i want to know who is a hacker in here.because i have a lot of questions.plz,tell me.thanks. :D

What's your definition of a 'hacker'?

P.S. I read somewhere that true hackers are named by the community.

'Hacking' itself is not a seperate entity from having knowledge of computers and programming. A 'hacker' is one who is experienced in the field of computers, specifically networking and programming.

Often, when one refers to a 'hacker', they mean 'cracking', which involves the use of software to steal passwords, gain access to sensitive information, etc.,

What are your questions  :P   :D
Not that I'll be able to answer them :rotfl:. I've got some basic knowledge.

Define what you really want to mean,or imply.I know that some "Hackers"are contributing to the "National Security"
of my nation,"United States Of America." I think anyone of us should educate ourselves that a hacker does not really mean a
'BAD"computer literate.It depends on the activities of such individual.

                                                    Respectfully,
                                                  A.Eddington

Why do you want to know if there are any hackers in the forum? :?:
And if you have questions about hacking, just search it with google.I'm sure you know advanced google searching.I am a Hacker and i think there are more other hackers in here.So if you have questions just ask!!!!!!
I will be posting a Hacking guide in this forum, so stay tuned!!! :D

Hacking and cracking are not the same thing. Cracking involves using software to override security. Hacking involves taking code, changing, or "hacking," it, and using it as one's own program.

In its [i]original[/i] sense as applied to computers (the really original use is said to go back to MIT, and you can Google that yourself), though there has been more than a little debate, hacking refers to finding a solution to a problem (often in a program).  Usually, it was something brilliant/original but inelegant.

Cracker--and I'll disagree with Eric on this--is the more accepted term in the connotation you are using.   However, I'll be the first to admit, I've never cared much about the terminology.

I will say that if you want to learn about hacking, you don't need a guide... you need to learn about networking, OS internals, etc.  It really all depends on WHAT, exactly, you are trying to do.  Most methods of 'cracking/hacking/whatever' are built on a pyrimidal design, beginning with local (potentially hardware driven) strategies that move on up.  The next area of attack involves networking, specifically the existence and exploitation of a LAN.  After that, there is OS backend and frontend.  The final, top level usually involves remote, web-based attacks (starting with the server and moving up to service, and eventually application 'hacks').

I can't remember a link off the top of my head, but there are several online series that detail attack theory and structure, as such.

Edit: [url=http://www.antionline.com]Here[/url] is a link to a pretty nifty security portal that I follow now and then.


a,b,c are all positive real numbers and a+b+c=1. Prove this:
\[ \sum_{cyc}\sqrt{a^2+abc}\geq\frac2{3}\sum_{cyc}\sqrt{a}\]

[quote="gus135791"]a,b,c are all positive real numbers and a+b+c=1. Prove this:
\[ \sum_{cyc}\sqrt {a^2 + abc}\geq\frac2{3}\sum_{cyc}\sqrt{a}
\]
[/quote]
$ \sum_{cyc}\sqrt {a^2 + abc}\geq1$ is true too. :wink:


Im not entirely sure of the source of this question, but I'm sure ive definitely seen it somewhere before.

Let ABCD be a square such that a point P inside this square, AP=1, BP=2, CP=3.  Find $\angle APB$.  Its not too difficult, but im having trouble finding an elegant solution...

I can calculate DP very easily, but then I'm stuck trying to find anything past that.  How do you solve for the angle in a non-elegant way?

It is not easy !

 Take a right - isosceles  triangle PBE with$BP = BE$ and E out of te square. Then triangle ABE = triangle BPC , so AE = PC = 3. But from converse of Pythagoras, triangle APE is right at P(please  do it). and angle  $APB = 90 + 45 = 135$

[u]Babis[/u]

Im not quite getting how triangle ABE = triangle BPC

The way i solved this problem was to say that triangle APB's area must be equal to any other triangle, as long as they have the same height and base.  So drawing a line from P to BC (such that this line intersects BC perpendicularly at a point, say X), then area AXB= area APB.  I used sine and cosine laws (with a dash of pythagoras) to solve the rest.  But this is quite inelegant.

Yes , I can understand you. You must make the right figure. Points $E$  and $C$ are on oposite sides with respect to line $AB$ . But then

   $BA = BC , BE = BP , \angle ABE = \angle PBC = 90 - \angle ABP$. Then we' ll  have $AE= PC = 3 , AP = 1 , PE = \sqrt8$. But now for the sides of $\triangle PAE$ we can have the Pythagorian relation....

 I hope , I have helped you. I ' ll read carefully your solution after two moments.

 [u]Babis[/u]


function
calculus
derivative
algebra
functional equation
calculus computations

The question is:
Find all continuous functions satisfying:
$ f(x + y) = f(x) + f(y) + xy(x + y) ; x, y \in \mathbb R.$

[quote="What I did is :"]
Let $ y=x$.
Then,
$ f(2x) = 2f(x) + x^2(2x)$
$ 2f'(2x)= 2f'(x) + 3x^2$
$ 2f''(2x)= f''(x) + 6x$
$ 4f'''(2x)= f'''(x) + 6$

Therefore:
$ 4f'''(0)= f'''(0) + 6$
$ f'''(0) = 2$
Now integrating from $ 0$ to $ x$, I got $ f(x) = \frac {x^3}{3} + cx^2$
The correct answer is ${ \color{red}{f(x) = \frac {x^3}{3} + cx}}$ where $ \color{red}c = f'(0)$
[b]Where did I possibly go wrong???[/b] :huh:

[/quote]

You cannot integrate from $ 0$ to $ x$ if you only have information about the third derivative at the single point $ 0$.  Besides, the problem asks to find [i]all continuous[/i] solutions of the functional equation, so any approach that assumes differentiability a priori cannot alone demonstrate uniqueness.

Here's one approach: Assume $ f_1$ and $ f_2$ are two solutions of the functional equation and note that $ f_3 \equal{} f_1 \minus{} f_2$ satisfies $ f_3(x \plus{} y) \equal{} f_3(x) \plus{} f_3(y)$.  This is Cauchy's functional equation, whose only continuous solutions are $ f_3(x) \equal{} cx$ ([url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]proof[/url]).  This implies that all solutions of the original functional equation are of the form $ f(x) \equal{} g(x) \plus{} cx$ for some function $ g$.  Since $ \frac {x^3}{3}$ is, by inspection, a particular solution of the equation, the general solution must be $ f(x) \equal{} \frac {x^3}{3} \plus{} cx$.


probability
Asymptote
floor function
ceiling function
algebra
polynomial

[quote="krsattack"]Say we have an n x 3 grid of points.  For example, if n=12, we have:
 . . . . . . . . . . . .
 . . . . . . . . . . . .
 . . . . . . . . . . . .
If we choose any three distinct points, find the probability, in terms of n, that those three points are collinear.[/quote]

This problem is so painful... I wonder if there is a better way to do it (there probably is).

[hide]Consider the number of collinear triples of points.
Let's start with the horizontal ones. Since there are $ n$ on each row, we have $ 3\binom{n}{3}$ ways, since there are 3 rows and we need to pick 3 out of $ n$ from each row.

Now, we count the number of vertical ones. This is easy: $ n$.

Then, we do the diagonals. This is relatively pretty easy: $ (n\minus{}2)\plus{}(n\minus{}2) \equal{} 2n\minus{}4$ (the $ n\minus{}2$s come from each direction).

Now we do the weird ones. By weird one, I mean one of them that hasn't fell into one of the three categories already mentioned, like these:

[asy]import cse5; D((0,0)--(4,2),linewidth(2)); D((0,0)--(6,2),linewidth(2)); D((0,0)--(8,2),linewidth(2)); D((0,0)--(10,2),linewidth(2)); for(int i=0; i<12; ++i) for(int j=0; j<3; ++j) D((i,j));[/asy]

These are also easy to count:
$ 2(n\minus{}4)\plus{}2(n\minus{}6)\plus{}2(n\minus{}8)\plus{}\cdots \equal{}\begin{cases}(n\minus{}4)(n\minus{}2)/2 &\text{if }n\text{ is even}\\ (n\minus{}3)^{2}/2 &\text{if }n\text{ is odd}\end{cases}$

We can add them up, and divide by $ \binom{3n}{3}$, the total number of ways, to get:

\[ \begin{cases}\frac{3\binom{n}{3}\plus{}3n\minus{}4\plus{}(n\minus{}4)(n\minus{}2)/2}{\binom{3n}{3}}&\text{if }n\text{ is even}\\ \frac{3\binom{n}{3}\plus{}3n\minus{}4\plus{}(n\minus{}3)^{2}/2}{\binom{3n}{3}}&\text{if }n\text{ is odd}\end{cases}\]

I really don't know how to simplify that...[/hide]

Very Good!!!  My solution is a little different though.  And to simplify your solution you can write things like $ \binom{n}{3}$ as $ \frac{n(n\minus{}1)(n\minus{}2)}{6}$.

This is the way I did it...[hide]First, I found the term $ 3\binom{n}{3}$ the same way as vishalarul did.  However, the rest of the lines I consider together instead of basing them on slant.

Consider we pick any point on the top row.  Then, three collinear points exist only if we choose a point on the bottom row whose...let's call it a horizontal position...is a multiple of two from the point we chose.

We say that every point on the top row that is two away from the top left point has odd parity while the rest have even parity.  We do the same for the bottom row.  There are $ \lfloor\frac{n}{2}\rfloor$ even points and $ \lceil\frac{n}{2}\rceil$ odd points.  Every odd point on the top row must match up with an odd point on the bottom while every even point must match with an even point on the bottom.  Thus, we have $ (\lfloor n/2\rfloor)^{2}\plus{}(\lceil n/2\rceil)^{2}$ non-horizontal points.  Thus, our total answer is:

$ \boxed{\frac{(\lfloor n/2\rfloor)^{2}\plus{}(\lceil n/2\rceil)^{2}\plus{}3\binom{n}{3}}{\binom{3n}{3}}}$
I leave this to someone else to simplify into polynomials in the numerator and denominator.[/hide]

I had seen this question on a local competition for the 3x3 case, and figured I might be able to generalize it, which is why I found this solution.  This might be a big pain, but is it possible to generalize this to m x n?  I haven't tried it, although it might be interesting to look at...

[quote="krsattack"]$ \frac{(\lfloor n/2\rfloor)^{2}\plus{}(\lceil n/2\rceil)^{2}\plus{}3\binom{n}{3}}{\binom{3n}{3}}$[/quote]
$ \binom{n}{m}\equal{}\frac{n!}{m!(n\minus{}m)!}$

Based entirely on these statements (because I don't know how to solve the problem myself) we have:
$ \frac{(\lfloor n/2\rfloor)^{2}\plus{}(\lceil n/2\rceil)^{2}\plus{}3\frac{n!}{3!(n\minus{}3)!}}{\frac{(3n)!}{3!(3n\minus{}3)!}}\rightarrow\frac{(\lfloor n/2\rfloor)^{2}\plus{}(\lceil n/2\rceil)^{2}\plus{}3\frac{n(n\minus{}1)(n\minus{}2)}{6}}{\frac{3n(3n\minus{}1)(3n\minus{}2)}{6}}\rightarrow\frac{(\lfloor n/2\rfloor)^{2}\plus{}(\lceil n/2\rceil)^{2}\plus{}\frac{n(n\minus{}1)(n\minus{}2)}{2}}{\frac{n(3n\minus{}1)(3n\minus{}2)}{2}}\rightarrow\frac{\left((\lfloor n/2\rfloor)^{2}\plus{}(\lceil n/2\rceil)^{2}\plus{}\frac{n(n\minus{}1)(n\minus{}2)}{2}\right)2}{\left(\frac{n(3n\minus{}1)(3n\minus{}2)}{2}\right)2}\rightarrow\frac{\left((\lfloor n/2\rfloor)^{2}\plus{}(\lceil n/2\rceil)^{2}\right)2\plus{}n(n\minus{}1)(n\minus{}2)}{n(3n\minus{}1)(3n\minus{}2)}\rightarrow\boxed{\frac{\left((\lfloor n/2\rfloor)^{2}\plus{}(\lceil n/2\rceil)^{2}\right)2\plus{}n(n^{2}\minus{}3n\plus{}2)}{n(9n^{2}\minus{}9n\plus{}2)}}$
Doesn't seem to get any simpler than that krsattack, unless there is a way to simplify ceiling and floor that I don't know.

[quote="Kyron"]$ \boxed{\frac{\left((\lfloor n/2\rfloor)^{2}\plus{}(\lceil n/2\rceil)^{2}\right)2\plus{}n(n^{2}\minus{}3n\plus{}2)}{n(9n^{2}\minus{}9n\plus{}2)}}$
Doesn't seem to get any simpler than that krsattack, unless there is a way to simplify ceiling and floor that I don't know.[/quote]

I should have been more specific... Try to break it up into the even and odd case.  This is the most simplified general answer, but if we want to simplify it even further, we know that if n is even, then:
$ \lfloor n/2\rfloor \equal{}\lceil n/2\rceil \equal{} n/2$

And if n is odd, then:
$ \lfloor n/2\rfloor \equal{}\frac{n\minus{}1}{2}$ and $ \lceil n/2\rceil \equal{}\frac{n\plus{}1}{2}$

Of course, it might have been easier just to simplify straight from vishalarul's answer.

What I'm more concerned about is whether there is a general answer for the $ m$ x $ n$ case.  I have not worked it out, but if anyone is willing to try...


function trigonometry calculus derivative

I got stuck with the equation : 2cos(x) = x^2.

 2cos(x) = e^ix + e^-ix

If x^2 = e^ix + e^-ix

how can i continue?

