c)^2(a \plus{} b^2)}
\]
[/hide][/quote][quote="arqady"]$ \sum_{cyc}\frac {1}{a \plus{} b^2}\geq\frac {144(a \plus{} b \plus{} c)^2}{\sum(2a \plus{} 3b \plus{} 7c)^2(a^2 \plus{} b^2 \plus{} ab \plus{} ac)}.$

Id est, it remains to prove that

$ 288(a \plus{} b \plus{} c)^4\geq13\sum_{cyc}(2a \plus{} 3b \plus{} 7c)^2(a^2 \plus{} b^2 \plus{} ab \plus{} ac).$
But[/quote]$ G(a,b,c)\equiv 288(a \plus{} b \plus{} c)^4 \minus{} 13\sum_{cyc}(2a \plus{} 3b \plus{} 7c)^2(a^2 \plus{} b^2 \plus{} ab \plus{} ac)$

$ \ne 119\sum_{sym}{a^4} \minus{} 187\sum_{cyc}{a^3b} \plus{} 411\sum_{cyc}{ab^3} \minus{} 235\sum_{sym}{b^2c^2} \plus{} 180\sum_{sym}{a^2bc}.$[quote="morheli"]i still cant get it after seeing the hint, can anyone tell me how it works?[/quote]$ G(a,b,c) \equal{} 119\sum_{sym}{a^4} \minus{} 239\sum_{cyc}{a^3b} \plus{} 463\sum_{cyc}{ab^3} \minus{} 235\sum_{sym}{b^2c^2} \plus{} 180\sum_{sym}{a^2bc}$

$ \equal{} G(a,a \plus{} s,a \plus{} t) \equal{} 864a^4 \plus{} 1152(s \plus{} t)a^3 \plus{} 8(137s^2 \plus{} 79st \plus{} 137t^2)a^2$

$ \plus{} (700s^3 \minus{} 1007s^2t \plus{} 1099st^2 \plus{} 700t^3)a$

$ \plus{} \frac {(5s\minus{}9t)^2(86751s^2\plus{}133925st\plus{}26775t^2)\plus{}103690s^3t\plus{}74169s^2t^2}{18225}> 0,$

which is obviously true for  $ a \equal{} \min\{a,b,c\}.$

See also here : http://www.mathlinks.ro/viewtopic.php?t=88792

hey....can you kindly share your private techniques???


function
integration
induction
calculus
derivative
inequalities
real analysis

Let $a,b\in(0,1)$ and  $f:[0,1]\to\rr$ a continuous function, such that
\[\int_0^x f(t){\rm d}t=\int_0^{ax}f(t){\rm
d}t+\int_0^{bx}f(t){\rm d}t,\>\>\hbox{for any}\>\>x\in[0,1].\]

 (i)  Prove that $a+b<1$ implies $f=0$.

(ii) Prove that $a+b=1$ implies $f$ is constant.
I particularly liked this problem.

Let $\displaystyle F(x)= \int^x_0  f(t) dt$. Then $F$ is derivable and $F(x)=F(ax)+F(bx)$. An easy induction shows that $\displaystyle F(x)= \sum_{k=0}^n { n\choose k} F(a^{n-k}b^kx)$. Because the latter expression tends to 0 for $a+b<1$ and to $f(0)$ for $a+b=1$, when $n$ tends to infinity, we have either $F(x)=0$ or $F(x)=f(0)x$, from where the conclusion follows.

