Given $ a, b, c > 0$ and $ n$ is an integer satisfy $ n \ge 4$. Prove that:
\[ \frac{ab^{n\minus{}1}}{c^n}\plus{}\frac{bc^{n\minus{}1}}{a^n}\plus{}\frac{ca^{n\minus{}1}}{b^n} \ge \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\]
 :)

[quote="nguoivn"]Given $ a, b, c > 0$ and $ n$ is an integer satisfy $ n \ge 4$. Prove that:
\[ \frac {ab^{n \minus{} 1}}{c^n} \plus{} \frac {bc^{n \minus{} 1}}{a^n} \plus{} \frac {ca^{n \minus{} 1}}{b^n} \ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\]
:)[/quote]

we have

$ \sum_{cyc}\left(\frac {ab^{n \minus{} 1}}{c^n} \plus{} \frac {c}{a}\right)\ge 2\sum_{cyc}\left(\frac {b^{\frac {n \minus{} 1}{2}}}{c^{\frac {n \minus{} 1}{2}}}\right)\ge \frac {2}{3}(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})(\sum_{cyc}\left(\frac {b^{\frac {n \minus{} 1}{2} \minus{} 1}}{c^{\frac {n \minus{} 1}{2} \minus{} 1}}\right) \ge 2 \text{RHS}$

Done

or we can use

$ \left(\frac {ab^{n \minus{} 1}}{c^n} \plus{}\frac{c}{a} \plus{}n \minus{} 3 \plus{} \frac {b}{c}\right)\ge n.\frac {b}{c}$


number theory
relatively prime
prime numbers

The largest prime is said to have been found. But how can it exist?

Let P1, P2,.....,Pn be the set of all the primes.

Then (P1.P2....Pn)+1 will not be divisible by any of the primes.

This means that this itself is prime.

But this is greater than Pn.

Thus, we reach a contradiction.

Where did you hear that?  In all likelihood, whatever source you read meant the largest [i]known[/i] prime.

From Elementary Number Theory by William Stein
[quote]A Mersenne prime is a prime of the form $ 2q \minus{} 1$. The largest known prime as of July $ 2004$ is the Mersenne prime $ p \equal{} 2^{24036583} \minus{} 1$, which has $ 7235733$ decimal digits.[/quote]

If you want to know how big can be such a number ,you can see that in this link : [url]http://www.math.utah.edu/~pa/math/largeprime.html[/url]

Impressive  :D
Is there anyone can memorize it  :P

Because there is four topics about the same thing "JRav" seems not to notice this topic so I would like to group everything in one place.
[quote="JRav"]Indeed, there are an infinite number of primes.  What I'm going to guess you're referring to is that some group of researchers have found a prime larger than all others previously discovered.  It's not the largest prime (there is none), just the largest found so far.[/quote]

There is no such thing as a largest prime.  Your proof that there are infinitely many primes is completely rigorous, and rigorous proofs tend to hold true...

As many of the people here have replied, you probably misread whatever you read and it actually said "largest known prime."

It might be best to note that in Euclid's proof of the infinitude of the primes, the number (p0*p1*...*pn-1*pn) + 1, while guaranteed to be relatively prime to all numbers p0, p1 ..., pn, may not itself be prime.

As another proof of the infinitude of the primes, Furstenberg's illustrates the many ways in which it can be shown that the primes have no limit.

http://primes.utm.edu/notes/proofs/infinite/topproof.html

Heuristics arguments also hold in showing the infinitude of the primes (as well as the number of primes less than n).

[quote="cstan"]It might be best to note that in Euclid's proof of the infinitude of the primes, the number (p0*p1*...*pn-1*pn) + 1, while guaranteed to be relatively prime to all numbers p0, p1 ..., pn, may not itself be prime.[/quote]
That's completely a matter of how you phrase the contradiction.  [i]In the proof[/i], the contradiction is that $ p_0 p_1 ... p_n \plus{} 1$ is not divisible by any smaller prime (by assumption), which is the definition of primality.  On the other hand, it also can't itself be prime (by assumption).  [i]Outside the proof[/i], either of these conclusions may or may  not hold.