$ x^{2}\equal{} 2cosx$
$ f(x)\equal{}x^{2}\minus{}2cosx$
$ f(x)\equal{}f(\minus{}x)$
Consider $ f(x)$ on $ [0,\infty)$
$ f'(x)\equal{}2x\plus{}2sin x>0$
$ f'(x)$ monotone on $ [0,\infty)$ ,$ f(0)f(2)<0$ so it has unique solution on $ R^{\plus{}}$

since both functions are symmetric we can focus on $ x\geq 0$. Obviously, we can reduce to $ 0<x<\sqrt{2}$. Consider $ f(x)\equal{}x^{2}\minus{}2\cos x$, then $ f(0)\equal{}\minus{}2,f(1)<0,f(\sqrt{2})>0$ and for the derivative we have $ \forall 0<x<\sqrt{2}<\frac{\pi}{2}:\,f'(x)\equal{}2(x\plus{}\sin x)>0$, i.e. there is exactly one solution in the open interval $ ]1,\sqrt{2}[$


i encounter a junior science question in the schoolwork of a young kid
, it is "does solid atom has more energy or liquid atom"?

the question was simple like that, doesn't specify what kind of energy and doesn't have any more information or  condition provided, I told him that it should be liquid atom because it can move around while atoms in solid could not, so that it will have at least some kinetic energy...

i asked my friend who is very good at chemistry, but he then doubted my choices, and he thought for a while, then he didn't come into any conclusions with me

so was my answer to the kid right or wrong? Any one good at chemistry could explain me? Thanks! much!

Although I am not much sure about this,but here's what i think...
Solid atoms are more tightly bound and have strong intermolecular forces than liquid atoms.Hence they have more potential energy stored in them which can be converted into other forms.So I think solid atoms should have more energy :wink:

Of course, solid tungsten atoms at 5000K will have less energy than liquid nitrogen at 70K.

Sorry, that post was intended to be sarcastic. Since it is a question meant to be answered by a child, we deduce that the answer is "liquid" on the assumptions that:
-The "solid" and "liquid" in question are derived from the same substance.
-The term "energy" refers to internal energy, which for all practical purposes is directly proportional to thermodynamic temperature (also accounting for latent energy from phase changes) in this context.
-The solid and liquid are under the same pressure.
-The substance is not some sort of exotic material that changes phase from liquid to solid upon addition of heat.

Looking at this more generally, however, yields ambiguity en masse.

First we address the meaning of "energy". Is energy defined as thermodynamic free energy, or simply the sum of all energy within the system, including non-useful energy? Does the energy account for mass-energy equivalence, and therefore the masses of the liquid and solid samples? What about potential mechanical energy? If energy accounts for potential energy, must the substances be able to release this energy spontaneously? Are we comparing total energy, or energy density per mole/kilogram/liter/etc.? ...

Then, the terms "solid" and "liquid". Do they have the same mass? Are they the same substance? Are they under the same pressure? What about thermal differences between the samples? Are they chemically active? Is the "liquid" a pure substance or a homogeneous solution? Will the samples be sufficiently large that when isolated, they form stellar bodies capable of nuclear fusion? [i]ad infinitum...[/i]

In any case, this question is ambiguous, but from the context in which it was presented, "liquid" is most likely the correct answer.

(P.S. Strong chemical bonds do not mean high potential energy; instead, WEAKER, unstable bonds in a substance usually equates to higher potential energy. Keep in mind that bond strength directly correlates with the energy released when it is formed, and the energy required to break it.

Therefore, a substance with strong bonds will require large amounts of energy to break said bonds and participate in further chemical reactions, resulting in a large quantity being subtracted from the yield.

Additionally, intermolecular bonds are not chemical bonds. They are mere intermolecular forces, although the same logic applies: Energy is required to break bonds. By transferring energy to a solid it melts and becomes a liquid eventually, the energy having gone into weakening the intermolecular forces.

Think about it: Would you say that a block of ice has more energy than a cup of hot water, merely because the ice has an organized lattice of water molecules bound with relatively strong hydrogen bonds? Of course not.)

[quote="Aashay THE ROKR"]How come?Can you please elaborate?[/quote]
hello,
he already elaborate too much. i don't think there is need for more information.


Consider real numbers x,y,z contained in the interval $ [\minus{}1,1]$ satisying x+y+z=0.
Prove that: $ \sqrt{1\plus{}x\plus{}\frac{y^{2}}{6}}\plus{}\sqrt{1\plus{}y\plus{}\frac{z^{2}}{6}}\plus{}\sqrt{1\plus{}z\plus{}\frac{x^{2}}{6}}\leq 3$

:lol:  I find this solution while I'm taking politics test

I use the inequality sqrt(a+b)<=sqrt(a)+sqrt(b) for a,b>=0

because x,y,z>=-1

so
sqrt(1+x+y^2/6)<=sqrt(1+x)+y*sqrt(6)/6
sqrt(1+y+z^2/6)<=sqrt(1+y)+z*sqrt(6)/6
sqrt(1+z+x^2/6)<=sqrt(1+z)+x*sqrt(6)/6

add them we get

sqrt(1+x+y^2/6)+sqrt(1+y+z^2/6)+sqrt(1+z+x^2/6)<=sqrt(1+x)+sqrt(1+y)+sqrt(1+z)<=sqrt(3*(1+x+1+y+1+z))=3

Sorry for my poor English

[quote="Dumel"]you are wrong, it's not a good solution[/quote]
Yes, you're right! I'm sorry!

[b]snowingforest[/b]'s one is also faulty: $ \sqrt {y^2}\equal{}|y|$, not $ y$.


In order to join the math club, potetial membesr must guess a secret whole number from 1 to 50, by asking yes or no questions only. Matt E. Matics, an excellent logician, wanted to join, and he tried the following four questions:

1. Is the number greater than 25?
2. Is the number evenly divisible by 2?
3. Is the number evenly divisible by 3?
4. Is the number evenly divisible by 5?

After he was told the answers, he did some figuring and said, "I still dont have enough information. Is the number a perfect square?" When he got the reply "NO," Matt knew what the number was. Can you determine what the number was?
 :wallbash_red:

I think that the number is 49. The answers are no, no, no, no, yes.

[quote="tennis429"]
After he was told the answers, he did some figuring and said, "I still dont have enough information. Is the number a perfect square?" [b]When he got the reply "NO,"[/b] Matt knew what the number was. Can you determine what the number was?
[/quote]

13. All 5 answers were no, that's why the last one was capitalized. Additionally, since he didn't know after the first four questions, it could have still been 1 or 13.
Edit: nvm I'm wrong.

Sorry, I don`t read the question very well.
I think that the number is 5. The answers are no, no, no, yes, no.

Um, I just posted a solution, and $ 5\not>25$, so the answers are no,no,no,yes,no. :wink:

so which one is right? all these possibilities are possible i guess.... :fool:

[quote="math154"]The perfect square condition guarantees that before he asked the last question, there was at least one plausible perfect square answers and, most importantly, [b]one non-square possibility[/b].

If one of the possible perfect square answers was $ 1$, then the answers to the first four would be "no,no,no,no." However, there are many (more than one) prime numbers that also satisfy "no,no,no,no," so it couldn't have been $ 1$.

$ 49$ is similar: there are many (more than one) prime numbers that also satisfy "yes,no,no,no," so it couldn't have been $ 49$.

For $ 4$ or $ 16$, it would be "no,yes,no,no." However, both $ 2$ and $ 8$ satisfy this as well, so it couldn't have been $ 4$ or $ 16$.

For $ 9$, it would be "no,no,yes,no." But $ 3$ and $ 21$ both satisfy this as well, so it couldn't have been $ 9$.

For $ 36$, it would be "yes,yes,yes,no". Both 42 and 48 satisfy this, so it couldn't have been $ 36$.

For $ 25$, it would be "no,no,no,yes." The only non-square possibility after he heard this would be $ 5$. So the number must have been $ \boxed5$, after he eliminated $ 25$ from the choices $ 5,25$.[/quote]

Come on...read the whole thread before posting, especially if you asked the question. The answer is $ 5$.

wat the hell, the last question said "is the number evenly divisible by 5?" and he got no. so hoow can it be 5? :moose:

Did you read math154's post? In the case where the only possibilities are 25 and 5, it is no,no,no,yes.

Yes; the last question is "is it a perfect square?," whose answer is "no."


How many ways are there to go from one letter to one below it in this triangle and spell the word MATH?
[code]
     M
    A A
   T T T
 H H H H
[/code]
Note: The M is directly above the center T and the A's are directly above the center H's, but the post turned out differently.

That's far too easy but only because I had a hard time constructing the triangle. What is the general algorithm for solving this type of problem with longer words?

Alternately, how many ways are there to start at the upper left corner in the rectangle and end at the lower right corner, only going down or to the right? (In terms of v and h)

________________________
|   ...v vertical roads...             |
|_______________________|
|   ...h horizontal roads...        |
|_______________________|

Also note that the roads are supposed to be straight.

[quote="mathnerd314"]How many ways are there to go from one letter to one below it in this triangle and spell the word MATH?

     M
    A A
   T T T
 H H H H

Note: The M is directly above the center T and the A's are directly above the center H's, but the post turned out differently.

That's far too easy but only because I had a hard time constructing the triangle. What is the general algorithm for solving this type of problem with longer words?

Alternately, how many ways are there to start at the upper left corner in the rectangle and end at the lower right corner, only going down or to the right? (In terms of v and h)

________________________
|   ...v vertical roads...             |
|_______________________|
|   ...h horizontal roads...        |
|_______________________|

Also note that the roads are supposed to be straight.[/quote]

[hide]The way to generalize about things where there's a triangle with the number of letters in each row increasing by 1 (like you example) is this:

If there are $n$ rows, then there are $2^{n-1}$ ways.

So the answer to your question would be $2^{4-1}=8$[/hide]

If I understand your second problem right, you must go v units down and h units across no matter what route you go, so it's really a matter of how you go these routes. Thus, the number of paths is ${{v+h}\choose{v}}$

[quote="mathnerd314"]
Alternately, how many ways are there to start at the upper left corner in the rectangle and end at the lower right corner, only going down or to the right? (In terms of v and h)
[/quote]
i did a problem like this in my Mathcounts practice!!!

Yeah, so did I.

Thanks, guys.

But why do those formulas work?

Well, I sort of explained the second one, but for the first,

Start with letter M. There are two ways to get to A. For every letter A, there are two ways to get to letter T. So from M to T, there are 4 possible ways. Finally, there are two ways to get to letter H from any letter T. There are 8 possible ways total.

There were $2^1$ ways to get to the 2nd row, and $2^2$ ways to get to the third row. To the fourth row, there needed $2^3$ ways. Do you see the pattern?

Oh, I feel so stupid now. Thanks.

But I still don't understand the second one (rectangles). I'm slow. :( :roll:

Well, let's consider a 3 by 3 grid. How many ways can you go from bottom left to the upper right vertex in just 6 lengths?

It doesn't matter how you move, you will always move 3 spaces up, and three spaces right, so now how many ways can you arrange these three moves up and three moves right.

R R R U U U.

R represents the moves right, and U represents the moves up. Arrange the R's and U's how you want. You will still get to B. How many ways can you arrange the RRRUUU thing above? There are 6 choose 3 ways. Now can you generalize it for a v-by-h grid?

Oh. I understand now.

These formulas should save me some time in Mathcounts.

Any way to generalize it to other figures?


[quote="onlylove_math"][quote="Allnames"]Let$ a,b,c\geq 0,a \plus{} b \plus{} c \equal{} 1$
Prove that
$ \frac {a}{\sqrt {5a \plus{} 6b^2}} \plus{} \frac {b}{\sqrt {5b \plus{} 6c^2}} \plus{} \frac {c}{\sqrt {5c \plus{} 6a^2}}\leq \frac {3}{\sqrt {21}}$[/quote]
Need prove $ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{5(\sum ab^2)(a \plus{} b \plus{} c) \plus{} 6\sum b^4} \ge\ \frac {1}{7}$[/quote]
setting $ p \equal{} a \plus{} b \plus{} c \equal{} 1,p \equal{} \sum ab,r \equal{} abc$
we used lemma below
$ a \plus{} b \plus{} c \equal{} 1\Rightarrow \sum ab^2\le \frac {4}{27} \minus{} r$
$ Ineq \Leftrightarrow f(r)\equal{}\frac {7}{27} \minus{} 4q \plus{} 16q^2 \minus{} 19r\ge 0$
we have $ r\le \frac{q^2(1\minus{}q)}{2(2\minus{}3q)}$
$ f'(r)<0 \Rightarrow f(r)\ge f(\frac{q^2(1\minus{}q)}{2(2\minus{}3q)})\equal{}\frac{(3q\minus{}1)(693q^2\minus{}390q\plus{}28)}{54(3q\minus{}2)}\ge 0$

[quote="bunhiacovski"][quote="onlylove_math"][quote="Allnames"]Let$ a,b,c\geq 0,a \plus{} b \plus{} c \equal{} 1$
Prove that
$ \frac {a}{\sqrt {5a \plus{} 6b^2}} \plus{} \frac {b}{\sqrt {5b \plus{} 6c^2}} \plus{} \frac {c}{\sqrt {5c \plus{} 6a^2}}\leq \frac {3}{\sqrt {21}}$[/quote]
Need prove $ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{5(\sum ab^2)(a \plus{} b \plus{} c) \plus{} 6\sum b^4} \ge\ \frac {1}{7}$[/quote]
$ \frac {1}{27}(144q \minus{} 7)(3q \minus{} 1)\ge 0$[/quote]
Really?I think it wrong  :wink: . Try $ a\equal{}b\equal{}0,3;c\equal{}0,4$

[quote="onlylove_math"]
Really?I think it wrong  :wink: . Try $ a \equal{} b \equal{} 0,3;c \equal{} 0,4$[/quote]
sorry, I edited  :wink:

