logarithms
vector
function
real analysis
real analysis unsolved

if : $ f\left( x \plus{} y\right) \equal{} f\left( x\right) . f\left( y\right)$ and $ f\left( 1\right) \equal{}$  10, then : $ f\left( x\right)$ >0

ps. e and  log are undefine now!!!

Same story (just take $ \log f$). As long as you do not assume any regularity, you can just consider $ \mathbb R$ as a vector space over $ \mathbb Q$, construct the Hamel basis and let $ f(x)$ be the coefficient of $ x$ at some basis vector (for the additive case) and $ e^\text{that}$ for the multiplicative case.

Ah, you changed the problem again :roll: Then, certainly, what I said doesn't make much sense any more. The new version is trivial unless $ f$ is allowed to take complex values. Since $ 0\ne f(1)\equal{}f(x)\cdot f(1\minus{}x)$, we conclude that $ f(x)\ne 0$ for all $ x$. Then $ f(x)\equal{}f(x/2)^2>0$ since the square of any non-zero real number is positive. If $ f$ is allowed to take complex values, it is still false (take $ 10^{g(x)}$ where $ g(x)$ is the function given by alekk only change $ a\plus{}b$ to $ a\plus{}ib$)

Sir fedja:

you  are so kind , to forgive for my rudeness !

apologize to you again!


In how many ways can be distributed $ n$ distinct letters (adresessed) in $ N$ mailboxes such that there are $ k$ letters:
a) In a fixed mailbox
b) In one of the mailboxes.

if the mailboxes are distinct:

a) $ \binom{n}{k}$ ways to pick the $ k$ letters to put into the chosen mailbox, and then $ N\minus{}1$ choices for each of the other $ n\minus{}k$ letters. so the answer in this case is $ (N\minus{}1)^{n\minus{}k}\binom{n}{k}$.

b) this one is a bit tougher, but you can use the idea in part (a) and PIE to get the answer.

a) is correct.
for b) I think we have to multiply the result from a) with $ n$, because we can choose a mailbox to contain $ k$ letters in $ \binom {n}{1}$ ways. Why using PIE?

you mean $ N$, not $ n$, right?

in (b), the mailbox is not fixed, and there may be more than one such mailbox. doing it the way you describe over-counts all distributions where at least two mailboxes get exactly $ k$ letters.

Indeed is $ N$ ot $ n$ :) Sorry.

And you are right for the b) case also. I'll think more about it...


exterior angle
geometry
angle bisector
geometry unsolved

Given a [b]triangle ABC[/b].The bisector of exterior angle of [b]angle A[/b]  intersects [b]side BC[/b] in point [b]D.[/b]
  [b]To show: [/b]
                     [b]AB/AC=BD/CD [/b]

Does not belong in the Olympiad section...

This is the [url=http://planetmath.org/?op=getobj&from=objects&id=6488]Angle Bisector Theorem[/url].

The diagram in the link you gave does not match which i have


A tank contains 200 fish and $ 99 \%$ of them are tetras. How many tetras should be removed so that exactly $ 98 \%$ of the fish in the tank are tetras?

Since 99% of the fish are tetras, 2 of them represent other fish. To get 2% other fish, we must remove 198-98=100 tetras.

Isn't there something similaar to this on the Algebra Challenge set?

[quote="dragon96"]
Isn't there something similaar to this on the Algebra Challenge set?[/quote]
I don't think so.


did anyone get # 25 on the 2007 amc 10a, if so how do you do it.

This has been posted before..

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=amc+digits+sum&t=132743&sid=e8bac0c304cb7dc2fd5b3696fadeb8c4

More to the point, if you want to know the solution to any of the last 5 questions on any AMC 10 or 12 from 2004 or later or any AIME question in the same time frame, simply look through the math jam transcripts to find the appropriate one.

did you know that I took a random guess on that one and still got it right?

i did ennie minnie mini moe...
(did I spell all of those correctly?)

No offense, but that post was basically spam.


Let a,b,c >0 such that a+b+c+1=4abc. Prove that
$ \frac{1}{{a^4  \plus{} b \plus{} c}} \plus{} \frac{1}{{b^4  \plus{} c \plus{} a}} \plus{} \frac{1}{{c^4  \plus{} a \plus{} b}} \le \frac{3}{{a \plus{} b \plus{} c}}$

[quote="tronghieu"]Let a,b,c >0 such that a+b+c+1=4abc. Prove that
$ \frac {1}{{a^4 \plus{} b \plus{} c}} \plus{} \frac {1}{{b^4 \plus{} c \plus{} a}} \plus{} \frac {1}{{c^4 \plus{} a \plus{} b}} \le \frac {3}{{a \plus{} b \plus{} c}}$[/quote]
The inequality comes from:
If $ a,b,c>0$ such that $ a\plus{}b\plus{}c \ge 3$, then
$ \frac {1}{{a^4 \plus{} b \plus{} c}} \plus{} \frac {1}{{b^4 \plus{} c \plus{} a}} \plus{} \frac {1}{{c^4 \plus{} a \plus{} b}} \le \frac {3}{{a \plus{} b \plus{} c}}.$
This is a very known result. ;)

[quote="tronghieu"]Let a,b,c >0 such that a+b+c+1=4abc. Prove that
$ \frac {1}{{a^4 \plus{} b \plus{} c}} \plus{} \frac {1}{{b^4 \plus{} c \plus{} a}} \plus{} \frac {1}{{c^4 \plus{} a \plus{} b}} \le \frac {3}{{a \plus{} b \plus{} c}}$[/quote]
We have : $ a^4 \plus{} b \plus{} c \equal{} \frac {(a^2)^2}{1} \plus{} \frac {(b^2)^2}{b^3} \plus{} \frac {(c^2)^2}{c^3}\geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^{2}}{1 \plus{} b^3 \plus{} c^3}$
$ \Leftrightarrow \sum\frac {1}{a^4 \plus{} b \plus{} c} \leq \frac {3 \plus{} 2(a^3 \plus{} b^3 \plus{} c^3)}{(a^2 \plus{} b^2 \plus{} c^2)^{2}}$
We have to prove that :
$ \frac {3 \plus{} 2(a^3 \plus{} b^3 \plus{} c^3)}{(a^2 \plus{} b^2 \plus{} c^2)^{2}} \leq \frac {3}{a \plus{} b \plus{} c}$ (*)
with : $ a,b,c > 0$ and $ a \plus{} b \plus{} c \plus{} 1 \equal{} 4abc$
$ a\plus{}b\plus{}c\equal{}p ;ab\plus{}bc\plus{}ca\equal{}q;abc\equal{}r$
We have : $ 4abc\equal{}a\plus{}b\plus{}c\plus{}1 \geq 4\sqrt[4]{abc} \Leftrightarrow abc\equal{}r \geq 1$
$ (*)\Leftrightarrow (p^2\minus{}3q)^{2}\plus{}(q^2\minus{}3p)\plus{}2(q^2\minus{}3pr) \geq 0$
We have: $ (p^2\minus{}3q)^{2}\plus{}(q^2\minus{}3p)\plus{}2(q^2\minus{}3pr) \geq (p^2\minus{}3q)^{2}\plus{}3(q^2\minus{}3pr) \geq 0$
We have done :D

[quote="mai quoc thang"][quote="tronghieu"]Let a,b,c >0 such that a+b+c+1=4abc. Prove that
$ \frac {1}{{a^4 \plus{} b \plus{} c}} \plus{} \frac {1}{{b^4 \plus{} c \plus{} a}} \plus{} \frac {1}{{c^4 \plus{} a \plus{} b}} \le \frac {3}{{a \plus{} b \plus{} c}}$[/quote]
We have : $ a^4 \plus{} b \plus{} c \equal{} \frac {(a^2)^2}{1} \plus{} \frac {(b^2)^2}{b^3} \plus{} \frac {(c^2)^2}{c^3}\geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^{2}}{1 \plus{} b^3 \plus{} c^3}$
$ \Leftrightarrow \sum\frac {1}{a^4 \plus{} b \plus{} c} \leq \frac {3 \plus{} 2(a^3 \plus{} b^3 \plus{} c^3)}{(a^2 \plus{} b^2 \plus{} c^2)^{2}}$
We have to prove that :
$ \frac {3 \plus{} 2(a^3 \plus{} b^3 \plus{} c^3)}{(a^2 \plus{} b^2 \plus{} c^2)^{2}} \leq \frac {3}{a \plus{} b \plus{} c}$ (*)
with : $ a,b,c > 0$ and $ a \plus{} b \plus{} c \plus{} 1 \equal{} 4abc$
$ a \plus{} b \plus{} c \equal{} p ;ab \plus{} bc \plus{} ca \equal{} q;abc \equal{} r$
We have : $ 4abc \equal{} a \plus{} b \plus{} c \plus{} 1 \geq 4\sqrt [4]{abc} \Leftrightarrow abc \equal{} r \geq 1$
$ (*)\Leftrightarrow (p^2 \minus{} 3q)^{2} \plus{} (q^2 \minus{} 3p) \plus{} 2(q^2 \minus{} 3pr) \geq 0$
We have: $ (p^2 \minus{} 3q)^{2} \plus{} (q^2 \minus{} 3p) \plus{} 2(q^2 \minus{} 3pr) \geq (p^2 \minus{} 3q)^{2} \plus{} 3(q^2 \minus{} 3pr) \geq 0$
We have done :D[/quote]
I don't think that it's nice solution althogh I have known it before. I think not easy to see that $ a^4 \plus{} b \plus{} c \equal{} \frac {(a^2)^2}{1} \plus{} \frac {(b^2)^2}{b^3} \plus{} \frac {(c^2)^2}{c^3}\geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^{2}}{1 \plus{} b^3 \plus{} c^3}$
So I expect another solution.
To can_hang: Can you post your solution for your inequality. I hope It won't same as mai quoc thang. ( sorry about my bad English )

[quote="tronghieu"]Nobody help me. Can_hang, can you post your solution?  :wink:[/quote]
Of course that my solution is different from Mai Quoc Thang's one. Please read my hint carefully before stating that:
[quote="tronghieu"]To can_hang: Can you post your solution for your inequality. I hope It won't same as mai quoc thang. ( sorry about my bad English)[/quote]
;)
Anyways, I dont want to post my solution here. All I can say is "there is a nice solution with Cauchy Schwarz". And I think you guy, the one usually...(spam) will find your own solution as you always states (Please reread your post) ;)!

[quote="vuductuong"]This is a problem in 30/4 contest  ??[/quote]
Yes. It's true.  :lol:


[quote="Allnames"]In an acute triangle $ ABC$ segments $ BE$ and $ CF$ are altitudes. Circle $ (O_1)$ passing through the point $ A$  and tangent to the line $ BC$ at the points $ P$, Circle $ (O_2)$ passing through the point $ F$ and tangent to the line $ BC$ at the points $ Q$,so  that $ B$ lies between $ C$ and $ Q$. Prove that lines $ PE$ and $ QF$ intersect on the circumcircle of triangle $ AEF$.
I hope it is mine. I am pondering whether  it is generalization of that nice one, isn't it ?.[/quote]
Is it true?  :huh:  :?:


hello, $ x\equal{}0$ or $ x\equal{}1$ are the only solutions.
Sonnhard.

[quote="phuong"]Solve: $ 2^{2^x} \plus{} 3^{2^x} \equal{} 2^x \plus{} 3^{x \plus{} 1} \plus{} x \plus{} 1$[/quote]

We have $ 2^{2^x}\minus{}2^x\minus{}x>1 for x>1$ and equality for x=1.Also$ 3^{2^x} > 3^{x \plus{} 1}$for x>1 and equality for x=1. So $ x \le 1$.


USAMTS
algebra
polynomial
induction
binomial coefficients

This problem is related to #1 on the past USAMTS math jam, the one involving polynomials in which one must create a new, but related polynomial to solve the problem. Here is a similar problem, but a bit more difficult because the new polynomial is harder to find. How would I do this one?

If f(x) is a polynomial of degree n such that f(k) = 2^k for k=0, 1, ...n, find f(n+1).

Where did this problem come from? If you made it up yourself, it may be incredibly difficult.

No, it was an IMO proposal, 1998 I believe.

Lagrange polynomial works perfectly here:
Lagrange thm:

The only polynomial of degree <=n satisfying f(a_i)=c_i for all i from 0 to is this:
f(x)=
Sum {i=0 to n} c_i*(x-a_0)..(x-a_(i-1))(x-a_(i+1))..(x-a_n)/(a_i-a_0)..(a_i-a_(i-1))(a_i-a_(i+1))..(a_i-a_n). We can prove this by observing that this polynomial checks the identities f(a_i)=c_i and it's unique because we can write all those identities as a system with n+1 eqns and at most n+1 unknown terms (the coefficients of the polynomial).

So the polynomial we're looking for is of this form and a_i=i and c_i=2^i, so all we have to do is calculate this expression for x=n+1. This isn't hard at all (we get some binomial coefficients and stuff). In the end, the result is 2^(n+1)-1. This can even be checked for small values of n.

I found a polynomial, but I'm stuck at that point.  Maybe there's a better polynomial than the following: (I haven't had any luck getting spoiler tags to work, so I'm just going to leave lots of space below)

f(x+1) - 2f(x) = cx(x-1)(x-2)(x-3)...(x-(n-1))
I don't see how to solve for c from here, and I'm pretty sure that you need c in order to find f(n+1).  On the other hand I have an educated guess for the answer (for what that's worth...)

2^(n+1)-1 was my guess, based on trying the cases n=0, 1, 2, and 3.

JBL said (jokingly) that every problem involving an integer n can be solved by induction, and this problem is no exception.  Let g(x) = f(x+1) - f(x) be the first difference of f.  Then g satisfies the hypothesis for n - 1, and so by induction g(n) = 2^n - 1.  Then f(n+1) = f(n) + g(n) = 2^n + 2^n - 1 = 2^{n+1} - 1.

We didn't say that f(n+1) = 2^{n+1} - 1 is true for every value of n.  It's just true for the specific value of n in the problem (the degree of f).
In other words, we are trying to prove the following property P(n): "For every polynomial f of degree n such that f(k) = 2^k for k in 0..n, we have f(n+1) = 2^{n+1} - 1".  My proof showed that P(n) is true by induction on n.

[hide]

the nth difference of the nth degree polynomial is a constant

a bit induction shows that the nth difference is the equation

have coefficients (-1)^i*nCi*f(i)

proceed

[/hide]


geometry
MATHCOUNTS
trapezoid
rotation
geometric transformation
reflection
homothety

Let $ABCD$ be a quadrilateral with $AB\parallel CD$ and $AC \perp BD$. Let $O$ be the intersection of $AC$ and $BD$. On the rays $(OA$ and $(OB$ we consider the points $M$ and $N$ respectively such that $\angle ANC = \angle BMD = 90^\circ$. We denote with $E$ the midpoint of the segment $MN$. Prove that

a) $\triangle OMN \sim \triangle OBA$;
b) $OE \perp AB$.

[i]Claudiu-Stefan Popa[/i]

Awww... it's not [i]that[/i] bad... but way harder than our MathCounts, and that includes 8th graders.