Just continuing from the other posts above, you can alternatively prove that for any $n$, there necessarily exists a prime $p$ such that $n<p<n!$.  If it holds for all $n$, you've got an infinite number of $p$'s.
Thanks hasan4444 for moving my comment into this thread.
[url=http://en.wikinews.org/wiki/Two_largest_known_prime_numbers_discovered_just_two_weeks_apart,_one_qualifies_for_$100k_prize]Here[/url] is a link about the largest discovered primes.


trigonometry
limit
calculus
derivative
calculus computations

Apostol's text asks to prove this limit. Can anyone give me any hints on how to go about it?

\[ \lim_{x\rightarrow 0} \frac{\sin x \minus{}\sin \alpha}{x\minus{}\alpha} \equal{} \cos \alpha\]

You've certainly made a mistake, because that limit is clearly $ \frac{\sin\alpha}{\alpha}$. If you're asking about how to derive the standard expression for the derivative of $ \sin$, it's a tricky problem which depends strongly on the definitions you use.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=165805]Two[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=120492]threads[/url] to look at.

I have not made a mistake. Others who own Apostol can verify that. Page 138 number 19. I do however understand your answer and I thought that was obvious but $ \displaystyle \cos a$ seems like a rather large mistake.

Maybe it's a typo (not you emjay285, I meant in the book)!? If you put $ x\rightarrow\alpha$ instead of $ x\rightarrow0$, you get $ \cos\alpha$ for the answer! And this seems fairly logical.. At least to me! :)

you're right that is logical.  I didnt even think of that  :blush:

That's because you all got wrapped up in limits and calculations and stuff! :) Things are simple sometimes.. ;)


integration
trigonometry
function
real analysis
real analysis unsolved

Consider $ I \equal{} \int_{0}^{\infty}\frac {\sin{(xt)}}{t}dt$

Show that $ I$ converges uniformly for $ x$ on any compact interval, $ C$, with $ 0\not\in C$, but not when $ 0\in C$.

Following alek's substitution:

$ du \equal{} tdt \iff dt \equal{} \frac {du}{t}$ thus;

$ \int_{0}^{\infty} \sin(u) \; du \equal{} \minus{} \frac {\cos(tx)}{t^{2}} \; \Bigg|_{0}^{\infty}$

Uhh, I suppose $ I$ converges uniformly since you can evaluate $ \minus{} \frac {\cos(tx)}{t^{2}} \; \Bigg|_{a}^{b}$ so long as neither a or b are 0, since that leads to an indeterminate form (division by 0).

I'm probably completely off base  :huh:

Yes, you are. You've done the substitution wrong; it's not something you can write down an antiderivative for.

I'm not sure about the premise: there doesn't seem to be a variable for it to converge uniformly in. Do you mean uniform continuity?

Although this goes beyond the original statement, it would be fair to ask it this way:

Consider the functions $ I_n(x)\equal{}\int_0^n\frac{\sin xt}{t}\,dt.$

As a sequence of functions, does this converge uniformly on $ [a,b]$ for $ a>0?$ Does it converge uniformly on $ [0,b]?$

The second question has a cheap answer: since each $ I_n$ is a continuous function but the limit is not continuous at $ 0,$ the convergence cannot be uniform on any interval containing zero.

alekk's suggestion would give us

$ I_n(x)\equal{}\int_0^{nt}\frac{\sin u}{u}\,du.$ Integrate this by parts to get

$ I_n(x)\equal{}\frac{1\minus{}\cos nt}{nt}\plus{}\int_0^{nt}\frac{1\minus{}\cos u}{u^2}\,du.$

Examine that for uniform convergence or the lack thereof.


function
logarithms
inequalities proposed
inequalities

[b]PRO-1[/b]: Assume $0<a<b<c$ , show that:$ a^bb^cc^dd^a>b^ac^bd^ca^d$
[b]PRO-2[/b]: Assume $a>b>c>d>e$ , show that:$ a^{e^b}\cdot b^{e^c}\cdot c^{e^d}\cdot d^{e^a}\ge b^{e^a}\cdot c^{e^b}\cdot d^{e^c}\cdot a^{e^d}$
NOTE: e such that: $\ln{e}=1$ :D