[quote="onlylove_math"][quote="Allnames"]Let$ a,b,c\geq 0,a \plus{} b \plus{} c \equal{} 1$
Prove that
$ \frac {a}{\sqrt {5a \plus{} 6b^2}} \plus{} \frac {b}{\sqrt {5b \plus{} 6c^2}} \plus{} \frac {c}{\sqrt {5c \plus{} 6a^2}}\leq \frac {3}{\sqrt {21}}$[/quote]
Assum $ c \equal{} max${a;b;c}
With C-S we have:
$ LHS^2 \le\ \frac {a}{5a \plus{} 6b^2} \plus{} \frac {b}{5b \plus{} 6c^2} \plus{} \frac {c}{5c \plus{} 6a^2}$
Need prove $ \frac {a}{5a \plus{} 6b^2} \plus{} \frac {b}{5b \plus{} 6c^2} \plus{} \frac {c}{5c \plus{} 6a^2} \le\ \frac {9}{21}$
$ \leftrightarrow \frac {b^2}{5a \plus{} 6b^2} \plus{} \frac {c^2}{5b \plus{} 6c^2} \plus{} \frac {a^2}{5c \plus{} 6a^2} \ge\ \frac {1}{7}$
$ \leftrightarrow \frac {b^4}{5ab^2 \plus{} 6b^4} \plus{} \frac {c^4}{5bc^2 \plus{} 6c^4} \plus{} \frac {a^4}{5ca^2 \plus{} 6a^4} \ge\ \frac {1}{7}$
And $ \frac {b^4}{5ab^2 \plus{} 6b^4} \plus{} \frac {c^4}{5bc^2 \plus{} 6c^4} \plus{} \frac {a^4}{5ca^2 \plus{} 6a^4}\ge\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{5\sum ab^2 \plus{} 6\sum b^4}$
Need prove $ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{5(\sum ab^2)(a \plus{} b \plus{} c) \plus{} 6\sum b^4} \ge\ \frac {1}{7}$
$ \leftrightarrow S_a(a \minus{} b)^2 \plus{} S_b(a \minus{} c)(b \minus{} c) \ge\ 0$
With $ S_a \equal{} (a \plus{} b)^2 \plus{} c^2 \plus{} 9c^2 \minus{} 5(ac \plus{} bc)$
       $ S_b \equal{} ab \plus{} (a \plus{} c)(b \plus{} c) \plus{} 9bc \minus{} 5(b^2 \plus{} bc)$
We easy $ S_a;S_b \ge\ 0$
We have done![/quote]
I don't think it is true my friend
Why $ S_a,S_b\geq 0$ it  is not cleary
My proof is quite hars,and i think you need detail it :roll:

[quote="Allnames"]Let$ a,b,c\geq 0,a \plus{} b \plus{} c \equal{} 1$
Prove that
$ \frac {a}{\sqrt {5a \plus{} 6b^2}} \plus{} \frac {b}{\sqrt {5b \plus{} 6c^2}} \plus{} \frac {c}{\sqrt {5c \plus{} 6a^2}}\leq \frac {3}{\sqrt {21}}$[/quote]
The following is still true
\[ \sum\frac{a}{\sqrt{a\plus{}kb^2}} \le \frac{3}{\sqrt{k\plus{}3}}\]
for all $ a,b,c \ge 0$ such that $ a\plus{}b\plus{}c\equal{}1$ and $ 0 \le k \le 2$
:)
And we can prove it by Cauchy Schwarz Inequality.
:)

[quote="Allnames"][quote="onlylove_math"][quote="Allnames"]Let$ a,b,c\geq 0,a \plus{} b \plus{} c \equal{} 1$
Prove that
$ \frac {a}{\sqrt {5a \plus{} 6b^2}} \plus{} \frac {b}{\sqrt {5b \plus{} 6c^2}} \plus{} \frac {c}{\sqrt {5c \plus{} 6a^2}}\leq \frac {3}{\sqrt {21}}$[/quote]
Assum $ c \equal{} min${a;b;c}
With C-S we have:
$ LHS^2 \le\ \frac {a}{5a \plus{} 6b^2} \plus{} \frac {b}{5b \plus{} 6c^2} \plus{} \frac {c}{5c \plus{} 6a^2}$
Need prove $ \frac {a}{5a \plus{} 6b^2} \plus{} \frac {b}{5b \plus{} 6c^2} \plus{} \frac {c}{5c \plus{} 6a^2} \le\ \frac {9}{21}$
$ \leftrightarrow \frac {b^2}{5a \plus{} 6b^2} \plus{} \frac {c^2}{5b \plus{} 6c^2} \plus{} \frac {a^2}{5c \plus{} 6a^2} \ge\ \frac {1}{7}$
$ \leftrightarrow \frac {b^4}{5ab^2 \plus{} 6b^4} \plus{} \frac {c^4}{5bc^2 \plus{} 6c^4} \plus{} \frac {a^4}{5ca^2 \plus{} 6a^4} \ge\ \frac {1}{7}$
And $ \frac {b^4}{5ab^2 \plus{} 6b^4} \plus{} \frac {c^4}{5bc^2 \plus{} 6c^4} \plus{} \frac {a^4}{5ca^2 \plus{} 6a^4}\ge\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{5\sum ab^2 \plus{} 6\sum b^4}$
Need prove $ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{5(\sum ab^2)(a \plus{} b \plus{} c) \plus{} 6\sum b^4} \ge\ \frac {1}{7}$
$ \leftrightarrow S_a(a \minus{} b)^2 \plus{} S_b(a \minus{} c)(b \minus{} c) \ge\ 0$
With $ S_a \equal{} (a \plus{} b)^2 \plus{} c^2 \plus{} 9c^2 \minus{} 5(ac \plus{} bc)$
       $ S_b \equal{} ab \plus{} (a \plus{} c)(b \plus{} c) \plus{} 9ab \minus{} 5(b^2 \plus{} bc)$
We easy $ S_a;S_b \ge\ 0$
We have done![/quote]
I don't think it is true my friend
Why $ S_a,S_b\geq 0$ it  is not cleary
My proof is quite hars,and i think you need detail it :roll:[/quote]
Oh my friend ; I think my proof clearly. And $ S_a,S_b\geq 0$ it allways true with $ c \equal{} min${$ a;b;c$}
Can U check careful ?
Thanks! :wink:

[quote="onlylove_math"][/quote][quote="Allnames"][quote="onlylove_math"][quote="Allnames"]Let$ a,b,c\geq 0,a + b + c = 1$
Prove that
$ \frac {a}{\sqrt {5a + 6b^2}} + \frac {b}{\sqrt {5b + 6c^2}} + \frac {c}{\sqrt {5c + 6a^2}}\leq \frac {3}{\sqrt {21}}$[/quote]
Assum $ c = max${a;b;c}
With C-S we have:
$ LHS^2 \le\ \frac {a}{5a + 6b^2} + \frac {b}{5b + 6c^2} + \frac {c}{5c + 6a^2}$
Need prove $ \frac {a}{5a + 6b^2} + \frac {b}{5b + 6c^2} + \frac {c}{5c + 6a^2} \le\ \frac {9}{21}$
$ \leftrightarrow \frac {b^2}{5a + 6b^2} + \frac {c^2}{5b + 6c^2} + \frac {a^2}{5c + 6a^2} \ge\ \frac {1}{7}$
$ \leftrightarrow \frac {b^4}{5ab^2 + 6b^4} + \frac {c^4}{5bc^2 + 6c^4} + \frac {a^4}{5ca^2 + 6a^4} \ge\ \frac {1}{7}$
And $ \frac {b^4}{5ab^2 + 6b^4} + \frac {c^4}{5bc^2 + 6c^4} + \frac {a^4}{5ca^2 + 6a^4}\ge\frac {(a^2 + b^2 + c^2)^2}{5\sum ab^2 + 6\sum b^4}$
Need prove $ \frac {(a^2 + b^2 + c^2)^2}{5(\sum ab^2)(a + b + c) + 6\sum b^4} \ge\ \frac {1}{7}$
$ \leftrightarrow S_a(a - b)^2 + S_b(a - c)(b - c) \ge\ 0$
With $ S_a = (a + b)^2 + c^2 + 9c^2 - 5(ac + bc)$
       $ S_b = ab + (a + c)(b + c) + 9bc - 5(b^2 + bc)$
We easy $ S_a;S_b \ge\ 0$
We have done![/quote]
I don't think it is true my friend
Why $ S_a,S_b\geq 0$ it  is not cleary
My proof is quite hars,and i think you need detail it :roll:[/quote][quote="onlylove_math"]
Oh my friend ; I think my proof clearly. And $ S_a,S_b\geq 0$ it allways true with $ c = max${$ a;b;c$}
Can U check careful ?
Thanks! :wink:[/quote]
Detail it,please
For example,What is your proof for $ S_b\geq 0$

[quote="Allnames"][/quote][quote="onlylove_math"][/quote][quote="Allnames"][quote="onlylove_math"][quote="Allnames"]Let$ a,b,c\geq 0,a + b + c = 1$
Prove that
$ \frac {a}{\sqrt {5a + 6b^2}} + \frac {b}{\sqrt {5b + 6c^2}} + \frac {c}{\sqrt {5c + 6a^2}}\leq \frac {3}{\sqrt {21}}$[/quote]
Assum $ c = min${a;b;c}
With C-S we have:
$ LHS^2 \le\ \frac {a}{5a + 6b^2} + \frac {b}{5b + 6c^2} + \frac {c}{5c + 6a^2}$
Need prove $ \frac {a}{5a + 6b^2} + \frac {b}{5b + 6c^2} + \frac {c}{5c + 6a^2} \le\ \frac {9}{21}$
$ \leftrightarrow \frac {b^2}{5a + 6b^2} + \frac {c^2}{5b + 6c^2} + \frac {a^2}{5c + 6a^2} \ge\ \frac {1}{7}$
$ \leftrightarrow \frac {b^4}{5ab^2 + 6b^4} + \frac {c^4}{5bc^2 + 6c^4} + \frac {a^4}{5ca^2 + 6a^4} \ge\ \frac {1}{7}$
And $ \frac {b^4}{5ab^2 + 6b^4} + \frac {c^4}{5bc^2 + 6c^4} + \frac {a^4}{5ca^2 + 6a^4}\ge\frac {(a^2 + b^2 + c^2)^2}{5\sum ab^2 + 6\sum b^4}$
Need prove $ \frac {(a^2 + b^2 + c^2)^2}{5(\sum ab^2)(a + b + c) + 6\sum b^4} \ge\ \frac {1}{7}$
$ \leftrightarrow S_a(a - b)^2 + S_b(a - c)(b - c) \ge\ 0$
With $ S_a = (a + b)^2 + c^2 + 9c^2 - 5(ac + bc)$
       $ S_b = ab + (a + c)(b + c) + 9ab - 5(b^2 + bc)$
We easy $ S_a;S_b \ge\ 0$
We have done![/quote]
I don't think it is true my friend
Why $ S_a,S_b\geq 0$ it  is not cleary
My proof is quite hars,and i think you need detail it :roll:[/quote][quote="onlylove_math"]
Oh my friend ; I think my proof clearly. And $ S_a,S_b\geq 0$ it allways true with $ c = min${$ a;b;c$}
Can U check careful ?
Thanks! :wink:[/quote][quote="Allnames"]
Detail it,please
For example,What is your proof for $ S_b\geq 0$[/quote]
Why don't U believe me? :(
$ S_a = (a + b)^2 + 9c^2 + c^2 - 5(ac + bc)$
    $ \ge\ 6c(a + b) + c^2 - 5(ac + bc)\ge\ 0$

$ S_b = ab + (a + c)(b + c) + 9ab - 5(b^2 + bc)$
       $ = ab + (a + c)(b + c) + 5(2ab - b^2 - bc) \ge\ 0$ by $ c = min${$ a;b;c$}
Are U ok? :wink:

[quote="onlylove_math"][quote="can_hang2007"][quote="Allnames"]Let$ a,b,c\geq 0,a \plus{} b \plus{} c \equal{} 1$
Prove that
$ \frac {a}{\sqrt {5a \plus{} 6b^2}} \plus{} \frac {b}{\sqrt {5b \plus{} 6c^2}} \plus{} \frac {c}{\sqrt {5c \plus{} 6a^2}}\leq \frac {3}{\sqrt {21}}$[/quote]
The following is still true

$ \sum\frac {a}{\sqrt {a \plus{} kb^2}} \le \frac {3}{\sqrt {k \plus{} 3}}$

for all $ a,b,c \ge 0$ such that $ a \plus{} b \plus{} c \equal{} 1$ and $ 0 \le k \le 2$
:)
And we can prove it by Cauchy Schwarz Inequality.
:)[/quote]
I think it also similar. And it 'll be solve by SS :wink:[/quote]
Are you sure? ;)

[quote="can_hang2007"][/quote][quote="onlylove_math"][quote="can_hang2007"][quote="Allnames"]Let$ a,b,c\geq 0,a + b + c = 1$
Prove that
$ \frac {a}{\sqrt {5a + 6b^2}} + \frac {b}{\sqrt {5b + 6c^2}} + \frac {c}{\sqrt {5c + 6a^2}}\leq \frac {3}{\sqrt {21}}$[/quote]
The following is still true

$ \sum\frac {a}{\sqrt {a + kb^2}} \le \frac {3}{\sqrt {k + 3}}$

for all $ a,b,c \ge 0$ such that $ a + b + c = 1$ and $ 0 \le k \le 2$
:)
And we can prove it by Cauchy Schwarz Inequality.
:)[/quote]
I think it also similar. And it 'll be solve by SS :wink:[/quote][quote="can_hang2007"]
Are you sure? ;)[/quote]
sorry if my idea is wrong. I 'll check again carefully  :)

sorry onlylovemath :roll: ,because my proof has to be  used $ \sum (2a^2\minus{}b^2\minus{}c^2\minus{}5ab\plus{}5bc)^2 \geq 0$ (it is true :P )
With the general problem that can_hang2007 has posted,i think we can used
$ \sum (2a^2\minus{}b^2\minus{}c^2\minus{}kab\plus{}kbc)^2 \geq 0$
Sorry if my ideal is wrong :P

[quote="Allnames"]
With the general problem that can_hang2007 has posted,i think we can used
$ \sum (2a^2 \minus{} b^2 \minus{} c^2 \minus{} kab \plus{} kbc)^2 \geq 0$
Sorry if my ideal is wrong :P[/quote]

Really? I think it don't similar your problem. If follow your way ; with $ 2\ge\ k\ge\ \frac {3}{2}$your proof is wrong ( Follow can_hang2007 's ideal ) :wink:
[hide="@ Allnames"] Sorry Allnames my proof have some mistakes . I have edited[/hide]
[hide="@ can_hang2007"] Sorry can_hang2007 by SS i only prove it with $ k \le\\frac {3}{2}$  [/hide]

[quote="Allnames"]i think your post Which after you edited is still wrong,check $ S_b\ge 0$ carefully[/quote]