[quote="mna851"]This is unrelated, but does anyone else find it odd that advanced American math students struggle with problems 7th grade Romanian students can solve.[/quote]

Unrelated too, I also noticed that, then, I am trying to discuss about that in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31971]this thread[/url], if you want to give your opinion  :D . By the way, Valentin (and also any Romanian who is either a 12th grader or already in college), I would like to know your opinion about the thread topic I am mentioning!

Probability1.01 I know the problem isn't that bad I can solve it and I'm a freshmen, but there is no chance I could have done it when I was in 7th grade.

As far as the stats go, this problem was solved by 20 contestants out of a pool of 97 students. Anyway let's stop the off-topic now and concentrate on solving this very nice problem :)

a) $\triangle OMN \sim \triangle OBA$;

Set OA=a, OB=b, and by similar triangles AOB and COD, set OC=ka, OD=kb. Then triangles AON and NOC are similar, so a/ON=ON/ka, i.e. $ON=a\sqrt k$, and similarly, $OM=b\sqrt k$. Then OM/ON=b/a=OB/OA, so the triangles are similar.

b) $OE \perp AB$.

Let OE and AB intersect at X. Then

<XOA = <EOM = <EMO (since MEO is isosceles)
=<XBO (by similar triangles in part a)
=90-<XAO (since AOB is right),

so triangle AXO is a right triangle.

[quote="mna851"]This is unrelated, but does anyone else find it odd that advanced American math students struggle with problems 7th grade Romanian students can solve.[/quote]

These problems look intimidating at first, but just stare hard at them for a couple minutes and solutions quickly emerge....

[quote="polymorphic"]I'm just curious if there's some way to do this only with angles i.e. angle chasing.[/quote]None that I know of.

The triangles $\triangle OMN \sim \triangle OBA$ are similar regardless of whether the trapezoid diagonals $AC, BD$ are perpendicular or not. Moreover, the angles $\measuredangle ANC = \measuredangle BMD$ have to be only equal, not necessarily right. From similarity of the triangles $\triangle OBA \sim \triangle ODC$, it follows that $\frac{OA}{OC} = \frac{OB}{OD}$. Rotate the the triangle $\triangle ACN$ by the angle $\measuredangle AOB$ around the point $O$ counter-clockwise and then reflect the rotated triangle in the line $BD$ into a congruent triangle $\triangle A'C'N'$. The corresponding sides of the triangles $\triangle A'C'N', \triangle BDM$ become parallel and the triangles themselves centrally similar with homothety center $O$ and homothety coefficient $h = \frac{OA}{OC} = \frac{OB}{OD}$. Consequently, the angles $\measuredangle MAN \equiv \measuredangle CAN = \measuredangle DBM \equiv \measuredangle NBM$ are equal. This means that the quadrilateral $ABNM$ is cyclic and the sums of its opposite angles are equal to $180^o$. Thus

$\measuredangle OMN = 180^o - \measuredangle AMN = \measuredangle ABN \equiv \measuredangle ABO$

$\measuredangle ONM = 180^o - \measuredangle BNM = \measuredangle BAM \equiv \measuredangle BAO$

As a result, the triangles $\triangle OMN \sim \triangle OBA$ are similar. This implies that the triangles $\triangle OMN \sim \triangle ODC$ are also similar and consequently, the quadrilateral $MCDN$ is cyclic. Until now, no right angles have been used. If the diagonals $AC \perp BD$ are perpendicular, the diagonals $MC \perp ND$ of the cyclic quadrilateral $MCDN$ are also perpendicular. By one of Brahmagupta's theorems, the diagonal intersection of a cyclic quadrilateral with perpendicular diagonals is identical with its anticenter - the intersection of its maltitudes, i.e., normals from midpoints of its sides to the opposite sides. Since $E$ is the midpoint of the side $MN$, the line $EO$ is one of the maltitudes and therefore, $OE \perp CD$. Since the trapezoid bases $AB \parallel CD$ are parallel, $OE \perp AB$ as well.


Circle P of radius $ 6$ is internally tangent to circle O of radius $ 12$. Two circles, Q and R, of equal radii, are internally tangent to circle O and externally tangent to circle P. What is the radius of circle Q?

Let O be the center of the outer circle, A the center of the circle radius 6 (which contains O), B and C the centers of the equal circles, P and Q the respective points of tangency to the outer circle, M the midpoint of BC, also the point of tangency to the two equal circles, and A,O, and M are collinear. Let the equal circles have radius x. O,B, and P are collinear, so $ OB \equal{} 12 \minus{} x$. $ AB \equal{} x \plus{} 6$. We have $ AO \plus{} OM \equal{} AM$, so by Pythag,
$ 6 \plus{} \sqrt {(12 \minus{} x)^2 \minus{} x^2} \equal{} \sqrt {(x \plus{} 6)^2 \minus{} x^2}$
$ 6 \plus{} \sqrt {144 \minus{} 24x} \equal{} \sqrt {12x \plus{} 36}$
Squaring,
$ 36 \plus{} 144 \minus{} 24x \plus{} 12 \sqrt {144 \minus{} 2x} \equal{} 12x \plus{} 36$
$ 144 \minus{} 24x \equal{} (3x \minus{} 12)^2$
$ x \equal{} \frac {70}{9}$


Can you tell me what I did wrong in LaTeX?

\documentclass{article}
\usepackage[left=2cm,top=1cm,right=3cm,nohead,nofoot]{geometry}
\usepackage{asymptote}
\title{akhil0422's Challament}
\begin{document}
\citation
\bibdata
\bibstyle
\maketile
\begin{enumerate}
\item What is the ratio of the area of square ABCD to square EFGH?
\begin{figure}[h]
\begin{asy}
\unitsize(1inch)
\draw((0,0)--(1,0)--(1,1)--(0,1));
\draw((2,2)--(2.25,2)--(2.25,2.25)--(2,2.25));
\label("$ A$",(0,0),SE);
\label("$ B$",(1,0),SW);
\label("$ C$",(1,1),NE);
\label("$ D$",(0,1),NW);
\label("$ x$",(0,0)--(1,0),S);
\label("$ E$",(2,2),SE);
\label("$ F$",(2.25,2),SW);
\label("$ G$",(2.25,2.25),NE);
\label("$ H$",(2,2.25),NW);
\label("$ \frac{x}{4}$",(2,2)--(2.25,2),S);
\end{asy}
\end{figure}
\end{document}

In LaTeX you only got one thing wrong which is that you had no \end{enumerate}. But your Asymptote code is all wrong. It doesn't use \ for commands so it should be unitsize, draw, label not \unitsize, \draw, \label. Also all lines of Asymptote code must end in ; and your first line doesn't have it.

It's a good idea to leave out the Asymptote code until the rest of your LaTeX code compiles without error. Then you can put the Asymptote code in.

For more help with Asymptote (as distinct from LaTeX) there's a [url=http://www.artofproblemsolving.com/Forum/index.php?f=519]special forum[/url] for it.


A while ago I posted the problem "A dice game for all you probability lovers".  However, I have a variation on the problem which I did not originally post.

First, imagine you have a six-sided die labeled 1-6 and somebody else has one labeled 2-7.  If you both roll your dice, what is the probabilty of you winning?

What if the other person has a 7 sided die 1-7?
What about 1-8?
What about 1-n?

Search for the post "A dice game for all you probability lovers" and read the rules.  Then imagine that your opponent has an n sided die and you only have an m sided die.  What is your strategy?

Ok here is my solution to this:

Warning: I keep switching from capitals to lower case.  forgive me!

My first step was to do a chart of the 2-7 and the 1-6 die.  It's easy to count the number of times "I" win.  It's obviously 10, and there are obviously 36 total outcomes.  That gives an easy answer of $\frac{5}{18}$  However, to put it in terms of N, where 1-N are the labels for the sides, and you have a 1-6 sided die, you're # of wins are always the same, however, the total number of outcomes changes.  Your probability of winning against an N sided die where N is greater than or equal to 6, is $\frac{15}{6(n - lower limit + 1)}$.  When N is less than 6, you can subtract the Triangular number of $6 - (n + 1)$.  This gives you $\frac{15 - \frac{(6-(n+1))(7-(n+1))}{2}}{6(n -m + 1)}$.  If the 1 in 1-N changes to a 2 or a 3 or to, let's say, M, then the triangular number resulting from the # of your wins decreases by M-1 terms.

End result:

Your probability of winning (in terms of whatever kind of dice the other guy has) is:

If m < n, and n > or = 6: edit: (m < 6)

$\frac{(6-m)(7-m)}{12(n-m+1)}$

I still am trying to figure out how to work out a decent definition for this!  I'll keep this work for now, but don't be surprised if something doesn't work for something or other.  I need a bit of time!

Also, if you have an "m" sided die, (i'll change this to x since I accidentally used an m already), replace all of the 6's with x's.  However, if this x sided die begins with a number that is not one, I do not want to write another formula for it!

Please, try not to double/triple post. Edit your last message instead.


abstract algebra
calculus
integration
algebra
function
domain
Ring Theory

Let $R$ be a noetherian integral domain. Show that the following are equivalent:
(i) $R$ has only finitely many prime ideals
(ii) $R$ has only finitely many prime ideals and $\mathrm{dim}(R) \leq 1$
(iii) There is a $f \in R$ such that $R_f$ is a field.

For the easy implications:

$i\to ii:$ Assume $R$ is not a field i.e. $dim_{Krull}R>0$.
Let $m$ be a maximal ideal of $R$. Then $R_m$ is a local noetherian domain and its Krull dimension is $ht(m)$.
By Krull's Hauptidealsatz, every nonzero element of $mR_m$ is contained in a prime ideal of height $1$. But then we get that $m$ is the union of the prime ideals of height $1$ contained in $m$. There are finitely many such ideals by assumption. By the Avoiding Lemma, $m$ is one of these ideals, hence $m$ is of height one. Since we can do this for all maximal ideals $m$ of $R$, we prove $dim_{Krull}R=1$.

$ii\to iii:$ Let $p_1,\ldots,p_n$ be the nonzero prime ideals of $R$. Since $R$ is a domain, $I=p_1\cdot\ldots\cdot p_n\neq (0)$. Let $f\in I$, $f\neq 0$.
$Spec R_f$ identifies with the Zariski open set $D(f)=\{p\in Spec(R)|\ f\not\in p\}$. But the only prime ideal of $R$ which does not contain $f$ is $(0)$, so $R_f$ is a field.

For $iii) \implies i)$, can we show that for infinite Spec, $\cap_{\mathfrak p \neq (0)} = (0)$? It looks like Krull's intersection theroem, i.e. can we find a prime ideal $\mathfrak p_n$ with $\mathfrak a^n \supseteq \mathfrak p_n \supsetneq (0)$ for all $n > 0$, $\mathfrak a \neq (1)$? EDIT: no, certainly not :(

amfulger's proof of $(i)\Rightarrow(ii)$ can be imitated to get a proof for $(iii)\Rightarrow(ii)$:

Because $R$ is a domain, $(iii)$ says basically that there is an $f\in R$ which belongs to all non-zero prime ideals. $(f)$ has a primary decomposition, and the minimal prime ideals which contain $f$ are precisely the minimal prime ideals which correspond to $(f)$ in this decomposition, hence there are finitely many of them. Coupled with our hypothesis $(iii)$, this shows that there are only finitely many minimal non-zero prime ideals of $R$. Now just repeat amfulger's argument. The conclusion will be that every non-zero prime ideal is one of these minimal non-zero ideals.


Let $ ABC$ be a triangle with $ Q,R\in BC$ so $ B\minus{}Q\minus{}R\minus{}C$ is in that order. Denote $ AQ\equal{}q$, $ AR\equal{}r$, $ BQ\equal{}y$, $ CR\equal{}z$, and $ QR\equal{}x$. Given $ \frac{r}{q}\equal{}\sqrt{\frac{b}{c}}$ and $ \angle QAR\equal{}\frac{1}{2}\cdot \angle A$, prove \[ x^2 bc\equal{}y^2b^2\plus{}z^2c^2\plus{}yz(b^2\plus{}c^2\minus{}a^2)\]

I's difficult than I thougt. Does it use Stewart' theorem?

The condition: $ \frac {r}{q} \equal{} \sqrt {\frac {b}{c}}$ plays an important role in the solution.

My solution is not as simple as Stewart's theorem...It may work I don't know.


[quote="pro_math"]Solve for $ x$:

$ \sqrt [3]{x \plus{} 6} \plus{} \sqrt {x \minus{} 1} \equal{} x^2 \minus{} 1$[/quote]
Let $ x \equal{} k^2 \plus{} 1$ when $ k \ge 0$
So $ \sqrt [3]{k^2 \plus{} 7} \equal{} k^4 \plus{} 2k^2 \minus{} k$
So $ k^2 \plus{} 7 \equal{} (k^4 \plus{} 2k^2 \minus{} k)^3 \equal{} k^{12} \plus{} 6k^{10} \minus{} 3k^9 \plus{} 12k^8 \minus{} 12k^7 \plus{} 11k^6 \minus{} 12k^5 \plus{} 6k^4 \minus{} k^3$ $ \iff$
$ k^{12} \plus{} 6k^{10} \minus{} 3k^9 \plus{} 12k^8 \minus{} 12k^7 \plus{} 11k^6 \minus{} 12k^5 \plus{} 6k^4 \minus{} k^3 \minus{} k^2 \minus{} 7 \equal{} 0$
But the only positive solution is $ k\equal{}1$. So $ x \equal{} 2$.

[quote="GioMott"]LaTeX tip: if you use $ a^10$ it won't appear correctly-use instead $ a^{10}$.
By the way, I don't think this is really useful: we get a horrific 12th degree equation which doesn't seem solvable analytically.
However, I can't find any clever way to solve this. :|

Bye
GM[/quote]
It's actually pretty easy to solve analytically if you guess the root $ k \equal{} 1$. Cause it's equivalent to:
$ (k\minus{}1)(k^{11} \plus{} k^{10} \plus{} 7k^9 \plus{} 4k^8 \plus{} 16k^7 \plus{} 4k^6 \plus{} 15k^5 \plus{} 3k^4 \plus{} 9k^3 \plus{} 8k^2 \plus{} 7k \plus{} 7)$ which has the only root $ k \equal{} 1$ cause $ k > 0$ :)

[quote="pro_math"]Solve for $ x$:

$ \sqrt [3]{x \plus{} 6} \plus{} \sqrt {x \minus{} 1} \equal{} x^2 \minus{} 1$[/quote]
If $ 1\leq x<2$ then $ \sqrt [3]{x \plus{} 6} \plus{} \sqrt {x \minus{} 1}>x\plus{}x\minus{}1\equal{}2x\minus{}1 >x^2 \minus{} 1.$
If $ x>2$ then $ \sqrt [3]{x \plus{} 6} \plus{} \sqrt {x \minus{} 1}<x\plus{}x\minus{}1\equal{}2x\minus{}1 <x^2 \minus{} 1.$
Thus, only $ 2$ can be a root of the equation.

This should be locked.

See here![url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1281130#1281130[/url]

pro_math, are you Vietnamese?

Japanese Communities Moderator

kunny


What is the sum of the least common multiple of 12 and 18 and the greatest common divisor of 12 and 18?