[quote="nttu"]It's solved on M&Y 1/05  :)[/quote]

yes,but only PRO-1 :D , and PRO-2 is mine.okie?


trigonometry
geometry
circumcircle
angle bisector
geometry proposed

In a triangle $\triangle ABC$ prove that the internal angle bisector of $A$ passes through the center of the 9-point circle if and only if

\[
\left( {2\cos A - 1} \right)\left( {\cos B - \cos C} \right) = 0
\]

Let us show that the internal angle bisector passes through the center of the nine point circle iff $AB=AC$ or $\angle{A}=60^{\circ}$. Clearly, it solves the problem.

Let ABC be an arbitrary triangle, H be its orthocenter, O be the center of the circumcircle, N be the center of the nine point circle.

It's obvious that if $AB=AC$ then N lies at the angle bisector of $A$. Let's prove the converse.

There are two well-known facts.

(1) The center of the nine point circle is a midpoint of the segment OH.

(2) The points O and H are isogonal conjugate.

Suppose that N lies at the internal angle bisector of A. From (2) we immediately obtain that AN is an angle bisector in the triangle OAH. And from (1) we obtain that AN is a median in the triangle OAH. And from the last two facts we have that the triangle OAH is isosceles (or the points O, H and A are collinear, but then $AB=AC$ and we are done). It means that $AO=AH$. But AO is a circumradius and AH is a circumdiameter of the triangle $H_BAH_C$ (where $H_B$ and $H_C$ are the feet of altitudes of the original triangle). But it is well-known that the triangles $BAC$ and $H_CAH_B$ are similar with coefficient $\cos{\alpha}$, where $\alpha=\angle{BAC}$. And hence $AH=2AO \cos{\alpha}$. Since $AO=AH$, we have $\cos{\alpha}=1/2$. It can be true only if $\alpha=60^{\circ}$. So, we are done.


A friend of mine came up with this question. We solved it different ways and got different answers. Any help?
$ \displaystyle\sum^\infty_{n\equal{}1}\dfrac{\cos{(n\cdot x)}}{3^n}$
(Where cos is in radians obviously)

Things very much like this have popped up on the forum on several occasions, but I'm not sure what a good search would be to try.  The easiest solution method starts by noting that [hide]$ \cos t \equal{} \textrm{Re}(e^{i t})$.[/hide]

[hide="Solution"]
$ \sum_{n\equal{}1}^{\infty}\frac{\cos nx}{3^n}\equal{}\text{Re}\sum_{n\equal{}1}^{\infty}\left(\frac{e^{ix}}{3}\right)^n$

The sum is $ \frac{e^{ix}{3}}{1\minus{}\frac{e^{ix}}{3}}\equal{}\frac{e^{ix}}{3\minus{}e^{ix}}\equal{}\frac{\cos x\plus{}i\sin x}{3\minus{}\cos x\minus{}i\sin x}$.

$ \frac{\cos x\plus{}i\sin x}{3\minus{}\cos x\minus{}i\sin x}\cdot\frac{3\minus{}\cos x\plus{}i\sin x}{3\minus{}\cos x\plus{}i\sin x}\equal{}\frac{3\cos x\plus{}3i\sin x\minus{}1}{10\minus{}6\cos x}$

The real part of that is, of course, $ \boxed{\frac{3\cos x\minus{}1}{10\minus{}6\cos x}}$.
[/hide]

Oops Im sorry, I didn't think to search before I posted.
Thanks for the solution!

[quote="grn_trtle"]

The sum is $ \frac {e^{ix}{3}}{1 \minus{} \frac {e^{ix}}{3}}$
[/quote]

This should be $ \frac {\frac{e^{ix}}{3}}{1 \minus{} \frac {e^{ix}}{3}}$.


function
inequalities
real analysis
real analysis unsolved

Hello!