[quote="Allnames"]Let$ a,b,c\geq 0,a \plus{} b \plus{} c \equal{} 1$
Prove that
$ \frac {a}{\sqrt {5a \plus{} 6b^2}} \plus{} \frac {b}{\sqrt {5b \plus{} 6c^2}} \plus{} \frac {c}{\sqrt {5c \plus{} 6a^2}}\leq \frac {3}{\sqrt {21}}$[/quote]
Sorry Allnames, Here is my full proof . I sure it not wrong
Assum $ c \equal{} max${$ a;b;c$}
With C-S we have:
$ LHS^2 \le\ \frac {a}{5a \plus{} 6b^2} \plus{} \frac {b}{5b \plus{} 6c^2} \plus{} \frac {c}{5c \plus{} 6a^2}$
Need prove $ \frac {a}{5a \plus{} 6b^2} \plus{} \frac {b}{5b \plus{} 6c^2} \plus{} \frac {c}{5c \plus{} 6a^2} \le\ \frac {9}{21}$
$ \leftrightarrow \frac {b^2}{5a \plus{} 6b^2} \plus{} \frac {c^2}{5b \plus{} 6c^2} \plus{} \frac {a^2}{5c \plus{} 6a^2} \ge\ \frac {1}{7}$
$ \leftrightarrow \frac {b^4}{5ab^2 \plus{} 6b^4} \plus{} \frac {c^4}{5bc^2 \plus{} 6c^4} \plus{} \frac {a^4}{5ca^2 \plus{} 6a^4} \ge\ \frac {1}{7}$
And $ \frac {b^4}{5ab^2 \plus{} 6b^4} \plus{} \frac {c^4}{5bc^2 \plus{} 6c^4} \plus{} \frac {a^4}{5ca^2 \plus{} 6a^4}\ge\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{5\sum ab^2 \plus{} 6\sum b^4}$
Need prove $ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{5(\sum ab^2)(a \plus{} b \plus{} c) \plus{} 6\sum b^4} \ge\ \frac {1}{7}$
$ \leftrightarrow S_a(a \minus{} b)^2 \plus{} S_b(a \minus{} c)(b \minus{} c) \ge\ 0$
With $ S_a \equal{} (a \plus{} b)^2 \plus{} 10c^2 \minus{} 5(ab \plus{} c^2)$
       $ S_b \equal{} ab \plus{} (a \plus{} c)(b \plus{} c) \plus{} 9ab \minus{} 5(ab\plus{}bc\minus{}c^2)$
We easy $ S_a;S_b \ge\ 0$ by $ c \equal{} max${$ a;b;c$}
We have done!


Determine a function f(x) such that
$\large f(t^2+t+1)=t$ for all $\large t \geq  0$

Let $x=t^2+t+1,x \ge 1$
Solve the equation:
$t^2+t+1-x=0$


Let P(z) = az^2 +bz+c.
Assume P(z) is real if and only if z is real.
Show that a =0.

$z=\lambda+i\mu$
$P(z)=a(\lambda^2-\mu^2)+b\lambda+c+(2a\lambda\mu+b\mu)i$
$P(z)$ is real if and only if $z$ is real (i.e. $\mu=0$), so $\mu(2a\lambda+b)=0\Leftrightarrow\mu=0$.  This implies that $2a\lambda+b\not=0$ for all $\lambda$.  If $a\not=0$, this in turn implies that $\lambda\not=-\frac{b}{2a}$; but $\lambda$ can take on all real values, so this is not true.  Then, assume $a=0$.  If $b\not=0$, the unequality will always hold.  Therefore, $P(z)$ is real if and only if $z$ is real is true if and only if $a=0$ and $b\not=0$.


[quote][color=green]It is given the triangle $ ABC$ and the points $ K,L$ on the sides of $ AB,AC$ such as $ \frac{KB}{AK}\plus{}\frac{LC}{AL}\equal{}1$
Prove that the segment $ LK$ passes thru the centroid of the triangle $ ABC$[/color][/quote]

[hide][quote="balan razvan"][color=green]It is given the triangle $ ABC$ and the points $ K,L$ on the sides of $ AB,AC$ such as $ \frac {KB}{AK} \plus{} \frac {LC}{AL} \equal{} 1 \ \ (1)$
Grove that the segment $ LK$ passes thru the centroid of the triangle $ ABC$[/color][/quote]

Let $ M$ be the midpoint of $ BC$. Let $ G\in AM \cap KL$

Let $ B',C' \in AG$ such that $ BB' \parallel CC\parallel KL$
From the equality of the triangles $ BMB'$ and $ CMC'$we find that $ M$ is the midpoint of $ B'C'$ ($ BB'CC'$ is a parallelogram)

Also we should notice that $ G$ is [b]not[/b] in the segment $ B'C'$.
[hide="Indeed"]$ K$ is between $ A$ and $ B$, so $ G$ is between $ A$ and $ B'$. Similarly $ G$ is between $ A$ and $ C'$
Consequently $ G$ is not in the segment $ B'C'$. This allows us to write below $ GM \equal{} \frac {GB' \plus{} GC'}{2}$
[/hide]
----------------------------------------------------------------------------------------------

[asy]pair A=(1.5,16), B=(0,0), C=(20,0), H, O, M, G;

G=centroid(A,B,C);
M=midpoint(B--C);

pair K = A+0.8*(B-A);
dot(K);
label("K", K, NW);

pair L = A+(4/7)*(C-A);
dot(L);
label("L", L, NE);

draw(K--L);
dot(A^^B^^C^^G^^M);

draw(A--M);
draw(A--B--C--A);

pair B1 = intersectionpoint(A--M, B--B+(L-K));
dot(B1);
label("B'", B1, E);
draw (B--B1);

pair C1 = intersectionpoint(A--M+(M-A), C--C+(K-L));
dot(C1);
label("C'", C1, S);
draw (C--C1);

draw(M--C1);

label("A",A,NE); label("B",B,SW); label("C",C,SE);
label("G",G,N);
label("M",M,SW);[/asy]

We have $ \frac {KB}{AK} \equal{} \frac {GB'}{AG}$ and $ \frac {LC}{AL} \equal{} \frac {GC'}{AG}$

From $ (1)$ we get $ \frac {GB'}{AG} \plus{} \frac {GC'}{AG} \equal{} 1 \equal{} > GB' \plus{} GC' \equal{} AG \equal{} > 2GM \equal{} AG$

This means that $ G$ divides the segment $ AM$ in a ratio $ 2: 1$. Also $ G$ is between $ A$ and $ M$, hence $ G$ is the centroid of $ \triangle ABC$[/hide]

[quote="balan razvan"][quote][color=green]It is given the triangle $ ABC$ and the points $ K,L$ on the sides of $ AB,AC$ such as $ \frac {KB}{AK} \plus{} \frac {LC}{AL} \equal{} 1$
Prove that the segment $ LK$ passes thru the centroid of the triangle $ ABC$[/color][/quote][/quote]

[hide="another hint"]If $ KL \parallel{} BC$ the conclusions is imediately.
If $ M\in KL \cap BC$and $ N\in KL\cap AA'$ where $ A'$ is the middle of $ BC$.And try the problem by aplying Menelaus[/hide]


Hey guys, i was wondering how do you find a slope of a line with only 2 coordinates and no graph paper?

Difference in y divided by difference in x

Or to actually explain it...

Given two points $ (x_1,y_1)$ and $ (x_2,y_2)$, the slope of the line connecting the two points is $ \frac{y_2\minus{}y_1}{x_2\minus{}x_1}$.

lol oops forgot to explain
basically rise (y) over run (x)
but keep them in order.

hello, let $ P_1(x_1,y_1)$ and $ P_2(x_2,y_2)$ then the slope is given by $ \frac{y_2\minus{}y_1}{x_2\minus{}x_1}$ if $ x_1\neq x_2$.
Sonnhard.

If $ x_1\equal{}x_2$, then one could think of the slope as "$ \infty$." More accurately, though, it is undefined.

The slope can also be defined as $ \frac{\text{rise}}{\text{run}}$ (rise over run)

[quote="007math"]The slope can also be defined as $ \frac {\text{rise}}{\text{run}}$ (rise over run)[/quote]

uh... isn't that what I said?

Here's an example. If you're given these two points:
(2,1) and (3,2) then you count off x1 being 2, y1 being 1. the x and y 1s are the first coordinate. The second coordinate would therefore be 3=x2 and 2=y2. So the formula is y2-y1/x2-x1. So you input the values of the variable in. You get 2-1/3-2=1/1=1. Slope=1


Prove that there are exist a prime number p sataifying the condition:
$p|2^{2^n}+1$ and $p\geq\ (n+1)2^{n+2}$

discussed before:

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=divisor&t=46125


[hide="problema 2 - solutia 2"]Folosim inegalitatea evidenta $a^{5}+b^{5}\geq a^{3}b^{2}+a^{2}b^{3}$(se verifica usor) in forma echivalenta $\frac{a^{5}}{b^{3}}\geq \frac{a^{3}}{b}+a^{2}-b^{2}$. Scriem si inegalitatile analoage si prin sumarea lor rezulta ca mai avem de justificat inegalitatea $\frac{a^{3}}{b}+\frac{b^{3}}{c}+\frac{c^{3}}{a}\geq a^{2}+b^{2}+c^{2}.$
Avem $\frac{a^{3}}{b}+\frac{b^{3}}{c}+\frac{c^{3}}{a}=\frac{a^{4}}{ab}+\frac{b^{4}}{bc}+\frac{c^{4}}{ca}\geq\frac{(a^{2}+b^{2}+c^{2})^{2}}{ab+bc+ca}\geq{a^{2}+b^{2}+c^{2}}$. Egalitatea are loc daca si numai daca $a=b=c$[/hide]
[hide="problema 2 - solutia 3"] Functia $f: \mathbb{R}\longrightarrow\mathbb{R}, f(x)=x^{3}$ este convexa pe $(0;\infty)$. Folosind inegalitatea lui Jensen,
$\frac{\frac{a^{5}}{b^{3}}+\frac{b^{5}}{c^{3}}+\frac{c^{5}}{a^{3}}}{a^{2}+b^{2}+c^{2}}=\frac{a^{2}(\frac{a}{b})^{3}+b^{2}(\frac{b}{c})^{3}+c^{2}(\frac{c}{a})^{3}}{a^{2}+b^{2}+c^{2}}\geq \Bigg(\frac{\frac{a^{3}}{b}+\frac{b^{3}}{c}+\frac{c^{3}}{a}}{a^{2}+b^{2}+c^{2}}\Bigg)^{3}\geq 1$[/hide]
[hide="problema 2 - solutia 4"]Avem inegalitatea $\frac{a^{5}}{b^{3}}\geq\frac{5}{2}a^{2}-\frac{3}{2}b^{2}$(usor de demonstrat) si inegalitatile analoage. Prin sumarea lor rezulta concluzia.[/hide]

Ideea comuna de rezolvare este aplicarea inegalitatii mediilor.

Spre exemplul 3. Aplicam inegalitatea AM-GM si avem: \[\frac{x^{3}}{(1+y)(1+z)}+\frac{1+y}{8}+\frac{1+z}{8}\geq\frac{3x}{4}.\] Scriind inca doua inegalitati similare, prin insumare obtinem solutia problemei (in final se mai tine cont ca $x+y+z\geq3\sqrt[3]{xyz}$). Celelalte probleme se rezolva asemanator.

[quote][hide="solutie 2"]$\sum_{cyc}\frac{a^{5}}{b^{3}}=\frac{1}{a^{3}b^{3}c^{3}}\sum_{cyc}a^{8}c^{3}$.
deci
ramane sa aratam ca
$\sum_{cyc}a^{8}c^{3}\geq\sum a^{5}b^{3}c^{3}$.
voi folosi inegalitatea mediilor (cu ponderi) :
$275a^{8}c^{3}+165b^{8}a^{3}+99c^{8}b^{3}\geq 539\sqrt[539]{a^{2695}b^{1617}c^{1617}}=539a^{5}b^{3}c^{3}$.
insumand ciclic aceasta inegalitate, voi obtine chiar inegalitatea dorita.[/hide][/quote]

Maamaaa ce macel a fost cu ponderile alea:))

[quote="pohoatza"][quote]
voi folosi inegalitatea mediilor (cu ponderi) :
$275a^{8}c^{3}+165b^{8}a^{3}+99c^{8}b^{3}\geq 539\sqrt[539]{a^{2695}b^{1617}c^{1617}}=539a^{5}b^{3}c^{3}$.
insumand ciclic aceasta inegalitate, voi obtine chiar inegalitatea dorita.[/quote]
Maamaaa ce macel a fost cu ponderile alea:))[/quote]

Se poate si un pic mai bine (cu Mathematica): \[25 a^{8}c^{3}+15 b^{8}a^{3}+9 c^{8}b^{3}\geq 49 \sqrt[49]{ a^{245}b^{147}c^{147}}= 49 a^{5}b^{3}c^{3}.\] (i.e. impartim la $11$ inegalitatea lui maky.)


Determine all functions f : R -> R which satisfy

f(x + y) + f(f(x) - y) = 2f(f(x)) + 2y

for all real numbers x and y.

I really enjoyed this one:

[hide]

We make x=0 to get f(y)+f(f(0)-y)=2f(f(0))+2y^2 (*). In (*) we make the transformation y->f(0)-y and we find f(f(0)-y)+f(y)=2f(f(0))+2(f(0)-y)^2 (**). From (*) and (**) we get y^2=(f(0)-y)^2 iff (f(0))^2=2f(0)*y for all real y iff f(0)=0. We use this with (*) to get f(x)+f(-x)=2x^2 for all real x (***).

Now, in the first equality we were given we make y=f(x) and we find f(f(x)+x^2)=2f(f(x))+2(f(x))^2 (#)

We now make y=-x^2 (also in the first relation we were given). We get f(f(x)+x^2)=2f(f(x))+2x^4 (##). From (#) and (##) we get (f(x))^2=x^4, so for all real x f(x) is x^2 or -x^2. Let's assume there exists x<>0 s.t f(x)=-x^2. We plug this in (***) to find f(-x)=3x^2, but f(-x) is either x^2 or -x^2 => contradiction because x<>0.

So the conclusion is: the only functon is f(x)=x^2. [/hide]

Yeah, I also enjoyed this one.