The prime factorization of these 2 numbers are as follows:
12 = {2,2,3}
18 = {3,2,3}

The greatest commen divisor is found by seeing that thhere is a  2 and a 3 in both of the numbers so $ 2\times3 = 6$

The least commen multiple is found by taking xor of the numbers and getting {2,2,3,3}. Thus $ 2\times2\times3\times\3 =36$

$ 36 + 6 = \boxed{42}$


function
LaTeX
trigonometry
calculus
calculus computations

Find the requested function value of sec x, if sin x = 14/15

x stands for theta. Im learning how to use latex


Let p be a prime number and n a positive integer.
Show that if $\frac{2}{3}n <p\leq n$ or $p>2n$, then ${2n\choose n}$ is not divisible by p.

This is quite easy using Legendre's formula about the exponent of $p$ in the prime decomposition of $n!$ and $(2n)!$....

Pierre.

I could be wrong, but isn't this a lemma in Erdos' proof of Bertrand's Postulate?

You're right.

And as Pierre said, it follows from the fact that, for $n > 4$, we have :
$v_{p}({{2n}\choose n}) = \sum_{k=1}^{\infty}\left(\left[\frac{2n}{p^{k}}\right]-2\left[\frac n{p^{k}}\right] \right) = 0$.

(And one checks manually the remaining cases.)

is there any simple ways of doing this question without using this kind of method?

anyone thanks to your suggested method, I finished the question.
Actually, it is cheating by using this kind of method in the chapter of my book.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Let R be an integral domain and let Q be the field of fractions of R [whose elements are denoted by (a,b) with a and b in R].

True or False: If p is irreducible in R then $ (p,1) \neq x^{2}$ for any x in Q.

I believe this is false.  For any field $ F$, consider the subring $ R = F+x^{2}F[x]$ of $ F[x]$.  $ x^{2}\in R$ is irreducible in $ R$ but $ x$ is in the field of fraction of $ R$ so that $ x^{2}\in R$ becomes a square in the field of fraction.


integration
trigonometry
complex analysis

$ \int_{0}^{\infty}\frac{\sin x}{x}dx$

can  someone do the complete solution of this using LAplace?

Thank you!!!

From that last link:

[quote="AndrewTom"] Therefore the limit as $ \alpha$ tends to $ 0$ of $ \int_{0}^{\infty} e^{ \minus{} \alpha x} \frac {\sin x}{x}\ dx \equal{} \int_{0}^{\infty} \frac {\sin x}{x}\ dx \equal{} \frac {\pi}{2}$.[/quote]
That is not an obvious claim at all. That limit is not justified by any of the usual limit theorems, such as the Dominated Convergence Theorem. It is true, but it requires a proof. What it is, as a theorem, is the continuous analogue of Abel's Theorem for the convergence of a power series at a boundary point.


Hello, i would like to put colours in some empty cells in my table. Can anybody help me please?

Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=colortbl]colortbl package[/url]. For example:
[code]\usepackage{colortbl}
...
\begin{tabular}{ccc}
1 & \cellcolor{blue} & 2 \\
3 & 4 & \cellcolor{red}
\end{tabular}[/code]


What is the value of $ (7\plus{}5)^2\minus{}(7\minus{}5)^2$?


Let be $ a,b,c,x,y,z>0$ such that $ x\plus{}y\plus{}z\equal{}\frac{1}{\sqrt{ab}}\plus{}\frac{1}{\sqrt{bc}}\plus{}\frac{1}{\sqrt{ca}}\equal{}1$ . Prove that :

$ \frac{1}{bcx}\plus{}\frac{1}{cay}\plus{}\frac{1}{abz}\ge 1$

[hide]For shortness, I'll write $ u\plus{}v\plus{}w\equal{}1$, to prove $ {u^2\over x}\plus{}{v^2\over y}\plus{}{w^2\over z}\geqslant 1$

$ \left({u^2\over x}\plus{}{v^2\over y}\plus{}{w^2\over z}\right)(x\plus{}y\plus{}z)\stackrel{\mathrm{Cauchy}}\geqslant(u\plus{}v\plus{}w)^2$ and the result follows directly.[/hide]


function
Support
algebra unsolved
algebra

For which a does there exist a nonconstant function
f: R -> R such that:
                               f(a(x+y))=f(x)+f(y) ?

I am not quite sure about this, can you use Cauchy's Equation if f is not continuous?

Setting x=y=0 gives f(0)=0.
Setting x=0 gives f(ay)=f(y). Similarly, f(ax)=f(x).
It follows that f(ax+ay)=f(ax)+f(ay). Define x'=ax and y'=ay yields f(x'+y')=f(x')+f(y'). Since f is well defined, by Cauchy's Equation, we have f(x')=cx' for some constant c. Note that since f is not a constant function, c\=0. Substituting, we get a2c(x+y)=ac(x+y), or, a=1.

[quote="al.M.V."]Note that since f is not a constant function, c\=0.[/quote]

And a\=0... ;)

[quote]
can you use Cauchy's Equation if f is not continuous?
[/quote]

No.

Pierre.[/quote]

[quote]No.

Pierre[/quote]

So my solution above is not valid? But even when proving Cauchy's Equation, if continuity is not explicitly given, we'd have to assume that the function is bounded below (or above) at some interval. In this problem, f is only defined for all reals, is that enough to support my solution, then?

Solutions to Cauchy equation are linear if you assume continuity, or that the functions is bounded on some interval with non-zero length, or if you assume mesurability on such interval.
Without any further hypothesis, there are many other solutions than the linear ones.

Pierre.

I will use al.M.V.'s achivement that $f$ is additive and $f(ax)=f(x)$.
If $a\neq 1$ then there are $u,v$ s.t. $u+v=0$ and $u+av=1$. We have \[f(x)=f((u+av)x)=f(ux)+f(avx)=f(ux)+f(vx)=f((u+v)x)=f(0).\]


To prove that  $ 7|n^3 \plus{} 3^n$ is corect only when $ 7|n^3.3^n\plus{}1$Thank you.

We have that $ n^3 \bmod{7} \in \{0, 1, 6\}$.

Therefore, if $ 7|n^3 \plus{} 3^n$, we have

$ n^3 \equiv 0 \pmod{7} \implies 3^n \equiv 0 \pmod{7}$
$ n^3 \equiv 1 \pmod{7} \implies 3^n \equiv 6 \pmod{7}$
$ n^3 \equiv 6 \pmod{7} \implies 3^n \equiv 1 \pmod{7}$

However, $ 3^n \equiv 0 \pmod{7}$ is clearly impossible, as $ 3^n$ only has factors of 3.

Therefore $ n^3 \cdot 3^n \equiv 6 \pmod{7}$.

Um, how does ffao's proof actually prove the thing?
As far as I can tell, this only proves that the relationship is true in one instance.
Am I missing something?
:?:

The chain of implication goes the other way.  In other words, ffao proves that the "one instance" is in fact the only possible instance.

I was going to say I didn't follow it, but I think I do now.  Just ffao has a typo: $ 3^n \equiv 1 \pmod{7}$ is perfectly possible. $ 3^n \equiv 0 \pmod{7}$ is impossible.  With this correction, he has proved $ 7|n^3\plus{}3^n$ implies $ 7|n^3 \cdot 3^n \plus{}1$.


no one interested in this issue? it very nice.

[quote="quykhtn-qa1"]let a,b,c be nonnegative real numbers such that a+b+c=3;  prove that:
 a^2/(a^2+2b+c) +b^2/(b^2+2c+a) +c^2/(c^2+2a+b) ≥ 3/4[/quote]
$ \sum_{cyc}\frac{a^2}{a^2\plus{}2b\plus{}c}\equal{}\sum_{cyc}\frac{a^4}{a^4\plus{}2a^2b\plus{}a^2c}\geq\frac{(a^2\plus{}b^2\plus{}c^2)^2}{\sum(a^4\plus{}2a^2b\plus{}a^2c)}.$
Id est, we need to prove that
$ 4(a^2\plus{}b^2\plus{}c^2)^2\geq3\sum_{cyc}(a^4\plus{}2a^2b\plus{}a^2c)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^4\plus{}8a^2b^2)\geq(a\plus{}b\plus{}c)\sum_{cyc}(2a^2b\plus{}a^2c)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^4\plus{}5a^2b^2\minus{}2a^3b\minus{}a^3c\minus{}3a^2bc)\geq0,$ which is true by  Vasc's inequality:
$ (a^2\plus{}b^2\plus{}c^2)^2\geq3(a^3b\plus{}b^3c\plus{}c^2a).$


[quote="NguyenDungTN"]Let $ a,b,c,d > 0$. Prove that:
$ \frac{a\plus{}2b}{a\plus{}2c}\plus{}\frac{b\plus{}2c}{b\plus{}2d}\plus{}\frac{c\plus{}2d}{c\plus{}2a}\plus{}\frac{d\plus{}2a}{d\plus{}2b}\ge 4$[/quote]
My problem very easy!
General:
$ \frac{a\plus{}kb}{a\plus{}kc}\plus{}\frac{b\plus{}kc}{b\plus{}kd}\plus{}\frac{c\plus{}kd}{c\plus{}ka}\plus{}\frac{d\plus{}ka}{d\plus{}kb}\ge 4$
For all $ k\ge 1$

[quote="stephen38"]Could you post your solution,NguyenDungTN?????? :maybe:[/quote]
Ok, stephen28! :)  :D
By Cauchy-Chwarz Inequality, we have:
$ \frac{a\plus{}kb}{a\plus{}kc}\plus{}\frac{b\plus{}kc}{b\plus{}kd}\plus{}\frac{c\plus{}kd}{c\plus{}ka}\plus{}\frac{d\plus{}ka}{d\plus{}kb}\ge\frac{(k\plus{}1)^{2}(a\plus{}b\plus{}c\plus{}d)^{2}}{\sum (a\plus{}kb)(a\plus{}kc)}\equal{}\frac{(k\plus{}1)^{2}(a\plus{}b\plus{}c\plus{}d)^{2}}{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\plus{}(k^{2}\plus{}k)(ab\plus{}bc\plus{}cd\plus{}da)\plus{}2k(ac\plus{}bd)}$
We will prove that:
$ \frac{(k\plus{}1)^{2}(a\plus{}b\plus{}c\plus{}d)^{2}}{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\plus{}(k^{2}\plus{}k)(ab\plus{}bc\plus{}cd\plus{}da)\plus{}2k(ac\plus{}bd)}\ge 4$
$ \Leftrightarrow (k\plus{}1)^{2}(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2})\plus{}2(k\plus{}1)^{2}(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd)\ge 4(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2})\plus{}4(k^{2}\plus{}k)(ab\plus{}bc\plus{}cd\plus{}da)\plus{}8k(ac\plus{}bd)$
$ \Leftrightarrow (k^{2}\plus{}2k\minus{}3)(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2})\plus{}2(k^{2}\minus{}2k\plus{}1)(ac\plus{}bd)\ge 2(k^{2}\minus{}1)(ab\plus{}bc\plus{}cd\plus{}da)$
we have:
$ (a\plus{}c)^{2}\plus{}(b\plus{}d)^{2}\ge 2(a\plus{}c)(b\plus{}d)$
And: $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\ge ab\plus{}bc\plus{}cd\plus{}da$
So:
$ (k^{2}\plus{}2k\minus{}3)(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2})\plus{}2(k^{2}\minus{}2k\plus{}1)(ac\plus{}bd)\ge (4k\minus{}4\plus{}2(k\minus{}1)^{2})(ab\plus{}bc\plus{}cd\plus{}da)\equal{}2(k^{2}\minus{}1)(ab\plus{}bc\plus{}cd\plus{}da)$
Equality holds if and only if $ a\equal{}b\equal{}c\equal{}d$ :wink:


Prove that 5n + 5m can be written as the sum of two squares (for positive integers m, n) if and only if m and n are of the same parity.

A triangle with sides of lenghts a, b, c has area S and a triangle with sides of lengths a + b, b + c, c + a has area T. Show that T :ge: 4S.

For the first one:

I used Fermat's thm:

n can be written as the sum of 2 squares iff any prime divisor of n of the form 4k+3 has an even exponent in n.

Let's assume m>n. If m and n are even then the number is already the sum of 2 squares. If m and n are odd then 5^m+5^n=5^n*(5^(m-n)+1). m-n is even, so 5^(m-n)+1 is the sum of 2 squares, and when we multiply it by 5^n we don't modify the exponents of the primes of the form 4k+3 which divide it, so by Fermat's thm 5^n*(5^(m-n)+1) can still be written as the sum of 2 squares. Conversely, if m and n have different parities then 5^m+5^n=5^n*(5^(m-n)+1)=5^n*(5^(2k+1)+1) = 2*3*5^n*(5^2k+..+5+1). It's obvious that 3 divides this number but 9 doesn't divide it, so, again by Fermat, it can't be written as sum of 2 squares (we don't even need Fermat for this one: 3|a^2+b^2=> a=b=0(mod 3) => 9|a^2+b^2).

I think we don't need Fermat for the first part either. Maybe there's an easy way to prove that 5(a^2+b^2) can be written as the sum of 2 squares. In fact, I'm sure of it, but I'm too lazy to think up another soln right now.

#2 [hide]is equivalent to proving that for 3 sides of a triangle a, b, c,

abc :ge: (a + b - c)(a - b + c)(-a + b + c).

I've verified this, but I don't know how to prove it (yet)...[/hide]

[hide]

replace a,b,c by u+v, v+w, u+w

abc >=(a + b - c)(a - b + c)(-a + b + c)

its equilavalent to

(u+v)(v+w)(u+w)>=8uvw

which is just a simple application of am-gm for each bracket!!

[/hide]

I SO knew I've seen that inequality before. *groan* and I'm doing the Tournament of Towns A-level paper today... watch me get none of the questions right. ;)


LaTeX
calculus
integration
modular arithmetic
number theory unsolved
number theory

I recently found a dyophantine equation which is apparently attributed to Henry Brocard. Please excuse the lack of Latex: 1+x+x^2+x^3=y^2. Does anybody know something about the way he solved this?

Suppose $ (x,y) \in \mathbb{Z}^{2}$ is a solution. Then $ (1+x)(1+x^{2}) = y^{2}$ and clearly $ y = 0$ iff $ x=-1$. So admit $ |y| \ne 0$. Then either $ \left\{\begin{array}{l}1+x = a^{2}\\ 1+x^{2}= b^{2}\end{array}\right.$ or $ \left\{\begin{array}{l}1+x = 2a^{2}\\ 1+x^{2}= 2b^{2}\end{array}\right.$, for some $ a,b\in \mathbb{N}_{\;\!0}= \{0, 1, \ldots\}$, according that $ x$ is either even or odd, respectively [just observe that $ \gcd(1+x,1+x^{2}) = \gcd(1+x,2x) \le 2$]. But now $ |x|^{2}< 1+x^{2}\le (|x|+1)^{2}$, with the equality holding iff $ x = 0$. So observing that $ y^{2}= 1$ iff $ y = \pm 1$, we put $ (x,y) = (0, \pm 1)$ and $ (x,y) = (-1,0)$ to one side and are left with sieving for integral solutions $ (a,b,x)$ to the second system such that $ x \equiv 1 \bmod 2$ and $ \gcd(a,b) = 1$.