Edit:

I'll repost the problem:

If $f(x,y)$ is a continuous function in $(x,y)$ and is defined on $M=C_{1}\times C_{2}$, where $C_{1}$ and $C_{2}$ are compact sets in the euclidean spaces $E^{m}$ and $E^{n}$, then $\Psi(x)$ is a continuous function in $x$, where $\Psi (x): = \mathop{\max }_{y \in C_{2}}f(x,y)$

/Edit

We have $f(x,y)$, $f$ is affine

Furthermore is f continuous on a compact (even convex) set $M$

$M: = C_{1}\times C_{2},x \in C_{1},y \in C_{2}$

The mapping $x \mapsto f(x,y)$ is affine

Let $\Psi (x): = \mathop{\max }_{y}f(x,y)$

 (The maximum of $f(x,y)$ upon $y$)

How do I show that $\Psi (x)$ is continuous on $M$?

$\Pi\Delta\Phi$

$C_{1}$ and $C_{2}$ are subsets of what? $\mathbb{R}?$ $\mathbb{R}^{m}$ and $\mathbb{R}^{n}?$ Some Banach space?

I will disregard convexity and linearity, and only assume that $C_{1}$ and $C_{2}$ are compact metric spaces. We know that $f$ is uniformly continuous on $M$. For any $\epsilon$ let $\delta$ be as in the definition of uniform continuity. Then $|\Psi(x)-\Psi(x')|\le \max_{y}|f(x,y)-f(x',y)|\le \delta$ whenever $|x-x'|<\epsilon$, as desired.

Hi!

$C_{1}$ and $C_{2}$ are indeed compact metric spaces.

I'll repost the problem:

If $f(x,y)$ is a continuous function in $(x,y)$ and is defined on $M=C_{1}\times C_{2}$, where $C_{1}$ and $C_{2}$ are compact sets in the euclidean spaces $E^{m}$ and $E^{n}$, then $\Psi(x)$ is a continuous function in $x$, where $\Psi (x): = \mathop{\max }_{y \in C_{2}}f(x,y)$

My idea is the following (it seems that linearity and convexity is not needed):

For every $\epsilon>0$ and $x' \in C_{1}$ we have to show that we can find a $\delta>0$ so that

$\left|{\Psi (x)-\Psi (x')}\right| < \varepsilon ,\forall x \in C_{1}$ with $\left|{x-x'}\right| < \delta$

As mlok said, from what is given we have that $f(x,y)$ is uniformly continuous, so that:

$\begin{gathered}\forall y,y' \in C_{2}: \hfill \\ \left|{f(x',y)-f(x,y)}\right| < \varepsilon \hfill \\ \end{gathered}$

AND

$\left|{f(x',y')-f(x,y')}\right| < \varepsilon$

if  $\left|{x-x'}\right| < \delta$ for $\delta>0$

Choose $y$ and $y'$ so that: $f(x',y') = \Psi(x')$ and $f(x,y)=\Psi(x)$ for some $x$ with $\left|{x-x'}\right| < \delta$ for $\delta>0$

Okay, now comes the problem:

If I can show that: $f(x',y) \leqslant f(x',y')$ and $f(x,y') \leqslant f(x,y)$ I would be basically finished.

How do I show this, or, why is it true?

$\Pi\Delta\Phi$

Your approach is correct through "Choose $y$ ...". However, you don't need $y'$, because there is no way to tell if $|y-y'|$ is small. Instead, observe that $\Psi(x)-\Psi(x')=f(x,y)-\Psi(x')\le f(x,y)-f(x',y)<\delta$. Since $x$ and $x'$ are interchangeable, you also have $\Psi(x')-\Psi(x)<\delta$, hence $|\Psi(x)-\Psi(x')|<\delta$.

Hi!

Nice proof

[quote="mlok"]Since $x$ and $x'$ are interchangeable[/quote]

Is this always true?