Here are three other functional equations :

(1) Find all f : R -> R such that for all reals x, y :
x + y + 2f(xy) = f(x + y)(f(x) + f(y))

(2) Find all f : R -> R such that for all reals x, y :
f(x - y) = (x - y)(f(x) + f(y)).

(3) Find all f : R -> R such that for all reals x, y :
f(xy) = max{f(x), y} + min{f(y), x}.

The last one is rather trivial ... :)

Arne wrote:(3) Find all f : R -> R such that for all reals x, y :
f(xy) = max{f(x), y} + min{f(y), x}.

[hide]Take x = y = 1.  Then f(1) = max(f(1), 1) + min(f(1), 1).  No matter the relation between f(1) and 1, we have f(1) = f(1) + 1.  Thus, there are no functions from R -> R satisfying that relation.[/hide]

Arne wrote:(1) Find all f : R -> R such that for all reals x, y :
x + y + 2f(xy) = f(x + y)(f(x) + f(y))

[hide]We plug in x=y=0 to find f(0)=0 or 1. We plug in y=-x to get either -x:^2:=f(-x:^2:) or 2x:^2:+2f(-x:^2:)=f(x)+f(-x). The first one gives us that for all t<0, f(t)=t. Now, if xy<0, wlog x>0 and y<0. We have then x:^2:+2xy+y:^2:=f(x+y)*(f(x)+y) Now, if x+y<0, we have x=f(x). This means for any t>0, as long as we can find a z such that z<0 and t+z<0, f(t)=t. Since we can always find such a z, f(t)=t for all t.

Now we consider the other case, f(0)=1. We plug in y=0 to get x:^2:+2=f(x)*(f(x)+1). We add 1/4 to both sides to get x:^2:+9/4=(f(x)+.5):^2: This means f(x)=-1/2:pm::sqrt:(x:^2:+9/4). For x=0, we take the positive square root. So if it's continuous, I think we always take the positive square root.

Anyways, plugging this function into the equation yields x:^2:+y:^2:-1:pm::sqrt:(4x:^2:y:^2:+9)=-1/2:pm::sqrt:((x+y):^2:+9/4)*(-1/2:pm::sqrt:(x:^2:+9/4)-1/2:pm::sqrt:(y:^2:+9/4)). Um don't think that's always true.  [/hide]

2) Find all f : R -> R such that for all reals x, y :

f(x - y) = (x - y)(f(x) + f(y)).

[hide]

x=y => f(0)=0

y=0

f(x^2)=xf(x)

f(x^2)=-xf(-x)=xf(x)

f(-x)=-f(x)

so we consider all x>0, under the assumption that there does not exist an x other than 0 such that f(x)=0...

f(x^2)=xf(x)=x(x^(1/2)*f(x^1/2))=...=x^(1+1/2+1/4+1/8+...)=x^2

[/hide]i was told i could not do this because we dont even know if f is continuous... but does this matter at all?[hide]

obviously i am missing some cases as seen from grobber's proof two posts down...

[/hide]

Arne wrote:(2) Find all f : R -> R such that for all reals x, y :
f(x - y) = (x - y)(f(x) + f(y)).

[hide]x=y=0 gives f(0)=0. y=0 gives f(x:^2:)=xf(x). x=0 gives f(-y:^2:)=-yf(y). Thus f is odd. x=1, y=0 gives f(1)=1*f(1)=c. As f(1)=c, f(-1)=-c, and y=1, -1 gives (x-1)(f(x)+c)=(x+1)(f(x)-c) => cx-f(x)=f(x)-cx => f(x)=cx. [/hide]

If I'm not mistaken, I'm the one who told bugzpodder that thing abt continuity (on Mathlinks; this problem was discussed there too). I don't know if it's true, but I'm beginning to think you're right.  Anyway, here's my soln:

[hide]I proved that f is odd in the same way bugzpodder did. What I did afterwards is this: f(x^2-y^2)=(x-y)(f(x)+f(y))=(x+y)(f(x)-f(y)), which we obtain by replacing y with -y in the initial equality. This means that xf(y)=y(f(x). For x,y<>0 we get f(x)/x=f(y)/y, so the functions are f(x)=ax for a real a.[/hide]

yes you are... and your proof is definately nice!

You can't take that last step -- f(x)^2 = x^4 does not imply f(x) = x^2 or f(x) = - x^2.  It does imply that for any given x, f(x) = x^2 or -x^2.  You need that same last paragraph that grobber had in order to nail down what the function can be specifically.


Scores on the Scholarly Ability Test are calculated by subtracting one-fourth of the number of questions answered incorrectly from the number of questions answered correctly. What is the score of a person who answered 38 questions correctly and 12 incorrectly on a 50 question test.

The 38 correct= 38*1 pt each=38 pts.
Subtract 12*1/4 for each wrong=3 pts.
38-3=35 pts
Scholarly Ability Test=SAT?
I'm jealous. :)


Prove: $n|\phi(2^n - 1)$,

where $\phi$ is Euler's totient function.

In the multiplicative group of residues modulo $2^n-1$ which are coprime with $2^n-1$, $2$ has order $n$, so $n$ divides the order of the group, which is precisely $\varphi(2^n-1)$.

We can write was grobber said in a slightly different way: using Euler's theorem we have $ 2^n-1| 2^{\phi(2^n-1)}-1$. Write $\phi(2^n-1)=nq+r$. It follows easily that $ 2^n-1|2^r-1$ which is possible only if $r=0$.


integration
inequalities
real analysis
real analysis unsolved

Hello, I have not read mathlinks since one month ! I was preparing for second concours of ENS Lyon and I'm admit. This is the oral exercise that I have done.

$f,g$ continuous, from $[a;b]$ to $\mathbb{R}^{+*}$. $I_n$ is defined as
\[I_n=\int_a^b f(x)\times g(x)^n dx\]
Show that $\sqrt[n]{I_n}$ and $\frac{I_{n+1}}{I_n}$ converges to maximum of $g$.

[quote="Rodolphe2005"]I was preparing for second concours of ENS Lyon and I'm admit.[/quote]
Congratulation

He gave me indication that $g$ is monotonous by "morceaux" in french (maybe parts in english)

It is easy problem comparing to the monsters given usually in oral examinations. Congratulations, Rodolphe. Hope to have the same success next year... :)

Let $a_1=I_1$ and, for $n\ge 2,\ a_n=\frac{I_n}{I_{n-1}}$. The first thing we have to show is that $\sqrt[n]{a_1a_2\ldots a_n}$ converges to $M=\max_{[a,b]}g$, and the second part is proving that $a_n$ converges to $M$. This second result is stronger, but it follows immediately from the first one once we prove that $a_n$ is increasing.

From the Cauchy-Schwartz inequality we immediately get $\int^b_afg^{n+1}dx\cdot\int^b_afg^{n-1}dx\ge\left[\int^b_afg^ndx\right]^2$, which is equivalent to $a_{n+1}\ge a_n$, so $a_n$ is increasing, as desired.

All we need to do now is prove that $\sqrt[n]{I_n}\to M$. Given any $\varepsilon>0$, let $I$ be a subinterval of $[a,b]$ on which $g$ takes values between $M-\frac\varepsilon 2$ and $M$. We then have $\left(M-\frac\varepsilon 2\right)\sqrt[n]{\int_Ifdx}\le\sqrt[n]{I_n}\le M\sqrt[n]{\int^b_afdx}$, so when $n$ is large, $\sqrt[n]{I_n}$ lies between $M-\varepsilon$ and $M$.

nice solution. I've done the same thing but I've not use sequence $(a_n)$ but proved $\frac{I_{n+1}}{I_n}$ converges to $M$ and that was harder and longer.

Indeed, it is not an easy task to prove the convergence of $\frac{I_{n+1}}{I_n}$ without using the convergence of $\sqrt[n]{I_n}$.  :)


geometry
3D geometry
number theory unsolved
number theory

[quote="al.M.V."]correct me if im wrong, but i seem to remember that no 3 cubic powers can form an arithmetic progression... :)[/quote]ok, thank you.  does anybody know a proof of this?  i've only found references to Mordell, "Diophantine Equations"...

Almv you are wrong. Take -a^3, 0, a^3 and you've got three cubes in AP.
The proof is not very difficlt. You use infinite descent and unique factorisation ring. First u do some Wlog assumptions, then use factorisation and a complex ring Z[sqrt(-3)] and its unique factoriaxtion like the Gaussian integers to bash it out. Then all solns are of form a=-b.

Bomb

Here's the soln in full:
Show that an equation of the form a^3 + b^3 = 2z^3
has no solutions other than those with a=+-b.  To prove this, assume
a,b,z are the solution with the absolutely smallest value of z.

Since a,b are coprime with distict magnitudes, and the sum of their
cubes is even, they must both be odd.  This implies the existence of
coprime integers u,v such that a=u+v and b=u-v.  Substituting into
a^3 + b^3 = 2z^3 and simplifying gives

                      u (u^2 + 3v^2)  =  z^3

which proves that  u  and  u^2 + 3v^2  must both be cubes.  Thus, we
have integers r,s such that

             u = r^3        and           u^2 + 3v^2 = s^3

The right hand equation can be factorised as

              (u + v sqr[-3]) (u - v sqr[-3])  =  s^3

We now take advantage of the fact that the ring of numbers of the form  a + b sqr(-3)
possesses unique factorization (where a,b are actually half-integers),
and assert that each of the (coprime) factors on the left is a cube.
Thus, we have coprime (half)integers g,h such that

     u + v sqr[-3]  =  (g + h sqr[-3])^3

                    =  g (g^2 - 9h^2)  +  h (3g^2 - 3h^2) sqr[-3]

from which it follows that

         u = g (g^2 - 9h^2)           v = 3h (g^2 - h^2)

Recalling that u = r^3, the left hand equation factors as

               r^3  =  g (g+3h) (g-3h)

(Notice that if g and h are really half-integers we can clear all
fractions by multiplying through by 2^3, which still gives a cube on
the left side.)  Since g,h are coprime, each of the above three factors
are coprime and therefore each must be cube.  Thus, we have integers
m,n,q such that

        g = m^3        g+3h = n^3       g-3h = q^3

which imply that  n^3 + q^3 = 2m^3.  But clearly m < g < r < z, so
m < z, which contradicts our assumption that z is the smallest integer
such that the double of its cube equals the sum of two cubes.  Therefore,
no non-trivial integer solutions of A^3 + B^3 = 2C^3 exist, as required.

Bomb

[quote="bomb"]Almv you are wrong. Take -a^3, 0, a^3 and you've got three cubes in AP.
[/quote]

Fair enough, I've edited my post, but I was discounting trivial cases. ;)

[quote="bomb"]                      u (u^2 + 3v^2)  =  z^3
which proves that  u  and  u^2 + 3v^2  must both be cubes.[/quote]
Not true... $\gcd (u, u^2+3v^2) = 1$ or $3$, so $u^2 + 3v^2 = s^3$ or $3s^3$ for some integer $s$. But this is trivial and can easily be patched up. The case when $\gcd (u, u^2+3v^2) = 1$ is already covered by [i]Bomb[/i], and for the case when $\gcd (u, u^2+3v^2) = 3$, by letting $u = 3u_0$, it reduces to $3{u_0}^2 + v^2 = s^3$, which is essentially the same as the first case.

[quote="al.M.V."]correct me if im wrong, but i seem to remember that no 3 positive distinct cubic powers can form an arithmetic progression... :)[/quote]

this is true in the case these three integers are prime

[quote="qwertyyy"]Find all solutions of $x^3 + y^3 = 2z^3$, x,y,z integers.[/quote]
I am intereesting about this problem too. I know one approach. Write me vluzgin@hotmail.com

whats wrong with $(1,1,1)$?

Or is that counted as a trivial solution?

[quote="bomb"]Here's the soln in full:
Show that an equation of the form a^3 + b^3 = 2z^3
has no solutions other than those with a=+-b.  To prove this, assume
a,b,z are the solution with the absolutely smallest value of z.

Since a,b are coprime with distict magnitudes, and the sum of their
cubes is even, they must both be odd.  This implies the existence of
coprime integers u,v such that a=u+v and b=u-v.  Substituting into
a^3 + b^3 = 2z^3 and simplifying gives

                      u (u^2 + 3v^2)  =  z^3

which proves that  u  and  u^2 + 3v^2  must both be cubes.  Thus, we
have integers r,s such that

             u = r^3        and           u^2 + 3v^2 = s^3

The right hand equation can be factorised as

              (u + v sqr[-3]) (u - v sqr[-3])  =  s^3

We now take advantage of the fact that the ring of numbers of the form  a + b sqr(-3)
possesses unique factorization (where a,b are actually half-integers),
and assert that each of the (coprime) factors on the left is a cube.
Thus, we have coprime (half)integers g,h such that

     u + v sqr[-3]  =  (g + h sqr[-3])^3

                    =  g (g^2 - 9h^2)  +  h (3g^2 - 3h^2) sqr[-3]

I think you are wrong here.

Because    of unit

[/quote]

[quote="BOMB"]

              (u + v sqr[-3]) (u - v sqr[-3])  =  s^3

We now take advantage of the fact that the ring of numbers of the form  a + b sqr(-3)
possesses unique factorization (where a,b are actually half-integers),
and assert that each of the (coprime) factors on the left is a cube.
Thus, we have coprime (half)integers g,h such that

     u + v sqr[-3]  =  (g + h sqr[-3])^3

                    =  g (g^2 - 9h^2)  +  h (3g^2 - 3h^2) sqr[-3]
[/quote][/quote]

I think you are wrong here.

Because    of unit

u+vsqr(-3)  can be  (1+sqr(-3))/2  * (g + h sqr[-3])^3


function
real analysis
real analysis unsolved

$f: \mathbb{R}^{n}\to \mathbb{R}$ a function. For $x \in \mathbb{R}^{n}$, one can define $Lip(f,x) = \limsup_{y \to x}\frac{|f(y)-f(x)|}{|x-y|}$. Let $L = \{x : L(f,x) < \infty \}$ be the set of point where ths quantity is finite.
Prove that for almost every $x \in L$, $f$ is differentiable at $x$.