Then plainly $ a^{4}+(a^{2}-1)^{2}= b^{2}$ and hence $ (a^{2},a^{2}-1,b)$ is a primitive Pythagorean tern. Consequently, there must exist $ u,v\in\mathbb{N}_{\;\!0}$ such that $ \gcd(u,v) = 1$ and $ \left\{\begin{array}{l}a^{2}= 2uv+v^{2}\\ a^{2}-1 = 2u^{2}+2uv \end{array}\right.$ or $ \left\{\begin{array}{l}a^{2}= 2u^{2}+2uv \\ a^{2}-1 = 2uv+v^{2}\end{array}\right.$. In the first case $ 2u^{2}+1 = v^{2}$, which is a famous equation, giving $ (u,v) = (0,1)$ as a unique solution; whereas in the second one $ 2u^{2}= v^{2}+1$, that is famous too and yields $ (u,v) = (1,1)$. Thus going back to the origin, we get $ (a,b) = (1,1)$ or $ (a,b) = (2,5)$, so finally $ (x,y) = (1,\pm 2)$ or $ (x,y)=(7,\pm 20)$. Now check these and you're done. :wink:

[quote="s.tringali"]In the first case $ 2u^{2}+1 = v^{2}$, which is a famous equation, giving $ (u,v) = (0,1)$ as a unique solution[/quote]
$ u=2, v=3$ (and infinitely more, see Pell equations).

[quote="s.tringali"]whereas in the second one $ 2u^{2}= v^{2}+1$, that is famous too and yields $ (u,v) = (1,1)$.[/quote]
$ u=5, v=7$, same like the case before.

I think there is something wrong when you solve the second equation, derived from the initiall one. Those are not the Phytagorean triplets. From the way you wrote you find as ZetaX confirmed infinite solutions, from Pell, but there is a mystake. In a book wrote by professor Cucurezeanu, there is an indication. The author says that the second ecuation can be turned into one of the following form $ x^{4}-y^{4}$= $ z^{2}$....again famous

So I don't think we've solved it...any other ideas? :maybe:

As s.tringali has noted we have $ x+1=2u^{2}\ \ 1+x^{2}=2v^{2},\ \ (u,v)=1$ and $ (u^{2})^{2}+(u^{2}-1)^{2}=v^{2}.$ Therefore $ (u^{2},u^{2}-1,v)$ is a primitive Pythagorean triple ( here we have to notice that the case $ u=1$ yields $ x=1,y=\pm 2$). Now let's suppose that $ x>1.$
[b]Case 1.[/b] Let's consider $ u-odd.$ Then there exist $ n,q\in \mathbb{N}$ with different parity such that $ (n,q)=1$ and \[ u^{2}=n^{2}-q^{2},\ \ u^{2}-1=2nq,\ \ v=n^{2}+q^{2}.\] Now we have \[ 1=u^{2}-(u^{2}-1)=n^{2}-q^{2}-2nq=(n-1)^{2}-2q^{2},\] \[ (q^{2}+1)^{2}=q^{4}+2q^{2}+1=q^{4}+(n-q)^{2}.\] Now because of the fact that $ (n+q,n-q)=(n-q,2q)=1$ and $ u^{2}=n^{2}-q^{2}$ we obtain that $ n-q=t^{2}.$ Thus $ (q^{2}+1)^{2}=q^{4}+t^{4}$ which is a contradition because the equation \[ x^{4}+y^{4}=z^{2}\] doesn't have a solution in integers with $ z>0.$
[b]Case 2.[/b] Now let's suppose that $ u-even.$ Thus \[ u^{2}-1=n^{2}-q^{2},\ \ u^{2}=2nq,\ \ v=n^{2}+q^{2}\] for $ (n,q)=1.$ We see that $ n$ is even because if doesn't we would have $ n^{2}\equiv 1\pmod{4},\ q^{2}\equiv 0\pmod{4}$ and $ n^{2}-q^{2}\equiv 1\mod{4}$ which is a contradiction with $ u^{2}-1\equiv-1\pmod{4}.$ Therefore we have \[ n=2t^{2}.\] Now $ 1=u^{2}-(u^{2}-1)=(n+q)^{2}-2n^{2}.$ Therefore \[ 2n^{2}=8t^{4}=(n+q)^{2}-1,\] \[ 8t^{4}=(n+q-1)(n+q+1),\] \[ 2t^{4}=\frac{n+q-1}{2}\frac{n+q+1}{2},\] \[ 2t^{2}=a(a+1).\] Since $ (a,a+1)=1$ we have that one of these numbers is $ b^{4}$ and the other is $ 2c^{4}.$ In both cases we obtain $ b^{4}-2c^{4}=\pm 1.$
First let's consider the case with "+". If $ d=c^{2}$ we have $ b^{4}-2d^{2}=1$ and therefore \[ (d^{2}+1)^{2}=d^{4}+2d^{2}+1=d^{4}+b^{4}\] which  is a contradiction. In the case with "-" and $ d=c^{2}$ we have $ b^{4}-2d^{2}=-1$ and $ (d^{2}-1)^{2}=d^{4}-2d^{2}+1=d^{4}-b^{4}.$ It's clear that $ b\not = 0$ ($ b^{4}=a$ or $ b^{4}=a+1$). Since the equation \[ z^{2}=x^{4}-y^{4}\] is solvable iff $ y=0$ or $ z=0$ we obtain that $ z=0$ and $ b^{4}=d^{4}.$ Since $ (c,d)=1$ we get $ b=\pm 1$ and $ d=\pm 1.$ Now we get $ x=7,y=\pm 20.$

Other similar equation is \[ 1+x+x^{2}+x^{3}+x^{4}=y^{2}.\] I know a solution using base-x representation of $ y$ so I believe that there is a solution of the above equation using such argument.


Given n boxes of identical apperance, n-1 of which are normal and the other is defective.  There are 12 bottles of beer in each box.The normal bottles are of equal weight, so are the defective ones, but the defective bottle is lighter than the normal one and we don't know their actual weights. We also have at hand a specially designed two-pan balance such that when bottles are put on both sides of the balance, we can tell at once not only which side is lighter but also the weight difference between two sides. We can use this balance 3 times. In order to guarantee to pick out the defective box, what's the maximum of n?


A small sphere is moving at a constant speed in a vertical circle. Below is a list of quantities that could be used to describe some aspect of the motion of the sphere.
         I - kinetic energy
        II - potential energy
       III - momentum
Which of these quantities will change as this sphere moves around the circle?
 A. I and II only     B. I and III only     C. II only     D. III only      E. II and III only

The answer is E, but I don't know why the kinetic energy doesn't change.

If the speed is $ v$ and the mass is $ m$, the kinetic energy is always $ \frac{1}{2}mv^2$.  We are given that the speed is constant.

Another way to put it is that kinetic energy is directionless.  So we can change the direction of our motion while maintaining the same [i]amount[/i] of kinetic energy.

Ah, I didn't notice the "at a constant speed", even though I typed them. :wink:


geometry
perimeter
trigonometry
parallelogram
angle bisector
geometry proposed

Let the parallel cuts $BC$ at $K$ and the angle bisector cuts $AC$ at $N$. We have to prove that $KC$=$\frac{a+c}{2}$.
\[KC=\frac{a+c}{2} \Leftrightarrow\]
\[MC \frac{\sin \angle KMC}{\sin \angle MKC} = \frac{a+c}{2}\Leftrightarrow\]
\[\frac{\sin \angle HNC}{\sin (180^{\circ}- C-\angle HNC)} = \frac{a+c}{b} \Leftrightarrow\]
\[\frac{\sin(90^{\circ}+\frac{ A - C}{2})}{\sin (90^{\circ}+\frac{A + C}{2})}=\frac{\sin A+\sin C}{\sin B}\Leftrightarrow\]
\[\frac{\cos\frac{A-C}{2}}{\sin\frac{B}{2}}=\frac{\sin A + \sin C}{2\sin\frac{B}{2}\cos\frac{B}{2}} \Leftrightarrow\]
\[2\sin\frac{A+C}{2}\cos\frac{A-C}{2} = \sin A + \sin C.\]
But it is the compound angle formula.

I suppose AB>BC (if AB=BC it is evidently). I note $U\in AB\cap l$ and the point S where the bisector of the angle $\widehat{ABC}$ meets the line AC. It can be demonstrated easily that MU || BS (with equalities between angles) . From $AUM\sim ABS$ results:

$\frac{c}{AU}=\frac{AS}{AM}=\frac{bc}{a+c}\cdot \frac{2}{b}=\frac{2c}{a+c}\Longrightarrow AU=\frac{a+c}{2},\ BU=\frac{c-a}{2}$.

Thus$,\ CB+BU=a+\frac{c-a}{2}=\frac{a+c}{2}=AU,\ i.e.\ MC+CB+BU=MA+AU$.

[quote="Jutaro"][color=darkred]Let $ABC$ be a triangle, $H$ the orthocenter and $M$ the midpoint of $AC$. Let $\ell$ be the parallel through $M$ to the bisector of $\angle AHC$. Prove that $\ell$ divides the triangle in two parts of equal perimeters.

[i]Pedro Marrone, Panamá[/i][/color] [/quote]

[color=darkblue][b]Proof.[/b] Construct the parallelogram $AUCD.$
Assume that: $AB>AC,$ and denote: $S$ be the foot of angle bisector of angle $\widehat{ABC},\, U=AB\cap l,\,V=BC\cap l$
Thus, easy show that: $BS\parallel MV\implies\widehat{BVU}=\widehat{CBS}=\widehat{SBA}=\widehat{MUA}=\widehat{BUV}=\widehat{MDC}$
$\implies\begin{cases}BU=BV\\ CV=CD\end{cases}\implies CB+BU=AU$
$\implies MC+CB+BU=MA+AU$[/color]

I will use the attached diagram;

First note that $MK$ is common in both parts, similarly with $MC$ that is equal to $MA$, so, the real problem is to show that $BK+AB=KC$, some angle chasing with the parallel will show that $\angle BKY=\angle BYK$, meaning that $BK=BY$, now for the conclusion, we draw the parallel to $MY$ through $C$, then it is clear that $KC=YD$ and that $Y$ is the midpoint of $AD$, so $AB+BY=YD$ but with substitutions we have that $AB+BY=AB+BK$ so we are done


Hi,

I was going through Art of Problem Solving Vol. 1, and I think there is a mistake in Chapter 5, Using the Integers. On page 44, it gives us the summary of what we know.

It says:

Let's summarize what we can do with congruences...If $ a \equiv b \pmod{m}$ and $ p \equiv q \pmod{m}$...we have:
.
.
.
$ 5)\ (a\plus{}b) \equiv (p\plus{}q) \pmod{m}$
$ 6)\ ab \equiv pq \pmod{m}$

That is exactly what the book says. They should instead be:

$ 5)\ (a\plus{}p) \equiv (b\plus{}q) \pmod{m}$
$ 6)\ ap \equiv bq \pmod{m}$

If this mistake could be corrected in the future editions, that would be appreciated.

(I hope I am not repeating a post, i.e. this has already been reported. And I also hope that I am posting this is the right section.)

Regards,
limac

I believe this was already addressed in another topic, and you have a much older version of the book.

On a related topic, on what I believe to be the latest edition, in Chapter 0 (Prove It!), it states the pigeonhole principle as:

"The Pigeonhole Principle states that given kn+1 objects which are placed in n boxes, there must be some box with k+1 objects. The principle..."

I don't think this is correct... I might be making a mistake, but what about, for example, the set

$ \{1,1,1,1,\hdots,1,(k\minus{}1)n\plus{}2\}$. This doesn't contain k+1 objects...

Unless I'm very mistaken, I [i]think[/i] that it should be [i]at least[/i] k+1 objects.

If I'm wrong, tell me!

Ok...So while we are at it, I think there is a/another mistake with the Solution Manual for AoPS Volume 1: the Basics. On page 31, it shows the proof of why the sum of three consecutive squares cannot equate to another perfect square.
This is the actual question by the way:

$ 109.$ Prove that it is impossible for three consecutive squares to sum to another perfect square.

On pg. 31(Solution Manual for AoPS: Vol 1: the Basics, pg.31), it says:

109) Let the squares be $ (x\minus{}1)^2,$ $ x^2,$ and $ (x\plus{}1)^2.$ The sum of these is $ 3x^2\plus{}2.$This number is congruent to $ 2\pmod{3}...$

That should be $ 3x^2\plus{}2x,$ instead of $ 3x^2 \plus{} 2,$ therefore, it shouldn't be $ 2\pmod{3}$, but should instead be $ 2x\pmod{3}$. And that , I think, changes the whole rest of the solution.

Tell me if I am mistaking. I have the 7th edition by the way(I bought it not that long ago...maybe a couple of months ago).

Regards,
limac

Ah, yikes, I see where I am mistaking. I kept on simplifying $ (x\minus{}1)^2$ as $ (x\minus{}1)(x\plus{}1)$ giving me $ x^2  \minus{} 1$ instead of $ x^2 \minus{}2x \plus{} 1$ lol :)

Thank you for your "confusion clearer"!

Regards,
limac

[quote="BOGTRO"]On a related topic, on what I believe to be the latest edition, in Chapter 0 (Prove It!), it states the pigeonhole principle as:

"The Pigeonhole Principle states that given kn+1 objects which are placed in n boxes, there must be some box with k+1 objects. The principle..."

I don't think this is correct... I might be making a mistake, but what about, for example, the set

$ \{1,1,1,1,\hdots,1,(k \minus{} 1)n \plus{} 2\}$. This doesn't contain k+1 objects...

Unless I'm very mistaken, I [i]think[/i] that it should be [i]at least[/i] k+1 objects.

If I'm wrong, tell me![/quote]

a box that has at least k+1 objects

has k+1 objects (and maybe some more...)

at least that's what's i know...


Let $ a,b,c\ge 0$ .Prove that:
$ \frac 1{x^2\plus{}xy\plus{}y^2}\plus{}\frac 1{y^2\plus{}yz\plus{}z^2}\plus{}\frac 1{z^2\plus{}zx\plus{}x^2}\ge \frac 9{(x\plus{}y\plus{}z)^2}$

To Consult Book

VASILE CIRTOAJE
ALGEBRAIC INEQUALITIES
GIL Publishing House
Pg. 231 & pg 373
 :)

Alberto I don't have this book.Can you give me a copy?I'm from Vietnam so it's very hard to get this book.


integration
logarithms
calculus
derivative
calculus computations

Can't quite figure this one out, I'd prefer solving it using an integrating factor.

$ \frac{dy}{dx} \minus{} \frac{2y}{x} \equal{} x^2$

thanks!

For differential equations of the form $ y' + p y = q$ we can indeed use the method of the integrating factor (multiplying by $ M = e^{\int p \, \,dx}$).

For this problem, $ p(x) = - \frac2x$, $ q(x) = x^2$.

We have that $ \int p(x) \, dx = - 2 \ln |x|$ hence $ M(x) = \frac1{x^2}$.

Our solution then becomes

$ y(x) = \frac{ \int q(x) M(x) \, dx + C}{M(x)} = x^3 + Cx^2$.

Where exactly did your problem with this exercise come from? It is a rather straightforward application of the technique. Or perhaps it is the technique itself that is somewhat unclear?