[quote="mlok"]However, you don't need $y'$, because there is no way to tell if $|y-y'|$ is small[/quote]

Ummm...
Check this out (the proof for lemma 1):

[img]http://img521.imageshack.us/img521/6133/77295079sw9.jpg[/img]

[img]http://img524.imageshack.us/img524/6457/67605287xt1.jpg[/img]

[img]http://img161.imageshack.us/img161/6024/32387299nz1.jpg[/img]

This is from SIAM Review, Vol. 7, No. 2. (Apr., 1965), pp. 181-188
You see the step where he gets two inequalities (Step 13)?  I don't understand how he gets this result.
Thanks for your posts.

$\Pi\Delta\Phi$

The inequalities (12) and (13) hold by the definition of $\psi$. This is essentially the same proof that I gave -- it would be hard to come up with something original here. The only difference is that I did not bother to write the second part, where the roles of $x$ and $x'$ are exchanged.
[quote][quote]Since x and x' are interchangeable[/quote]
Is this always true?[/quote] No. This is only true in a particular context.
[My standard reply to that question]


Sarah's mom wants to bake 4 meatballs, but only three of them fit in her pan at a time. It takes 5 minutes to bake one side of a meatball, and 5 minutes to bake the other side. Without cutting the meatballs, can all 4 of them be baked in less than 20 minutes?

Solution: [hide] First 3 meatballs: 5 + 5 = 10, Second meatball: 5 + 5 = 10

Therefore, she cannot bake all the meatballs in less than 20 minute, but in exactly 20 minutes[/hide]

That's what I thought, but I have a feeling she can...
what is the longest period of time each meatball can stay if every meatball can have an equal time in the pan?

bake first 3 for 5 minutes (one side)

take out first, put in fourth. bake for five minutes

take out second and third. Put in first. Bake for 5 minutes.

This takes 15 minutes so yes.

Well must it be baked for five minutes continously, or cummulative?

realistically, i dont think that cumulative would work, because it would cool of, and then at least some of the time would be spent reheating, so the meatball would be just barely undercooked...but maybe im wrong...

i know I know, but this is an imaginative math world, that's why problem solving is an art in some ways, not only do you have to think of ideas, but you also have to imagine the situation  :lol:

i think you contradiced yourself...

first you said its imaginitive, then u said that you have to imagine the situation...which is right?

no...
which side r u on?

contiuously or cumulative?

each side has to be baked continuously for 5 minutes but you can cook the two sides separately

Then it's easy.

Label the bottom half of a meatball with a -1 suffix, and the top half with a -2 suffix.

First five minutes:

1-1
2-1
3-1

Next five minutes

2-2
3-2
4-2

2 and 3 are done.

Last five minutes

1-4
4-1

yeah but it wasn't really that hard of a question anyway

Thank you everyone, that problem just asked if she could cook the meatballs in less than 20 minutes, problem solved  :10: !


Is it true that if $x,y,z$ are three positive reals such that $xyz=1$, then
$\frac{1}{2(x+y)+1}+\frac{1}{2(y+z)+1}+\frac{1}{2(z+x)+1}\le \frac{1}{2}$

No, zhaobin is right  :(

I saw this inequality in my notebook, maybe there is a mistake and the sign is reversed. But anyway since i`m not sure just leave this topic.
Appologize to every one who saw this

But if you can prove it with $\le 1$, phamduyhiep, I ll be very happy to see the solution.

:o
That was USAMO. I didn`t saw that..
Thanks anyway
Must go to school. Goodbye for now   :bye:

[quote="zhaobin"]it is easy to prove:
$\frac{1}{2(x+y)+1}+\frac{1}{2(y+z)+1}+\frac{1}{2(z+x)+1}\le 1$
because:
$\frac{1}{2(x+y)+1}+\frac{1}{2(y+z)+1}+\frac{1}{2(z+x)+1}\le \frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}$
and let:$x=a^3,y=b^3,z=c^3$
then we have
$\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}= \frac{abc}{a^3+b^3+abc}+\frac{abc}{b^3+c^3+abc}+\frac{abc}{c^3+a^3+abc}$
use $a^3+b^3 \ge ab(a^2+b^2)$ and similar.
then we have:
$\frac{1}{2(x+y)+1}+\frac{1}{2(y+z)+1}+\frac{1}{2(z+x)+1}\le \frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1} \le 1$
in other hand:let $x=y \to 0$,we will see that 1 is the best constant :)[/quote]
Yes!It's my solution.Thanks,zhaobin!