The proof is long if one starts from scratch, so let's assume that we already know Rademacher's theorem: a Lipschitz function is differentiable a.e. Our $f$ is not Lipschitz, but we can define its Lipschitz majorants and minorants, which are differentiable a.e. At a.e. points of the set $L$ the majorant and minorant squeeze $f$ hard enough to make it differentiable, too. This idea leads to a simple proof (due to Jan Maly) which is reproduced on pp.23-24 of [url=http://www.math.jyu.fi/research/reports/rep100.pdf]Lectures on Lipschitz Analysis[/url] by Juha Heinonen.


Prove that for each $ a\in\mathbb N$, there are infinitely many natural $ n$, such that
\[ n\mid a^{n \minus{} a \plus{} 1} \minus{} 1.
\]

hallo,
$n=(a-1)p$ for $p \in \mathbb{P}$.

$ a^{n \minus{} a \plus{} 1} \minus{} 1 \equal{} a^{ap \minus{} a \minus{} p \plus{} 1} \minus{} 1 \equal{} a^{(a \minus{} 1)(p \minus{} 1)} \minus{} 1 \equiv 0 \bmod p, (a \minus{} 1)$.  But we need the condition $ p \not | a(a \minus{} 1)$ to conclude.

I don't understand why the condition for p not dividing a(a-1) is included...
also, is it congruent to 0 mod p and congruent to 0 mod a-1?  why isn't it just a? :S

$ a \minus{} 1 | a^m \minus{} 1$ for all $ m$.  We need $ p \not | a$ or else $ p$ can't divide a number of the form $ a^m \minus{} 1$, and we need $ p \not | a \minus{} 1$ to conclude that divisibility by $ p$ and $ a \minus{} 1$ implies divisibility by $ p(a \minus{} 1) \equal{} n$.

You're wonderful :) Thank you.

Hopefully you'll have a good afternoon also :D

We can take $ n\equal{}a^{p}\minus{}1$ where $ p$ is a arbitrary prime number . then :
$ n\minus{}a\plus{}1\equal{}a^{p}\minus{}a\equal{}pk$ so , $ a^{p}\minus{}1|a^{n\minus{}a\plus{}1}\minus{}1$


LaTeX
Asymptote
geometry
ratio
number theory
prime numbers
irrational number

I'm not sure how to use latex (or post in the right category on the forum) so bear with me...

I was pondering how to come up with a formula to get all the prime numbers (as futile as that may be) when I came up with this following proof:

First, proof of Irrational*Rational=Irrational (I*R=I)

If Irrational*Rational=Rational, Irrational=Rational/Rational. Since rational/rational=rational (by definition of rational numbers), the assumptions must be false and therefore Iratioanl*Rational must be Irrational ([b]Irrational * Rational=Irrational[/b])

1/Rational=?

If 1/Rational=Irrational, Irrational*Rational=1 (which is a Rational (number)). Contradition. (three dots/therefore) [b]1/Rational=Rational[/b]
Same goes for Irrational: [b]1/ Irrational=Irrational[/b]

Now for the Irrational*Irrational=Rational

Start with Irrational*Rational=Irrational (proven above)

Irrational*Rational=Irrational
Irrational*(1/Irrational)=1/Rational (dividing both sides by Rational*Irrational/the Irrationals have switched sides)
[b]Irrational*Irrational=Rational[/b] (1/Irrational=Irrational and 1/Rational=Rational)
Q.E.D.

Using this proof you can now state sqrt2*sqrt3=rational number or (pi)^2=rational number.

Feel free to nominate me for the Fields medal :lol:

posted by Go (pseudonym Tare) @8/8/2007

I just realized if you square sqrt(pi), you get (pi) which is irrational. But if

rational*rational=rational
rational*irrational=irrational
irrational*irrational=rational

then sqrt(pi) (or any sqrt(irrational)) is neither rational nor irrational...strange

You didn't prove anything for all pairs of irrational numbers, you only proved it for certain pairs.

$ \sqrt{2}\times \sqrt{3}= \sqrt{2 \times 3}= \sqrt{6}$

All irrational, hence your claim is disproven.

Also, the reciprocal of a rational number is always rational. Since any rational number can be written as a fraction with an integer denominator and numerator, we have:

$ \forall a,b \in \mathbf{Z}$

$ \textrm{rational value}= \frac{a}{b}$

$ (\textrm{rational value})^{-1}= \left(\frac{a}{b}\right)^{-1}$

$ \frac{1}{\textrm{rational value}}= \frac{b}{a}$

And since b and a are integers, our initial conditions are still satisfied.

As for learning latex, check "latex and asymptote" in the site & general information area of the forum.

Finally, I'm not sure what most of your conclusions are based on. That probably explains why most of them are wrong (no offense).

I based everything on two facts.
1.) rational/rational=rational
2.) rational numbers and irrational numbers are mutually exclusive and together form the group "real numbers"

I'm not sure where I went wrong with the proof...

Altheman I'm not sure what you mean by certain pair versus all pair-the proof should work for all pair (theoretically at least)

Lastly [i]you cannot prove that sqrt of a non-square number is irrational[/i] here's my counterproof for the famous "sqrt(2) is irrational proof"

The last proof had some faults in it which may have caused some conclusion. Here's an updated version of it.

If we take the set of real numbers and say that that set could be divided into two sets, rationals and irrationals, the proof that I wrote above will stand. However as I added in the second post, if the proof was correct then
rational*rational=rational
irrational*rational=irrational
irrational*irrational=rational.
However sqrt (irrational) * sqrt(irrational)=irrational. Since you cannot multiply two numbers together and get an irrational number, the assumption ("Real numbers can be divided into two (non-exclusive) groups rational number and irrational number") must be [i]false[/i].

[i]Since[/i] the proof of "sqrt (non-square number)=irrational" is based on reductio ad absurdum (rational numbers and irrational numbers compose the whole set of real numbers) and I have just proven that that is not the case (there is a [i]third[/i] category of real numbers that is neither rational nor irrational), you [i]cannot say that sqrt (non-square number) is irrational[/i] using this proof. Therefore your claim that sqrt(6) is irrational cannot be for certain (for that matter, sqrt 2 and sqrt 3 may not be irrational either)

If there is no logical fault for the original proof, then that proof should go on to conclude that "Real numberrs contain more than three groups of numbers, rationals, irrationals, and some other number (s-o-ns))

ok so irrational*irrational may or may not be rational, but I still proved that there are numbers which are neither rational nor irrational. Si?

I think you're out of your mind.

[quote]
irrational*irrational=rational
[/quote]

WHY????????????? Give me one good reason why this is true in the general case.

[quote]
here's my counterproof for the famous "sqrt(2) is irrational proof"
[/quote]

:rotfl: do you honestly believe you can disprove such a widely accepted proof?

[quote]
Since you cannot multiply two numbers together and get an irrational number...[/quote]

$ e \times \pi$ is irrational. disproven.

[quote]
there is a third category of real numbers that is neither rational nor irrational
[/quote]

This is getting downright amusing. Now what on earth would be the property of these numbers that fit into your magical new category of numbers?

[quote]
Real numberrs contain more than three groups of numbers, rationals, irrationals, and some other number (s-o-ns))
[/quote]

I'm sure "some other number" is a nice title for your new group of numbers.

The only numbers that are neither rational or irrational are complex numbers such as $ i^{2}=-1$. I reccomend that you carefully read the following sentence:

a rational number can be written as a fraction with integer values for the denominator and numerator. Otherwise (with the exception of complex numbers), the given number is irrational.

This means that $ \frac{2\sqrt{3}}{5}$ is irrational and not part of some fabricated new set of numbers since $ 2\sqrt{3}$ is irrational (it is approximately $ 3.464101615$). I'm not sure how else to convince you that you're claim is ridiculous, other than to inform yourself better about what characterizes a real number, a rational number and an irrational number.

[quote="Prime factorization"]
The only numbers that are neither rational or irrational are complex numbers such as $ i^{2}=-1$.
[/quote]

So there is no more future arguement about that, stating the obvious $ \mathbb{C}\not\in \mathbb{R}$.

Sorta irrelavent, but the sqrt2 is irrational because:

If $ \sqrt{2}$ is rational, then it can be written as a $ \frac{p}{q}$ that is simplified

$ \sqrt{2}=\frac{p}{q}$
Square both sides
$ 2=\frac{p^{2}}{q^{2}}$
Multiply $ q^{2}$
$ 2q^{2}=p^{2}$
Therefore, p is even. Since all even squares are a multiple of 4 (or 2*2),q must also be even, However, that violates our assumption that p/q is reduced. Therefore, $ \sqrt{2}$ is $ \text{IRRATIONAL}$

P.S. $ This_{is^{totally_{irrelavent^{!!}}}}$

You probably should have thought a bit about my comment before posting that post.

Your proof is bad because you are not clear about your theorems. At least you use the following theorem in this manner at the later parts of your posts. I believe you are making the statement: for any $ I_{1}$; $ I_{2}$, there exists $ Q_{1}$ such that $ I_{1}*Q_{1}=I_{2}$. Clearly you have not proven this proposition. [I am going to add that $ Q_{1}\ne 0$.]

Lets walk through your proof: So you say that $ I_{1}*Q_{1}=$ irrational or rational.

It is clear enough that $ I_{1}*Q_{1}=I_{3}$ where $ I_{3}$ is irrational. But $ I_{3}$ is not an arbitrary irrational number; you didn't prove it for any irrational number $ I_{2}$. That is to say $ I_{3}$ is dependent on $ I_{1}$ since your proof is constructive.

For your proof to be valid, you would have to start with $ I_{1}$; $ I_{2}$ and show that $ Q_{1}$ exists. I think the origin of this problem in your proof is that you did not define your theorems carefully.

Of course the rest of your 'results' become nonsense once you realize that the '$ I_{1}*Q_{1}=I_{2}$' proposition is incorrect.

You've obscured a lot of what you're doing by not writing your quantifiers.  Allow me to explain your error, thereby duplicating some of the things that Altheman has written.

[quote="Tare"]Irrational*Rational=Irrational
Irrational*(1/Irrational)=1/Rational (dividing both sides by Rational*Irrational/the Irrationals have switched sides)
[b]Irrational*Irrational=Rational[/b] (1/Irrational=Irrational and 1/Rational=Rational)
Q.E.D.[/quote]

Your initial statement is, "The product of any rational number and any irrational number is irrational."  This is not actually true (the rational number could be 0), but it's close enough.  What's being quantified over?  We have universal quantifiers over both variables on the left.  By the third line, however, you're claiming "The product of any two irrational numbers is rational."  When you thought you were being clever with your arithmetic, what you actually did was cheat and start applying the universal quantifiers to the two irrational numbers, something without justification.  The actual result you can claim from your manipulations is, "If were have two irrational numbers whose ratio is rational, multiplying one by the reciprocal of the other yields a rational number."  This is rather less impressive.

It's quite easy to find pairs of irrational numbers which multiply to other irrational numbers (although why PrimeFactorization thought $ \pi \cdot e$ was a good example, I have no idea) or to prove nonconstructively that such pairs exist with a cardinality argument.

As for your grander claims: the irrational numbers are defined exactly as those real numbers which are not rational.  It therefore follows [i]from the definition[/i] that there are no real numbers which are neither rational or irrational.  If your claim that the product of two irrationals always yields a rational were true (which it isn't), your derivation of the result that there must be numbers which are neither rational or irrational would be correct.  Because your product result is false, however, this derivation is meaningless: any statement can be derived starting with a false premise.

I * I=I
SQRT(3) * SQRT(5)=SQRT(15)
I                 I             I
OR
I * I=R
SQRT(3)  * SQRT(12)=SQRT(36)=6
I                 I              R

hm.

So yeah, my pseudoproof was wrong then :(

I'm glad to see everyone's alive and well though :) JBL, forlorn...

Yeah.

yellowcakes.

[quote="Prime factorization"]The only numbers that are neither rational or irrational are complex numbers such as $ i^{2}=-1$. I reccomend that you carefully read the following sentence:

a rational number can be written as a fraction with integer values for the denominator and numerator. Otherwise (with the exception of complex numbers), the given number is irrational.[/quote]

There is no third set, even in $ \mathbb C$.
$ i$ is irrational as it is not rational. There is no reason not to apply the definition from $ \mathbb R$ to $ \mathbb C$, too.


Shalah has two bottles of ketchup. The larger bottle is 1/2 full and the smaller bottle is empty but holds 12 ounces. When the contents of the smaller bottle are dumped into the smaller, the smaller is 3/4 full. How many ounces of ketchup does the larger bottle hold?

if the smaller bottle is $ \frac{3}{4}$ full when its full is $ 12 oz.$, that means it now holds $ (\frac{3}{4})(12 oz.) \equal{} 9 oz.$ since this is $ \frac{1}{2}$ the total amount the larger bottle holds, the larger bottle holds $ (2)(9 oz.)$ which is $ 18 oz.$


1/ Person A is drawing a pattern on a large piece of paper. He starts by drawing a segment of length 1 cm, then turns $ 138^{\circ}$ to the left and draws a segment of length 2 cm, and again turns $ 138^{\circ}$ to the left. In this manner, he draws segments of length 1 cm, 2 cm, 1 cm, 2 cm, and so on with each segment drawn following a $ 138^{\circ}$ turn. The first four steps are shown in the diagram. Person A discovers that the pattern eventually starts to retrace itself exactly. What is the total length, in centimetres, of the pattern before it starts to retrace?

2/ A quadruple (a, b, c, d) of positive integers such that
•$ a\le b\le c\le d$,
• a, b, c, d are consecutive terms in an arithmetic sequence,
• $ a^3 \plus{} b^3 \plus{} c^3 \equal{} d^3$, and
• at least one of a, b, c, d between 1867 and 2008 inclusive.
is called a quixotic quadruple. How many quixotic quadruples are there?

[hide="Problem 2"]
Let:
\[ a\equal{}a\]
\[ b\equal{}a\plus{}x\]
\[ c\equal{}a\plus{}2x\]
\[ d\equal{}a\plus{}3x\]

Since we know that a,b,c and d form an increasing arithmetic sequence, we know that a and x are both positive integers. They are obviously not equal since that wouldn't allow the first, third and fourth conditions simultaneously.