I just don't fully understand this type of differentiation, its quite new to me. thanks for your help

hello, at first you can solve the homogeneous part of your equation
$ \frac{dy}{dx}\equal{}2\frac{dx}{x}$ with the solution $ y_h(x)\equal{}Cx^2$, now you must find a particular solution of the inhomogeneous part of your equation, this is $ x^3$ so we get
$ y(x)\equal{}x^3\plus{}Cx^2$ as the solution.
Sonnhard.


10)  There is snow on the ground.  This is a huge difference from Atlanta, where you could probably walk in your shorts and t-shirt at this time of the year.  It was funny to see the air traffic controller guy all bundled up in his puffy little snow suit.

9)  There is an election.  What???  I found this out when I picked up the National Post this morning (it's been a while since I last read that paper).  Hahaha, how ignorant am I.

8)  Nothing is crunk here.  Enough said.

7)  Canadians apologize to you when YOU bump into THEM.  There was a lot of this at the airport.

6)  My roomate before I boarded the plane:  "Where's Windsor?"  My Canadian buddies:  "Who's Georgia?"

5)  Hockey's in the newspaper.  Lol.

4)  Hockey's on TV...LOL.

3)  Kids play hockey outside.  ROFL. :rotfl:

2)  A deer runs out in front of your car on the middle of the road.  That was so weird when it happened...

And the number 1 thing I've noticed after half a year away from Canada is...

1)  Liberals.

[quote="lowest_reynolds"]
And the number 1 thing I've noticed after half a year away from Canada is...

1)  Liberals.[/quote]
The Liberals have been in power too long; they haven't helped Canada as much as I have expected them to. So let's go: New Democrat Party! ^_^

[quote="lowest_reynolds"]10)  There is snow on the ground.  This is a huge difference from Atlanta, where you could probably walk in your shorts and t-shirt at this time of the year.  It was funny to see the air traffic controller guy all bundled up in his puffy little snow suit.
[/quote]
It's quite opposite in Edmonton. It has been too warm here, so snow melted down in December.

[quote="G-UNIT"]NUMBER 1 THING I'VE LEARNED UPON SEEING THIS TOPIC

  Modern society is filled with sad, sad individuals who have WAY too much time on their hands.[/quote]

Is that comment supposed to sound rude??  :?

Anyways, welcome back to Canada!!

What an overwhelming misinterpretation; my comment was suppose to be pleasant and comforting.

[quote="blahblahblah"]You're not witty.[/quote]

-

[quote="blahblahblah"][quote="G-UNIT"]I find that comment highly offensive, considering it came from one whose name is representative of that which he speaks.[/quote]

Sorry, try again.[/quote]

pwnsauce!  holla

[quote="G-UNIT"]What an overwhelming misinterpretation; my comment was suppose to be [b]hurtful and conceited[/b] [/quote]

FYP

Okay, that's it! You people need to learn some public manners.

 :police:


geometry
integration
LaTeX
logarithms
calculus
derivative
calculus computations

Find the area between the curves of
y= 1/x
y=1/x^2
x=2

Thanks a lot, i was pretty sure that was how you did it but its been a while. Also how do you write with the symbols?

So is this basically the rest?

$\int^{2}_{1}x^{-}1-x^{-}2 dx$

then its $\log(x)+1/x$ from 2 to 1
then you do top - bottom
and have $\log(2)+1/2-\log(1)+1$ ?

When you want to group things in $\text{\LaTeX},$ use brackets, {}.

For instance, (dollar) x^-2 (dollar) becomes $x^{-}2$ which is no good.

What you really want is (dollar) x^{-2} (dollar) or $x^{-2}.$

thank you very much for that tip.
back to the problem, is what i had right?

Sign error on the evaluation.

$\left.\left(\ln(x)+\frac1x\right)\right|_{1}^{2}=\left(\ln 2+\frac12\right)-\left(\ln 1+1\right)=\ln2-\frac12\approx .193.$

It's an area, so it should be positive, but the curves aren't all that far apart, so the number should be fairly small.

Ya i was wondering why it was positive. So even though when evaluated you get a negative number you just make it positive?

[quote="orangedude654"]So even though when evaluated you get a negative number you just make it positive?[/quote]
Don't you dare do that. The fact that an area must be positive falls into the realm of "plausibility checks." Such checks are numerous, and the more of them you employ, the better. If you evaluate an area and get a negative number, then you've made a mistake, and you're now in de-bugging mode. Find and fix the mistake - don't just say "oh, well" and flip a sign around.

Ya i didnt think so. See the problem is when one uses log instead of ln you get a negative number? I thought the derivative of 1/x= log(x)?

The antiderivative of $\frac1x$ is the natural logarithm. It is fairly common among mathematicians to regard the natural logarithm as "the" logarithm and to use the notation $\log x$ for it - indeed, you will see many posters on this board using that notation. However, that's not the way calculator keys are labeled. The button on a calculator labeled $\log$ gives the base 10 or common logarithm. So enter $\log 2$ on that calculator and you will get $\log_{10}2\approx.30103,$ which is the wrong number to use in this problem. The button that gives the natural, or base $e$ logarithm is $\ln.$ Use that to get $\ln 2\approx .693.$

(I did that without a calculator, since I have both of those numbers memorized - $\log_{10}2$ from long-ago familiarity with the tables of common logarithms I used before calculators became common, $\ln 2$ because of its association with half life and doubling time computations.)


algebra
polynomial
calculus
derivative
function
algebra solved

Let f be a polynomial with real coefficients. If f has all its roots real prove that a*f+f' also has all roots real for every real a. (f' is the derivative polynomial)

Hi Grobber,

Suppose f is a polynomial  of degree n
Let x_1<...<x_r  distincts roots of f with multiplicity m_1,...,m_r
The polynomial f'-af have x_k roots  with multiplicity m_k- 1
and it has deg (f) - r  roots.

But f '/f  = sum_{k =1 to r} m_k/(x-x_k) tends  to
+oo when x-->x_k , x>x_k
 -oo when x-->x_(k+1), x <x_(k+1)

f'/f is continous on, so f'/f reach a on ]x_k ; x_(k+1)[ this give r-1 roots

finally f ' - af has at  least deg(f) - r + (r-1)= deg f -1 real roots, and
all its roots are real

Thanks, I did sth similar, but I used this function: g(x)=e^ax*f(x). This function only has the roots of f, since e^ax different from 0 even for complex x, since e^it=cost+isint=0 iff cost=sint=0, false. g'(x)=af(x)+f'(x) and then I did sth similar with orders of multiplicity and stuff like that. I'm not sure this is correct though  :D


[quote="Soarer"]Let $a+b+c=1$, we have to prove
$\sum \frac{1}{\sqrt{a^{2}+a}}\ge \frac{9}{2}$
[/quote]

sorry, but we have there $2a^{2}+ab+bc$ which is not $a^{2}+a$.

[quote="tripleth"]Let $a,b,c \geq 0$. Prove:

$\frac1{\sqrt{2a^{2}+ab+bc}}+\frac{1}{\sqrt{2b^{2}+bc+ca}}+\frac1{\sqrt{2c^{2}+ca+ab}}\geq \frac{9}{2(a+b+c)}.$[/quote]
Ugly but poof:
$\sum_{cyc}\frac1{\sqrt{2a^{2}+ab+bc}}=\sum_{cyc}\frac{2a+b+c}{2\cdot\frac{2a+b+c}{2}\sqrt{2a^{2}+ab+bc}}\geq$
$\geq\sum_{cyc}\frac{2a+b+c}{\left(\frac{2a+b+c}{2}\right)^2+2a^{2}+ab+bc}.$
But $\sum_{cyc}\frac{2a+b+c}{\left(\frac{2a+b+c}{2}\right)^2+2a^{2}+ab+bc}\geq\frac{9}{2(a+b+c)}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(100a^6+600a^5b+588a^5c+1123a^4b^2-357a^4c^2-1842a^3b^3+$
$+1090a^4bc-1414a^3b^2c+1330a^3c^2b-1218a^2b^2c^2)\geq0,$ which is easy.
Maybe there is something nicer?

@ pohoatza (who wrote the above in 2006!):
Since the inequality was homogeneous, Soarer assumed $a+b+c=1$, and thus $2a^2 + ab + bc = a^2 + a(a+b+c) = a^2 + a$ !!

We have 2a^2+ab+bc. But what you have is 2a^2+ab+ac!!!

in Vo Quoc Ba Can's book the inequality can be proved using Holder:
$$(\sum_{cyc}\frac{1}{\sqrt{2a^2+ab+bc}})^2\ge \frac{27(a+b+c)^3}{\sum_{cyc}(b+2c)^3(2a^2+ab+bc)}$$
but I don't know how the expression $(b+2c)^3$ comes out.

Hmmm what's that book... may I have a link?

http://diendantoanhoc.net/forum/index.php?/topic/57532-cac-chuyen-d%E1%BB%81-bdt-c%E1%BB%A7a-anh-c%E1%BA%A9n/#entry531466

the third one (sorry it's in Vietnamese,but not difficult to understand)


[quote="tranthanhnam"]$\left\{\begin{array}{l}1-2x=3\\1+2x+12y=0\\x^2+z=1\end{array}\right.$[/quote]

Please stop spamming.

  Darij

[quote="tranthanhnam"]$\left\{\begin{array}{l}1-2x=3\\1+2x+12y=0\\x^2+z=1\end{array}\right.$[/quote]

1)-do you think it's an exercice of geometry  :!:
2)-it's very easy (and not nice)
3)-[hide] $x=-1$ , $y=\frac{1}{12}$ , $z=0$[/hide]

problem of the year  :roll:  next time..


let $\triangle ABC$ be a triangle and $\omega$ a circle with center $O$ that is tangent to $AB$ in $K$ and to $AC$ in $L$, such that $O\notin BC$. $M$ is a point on $AO$ such that $O\in(AM)$ and $\angle KAL=\angle KLM$. the circle with diameter $CO$ is intersected again by the line $KL$ in $P$ and by the line $LM$ in $Q$. the circle with diameter $BO$ is intersected again by the line $KL$ in $S$ and by the line $KM$ in $T$. the lines $BO$ and $ST$ intersect in $U$, the lines $CO$ and $PQ$ intersect in $V$, and the lines $PQ$ and $ST$ intersect in $W$.
show that $O$ lies on the circumcircle of the triangle $UVW$.

[img]8295[/img]

What a messy problem!  :D

ML is a tangent to (AKL). Therefore $\angle MLO = \angle LKO = \angle OLK$ and $\angle QCO = \angle QLO = \angle MLO = \angle OLK = \angle OLP = \angle OCP$, using directed angles, then O is the midpoint of arc PQ, and $\angle OVQ = \angle OVW = 90$. We prove also that $\angle OUW=90$. Therefore $OW$ is the diametre of the circle $(UVW)$.

Did you invent it?  :)

[quote="edriv"]

Did you invent it?  :)[/quote]

I think so, seeing that he posted also at

http://www.mathlinks.ro/Forum/viewtopic.php?t=140630


geometry
circumcircle
trigonometry
similar triangles
trig identities
Law of Sines

[quote="Geometry Revisited"]Let $p$ and $q$ be the radii of two circles through $A$, touching $BC$ at $B$ and $C$, respectively. Then $pq = R^2$.[/quote]

I think I'm having a little trouble interpreting this problem. Could someone please clarify how exactly the circles are interacting with the triangle?

touch = are tangent to... I got confused on that for a while too

Since $BC$ is tangent to $\odot D$, thus $\angle{ABC}= \frac{1}{2} \angle{ADB} = \angle{ODB}$. Because $O$ is the circumcentre of $\triangle{ABC}$, thus $\angle{EOC}= \frac{1}{2} \angle{AOC} = \frac{1}{2} \cdot 2\angle{B}= \angle{B}$, thus $\angle{EOC}=\angle{ODB}$.

Since $\angle{OBC}=\angle{OCB}$, $\angle{DBC}=\angle{ECB}$, so $\angle{OBD}=\angle{ECO}$, thus $\triangle{ECO} \sim \triangle{OBD}$, thus $p: R=R: q \Longrightarrow R^2=pq$.

Wow, what an honor it is to be given a nice solution AND have a picture named after me. :)

I solved it using similar triangles and the Extended Law of Sines.

Extending BD on your diagram to a point P on circle D and CE to a point Q on circle E, respectively, we have right triangles PAB and QAC.

Now, $\angle B = \angle BPA$ and $\angle C = \angle CQA$. So $\sin B = \frac{c}{2p}$ and $\sin C = \frac{b}{2q}$.

Multiplying together, we have $4pq = \frac{c}{\sin B} \cdot \frac{b}{\sin C}$. And from the Extended Law of Sines, this is $4R^2$. Dividing by 4, we have $pq = R^2$ as desired.


Well this year [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=98]Tuymaada Yakut Olympiad[/url] seems to have almost finished.

I know some Romanian results: Ilie Livia Gold medal, Lucian Turea (maky) Silver medal and Mihai Dumitrescu Silver medal.

The questions will be coming in soon also.

i will post the problems :)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=576231#576231]Problem 1[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=576232#576232]Problem 2[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=576233#576233]Problem 3[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=576234#576234]Problem 4[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=576235#576235]Problem 5[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=576237#576237]Problem 6[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=576236#576236]Problem 7[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=576239#576239]Problem 8[/url]

there were two days, problems 1-4 are from first day, and problems 5-8 are from second day. the work time was 5 hours, each day.

the romanian team obtained good results

livia ilie               - gold - 7 7 5 5 7 7 1 3 - 42 pts
mihai dumitrescu - silver - 7 6 2 2 7 7 5 0 - 38 pts
lucian turea (me) - silver - 7 7 7 0 0 7 5 0 - 33 pts

unfortunately i misread the statement of problem 5, the easiest of all :( and i missed 7 points, but the medal would have been the same.
the best score was 50 - 7 7 7 7 7 7 7 1 , and was obtained by a student from moscow.

This is the third time I heard the word "Tuymaada". Before, I don't know this contest.
Another one is QEDMO! What's it?


algebra
polynomial
inequalities proposed
inequalities

[hide="Solution"] When $x+y+z = 0$ we know that $x^{3}+y^{3}+z^{3}= 3xyz$, so we know that the product of $x, y, z$ is positive.

We conclude that two of $x, y, z$ are negative.

An equivalent construction of the problem, then:  given two positive reals $a, b$ (such that WLOG $x =-a, y =-b, z = a+b$), we know (from the above argument) that

$(a+b)^{3}-a^{3}-b^{3}> 0$

(This is obvious by expansion.)  We want to show that

$(a+b)^{5}-a^{5}-b^{5}> 0$

(Also obvious by expansion.)  QED.

In fact, we can conclude that $x^{2k+1}+y^{2k+1}+z^{2k+1}> 0$ for $k \in \mathbb{N}$. [/hide]

[quote="Ottem"]Prove the following, for real numbers $x,y,z$ such that $x+y+z=0$:
$x^{3}+y^{3}+z^{3}> 0 \Longleftrightarrow x^{5}+y^{5}+z^{5}> 0.$[/quote]
Ugly solution:
Suppose wlog $x\le y\le z$, thus $z>0$. If $y>0$, we would have $y^{n}+z^{n}>(y+z)^{n}$ for $n=3$ or $5$, contradiction. So let $a: =-x$, $b: =-y$ and we have to prove $(a+b)^{3}>a^{3}+b^{3}\Longleftrightarrow (a+b)^{5}>a^{5}+b^{5}$ for $a,b\ge0$. Equivalently $a+b\ge0\Longleftrightarrow (a+b)(a^{2}+ab+b^{2})\ge0$ which is obvious.