Let us write a full solution to this one.
Suppose $a+b=1$. Take any  $\varepsilon >0$.  Since $F$ is derivable and its derivative in 0 is $f(0)$, there exists $p>0$ such  that if $|x|<p$ we have
\[ f(0)- \varepsilon < \frac {F(x)}x < f(0)+ \varepsilon . \]

 Take $n_0$ such that for all $n>n_0$ and all $x \in [0,1]$ in we have $a^kb^(n-k)x<p$ (this is clearly possible, since $a^kb^{n-k}x < \max{a,b}^n$ and $a,b<1$) for any $k=\overline{0,n}$. Then
\[ f(0)- \varepsilon < \frac {F(a^{n-k}b^kx)}{a^{n-k}b^kx}< f(0)+ \varepsilon . \]
Let us fix an $x$ and multiply these two inequalities with $\displaystyle {n\choose k}$ and use  that \[ \displaystyle F(x)=\sum_{k=0}^n {n\choose k } F(a^{n-k}b^kx).\]

We obtain that $f(0)- \varepsilon <F(x)<f(0)+ \varepsilon$, for all $x$ and for any $\varepsilon>0$, thus by making $\varepsilon \to 0$ we obtain $F(x)=f(0)$ for all $x$.

It's a nice approach. I tried using F(x)=F(ax)+F(bx) too, but got nowhere, so I proved that f(x)=af(ax)+bf(bx) instead. The result follows easily.


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \sum \sqrt{\frac{a^2\plus{}bc}{b^2\plus{}bc\plus{}c^2}} \ge \sqrt{6}\]
:)

[quote="can_hang2007"]Let $ x,y,z$ be nonnegative real numbers such that $ yz \plus{} zx \plus{} xy > 0$. Prove that

$ \sqrt {\frac {x^2 \plus{} yz}{y^2 \plus{} yz \plus{} z^2}} \plus{} \sqrt {\frac {y^2 \plus{} zx}{z^2 \plus{} zx \plus{} x^2}} \plus{} \sqrt {\frac {z^2 \plus{} xy}{x^2 \plus{} xy \plus{} y^2}} \ge \sqrt {6},$

with equality if and only if $ x \equal{} y \equal{} z.$[/quote]$ 6(x^2 \plus{} yz)(x^2 \plus{} y^2 \plus{} z^2 \plus{} yz \plus{} zx \plus{} xy)^2 \minus{} (y^2 \plus{} yz \plus{} z^2)(6x^2 \plus{} 4yz \plus{} zx \plus{} xy)^2$

$ \equal{} 6x^6 \plus{} 12x^5(y \plus{} z) \minus{} 6x^4(3y^2 \plus{} yz \plus{} 3z^2) \plus{} 12x^3yz(y \plus{} z)$

$ \plus{} x^2(y^2 \minus{} 5yz \plus{} z^2)(5y^2 \plus{} 4yz \plus{} 5z^2) \plus{} 4xyz(y \plus{} z)(y^2 \plus{} yz \plus{} z^2)$

$ \plus{} 2yz(y^2 \plus{} yz \plus{} z^2)(3y^2 \minus{} 5yz \plus{} 3z^2)$

$ \equal{} \frac {(x \minus{} y)^2(52x^3y \plus{} 40x^2y^2 \plus{} 103xy^2z \plus{} 5xz^3 \plus{} 24y^3z)}{8}$

$ \plus{} \frac {(x \minus{} z)^2(52x^3z \plus{} 40x^2z^2 \plus{} 103xz^2y \plus{} 5xy^3 \plus{} 24z^3y)}{8}$

$ \plus{} \frac {x(y \plus{} z \minus{} 2x)^2[12x^3 \plus{} 23x^2(y \plus{} z) \plus{} 28xyz \plus{} 9yz(y \plus{} z)]}{8}$

$ \plus{} yz(y \minus{} z)^2[(y \plus{} z \minus{} 2x)^2 \plus{} 2(y^2 \plus{} z^2)]\geq0\Longrightarrow$

$ \sum{\sqrt {\frac {6(x^2 \plus{} yz)}{y^2 \plus{} yz \plus{} z^2}} }\geq\sum{\frac {6x^2 \plus{} 4yz \plus{} zx \plus{} xy}{x^2 \plus{} y^2 \plus{} z^2 \plus{} yz \plus{} zx \plus{} xy}} \equal{} 6.$

See also here: http://www.mathlinks.ro/Forum/viewtopic.php?t=202843