What do you guys think of the equation editor in word 2007?

I mean sure its probably not as good as LaTex if you've been using it for years and remember all the complex codes and stuff.

But for someone like me who started typing these equations with mathtype it seems like an improvement.

And the new equation editor has latex like recognition capabilities so you can type on the keyboard without stopping anyway. just need to remember a new set of strings (and i think some of them were taken right off latex anyway)

any opinions? im considering whether or not i should get office 2007. ive only tried it.

Could you please post a screenshot of some of those equations? I haven't tried out Word 2007 yet. Somehow I'm not optimistic, though...

1) If you're going to learn the codes for EE, why not just learn the codes for LaTeX?
2) It is not, in fact, very difficult to learn TeX.  You just have to keep using it, and asking questions when you can't figure things out.
3) EE makes image files.  Documents become hugely massive and burdensome.  This really sucks.
4) Everyone can read .pdf-s (and could read .dvi-s or .ps-s with one or two additional pieces of free software).  Not everyone can read Word documents.  Also, you should find out if this new EE will be backwards-compatible, i.e. if anyone without Word 2007 will be able to read your equations.

$\LaTeX$ still is better.  I don't want to re-learn a lot of commands...

And you can't use EE codes on this forum  :D

but dun u think its easier to see what ur getting than a gazillion codes on one page??

Like you have to intertwine text with latex code and that just makes it really confusing for me

After using $\LaTeX$ for some time, you get used to the coding and can see the final result in your mind's eye. You always have the option to preview your posts, or compile your code.
I find this to be a lot less cumbersome than using the Equation editor, which has me open a small window and switch between mouse and keyboard to input symbols, and then repeat for more equations later on.

I'm sorry I'm writing in this topic. I just didn't want to create another similar topic.

Now the question :

I have two old files on Word 2003 with MathType equations. Now, I want to convert the files in Word 2007 format and I prefer the mathematical equation to be in Cambria Math style (MS Equation Editor 3.0), not as a MathType equation. I've searched in the net, but there isn't useful information. So, does anybody know how to convert MathType symbols into Cambria style  :|  ?

I, for one, use Equation Editor all the time!  It's just so simple, so I love it.

I use $ LaTex$ on the forum, and subscript on word 2007 when I am doing challenge set problems or homework. It is just annoying having codes and the words for my proofs mixed up on a page.
@Karamata there is [b]never[/b] a good reason for reviving a 2 year old topic

Oh, isn't it better than starting a new topic? I think less topics is better. :blush:

By the way, I found an interesting program - GrindEq Math Utilities - which can resolve the problem with MathType equations. If somebody don't like MathType and prefer MS office 2007's equations, this will convert them in Cambria style. The problem is in that the program isn't free and a registration code is wanted after 10 conversions. Fortunately, 10 attempts were enough for me.


Find all functions $f: \mathbb{Q}\to \mathbb{Q}$ such that for all $x,y \in \mathbb{Q}$: \[f(x+y)+f(x-y)=2(f(x)+f(y)).\]