Ugly because it is not straightforward to generalize this proof for other pairs of odd numbers instead of $(3,5)$ :maybe:

Edit: oops, t0rajir0u was faster. But what about the "$\Longleftrightarrow$"?

[quote="spanferkel"]But what about the "$\Longleftrightarrow$"?[/quote]

Hmm.  Then I need to handle [hide="more cases..."] If $xyz > 0$ then my argument already does the job.  Otherwise, there are two cases.

[b]Case:[/b]  $xyz = 0$, so that one of $x, y, z$ is zero and the others are negatives of each other.  But then $x^{2k+1}+y^{2k+1}+z^{2k+1}= 0 \forall k \in \mathbb{N}$ so this case is irrelevant.

[b]Case:[/b]  $xyz < 0$.  WLOG $z$ is negative, so that $x+y =-z$.  But because $x, y > 0$ we conclude that

$x^{3}+y^{3}-(x+y)^{3}< 0, x^{5}+y^{5}-(x+y)^{5}< 0$

Again from familiar binomial expansions, so this case is also irrelevant.

Edit:  In other words, I showed that

$xyz > 0 \implies x^{2k+1}+y^{2k+1}+z^{2k+1}> 0$
$xyz = 0 \implies x^{2k+1}+y^{2k+1}+z^{2k+1}= 0$
$xyz < 0 \implies x^{2k+1}+y^{2k+1}+z^{2k+1}< 0$

Taken together, this means that whenever $x^{3}+y^{3}+z^{3}> 0$ only the first case can occur, in which case $x^{5}+y^{5}+z^{5}> 0$, and vice versa. [/hide]

[quote="t0rajir0u"]
Edit:  In other words, I showed that

$xyz > 0 \implies x^{2k+1}+y^{2k+1}+z^{2k+1}> 0$
$xyz = 0 \implies x^{2k+1}+y^{2k+1}+z^{2k+1}= 0$
$xyz < 0 \implies x^{2k+1}+y^{2k+1}+z^{2k+1}< 0$
[/quote]
You've shown this only for $k=1,2$, right?

Now, what about the generalization $x^{2k+1}+y^{2k+1}+z^{2k+1}> 0\Longleftrightarrow x^{2\ell+1}+y^{2\ell+1}+z^{2\ell+1}> 0$ for $k,\ell\in\mathbb N$? I'm sure that holds, but how to show it without expanding?  :)

(Edit: I mean you have already proved this if one of $k,\ell$ is equal to 1.)
(Edit again: I am stupid, by the above it's proven for all $k,\ell$, as for any $k$, you have shown that $x^{2k+1}+y^{2k+1}+z^{2k+1}> 0$ is equivalent to the $k=1$ case, so for all odd numbers the ineqs are equivaent  :blush:  :blush: )

Nice solutions spanferkel and t0rajir0u! My solution was exactly as yours spanferkel!  :)

[quote="spanferkel"][quote="Gibbenergy"]
 Use identity:
 $3(x^{5}+y^{5}+z^{5})$ $=3(x+y+z)^{5}+5(x^{2}+y^{2}+z^{2}+xy+yz+zx)(x^{3}+y^{3}+z^{3}-(x+y+z)^{3})$
 :)[/quote]
Found by hand?  :P[/quote]

Yeap. Use Bezourt's theorem, you can factor $x^{5}+y^{5}+z^{5}-(x+y+z)^{5}$ very fast. For $(x+y+z)^{3}-x^{3}-y^{3}-z^{3}$, it is from my memory.

[quote="spanferkel"][quote="Gibbenergy"]Use Bezourt's theorem.[/quote]
 :?:
[quote="Gibbenergy"] For $(x+y+z)^{3}-x^{3}-y^{3}-z^{3}$, it is from my memory.[/quote]Good memory!! :P And the factor $\frac53$ is amazing.[/quote]

Let $f(x)=(x+y+z)^{5}-x^{5}-y^{5}-z^{5}$
Since $f(-y)=f(-z)=0$ hence $f(x)$ must have a factor $(x+y)(x+z)$. Similarity, you can prove that $f(x)$ has a factor $y+z$
Now $f(x)= k (x+y)(x+z)(y+z) Q(x)$
Because $f(x)$ is symmetric and have 4 degrees with respect to x.Hence $Q(x)$  must be a symmetric polynomial which had second degree with respect to $x$. So $Q(x)= m(x^{2}+y^{2}+z^{2})+k(xy+yz+xz)+r(xyz)$. Plug some values of $x,y,z$ into both sides of eq to solve for these constant  and you find the identitity :D


Consider a list of numbers. The first one is $25$. Then, the next number is the sum of the squares of the digits. So, the second number is $2^2+5^2=29$, and the 3º one is $2^2+9^2=85$. Find the number in the 2005th position.

[hide]

Listing out the first couple of terms, we have: $25, 29, 85, 89, 145, 42, 20, 4, 16, 37, 58, 89$. So after the first $2$ terms, the sequence repeats every $9$ terms.

$(2005 - 2) \equiv 5 \bmod 9$. The $2005th$ term is then $20$.

[/hide]

[quote="eLeM"][quote="Adunakhor"]

Listing out the first couple of terms, we have: $25, 29, 85, 89, 145, 42, 20, 4, 16, 37, 58, 89$. So after the first $2$ terms, the sequence repeats every $9$ terms.

$(2005 - 2) \equiv 5 \bmod 9$. The $2005th$ term is then $20$.

[/quote]

It repeats every $8$ term, so $2005th$ term is $145$.[/quote]

Heh, didn't see the $58$  :oops:

[hide]This sequence repeats.
25,29,85,89,145,42,20,4,16,37,58,89,...
If the 4th term is 89, and the 12th term is 49, every 8n+4th term is 89.  Then the 2004th term is 89.  Therefore, the 2005th term is 145.[/hide]
Oops, I didn't notice Jose posted it twice.  Will a mod please delete this?

My problem-solving experience indicates that pretty much every recursive sequence question on contests that is based on some kind of function of digits begins to repeat after a few terms.

Please don't post problems more than once. It gets difficult to keep track of.

Follow-up:  Can anyone characterize all $n$ for which this process produces a periodic sequence?  :)

A sequence seems to repeat once a term's digits are repeated later in the sequence, where order of the digits does not matter. In this case, 85 was reversed into 58, which continued the loop.

It seems logical to me that this works for all starting numbers. If the sequence wasn't to repeat, each progressing term should increase infinitely, but this isn't possible. I think.

my gut is telling me I'm wrong...

Well... consider numbers that aren't representible as a sum of squares.  :P

[quote="Pocket Sand"]Wouldn't all numbers be representable as the sum of squares, so long as the starting number's number of digits wasn't restricted?

(on a side note, I think I'm lost)[/quote]

Well, think about $3$.  You can write $3 = 1^2 + 1^2 + 1^2$, but then you'd have to show that some number with only the digits $1, 1, 1, 0, 0, 0, 0 , 0, ...$ is representible as another sum of squares, and...

Actually, this problem is very ugly in the general.  :|

it seems to me that the function has to be periodic at some point because it generally decreases, say you have a n-digit number, the max value for the next is 81n, and since it is bounded, there are a limited number of values that it can take on, and eventually it will have to get back to the number already in the sequence, becoming periodic


Find all natural numbers $a$ such as there exits $2002$ natural numbers $x_{0},x_{1},x_{2},....,x_{2001}$ sastify: :$a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+......+a^{x_{2001}}$

[moderator: maybe this¿]


Let poistive real numbers $ a,b$ and $ a\ge b$. Prove that $ (2^a\plus{}\frac{1}{2^a})^b\le (2^b\plus{}\frac{1}{2^b})^a$

[quote="thaithuan_GC"]Let poistive real numbers $ a,b$ and $ a\ge b$. Prove that $ (2^a \plus{} \frac {1}{2^a})^b\le (2^b \plus{} \frac {1}{2^b})^a$[/quote]

This's the problem in contest to the university of Viet Nam, 2007.  :D
We can generaltion: $ (k^a \plus{} \frac {1}{k^a})^b\le (k^b \plus{} \frac {1}{k^b})^a$  for all $ a\ge b$ and $ k>1$.  :roll:


Let $ a,b,c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac 1{\sqrt {5 \minus{} 4a \plus{} 5a^2} } \plus{} \frac 1{\sqrt {5 \minus{} 4b \plus{} 5b^2}} \plus{} \frac 1{\sqrt {5 \minus{} 4c \plus{} 5c^2}} \le \sqrt {\frac {3}{2}},$

with equality if and only if $ a \equal{} b \equal{} c \equal{} 1$.

[hide="Remark."]$ (11 \minus{} 4a \plus{} 5a^2)^2 \minus{} 24(5 \minus{} 4a \plus{} 5a^2) \equal{} (1 \minus{} a)^2(1 \plus{} 5a)^2\geq0\Longrightarrow$

$ \frac {12}{11 \minus{} 4a \plus{} 5a^2}\leq\sqrt {\frac {6}{5 \minus{} 4a \plus{} 5a^2}}$.[/hide]
See : http://www.mathlinks.ro/Forum/viewtopic.php?t=223910

[quote=Ji Chen]Let $ a,b,c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac 1{\sqrt {5 \minus{} 4a \plus{} 5a^2} } \plus{} \frac 1{\sqrt {5 \minus{} 4b \plus{} 5b^2}} \plus{} \frac 1{\sqrt {5 \minus{} 4c \plus{} 5c^2}} \le \sqrt {\frac {3}{2}},$

with equality if and only if $ a \equal{} b \equal{} c \equal{} 1$.
[/quote]
It's true by Vasc's LCF Theorem.


algebra
polynomial
number theory proposed
number theory

Let $P(x,y),Q(x,y)$ be two-variable polynomials with the coefficients of integer.Supposed that when $a_n,b_n$ are determined for certain integers $a_0,\ b_0$ by $a_{n+1}=P(a_n,b_n),\ \ b_{n+1}=Q(a_n,b_n)\ \ (n=0,1,2,\cdots)$ there existed positive integer $k$ such that $(a_1,b_1)\neq (a_0,b_0)$ and $(a_k,b_k)=(a_0,b_0)$.Prove that the number of  the lattice points on the segment with end points of $(a_n,b_n)$ and $(a_{n+1},b_{n+1})$ is indepedent of $n$.

First of all, observe that if $ a,b,c,d \in \mathbb{Z} $ then the number of
lattice points in the segment with endpoints $ (a,b) $ and $ (c,d) $ is exactly
$ gcd(c-a,d-b) + 1 $.
 Now assume that $ d = gcd(a_{n+1} - a_{n} , b_{n+1} - b_{n}) $. Then
$ a_{n+1} \equiv a_{n} (mod \ d) $ and $ b_{n+1} \equiv b_{n} (mod \ d) $.
Therefore $ a_{n+2} = P(a_{n+1},b_{n+1}) \equiv P(a_{n},b_{n}) = a_{n+1} (mod \ d) $
and $ b_{n+2} = Q(a_{n+1},b_{n+1}) \equiv Q(a_{n},b_{n}) = b_{n+1} (mod \ d) $, therefore
$ d | a_{n+2} - a_{n+1} , b_{n+2} - b_{n+1} $. Therefore we get that
$ gcd(a_{n+1} - a_{n} , b_{n+1} - b_{n}) | gcd(a_{n+2} - a_{n+1} , b_{n+2} - b_{n+1}) $.
On the other hand, we know that the sequence of points $ (a_{n},b_{n}) $ is periodic,
so the only possibility is that $ gcd(a_{n+1} - a_{n} , b_{n+1} - b_{n}) $
is independent of $ n $.


please help, this questions seems easy but is really hard for me.

show that (Tn)^2 = Tn-1* Tn+1  -1 if n is even

Note: the n is subtracting the 1 and adding the one. Not Tn

fibonacci sequence- 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610 (first 15)

it works if you sub in an T2 or T4 becuase they are even but how do we prove for any even value not just from the fibonacci sequence

$F_{0}=0,F_{1}=1,F_{n+1}=F_{n}+F_{n-1}$ for $n\in\mathbb{N}$.
Then
$\left(\begin{array}{cc}F_{n+1}&F_{n}\\ F_{n}&F_{n-1}\end{array}\right)=\left(\begin{array}{cc}1&1\\ 1&0\end{array}\right)^{n}$ for all $n\in\mathbb{N}$. (easy induction)
Now take the [url=http://en.wikipedia.org/wiki/Determinant]determinant[/url].
$F_{n+1}F_{n-1}-F_{n}^{2}=\det \left(\begin{array}{cc}F_{n+1}&F_{n}\\ F_{n}&F_{n-1}\end{array}\right)=\det \left(\begin{array}{cc}1&1\\ 1&0\end{array}\right)^{n}=(-1)^{n}$.
This formula can also be shown easily by induction, but this is more natural.

we have not yet been taught induction, so my tutor siad there must be another way to prove this.

or is there no other way

$F_{n+2}F_{n}-F_{n+1}^{2}=(F_{n+1}+F_{n})F_{n}-F_{n+1}^{2}\\ \hspace*{1.15in}=F_{n}^{2}+F_{n+1}(F_{n}-F_{n+1})\\ \hspace*{1.15in}=F_{n}^{2}-F_{n+1}F_{n-1}$

So if we define $a_{n}: =F_{n+1}F_{n-1}-F_{n}^{2}$ for $n\in\mathbb{N}$,
then $a_{n+1}=-a_{n}$ and $a_{n+2}=-a_{n+1}=a_{n}$ for all $n\in\mathbb{N}$.
So $1=a_{2}=a_{4}=\cdots=a_{2n}$ for all $n\in\mathbb{N}$.
But the last step is technically itself an induction.
$F_{n}$ is defined recursively, so you need to handle it recursively somehow, i.e. with induction.
Better to learn what that is and get it over with.


logarithms
function
trigonometry
integration
calculus
calculus computations

How do I solve:

1: $ y' \equal{} xe^{ \minus{} y} \plus{} \frac {2}{x}$

2: $ y'' \equal{} \frac {(y')^{2}}{y} \plus{} \frac {y'}{x}$

Thanks.

[hide="1"]
Rewrite as $ y^\prime e^y\equal{}x\plus{}\frac{2}{x}e^y$. As the left hand side is $ \frac{d}{dx}[e^y]$, substitute $ z\equal{}e^y$ to get
\[ z^\prime \equal{}\frac{2}{x}z\plus{}x\]
(That substitution's somewhat motivated by the fact that a logarithm for $ y$ would produce similar types of functions for $ y^\prime$ and $ e^{\minus{}y}$, while algebraics, exponentials, and trig do not.)