Plugging in those values into the equation:
\[ a^3 \plus{} (a\plus{}x)^3 \plus{} (a\plus{}2x)^3 \equal{} (a\plus{}3x)^3\]
\[ 3a^3 \plus{} 9a^2x \plus{} 15ax^2 \plus{} 9x^3 \equal{} a^3 \plus{} 9a^2x \plus{} 27ax^2 \plus{} 27x^3\]
\[ 2a^3 \equal{} 12ax^2 \plus{} 18x^3\]
\[ a^3 \equal{} 6ax^2 \plus{} 9x^3\]
\[ a^3 \equal{} 3(2ax^2 \plus{} 3x^3)\]

Therefore, a must be divisible by 3. So let a = 3k. Plugging that into the equation, we see that:
\[ 27k^3 \equal{} 18kx^2 \plus{} 9x^3\]
\[ 3k^3 \equal{} 2kx^2 \plus{} x^3\]
\[ 3k^3 \equal{} x^2(2k \plus{} x)\]

We know that x is positive, so let k = xl since k must be divisible by x.
\[ 3x^3l^3 \equal{} 2lx^3 \plus{} x^3\]
\[ 3l^3 \equal{} 2l\plus{}1\]

Solving this equation, we see that the only integer solutions is l = 1. So, we know that k = x and a = 3x. So our quixotic quadruple is 3x,4x,5x,6x. That is our only possibility. For all positive x, this sequence satisfies conditions 1,2, and 3. Now we must worry about condition 4. So we must make at least one of them between 1867 and 2008.

The smallest possible x value is when 6x is greater than 1867 for the smallest possible value. The largest possible value of x is when 3x is smaller than 2008 for the largest possible value of x. So we know that x is between 312 and 669 inclusive. So there are 669-311 = 358 quixotic quadruples.

[/hide]


If two random variables  X and Y assume only two values each, and if Cov(X,Y) =0, prove that X and Y are independent.

Let the range of X be x_1 and x_2 and the range of Y be y_1 and y_2. Then we need to somehow show that

$ P(X \equal{} x_i \mbox{ and } Y \equal{} y_j) \equal{} P(X \equal{} x_i) P(Y \equal{} y_j)$.

[quote="roah"]Hint:
$ Cov(aX \plus{} b,cY \plus{} d) \equal{} acCov(X,Y)$,
so we may assume that $ X$ and $ Y$ take only values 0 and 1.
[hide="Rest of solution:"]
If $ Z$ takes only 0 and 1 then $ E(Z) \equal{} P(Z \equal{} 1)$.
If $ E(XY) \minus{} E(X)E(Y) \equal{} 0$ then $ P(X \equal{} 1\cap Y \equal{} 1) \minus{} P(X \equal{} 1)P(Y \equal{} 1) \equal{} 0$.
[/hide][/quote]

But how does that show, for example that

$ P(X \equal{} 1\cap Y \equal{} 1) \equal{} P(X \equal{} 1)P(Y \equal{} 0)$

The pair $ (1,1)$ is enough. Either use the linear transformation that maps $ 0$ to $ 1$ and vice versa for the first variable, or just express $ P(X\equal{}0)\equal{}1\minus{}P(X\equal{}1)$, $ P(X\equal{}0,Y\equal{}1)\equal{}P(Y\equal{}1)\minus{}P(X\equal{}Y\equal{}1)$.

And of course this falls apart if you allow three values. My favorite example for "uncorrelated does not imply independent": Let $ (X,Y)$ take on the value $ (1,0),(0,1),(\minus{}1,0)$ and $ (0,\minus{}1)$ each with probability $ \frac14.$ (That is, one step of a two-dimensional random walk.)

[b] To Kent Merryfield[/b]: What is the motivation for giving this example to show that uncorrelated does not imply independent?
Thanks.

Uncorrelated does not imply independent.
Imagine data thus ditributed:
(1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,2.5),(2,3),(2,3.5),(2,4)
The data clearly show no correlation (every increase in the x value does not correlate to a specific value increase in y).
Regardless, y is not independent of x. It is clear that, although there is no correlation (Covar(x,y)=0), as x increases, the error of y increases. In this way, x does not affect the change in the center value of y (as correlation implies), but it does affect y's distribution.


since most forums here have their own formula and theorem link and since we here just have one link
so i thought why not we start a post containin good collection of formulaes , reactions and other beautiful important things.
well obviously this can happen if you ([b]carcul,GregX007,Cancer...[/b].....and all others (including me :wink:) ) want.so i just want your opinion...so should we have one or is this the first and last post on that idea.
if yes just post on and i will[i] edit this later[/i] :wink:

i think we seriously should have one :)

I don't mind to make a contribution for that idea, but don't you think that is already very easy to make a google search (for example) when we want to find something?

yes that's true but the reason for making is why others made  :maybe: which i think is limited search u have to do in a space that's too small with a high probability of finding something really useful than searching 5-6 web pages and finally getting your answer  :)

sometimes it so happens that u r searching absolutely nothing but then a reaction say posted by X may be imp and u missed ..so basically this(if created  :wink: ) will be probably of much help for those (like me :D ) who on most cases don't exactly know what we are googling but just want some imp. reactions...protecting groups..name reactions... :)

It would be more correct to say that halogen addition to a double bond protects it, by converting it to vicinal dihalide. Then to remove the protecting group (the two halogens) and re-obtain the double bond we can add some zinc.

[u][b]MOLE CONCEPT[/b][/u]

1 mole = $ 6.02 x 10^{2}3$
$ moles$=$ \frac{mass}{f.wt}$
$ moles$=$ conc.Xvol.$

I think we seriously need a new moderator.

I agree with that. At least someone that could have an active participation and, for example, address the viability of this proposal from Pardesi.

We should create vote on it ( that is if we have enough people to vote on it).

i would like to become the mod lest someone else is interested

Have you tried to contact the current moderator? Perhaps he can provide some relevant information.

did u see when he last posted .it was in 2005  :o

[quote="pardesi"]actually i asked valentin about this but sadly no reply... :(
i think the entire arts and science section is poorly moderated...it's really necessary.[/quote]

How long ago did you ask?

As far as I can tell the only active posters of this forum are Carcul, pardesi, gregx007, and BanishedTraitor. We seriously need to recruit some more people.

pardesi and Carcul post/visit the most often (I think), so one of them should probably be the new moderator.

[quote="gregx007"]How long ago did you ask?[/quote]
say a week or two  ago

[url=http://www.mathlinks.ro/viewtopic.php?t=53756]interesting comments from carcul[/url]  :D

Funny, we talk about getting a new moderator but we don't bother doing it.....

thanks brandished for posting it there  :lol:

thanks for the reply Banished
when u heat $ \text{Beta\minus{}Ceto Acids},\text{Glycidic Acids}$ at about $ 100^{o}\text{C}$ u will get $ \text{CO}_{2}$

Carbon dioxide is the least interesting product that we obtain from those acids. What is the other product in each case? That is what matters in organic synthesis.

yes i read that now u get decarboxylated(alkane kind) product this gives ketone

From a beta-cetoacid we get a ketone. How about from a glycidic acid?

not yet read  :rotfl:
will tell that after i read about it  :)

[u][i][b]NERNST EQUATIONS[/b][/i][/u]

E = E0 – (0.0591 / n ) log10 ([PRODUCTS]/ [REACTANTS])

First, when writing an equation we should also explain the meaning of the symbols used. Second, that form of the Nernst equation is only valid at 25ºC. Third, "log" is already understood as the base 10 logarithm, there is no need to write "log10".

While this thread is a great idea, upon closer inspection its ratio of formulas/discussion is too low.  I'll start a new thread for the sticky.  Please reproduce the [i]revelant[/i] posts there.

Actually u can move them there being a moderator

Yes, it's possible.  However, I feel that if we want anything to compile anything serious that the posts should be a lot prettier and more condensed than they currently are.  The quality of the first few posts will determine the quality of later ones.  Rather than going through and editing each post without permission of the relevant authors, therefore, I ask you to reproduce the posts.

If enough people just want me to move old posts, I will do so; however, I feel that this would undermine the quality of the resulting thread.

For the collection of law thing, do you think we should post by topics? Like Week 1 post things on Orbitals, VB etc. Because I think we need a certain degree of organization.

That's an idea, actually.  Or alternatively, modify an earlier suggestion of yours and have the Collection point towards a bunch of other threads, one thread for each topic.

I moved the posts I thought were on topic to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179750]here[/url].  We'll probably need better organization of sorts; I rather like Banished's idea, but it would take work.

In fact greg why not post a sticky where we discuss ideas for the forum and then evaluate their applicability? I think this forum can be great and all we need is a bit of organization  :) .


Find all pairs of integers $ (m,n)$ such that the numbers $ A\equal{}n^2\plus{}2mn\plus{}3m^2\plus{}2,\ B\equal{}2n^2\plus{}3mn\plus{}m^2\plus{}2,\ C\equal{}3n^2\plus{}mn\plus{}2m^2\plus{}1$ have a common divisor greater than $ 1$.

suppose $ p$ is prime and $ p|A,B,C$.

$ A\minus{}B \equal{} 2m^2 \minus{} mn \minus{} n^2 \equal{} (m\minus{}n)(2m\plus{}n) \qquad(1)$

$ C\minus{}B \equal{} m^2 \minus{} 2mn \plus{} n^2 \minus{} 1 \equal{} (m\minus{}n)^2\minus{}1 \qquad(2)$

From $ (1)$, $ p|(m\minus{}n)$ or $ p|(2m\plus{}n)$ but clearly $ p\not | (m\minus{}n)$ because of $ (2)$

replacing $ n\equiv \minus{}2m \mod p$ in $ A$ and $ C$ gives $ 3m^2\plus{}2 \equiv 12m^2\plus{}1 \mod p$

but $ \gcd(3m^2\plus{}2,12m^2\plus{}1) \equal{} \gcd(3m^2\plus{}2,7)$ so the greatest common denominator is at most $ 7$

so $ 3m^2\plus{}1 \equiv 0 \mod 7 \Longrightarrow m \equiv 2,5 \mod 7 \Longrightarrow n \equiv 3,4 \mod 7$

hence $ (m,n) \equal{} (7k_1\plus{}2,7k_2\plus{}3) or (7k_1\plus{}5,7k_2\plus{}4)$


Does anyone know where I can get worked out solutions to last years national competition?

thanks!

I am almost positive that there is no solution document to the competiton. The only place  to get the answers is your coach or buy the 2005 competition set from the  MATHCOUNTS store on their site

I think there are the ones that only have the answer but not worked out although at nationals they do have it worked out for you right after the test, but we were to nervous to learn how to do the problems.


group theory
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

If $ R$ is a ring with unity $ 1\ne 0$ , then $ Char(R)\equal{}2$ whenever $ |U(R)|$ is odd .

Note that :

$ U(R)$ is a group of units of the ring $ R$.

 :)

Show that for all other characteristics, there is a multiplicative subgroup of even order.

I know I need to prove that, but how ??

tell me the sketch of proof at least .

Thx  :)

There is a canonical element $ x \neq 1$ such that $ x^2\equal{}1$.

Let me to show my work :

let $ Char(R)\equal{}n\ne 2$ , and assume $ U(R)\equal{}\{1,a_1,a_2,\cdots,a_{2k}\}$, where $ a_i\ne a_j$, for all i,j

now , $ a_1U(R)\equal{}\{a_1,{a_1}^2, \cdots , a_1a_{2k}\}$

Clearly , $ a_1a_i\ne a_1a_j$ where $ i\ne j$.

So, if $ {a_1}^2\equal{}1$ , then let $ x\equal{}a_1$.done ,

if not , there exist an element $ a_1a_j$ such that $ {a_1}^2a_1a_k\equal{}1$ . since $ a_1\in U(R)$ , there exist an element $ a_i\ne a_1$, so we get $ {a_i}^2\equal{}1$ , contradiction , we must have $ {a_1}^2\equal{}1$.

Let, $ x\equal{}a_1$, Is there something wrong in my previous work?  :huh:

if my work is true, then how I can get contradiction .

Sorry  :roll:

[quote="Hidden Scofield"]if not , there exist an element $ a_1a_j$ such that $ {a_1}^2a_1a_k \equal{} 1$ . since $ a_1\in U(R)$ , there exist an element $ a_i\ne a_1$, so we get $ {a_i}^2 \equal{} 1$ , contradiction , we must have $ {a_1}^2 \equal{} 1$.[/quote]
This makes absolutely no sense to me.

What's some element whose square is $ 1$?  It's a very common element, you've seen it before. :)

Could u write a full solution for the problem LydianRain please, I'm warping about it .

Thx  :)

Try it for $ \mathbb{Z}/n$ first, please. :)

I know $ M\equal{}\mathbb Z/n$ is ring with $ Ch(M)\equal{}n$.Also, $ U(M)$ is of even order always whenever $ n>2$ (the contrapositive of my question  :D )

I know the ring $ R$ should has a subring that's isomorphic to $ \mathbb Z_n$ or $ Z$.

Sorry LydianRain, I don't know what do u point for ? :oops:

First of all, proving the contropositive is perfectly alright, unless you're a constructivist.  Maybe that's what you're saying--it's unclear.

Now, this is not just you, it seems that many people have a serious problem with the way they approach math exercises--they don't try small cases!  An overwhelming percentage of questions become much easier when considered *in light of* some concrete examples.

We are looking for an element of order $ 2$ in $ U(R)$.  Until you know what this element is for the rings $ \mathbb{Z}$, $ \mathbb{Z}/3$, $ \mathbb{Z}/4$, $ \mathbb{Z}/5$, and $ \mathbb{Z}/6$, please don't try anything more complicated!  You won't learn anything by having somebody tell you the answer to this, trust us. :)

I trust you dude  :wink:

I think I got that element , it's $ a\equal{}(n\minus{}1).1$, where $ ch(R)\equal{}n$.

So, $ a^2\equal{}(n\minus{}1).1(n\minus{}1).1\equal{}(n\minus{}1)(n\minus{}1).1\equal{}(n^2\minus{}2n\plus{}1).1\equal{}(n(n.1)\minus{}2(n.1)\plus{}1.1)\equal{}0\plus{}0\plus{}1\equal{}1$.