[b]1.[/b] If we plug in $ x = a$ and $ y = 0$ we get $ 2f(a) = 2(f(a)+f(0))$ therefore $ f(0) = 0$.
[b]2.[/b] If we plug in $ x = 0$ and $ y = a$ we get $ f(a)+f(-a) = 2(f(a)+f(0))$ so $ f(a)+f(-a) = 2f(a)$ hence $ f(a) = f(-a)$, therefore $ f(x) = f(-x)$ for any $ x\in\mathbb{Q}$.
[b]3.[/b] Take any $ t\in\mathbb{N}$. Then, plugging into the original equation $ x = y =\frac{1}{t}$ we obtain $ f(\frac{2}{t})+f(0) = 2(f(\frac{1}{t})+f(\frac{1}{t}))$ hence $ f(\frac{2}{t}) = 4f(\frac{1}{t})$.
Let us now prove by mathematical induction that for any $ n\in\mathbb{N}$ have $ f(\frac{n}{t}) = n^{2}f(\frac{1}{t})$.
[b]Base Step[/b]: For $ n = 1,2$ the result is true -- we have proved $ f(\frac{2}{t}) = 4f(\frac{1}{t})$ in the first paragraph of 3, and also $ f(\frac{1}{t}) = 1^{2}f(\frac{1}{t})$
[b]Induction Step[/b]: Let us assume the result is true for $ n = k,k-1$. Then, let us prove the result for $ n = k+1$. If we plug into the original equation $ x =\frac{n}{t}$ and $ y =\frac{1}{t}$. Then we get:
$ f(\frac{k+1}{t})+f(\frac{k-1}{t}) = 2(f(\frac{k}{t})+f(\frac{1}{t}))$
$ \Leftrightarrow f(\frac{k+1}{t}) = 2f(\frac{k}{t})+2f(\frac{1}{t})-f(\frac{k-1}{t})$
$ \Leftrightarrow f(\frac{k+1}{t}) = 2k^{2}f(\frac{1}{t})+2f(\frac{1}{t})-(k-1)^{2}f(\frac{1}{t})$
$ \Leftrightarrow f(\frac{k+1}{t}) = 2k^{2}f(\frac{1}{t})+2f(\frac{1}{t})-(k-1)^{2}f(\frac{1}{t})$
$ \Leftrightarrow f(\frac{k+1}{t}) = (k+1)^{2}f(\frac{1}{t})$
so the result is true for $ n = k+1$ and by mathematical induction it follows that for any $ n,t\in\mathbb{N}$ have $ f(\frac{n}{t}) = n^{2}f(\frac{1}{t})$.
[b]4.[/b] Now, take any $ t\in\mathbb{N}$. From 3, we get $ f(1) = f(\frac{t}{t}) = t^{2}f(\frac{1}{t})$ hence $ f(\frac{1}{t}) =\frac{1}{t^{2}}f(1)$.
Therefore, combining this with 3 we get for any $ n,t\in\mathbb{N}$ ${ f(\frac{n}{t}) = (\frac{n}{t})^{2}}f(1)$, and therefore for any $ x\in\mathbb{Q}$ with $ x > 0$ we get $ f(x) = x^{2}f(1)$.

Finally because from 2, $ f(x) = f(-x)$, $ f(0) = 0$ and for any $ x\in\mathbb{Q}$ with $ x > 0$ we get $ f(x) = x^{2}f(1)$, then for any $ x\in\mathbb{Q}$ have $ f(x) = x^{2}f(1)$.

[b]Solution:[/b] $ f(x) = x^{2}*c$ where $ c$ is any rational number.

Plugging $x=y=0$ yields $f(0)=0.$ By induction $f(xy)=y^2f(x).$ Noting that $f$ is even we get $f(x)=cx^2,$ for some rational constant $c$, it clearly works.


The numbers $ x,y,z>0$ are so that $ x 3^{x\plus{}1}\plus{}y3^{y\plus{}1}\plus{}z3^{z\plus{}1}\equal{}(xy\plus{}yz\plus{}zx)^2$.Prove that $ x\equal{}y\equal{}z.$

NICE ,SKYLOVER
Let $ LHS\equal{}f(x).RHS\equal{}g(x)$
then $ f'(x)\equal{}g'(x0$,simialerly,with $ f(y).g(y)$
We have $ T(a)\equal{}a3^{a\plus{}1}$ is a crease function with$ a>0$
So if $ x>y thus f'(x)>f'(y)  thus g'(x)>g'(y) hence y>x$ (it is easy to find)
So $ x\equal{}y$,
Then $ x\equal{}y\equal{}z$
We have done :lol:

Here is a another solution:
[hide]We have $ 3^x\geq x^3$ and the others, so $ (xy\plus{}yz\plus{}zx)^2\geq 3(x^4\plus{}y^4\plus{}z^4)\geq (x^2\plus{}y^2\plus{}z^2)^2\geq\ (xy\plus{}yz\plus{}zx)^2$ .
Thus, we have the equality case and the conclusion follows.[/hide]