This is a linear differential equation: here we can choose the integrating factor $ z\equal{}x^2u$ so $ z^\prime \equal{}2xu\plus{}x^2u^\prime$. We get
\[ 2xu\plus{}x^2u^\prime \equal{}2xu\plus{}x\Rightarrow u^\prime \equal{}\frac{1}{x}\]
So there is a constant $ A$ such that $ u\equal{}A\plus{}\ln x$, giving $ z\equal{}x^2(A\plus{}\ln x)$ and finally $ \boxed{y\equal{}2\ln x\plus{}\ln (A\plus{}\ln x)}$.
[/hide]

[quote="RealTime"]How do I solve:

2: $ y'' \equal{} \frac {(y')^{2}}{y} \plus{} \frac {y'}{x}$

Thanks.[/quote]

hint :

rewrite it as...

$ \frac {y''}{y'} \equal{} \frac {y'}{y} \plus{} \frac {1}{x}$

$ \Rightarrow (\ln\,y')' \equal{} (\ln\,y)' \plus{} (\ln\,x)'$

$ \Rightarrow \ln\,y' \equal{} \ln\,y \plus{} \ln\,x \plus{} A$

$ \Rightarrow  y' \equal{} e^A\,x\,y$

rewrite it as...

$ \frac {y'}{y} \equal{}e^A\, x$

$ \Rightarrow ( \ln\,y )'  \equal{} \frac{e^A}{2}\; (x^2)'$

$ \Rightarrow  \ln\,y \equal{}  \frac{e^A}{2}\; x^2 \plus{} B$

$ \Rightarrow y \equal{} \boxed{  e^B\;e^{\left(\frac{e^A}{2}\; x^2\right)}}$

hello, the second equation:
let us consider
$ y''\plus{}f(y)y'^2\plus{}g(x)y'\equal{}0$. After dividing by $ y'$ you get an exact differential equation. The solution is giving by:
$ \ln(|{y'}|)\plus{}\int {f(y)dy}\plus{}\int{ g(x)dx}\equal{}C$.
Your equation has the solution $ y(x)\equal{}C_1\exp(C_2x^2)$.
Sonnhard.


Let $ ABC$ be an obtuse inscribed in a circle of radius $ 1$.  Prove that $ \triangle ABC$ can be covered by an isosceles right-angled triangle with hypotenuse of length $ \sqrt {2} \plus{} 1$.

Assume that $ \widehat{ACB}>90^0$ and $ \widehat{CAB}\leq 45^0$.
Contruct an $ E$-isosceles right-angled triangle $ AED$ with $ AD\equal{}\frac{\sqrt{2}\plus{}1}{2}AB$ and $ C,E$ are same side with $ AB$.
Then $ AD\leq \sqrt{2}\plus{}1$ because $ AB\leq 2$
We prove that: $ C$ is inside triangle $ AED$.
In fact: Call $ M$ is midpoint of $ AB$ and $ F$ is foot of peperdicular line from $ M$ to $ DE$.
Then $ MF\equal{}\frac{MD}{\sqrt{2}}\equal{}\frac{AD\minus{}AM}{\sqrt{2}}\equal{}\frac{AB}{2}\equal{}AM\equal{}MB$
It mean $ DE$ is tangent with $ (AFB)$.
But $ C$ is inside $ (AFB)$, it mean $ C$ is inside triangle $ AED$.


[quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
$ (a^2 \plus{} 2b^2 \plus{} 3c^2)(b^2 \plus{} 2c^2 \plus{} 3a^2)(c^2 \plus{} 2a^2 \plus{} 3b^2) \geq\ \frac {8}{9}(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^4$
 :)[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3 \equal{} > a^2 \plus{} b^2 \plus{} c^2 \geq 3$
We only need prove
$ (3 \plus{} b^2 \plus{} 2c^2)(3 \plus{} c^2 \plus{} 2a^2)(3 \plus{} a^2 \plus{} 2b^2) \geq \frac {8}{3}(ab \plus{} bc \plus{} ca)$
We have
$ 3 \plus{} b^2 \plus{} 2c^2 \geq 3 \sqrt[3]{(1 \plus{} b^2)(1 \plus{} c^2)^2}$ (by AM-GM)
$ \equal{} > LHS \geq (1 \plus{} a^2)(1 \plus{} b^2)(1 \plus{} c^2)$
And $ (1 \plus{} a^2)(b^2 \plus{} 1) \geq (a \plus{} b)^2$ (by CS)
$ \equal{} > (1 \plus{} a^2)(1 \plus{} b^2)(1 \plus{} c^2) \geq (a \plus{} b)(b \plus{} c)(c \plus{} a) \geq \frac {8}{9}(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \equal{} RHS$ (because $ a \plus{} b \plus{} c \equal{} 3$)
We are done! :)

Yes, it's also my proof, 2424  :lol:
And try with the similar:
$ (a^2 \plus{} 2b^2)(b^2 \plus{} 2c^2)(c^2 \plus{} 2a^2) \geq\ \frac {1}{3}(ab \plus{} bc \plus{} ca)^2(a \plus{} b \plus{} c)$
 :)

[quote="great math"]But your last inequality is not homogeneous
[hide="Nguoivn"]Anh nhan duoc file cua em chua? :) [/hide][/quote]
Sorry, I edited it.  :)
[quote="nguoivn"]Yes, it's also my proof, 2424  :lol:
And try with the similar:
$ (a^2 \plus{} 2b^2)(b^2 \plus{} 2c^2)(c^2 \plus{} 2a^2) \geq\ \frac {1}{3}(ab \plus{} bc \plus{} ca)^2(a \plus{} b \plus{} c)^2$
[/quote]
[hide]Anh nhan. dc file cua em rui. Co' vai` cho~ co the? chinh? ngan' gon hon nhung noi chung la` ok.  :) [/hide]

Oh no, it is not homogeneous again since the left hand side has a degree 6 and the right hand side yields a degree 5

the last ineq stated by ngoivn is  true and sovable by cauchy and am-gm :wink:

Very sorry, I edited again my problem. That's so silly  :blush:
PS: [hide]Dao. nay` dau` oc' ngo' ngan? that.![/hide]
[quote="nguoivn"]
$ (a^2 \plus{} 2b^2)(b^2 \plus{} 2c^2)(c^2 \plus{} 2a^2) \geq\ \frac {1}{3}(ab \plus{} bc \plus{} ca)^2(a \plus{} b \plus{} c)^2$
[/quote]

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
$ (a^2 \plus{} 2b^2 \plus{} 3c^2)(b^2 \plus{} 2c^2 \plus{} 3a^2)(c^2 \plus{} 2a^2 \plus{} 3b^2) \geq\ \frac {8}{9}(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^4$
 :)[/quote]
Nice result,nguoivn :) .We will prove the inequality than stronger:
$ (a^2\plus{}2b^2\plus{}3c^2)(b^2\plus{}2c^2\plus{}3a^2)(c^2\plus{}2a^2\plus{}3b^2) \ge \frac{8}{9}(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)^4$
We can assume: $ ab\plus{}bc\plus{}ca\equal{}3$,Put $ x\equal{}a^2\plus{}b^2\plus{}c^2$,we have:
$ LHS\equal{}(x\plus{}b^2\plus{}2c^2)(c\plus{}c^2\plus{}2a^2)(x\plus{}a^2\plus{}2b^2)$
$ \equal{}6x^3\plus{}3(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)x\plus{}(b^2\plus{}2c^2)(c^2\plus{}2a^2)(a^2\plus{}2b^2)$
$ \ge 6x^3\plus{}3(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)x\plus{}27a^2b^2c^2$ (by Am_Gm inequality)
$ \ge 6x^3\plus{}3(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)x\plus{}3x(4(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\minus{}x^2)$ (By Schur inequality)
$ \equal{}3x(x^2\plus{}5(a^2b^2\plus{}b^2c^2\plus{}c^2a^2))$
$ \ge 3x(x^2\plus{}15)$
On the other hand: $ (a\plus{}b\plus{}c)^4\equal{}(x\plus{}6)^2$
so,we only need to prove;
$ 27(x^2\plus{}15) \ge 8(x\plus{}6)^2$
$ <\equal{}> (x\minus{}3)(19x\minus{}39) \ge 0$
Which 's obvious true because $ x \ge 3$
Equality holds if and only if: $ a\equal{}b\equal{}c$
 :)

Nice proof, Quy  :)
A similar result:
$ (a^2 \plus{} 2b^2 \plus{} 3c^2)(b^2 \plus{} 2c^2 \plus{} 3a^2)(c^2 \plus{} 2a^2 \plus{} 3b^2) \geq$$ (ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}$
PS: [hide]Loi` giai? cua em thuc chat' la` Am-Gm cho dai luong $ (a^2\plus{}2b^2)(b^2\plus{}2c^2)(c^2\plus{}2a^2)$ va` su? dung. pqr. Nhung anh rat thjk cach' trinh` bay` cua em  :) [/hide]

[quote="nguoivn"]Nice proof, Quy  :)
A similar result:
$ (a^2 \plus{} 2b^2 \plus{} 3c^2)(b^2 \plus{} 2c^2 \plus{} 3a^2)(c^2 \plus{} 2a^2 \plus{} 3b^2) \geq$$ (ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}$
PS: [hide]Loi` giai? cua em thuc chat' la` Am-Gm cho dai luong $ (a^2 \plus{} 2b^2)(b^2 \plus{} 2c^2)(c^2 \plus{} 2a^2)$ va` su? dung. pqr. Nhung anh rat thjk cach' trinh` bay` cua em  :) [/hide][/quote]
It 's weaker than our inequality :)

Yes, you are right. I don't see carefully  :oops:


[quote="Lovasz"]Let $a,b,c,d \ge 0$. Prove that:
\[ \left(\frac{a+b+c+d}{4} \right)^5 \ge \frac{a^2b^2c+b^2c^2d+c^2d^2a+d^2a^2b}{4}.  \][/quote]
I think this is not true. Try $a=b=c=1,$ $d=0.$ ;)
Maybe are you mean to
$\left(\frac{a+b+c+d}{4} \right)^4 \ge \frac{a^2bc+b^2cd+c^2da+d^2ab}{4}$ $?$


Prove that for all real $r$ and $x, y, z \ge 0$ we have
$\sum_{cyc}x^{r}(y-z)^{2}(x-y)(x-z) \ge 0$

In fact, I think if $r$ is an even nonneg integer I think it holds for all $x, y, z$.

It's not really like Shur. Suppose that $a\ge b\ge c$, then the inequality is equivalent to \[x^{r}(y-z)-y^{r}(x-z)+z^{r}(x-y)\ge 0\]  \[\Leftrightarrow (x^{r}-y^{r})(y-z) \ge (y^{r}-z^{r})(x-y)\] By Lagrange, there exist $u,v$ such that $x\ge u\ge y\ge v\ge z$ and \[x^{r}-y^{r}=ru^{r-1},\ y^{r}-z^{r}=rv^{r-1}\] so we must have (with $u\ge v$) \[ru^{r-1}\ge rv^{r-1}\] or $r<0$ and $r\ge 1$. If $0<r<1$ then the sign is reversed.  :)

For all  integer r and x,y,z>0 we have

                 Sum_{cyc}(x^r*(y-z)^2*(x-y)*(x-z))>=0 .


AMC
USA(J)MO
USAMO
floor function
function
inequalities

4th USAMO 1975

1.  Show that for any non-negative reals $x, y$, $\lfloor 5x \rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y} \rfloor + \lfloor {x+3y} \rfloor$.

i have a solution for this but i wanted to see other ppl's

[hide]

$\lfloor 5\lfloor x \rfloor + 5\{x\} \rfloor + \lfloor 5\lfloor y \rfloor + 5\{y\}\rfloor \ge \lfloor 3x \rfloor + \lfloor y \rfloor + \lfloor x \rfloor + \lfloor 3y \rfloor$

$5\lfloor x \rfloor + \lfloor 5\{x\} \rfloor + 5\lfloor y\rfloor + \lfloor 5\{y\} \rfloor \ge 3\lfloor x\rfloor + \lfloor 3\{x\} \rfloor + \lfloor y \rfloor + \lfloor x \rfloor + 3\lfloor y \rfloor + \lfloor 3\{y\} \rfloor$

$\lfloor x \rfloor + \lfloor y \rfloor + \lfloor 5\{x\} \rfloor + \lfloor 5\{y\} \rfloor \ge \lfloor 3\{x\} \rfloor + \lfloor 3\{y\} \rfloor $

which is true since all are positive

[/hide]

i saw your answer and am not sure if all of the math is 100% correct

im talking about the part where you basically said that [x+3y]=[x]+[3y], and other cases like this one. because if x= 1/3 and y=2/9 then  [x+3y]=1 but [x]+[3y]=0  so i dont think that the way you did it is totally right. sorry  :blush:

oh oops

i thought that was a rule but i guess i didnt think that one through :blush:

does ne1 know a nice solution?

[color=blue][i]mod edit: Please be respectful of proper forum etiquette.[/i][/color]

yeah i was working on this problem and could only prove for cases where x=y and x or y = 0. but i was only able to prove part of it for x>y and y=x-k (i would use this also to prove for y>x and X=y-k).  :mad:  :oops:  :blush:

$\displaystyle  \lfloor 5x \rfloor = 5.\lfloor x \rfloor + k$ for  $ 0 \le k \le 4$ we constact this considering that $5.({\lfloor x \rfloor},9999999...)$  is the maximus  kind of multiplication between a integer and a real number,  then:
$\displaystyle  \lfloor 5y \rfloor = 5.\lfloor y \rfloor + k'$ for  $ 0 \le k' \le 4$

$\displaystyle \lfloor 3x \rfloor = 3.\lfloor x \rfloor + t$ for $0 \le t \le 2$

$\displaystyle \lfloor 3y \rfloor = 3.\lfloor y \rfloor + t'$ for $0 \le t' \le 2$

Analising the maximus kind of adition we got:
$\displaystyle \lfloor {3.x + y} \rfloor = 3.\lfloor x \rfloor + t + \lfloor y \rfloor + l$ for $0 \le l \le 1$

$\displaystyle \lfloor {3.y + x} \rfloor = 3.\lfloor y \rfloor + t' + \lfloor x \rfloor + l'$ for $0 \le l' \le 1$
Since was posted that $\displaystyle \lfloor 5x \rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y} \rfloor + \lfloor {x+3y} \rfloor$, we must prove that the result of  all possibility is true:

$\displaystyle \lfloor 5x \rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y} \rfloor + \lfloor {x+3y} \rfloor$
->$\lfloor x \rfloor + \lfloor y \rfloor \ge t + t' + l + l' - k - k'$ suming the minimus and maximus  of thats variables we got:

${-2} \le ({t + t' + l + l' - k -k'}) \le  0$ and how $\lfloor x \rfloor + \lfloor y \rfloor \ge 0$: considering that the greatest functions are positive integers the inequality is true

um i can barely follow what you are trying to say.
i am able to follw what u have done up until this point;

$\displaystyle \lfloor 5x \rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y} \rfloor + \lfloor {x+3y} \rfloor
->\lfloor x \rfloor + \lfloor y \rfloor \ge t + t' + l + l' - k - k'$ thus:


arr. i've seen this problem before and forgot how to do it. or maybe i'm breaking some of the exponent rules. the way i did it, i narrowed it down to $2^X=X$ which gives no solutions and can't be right.  :help: (help)

[quote="math92"]Find x if $2^{16^x}=16^{2^x}$[/quote]

Well, you can log both sides and take x out thus getting $x(log 2^{16})=x(log 16 ^2)$, which is not equal, thus no solutions.