Is that true ??  :)

Yes.  Sometimes we also call this element $ \minus{}1$ :)

Oh yes, since $ n.1\equal{}0$  then $ (n\minus{}1).1\equal{}\minus{}1$.

Now,I know we need to equip other elements in the group $ U(R)$ as pairs , so we get a contradiction.

Thus, what is the strategy to guarantee they equip as pairs ? how we could guarantee they are not canonical elements also?

I'm so thankful LydianRain  :)

Don't try to pair up elements--use Cauchy's theorem.


(Make that the center of mass of a quadrant of an astroid.)

This was my lecture example for today. As a computational hint, I was talking about curves given by parametric equations.

Consider the region in the first quadrant ($ x\ge0,y\ge 0$) bounded by the two axes and by the curve $ x^{2/3}\plus{}y^{2/3}\equal{}a^{2/3},$ where $ a$ is a positive constant. Assuming constant mass density, find the center of mass of this object.

do you mean that
\[ \begin{gathered}
  x_0  \equal{} \frac{1}
{m}\iint\limits_S {x\rho dS} \hfill \\
  y_0  \equal{} \frac{1}
{m}\iint\limits_S {y\rho dS} \hfill \\
  z_0  \equal{} \frac{1}
{m}\iint\limits_S {z\rho dS} \hfill \\
\end{gathered}
\]
where
\[ m \equal{} \iint\limits_S {\rho dS}.
\]

This is two-dimensional. Everything is in the $ xy$ plane; there's no $ z$ in the problem. Imagine a thin flat sheet of constant thickness, cut to fit that shape in the plane.


[quote="Ligouras"]Prove that
$ \sec \frac{A}{2}\plus{}\sec \frac{B}{2}\plus{}\sec \frac{C}{2}\geq\frac{8\sqrt{3}}{9}$[/quote]
$cos \frac{A}{2}cos\frac{B}{2}cos \frac{C}{2}\le \frac{3\sqrt{3}}{8}\implies $

$\sec \frac{A}{2}\plus{}\sec \frac{B}{2}\plus{}\sec \frac{C}{2}\geq \frac{3}{\sqrt[3]{cos \frac{A}{2}cos\frac{B}{2}cos \frac{C}{2}}}\ge 2\sqrt{3}.$


[hide]Okay.

$a^{2}+a+1=0$ implies that

$a^{3}-1=0$, where $a\neq1$

So $a^{2006}+\frac{1}{a^{2006}}=a^{2}+\frac{1}{a^{2}}$.

Also, we know that this is equal to $\frac{a^{4}+1}{a^{2}}$.

So this implies that $\frac{a+1}{a^{2}}$ which implies that $\frac{-a^{2}}{a^{2}}$, so the answer is $-1$

$-1$[/hide]

[quote="mahyar_math"]if $a^{2}+a+1=0$ find $a^{2006}+\frac{1}{a^{2006}}$[/quote]

Well, dividing the whole expression by a,
$\frac{a^{2}}{a}+\frac{a}{a}+\frac{1}{a}=\frac{0}{a}$
 $a+1+\frac{1}{a}=0$
$a+\frac{1}{a}=-1$
 $(a+\frac{1}{a})^{2006}=-1^{2006}$
$a^{2006}+\frac{1}{a^{2006}}+2=1$
therefore $a^{2006}+\frac{1}{a^{2006}}=-1$
I asked my math teacher the SAME ONE  the other day..it was in our  textbk so..i guess this is another way of doin it
This is the first time i'm using Latex..hope it works!

yea thats what i thought but i was sooo confused how he got that.

$a^{2}+a+1=0\implies a+\frac{1}{a}=-1$
Now just to show a new and different way to solve this problem, we can delve into complex numbers and trigonometry:
It is well known that $\frac{e^{i\theta}+e^{-i\theta}}{2}=\cos \theta$
[hide="derived"]$e^{i \theta}=\cos \theta+i\sin \theta$, now substitute $-\theta$ for $\theta$ and add the two equations, then divide by two[/hide]
Substitute for $\theta$: $\frac{e^{i\theta*n}+e^{-i\theta*n}}{2}=\cos \theta*n$
Notice that we have so far: $a+a^{-1}=-1$ and we are trying to find: $a^{2006}+a^{-2006}$... look familiar now?
let $a=e^{i\theta}$, so plugging into our formula, $\cos \theta =-\frac12$
Therefore $\theta=\frac23 \pi$ (well it can also be another value, which behaves almost exactly like this value... check it!)
Now $\cos 2006*\frac23 \pi$ is the same as $\cos 2*\frac23 \pi=-\frac12$ so plugging back into our other formula, $\frac{a^{2006}+a^{-2006}}{2}=-\frac12\implies a^{2006}+a^{-2006}=\boxed{-1}$
This might be the wrong level for this forum, but it gives an idea of what one can do with diferent types of math in "unrelated" situations...

ya..the step..i have no clue..i am a  bit shaky in this area... :)

which step? I'm not sure what you're talking about...

De Moivre's is too strong for this problem.  I would advise against using it, since the simple algebraic solution builds better problem-solving skills, is faster to write out, and applies to more general problems of this type.

[quote="srulikbd"]
soomething strange... if you say that $a^{3}=1$ you can't say that $a=\pm1$?
 :maybe:[/quote]

Where did I say $a=\pm 1$?  :maybe:

[quote="srulikbd"]soomething strange... if you say that $a^{3}=1$ you can't say that $a=\pm1$?[/quote]

I think you mean $a = 1$.  And no.  $a$ here is complex.  You'll find that $a = 1$ does not satisfy the original condition $a^{2}+a+1 = 0$.

[quote="srulikbd"]but I think that also the complex number doesn't fit the condition $a^{3}=1$, right?[/quote]
$a^{2}+a+1=0\Longrightarrow a=\frac{-1\pm i\sqrt3}{2}\Longrightarrow a^{3}=1$

[quote="srulikbd"]but I think that also the complex number doesn't fit the condition $a^{3}=1$, right?[/quote]

$a^{3}= 1 \Leftrightarrow a^{3}-1 = 0 \Leftrightarrow (a-1)(a^{2}+a+1) = 0$


Answer this question correctly and I will keep posting new ones!

4 boys and 4 girls are randomly seated in a row of 8 chairs. What's the probability that they are seated alternately.( boy-girl-boy..... or girl-boy-girl)

2006 Chapter Team:

24x24x2=$ \boxed {1152}$

Isn't the point of a marathon to get the people who get it right to post new problems? :wink:

EDIT: nvmd...  lol, this isn't from 2006. :roll:

[hide]
First note that there are only two ways to seat the little children if they are distinguished by only gender. Because there are $ 4!$ ways to seat four boys and $ 4!$ for four girls, while $ 8!$ for eight people.  The answer is $ \dfrac{2*4!4!}{8!} \equal{} \dfrac{1}{35}$
[/hide]

*okay, if it's a true marathon i'm okay with it.

Your answer was correct.
Fine, the [i]solver[/i] can post the new questions
Next 2:

1)what is the greatest power of two which divides 20!

2)If 3 segments are chosen randomly for 6 segments having length 2,3,5,6,7,10. Whats the probability that the 3 segments can form a triangle?

Answer to my 2nd question:

[hide] the answer to my 2nd question was: 8/(6C3) = 2/5 (I'm pretty sure thats right but I might have forgotten a case)[/hide]

[quote="distracted523"][hide="#1"]
Note that there are 10 multiples of 2 in the terms, 5 multiples of 4, 2 multiples of 8, and 1 multiple of 16. For each higher power of 2 there is an uncounted power of 2. Our answer is $ 10 \plus{} 5 \plus{} 2 \plus{} 1 \equal{} \fbox{18}$.
[/hide]

[hide="#2"]
Consider the case where the longest segment is 10. Then the sum of the other two sides must be > than 10.
7-6, 7-5, and 6-5 work.

Now count for longest 7:
6-5, 6-3, 6-2, 5-3 work.

Longest 6:
5-3, 5-2 work

Longest 5 and below:
none work

So 9 out of $ \dfrac{6!}{3!3!} \equal{} 20$ possibilities work.

Answer $ \dfrac{9}{20}$
[/hide]

Improvised question:
There are metal rods of length 1, 7, 18, and 34 inches. In how many ways can you use one or more (i.e. joined together) of the rods for a rod to be used for a gate? Assume that solder takes up no space.[/quote]

Meh, you people should read my post better. Anyone want to solve the problem?

I'm tired so I'd probably get it right--omg, did I just say right? I meant wrong. that's how tired i am.

seems like there will be many cases and I always undercount when they are a lot of cases so I'llget it wrong anyways. Post a problem that doesn't involve so much casework

I suck at counting which means I suck at probability :(

You guys are pretty thick then.

[hide="I have to post my own solution meh"]
This problem is in actuality very easy. There is no way for you to combine two or three rods to get a rod you already have (try it if you don't believe me). Therefore you just proceed as follows:

1 rod: 4 combs
2 rods: 6 combs
3 rods: 4 combs
4 rods: 1 comb

Answer $ \fbox{15}$
[/hide]

And don't post if you are going to lament on how you don't want to do it. :spam:

I ask for a lock, because someone revived the old one. We don't need this topic anymore.


geometry
geometric transformation
induction
combinatorics proposed
combinatorics

From each of $k$ points on a plane, a few rays are drawn. No two rays intersect. Prove that one can choose $k-1$ of the segments connecting these points such that they are disjoint from one another and from any of the rays, except possibly at the $k$ points.

Pierre.

I didn't understand problem.
I tryed to find it in LMO Book, but I didn't succeed.

What don't you understand?
I give you the statement as given in Crux.

As I said in another thread, this problem appears at the end of the book (according to the files that Orlando sent to me) with no solution (even if I found one   :P  ). I checked the files and, keeping in mind that I do not read russian, I would say that it is problem 30 (1975).

Pierre.

Ok. I have found it!
But it was really hard to understand english variant.

Could you give your own translation please?

Pierre.

Hmm... Problem is easy, isn't it? :)

Consider arbitrary point $A$ and ray $r_A$.
[b]Case 1.[/b] Whole line $l_A$ ($r_A\subset l_A$) doesn't intersect with other rays. If $l_A$ separates remaining points into to nonempty subsets then apply induction to both these parts (take into consideration that $A$ belongs to both parts!). If one of half-planes doesn't contain points from our set, then we can remove $r_A$ from consideration!

[b]Case 2.[/b] Line $l_A$ meets ray $r_B$ at point $C$. We assume that it is the nearest to $A$ such point. If angle $X$ contains points (except $A$) then apply induction (and again $A$ belongs to $X$ and $Y$ simultaneously). If angle $X$ is empty then we can exclude ray $r_A$ from consideration!

But how can you apply induction in case 2 if you are not sure that rays emanating from $Y$ do not interfer with segments between points from $X$ (and vice-cersa)?

Pierre.

:what?: Pierre! It is very strange to hear such question from you!
Rays from $X$ cannot interset bounding angle (and vice versa)!

I didn't understand what was $Y$, I thought it was bounded by $l_B$... :blush:

Pierre.

A three pages long one  :?  where I proved that it is always possible to draw some segments between points so that they form a connected planar graph with no edge intersected by any ray...

Anyway, it is the one published in Crux (2004, p.419-421) since nobody elxe seems to have answered to that problem  :(  .

Pierre.

You can send my solution to CRUX  :) Maybe it will be published as a remark to previous solved problem. :D


2.	Find an equation for the set of points P(x, y) equidistant from the point F(2, -3) and the line
x – 2y +5 = 0.  Name the curve and find its vertex

hmm so equidistant from point+line = parabola

and then... we find the line through (2,3) perpendicular to 2y=x+5,

which is of the form y=-2x+b... so b=7

so projection of (2,3) onto 2y=x+5 is when
2(-2x+7)=x+5 $ \Leftrightarrow$ 5x=9
so like y=17/5

and then midpoint = (19/10, 16/5) = answer

I might have done the algebra wrong, but you can see how it goes

there's probably a much 1337er solution but oh well

You only answered half the question  :( The answer's supposed to be in the form $ Ax^{2} \plus{} Bxy \plus{} Cy^{2} \plus{} Dx \plus{} Ey \plus{} F \equal{} 0$, absolutely no idea how to reach it, though.

Why not write it out as $ y \equal{} a(x \minus{} h)^2 \plus{} k$ where $ (h,k)$ is the vertex?

As for the work, the perpendicular line should instead be $ y \equal{} \minus{} 2x \plus{} \color{red}1$ in order to contain the foci which is $ (2, \minus{} 3)$ (notice the negative).

After some work I got the vertex as
[hide="Vertex"]$ \left(\frac {13}{10},\frac {13}{5}\right)$[/hide]
[b][Edit][/b] Thank you t0rajir0u.  :)
[hide="Answer in Expanded Form"]
$ (x \minus{} 2y \plus{} 5)^2 \equal{} 5(x \minus{} 2)^2 \plus{} 5(y \plus{} 3)^2$
$ \implies \boxed{4x^2 \plus{} 4xy \plus{} y^2 \minus{} 30x \plus{} 50y \plus{} 40 \equal{} 0}$[/hide]

[quote="Patterns_34"]Why not write it out as $ y \equal{} a(x \minus{} h)^2 \plus{} k$ where $ (h,k)$ is the vertex?[/quote]

Parabolas of this form have directrixes parallel to the $ x$-axis.  This is clearly not the case here.

[hide="Solution"] Here's a formula you might want to know.  The distance from a line $ ax \plus{} by \plus{} c \equal{} 0$ to a point $ (x_0, y_0)$ is

$ \frac {|ax_0 \plus{} by_0 \plus{} c|}{\sqrt {a^2 \plus{} b^2}}$.

What we want, then, is the set of points such that

$ \frac {|x \minus{} 2y \plus{} 5|}{\sqrt {5}} \equal{} \sqrt {(x \minus{} 2)^2 \plus{} (y \plus{} 3)^2} \implies$
$ (x \minus{} 2y \plus{} 5)^2 \equal{} 5(x \minus{} 2)^2 \plus{} 5(y \plus{} 3)^2$.

Now it's just a matter of expanding.  :) [/hide]