To compare the exponents, we must change the bases so that they are equal.
\begin{eqnarray*}
2^{16^x}&=&16^{2^x}\\
2^{16^x}&=&(2^4)^{2^x}\\
2^{16^x}&=&2^{4(2^x)}
\end{eqnarray*}
If $2^a=2^b$, then $a=b$, so $16^x=4(2^x)$.  Again we must change the bases so that they are equal.
\begin{eqnarray*}
16^x&=&4(2^x)\\
(2^4)^x&=&(2^2)(2^x)\\
2^{4x}&=&2^{2+x}\\
4x&=&2+x\\
3x&=&2\\
x&=&\frac {2}{3}
\end{eqnarray*}

[quote="Iversonfan2005"]Well, you can log both sides and take x out thus getting $x(log 2^{16})=x(log 16 ^2)$, which is not equal, thus no solutions.[/quote]

I'm not sure if you can log both sides of an equation, but regardless, there is one problem with your solution.  If $x\times a=x\times b$ where $a$ and $b$ are different, then $x=0$. ;)

Iversonfan2005, here's the problem with your solution.  It is indeed possible to log both sides of an equation.  However, you attempted to solve the equation using the incorrect identity $\log x^{y^z}=z \log x^y$.  The identity you were thinking of is actually $\log x^y=y\log x$.

[quote="Palytoxin"]so Iversonfan2005's equation should have been $16^x\log 2=2^x\log 16$ and so
\[\frac{16^x}{2^x}=\frac{\log 16}{\log 2}\]
\[8^x=\log_2 16=4\]
\[2^{3x}=2^2\]
\[x=\frac23\][/quote]

Argghh.. :mad: , I need to brush up on my Logarithm rules...


geometry
trapezoid
parallelogram
geometric transformation
reflection
geometry unsolved

Point $ A'$ on the boundary of a trapezoid $ ABCD$ is such that the line $ AA'$ bisects the area of the trapezoid. Points $ B',C',D'$ are similarly defined. Prove that the intersection points of the diagonals of the quadrilaterals $ ABCD$ and $ A'B'C'D'$ are symmetric with respect to the midpoint of the middle line of the trapezoid $ ABCD$.

[hide="Solution"]
Let the midpoints of $ AD$ and $ BC$ be $ M$ and $ N$ respectively, $ AC$ and $ BD$ meet at $ O_1$, let $ A'C'$ and $ B'D'$ meet at $ O_2$, $ A'C'$ and $ BD$ meet at $ Q$, and $ B'D'$ and $ AC$ meet at $ P$.  Furthermore, let $ AA'$ and $ BB'$ meet at $ O$ and $ A'C'$ meet at $ AB$ at $ C''$.  Now, notice that $ [AA'D] \equal{} \frac {[ABCD]}{2} \equal{} [CC'D]$, so $ [AA'C'] \equal{} [C'A'C]$.  This means that $ A'C'\parallel AC$.  Similarly, $ B'D'\parallel BD$.  Thus, $ O_1Q\parallel PO_2$ and $ O_2Q\parallel O_1P$, so $ O_1PO_2Q$ is a parallelogram.  This means that $ QP$ and $ O_1O_2$ bisect each other.  Now, notice that
\[ [AA'D] \plus{} [BB'C] \equal{} \frac {[ABCD]}{2} \plus{} \frac {[ABCD]}{2} \equal{} [ABCD] \equal{} [AA'D] \plus{} [BB'C] \minus{} [A'B'O] \plus{} [AOB]
\]
which implies that $ [A'OB'] \equal{} [AOB]$, so $ A'B\parallel AB'$.  Since $ AB\parallel A'B'$, we have that $ ABA'B'$ is a parallelogram.  Furthermore, $ [AOB'] \equal{} [BOA']$ since $ AB\parallel A'B'$ and $ [AA'D] \equal{} [BB'C]\implies [AOB'D] \equal{} [BOA'C]$.  This means that $ [AOB'D] \minus{} [AOB'] \equal{} [BOA'C] \minus{} [BOA']$, so $ [B'AD] \equal{} [A'BC]$. Since the heights from $ A$ and $ B$ to $ B'D$ and $ A'C$ respectively are equal because of the fact that $ AB\parallel DC$, we see that $ B'D \equal{} A'C$. Now, $ AC\parallel A'C''$ and $ AC''\parallel A'C$, so $ AC''A'C$ is a parallelogram. Thus, $ AC'' \equal{} A'C \equal{} B'D$ and $ AC''\parallel B'D$, so $ AC''DB'$ is a parallelogram.  This implies that $ C''D \equal{} AB' \equal{} A'B$, the last part becaues $ ABA'B'$ is a paralellogram.  Also, $ C''D\parallel AB'\parallel A'B$, so $ C''DA'B$ is a parallelogram.  This means that $ C''A'$ passes through the midpoint of $ BD$, which implies that $ Q$ is the midpoint of $ DB$.  Yet, $ M$ is the midpoint of $ AD$, so $ QM$ is a midline of $ \triangle ADB$, so $ MQ\parallel AB$.  Also, $ MN\parallel AB$ since $ MN$ is a midline of trapezoid $ ABCD$, so $ M$, $ Q$, and $ N$ are collinear.  Through the same reasoning, $ M$, $ P$, and $ N$ are collinear, so $ MQPN$ is a line.  From the fact that $ MQ$ and $ PN$ are midlines of $ \triangle ADB$ and $ \triangle ACB$ respectively gives that $ MQ \equal{} \frac {AB}{2} \equal{} PN$.  Thus, the midpoint of $ O_1O_2$ is the midpoint of $ QP$ because $ O_1QO_2P$ is a parallelogram and the midpoint of $ QP$ is the midpoint of $ MN$ since $ QM\equal{}PN$.  We conclude that the midpoint of $ O_1O_2$ is the midpoint of $ MN$, which means that the reflection of $ O_1$ about the midpoint of $ MN$ is $ O_2$. [/hide]

Although there are so many lines in the drawing however, this is not a difficult problem.

In my drawing $ AB\parallel CD$ and $ AB < CD$ and let $ K,\ L$ be, the intersection points of the sideline $ CD,$ from the line through the points $ A,\ B$ and parallel to $ BD,\ AC,$ respectively.

It is easy to show that the line segments $ BK,\ AL,$ pass through the points $ M,\ N$ respectively, as the midpoints of the side-segments $ AD$ and $ BC.$

Because of now, the points $ A',\ B',$ are the midpoints of the segments $ DL,\ CK$ respectively, we conclude that the line segments $ AA'$ and $ BB',$ pass through the midpoint $ O,$ of the segment $ MN$ and so, we have $ AO \equal{} OA'$ $ ,(1)$ and $ BO \equal{} OB'$ $ ,(2)$

It is easy to show that $ A'C'\parallel AC$ and $ B'D'\parallel BD,$ where $ C'\in AD$ and $ D'\in BC.$

Hence, because of $ (1),$ $ (2),$ we conclude that the points $ P\equiv AC\cap BD$ and $ P'\equiv A'C'\cap B'D',$ as the intersection points of the diagonals of the trapeziums $ ABCD$ and $ A'B'C'D'$ $ ($ easy to prove that $ C'D'\parallel CD$ $ )$ respectively, are symmetric each other, with respect to the point $ O$ and the proof is completed.

Kostas Vittas.

[quote="discredit"][quote="vittasko"]

Hence, because of $ (1),$ $ (2),$ we conclude that the points $ P\equiv AC\cap BD$ and $ P'\equiv A'C'\cap B'D',$ as the intersection points of the diagonals of the trapeziums $ ABCD$ and $ A'B'C'D'$ respectively, are symmetric each other, with respect to the point $ O$ and the proof is completed.

[/quote]

Well, I can't quite see how we can conclude that. Please give some more information...[/quote]
Because of $ OA \equal{} OA'$ and $ OB \equal{} OB',$ we have that $ ABA'B'$ is a parallelogram.

We consider now, the point $ P,$ as the intersection point of two arbitrary lines through vertices $ A$ and $ B.$

So, the line through the point $ A'$ and parallel to $ AC\equiv AP,$ intersects the line segment $ PO$ at one point, so be it $ P',$ such that $ PO \equal{} OP'$ ( easy to prove by [b][size=100]Thales theorem[/size][/b], because of $ OA \equal{} OA'$ ).

Similarly, the line through the point $ B'$ and Parallel to $ BD\equiv BP,$ intersects the line segment $ PO,$ at point $ P'$ too.

Kostas Vittas.

PS. As an interseting remark in this problem, we can say that the line segment $ POP',$ passes through the points $ E\equiv AD\cap BC$ and $ E'\equiv A'D'\cap B'C',$ easy to prove by [b][size=100]Desarques theorem[/size][/b].

[quote="vittasko"]
PS. As an interseting remark in this problem, we can say that the line segment $ POP',$ passes through the points $ E\equiv AD\cap BC$ and $ E'\equiv A'D'\cap B'C',$ easy to prove by [b][size=100]Desarques theorem[/size][/b].[/quote]

Again, sorry for my dumbness, but please show this  :blush:

[quote="vittasko"]  PS. As an interseting remark in this problem, we can say that the line segment $ POP',$ passes through the points $ E\equiv AD\cap BC$ and $ E'\equiv A'D'\cap B'C',$ easy to prove by [b][size=100]Desarques theorem[/size][/b].[/quote]
(See the schema [b][size=100]t=207439[/size][/b] ) It is easy to show that the triangles $ \bigtriangleup AA'C'$ and $ \bigtriangleup BB'D'$ are perspective, because of $ \infty \equiv AB\cap A'B'\cap C'D'$ $ ($ From $ AB\parallel A'B'\parallel C'D'$ $ ).$

So, by [b][size=100]Desarques theorem[/size][/b], we conclude that the points $ E\equiv AC'\cap BD'$ and $ O\equiv AA'\cap BB'$ and $ P'\equiv A'C'\cap B'D',$ are collinear.

$ \bullet$  We can say now, that the line connecting the point $ E,$ as the intersection point of the non-parallel sidelines $ AD,\ BC,$ of the trapezium $ ABCD,$ with the point $ P,$ as the intersection point of its diagonals $ AC,\ BD,$ bisects its bases $ AB,\ CD$ and also the segment $ C'D',$ because of $ C'D'\parallel AB\parallel CD.$

So, in the configuration of the trapezium $ A'B'C'D',$ this line connects the midpoint of its base $ C'D',$ with the point $ P',$ as the intersection point of its diagonals $ A'C',\ B'D'$ and then, it passes through the point $ E'\equiv A'D'\cap B'C',$ as well.

Hence, the line segment $ POP',$ passes through the points $ E,\ E'$ and the proof is completed.

Kostas Vittas.


geometry
trapezoid
ratio
similar triangles

Let $ABCD$ be a quadrilateral with $AB \parallel CD$, $AB > CD$. Prove that the line passing through $AC \cap BD$ and $AD \cap BC$ passes through the midpoints of $AB$ and $CD$.

============

Consider a triangle ABC with X on CB and Y on CA so that XY is parallel to AB.  (*)

-------------
CXY is similar to CBA, so that CX/CB = CY/CA.  It follows BX/XC = AY/YC. (**)

Suppose AX and BY intersect in J.  and CJ intersects AB at Z.  From our result (**) + Ceva, it follows AZ = ZB.

Finally, because our triangles are similar, the (collinear) line CJZ also hits the midpoint of XY.

So we have proved with the above configuration (*) that C, the midpoint of XY, the intersection of AX and BY, and the midpoint of AB are collinear.

============

Now putting $A',B',C',X',Y' = A,B,AD\cap BC,C,D$ we can appeal to the result above.

OK, here is another proof

Let $P  = AD \cap BC$ and $O = AC \cap BD$

From similar triangles $PDC,PAB$ we have:

$\frac{PD}{PA} = \frac{DC}{AB}$

Also, the triangles $OAB,OCD$ are similar ($\angle OAB = \angle OCD, \angle DOC = \angle AOB$)

So, $\frac{OC}{OA} = \frac{DC}{AB} \Rightarrow $

$\frac{OC}{OA} = \frac{PD}{PA}$

Take a point $F \in OA$, such that $OF=OC$ (since $CD<AB \Rightarrow OC<OA \Rightarrow  F$ is inside the segment $OA$)

Now, $\frac{OF}{OA} = \frac{OC}{OA} \Rightarrow $

$\frac{OF}{OA} = \frac{PD}{PA}$

Hence $DF \parallel PO $

In the triangle $CDF, \ O$ is the midpoint of $CF$ and $PO \parallel DF \Rightarrow  PO$ intersects $DC$ at its midpoint $H$.

Finally, from similar triangles $PDC,PAB$ we find that the line $PH$ is also median for $\triangle PAB$.

Let the intersection of $AC$ and $BD$ be $O$, and the intersection of $AD$ and $BC$ be $E$. Let $EO \cap AB = M$, use area ratios:
\[\frac{AM}{BM}=\frac{\triangle{AOE}}{\triangle{AOB}}\cdot \frac{\triangle{AOB}}{\triangle{EOB}}=\frac{EC}{CB} \cdot \frac{ED}{DC} = 1\]

By the way, in a contest in Beijing in the beginning of the 1980's, there was a problem exactly like this one. The original problem said: Given a segment and a line paralel to it. Use ruler to find the mid-point of the segment.

[hide="Another solution"]We will use a theorem (I don't remember this theorem's name):

[color=brown][size=100][b]Lemma[/b][/size]
Let $A,B,E$ be three collinear points and a point $P \not \in AB$.
Let a line $d \parallel AB.$
$d$ intersects the lines $PA,PB,PE$ at the points $A',B',E'$ respectively.

Then $\frac{EA}{EB}= \frac{E'A'}{E'B'}$[/color]

[hide="Proof of Lemma"]
$\triangle PAE \sim \triangle PA'E' \Rightarrow  \frac{PA}{PA'}= \frac{EA}{E'A'}$

$\triangle PBE \sim \triangle PB'E' \Rightarrow  \frac{PB}{PB'}= \frac{EB}{E'B'}$

But $\frac{PA}{PA'}= \frac{PB}{PB'}\Rightarrow  \frac{EA}{E'A'}= \frac{EB}{E'B'}\Rightarrow  \frac{EA}{EB}= \frac{E'A'}{E'B'}$[/hide]

So, if $E$ is the midpoint of $AB$ then the line $PE$ intersects the segment $A'B'$ on its midpoint $E'$
This means that the line $EE'$ passes through $P=AA' \cap BB'.$

Let $M,N$ be the midpoints of $AB,CD$ respectively.

$A'=C, B'=D \Rightarrow  MN$ passes through $AC \cap BD$

$A'=D, B'=C \Rightarrow  MN$ passes through $AD \cap BC$[/hide]


Where i can find Romanian MO 2008 includes everything (JBMO TST,IMAR,National Olympiad,Team Selection tests...) Please send me smth about this.Thank You

http://www.mathlinks.ro/Forum/resources.php?c=142
Check there
Daniel