[quote="skylover"]Here is a another solution:
[hide]We have $ 3^x\geq x^3$ and the others, so $ (xy \plus{} yz \plus{} zx)^2\geq 3(x^4 \plus{} y^4 \plus{} z^4)\geq (x^2 \plus{} y^2 \plus{} z^2)^2\geq\ (xy \plus{} yz \plus{} zx)^2$ .
Thus, we have the equality case and the conclusion follows.[/hide][/quote]
It is wrong,try $ x\equal{}2,8$, :wink:  :wink:  :wink:

Yeah, that ineq holds only for $ x\geq 3$..:D..

Oh no,your proof is still true for $ 0<x<e,and x\ge 3$.did you understand
But with minr,we dont need it :lol:


1/ Find $ x,y,z \in R$ such that:
$ \frac{4x^2}{4x^2\plus{}1}\equal{}y$
$ \frac{4y^2}{4y^2\plus{}1}\equal{}z$
$ \frac{4z^2}{4z^2\plus{}1}\equal{}x$
2/ Find all $ p,q$ are primes such that $ pq$ divides $ (5^p\minus{}2^p)(5^q\minus{}2^q)$
3/ Given a trapezoid $ ABCD (AB//CD). P$ is on the segment $ AB$ such that $ PC\equal{}PD. E$ is the intersection of AC and BD. Let $ O_1, O_2$ be the circumcentres of triangles $ ADP$ and $ BCP$ respectively. Prove that $ PE$ is perpendicular to $ O_1O_2$.
(Give me detail solutions, please  :) )

I think that #2 is PEN problem, but I'm really not sure...

[quote="Kyoshiro"]1/ Find $ x,y,z \in R$ such that:
$ \frac {4x^2}{4x^2 \plus{} 1} \equal{} y$
$ \frac {4y^2}{4y^2 \plus{} 1} \equal{} z$
$ \frac {4z^2}{4z^2 \plus{} 1} \equal{} x$
[/quote]
[hide]Multiplying these 3 inequalities we get: 64x^2y^2z^2/(4x^2+1)(4z^2+1)(4y^2+1)=yxz and considering that x,y,z>0
Applying AM-GM we have: 4x^2+1>=4x
                                    4y^2+1>=4y
                                    4z^2+1>=4z
Therefore, the LHS=<xyz=RHS
The equality occurs when x=y=z=1/2(since x,y,z>0). That's the only solution to this system of equation(Q.E.D)[/hide]

2 - it is similar to Bulgaria 1996 - IV Round problem
1 - Canadian Math Olympiad 1996 problem

[quote=">Aver<"]I think you forgot that $ x \equal{} y \equal{} z \equal{} 0$ is also true for all three equations ;)[/quote]
Oh yeah you're right :blush: . I left out that simple case.

Someone solves the #2 and #3 for me, please!  :(

one solution for #2 is $ p\equal{}q\equal{}3$ (because $ 3|5^x \minus{} 2^x$ for all positive integers $ x$)

you can see the solution of problem#2 here
[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1791960291&t=50838[/url]


linear algebra
matrix
algebra
polynomial

Let $ A \equal{} \begin{pmatrix} 7 \minus{} \lambda & \minus{} 12 & 6 \\
10 & \minus{} 19 \minus{} \lambda & 10 \\
12 & \minus{} 24 & 13 \minus{} \lambda\end{pmatrix}$. Find $ \operatorname{rank} A$

Of course, the rank depends on $ \lambda.$ For all but at most three values of $ \lambda,$ the rank is 3. The only $ \lambda$ for which the question is interesting are the roots of the characteristic polynomial, which is to say the eigenvalues. Do you know what those are?

[hide]All lines add up to 1, so 1 is an eigenvalue ($ \lambda_3 \equal{} 1$). The trace equals 1, so $ \lambda_1 \plus{} \lambda_2 \plus{} 1 \equal{} 1 \equal{} > \lambda_2 \equal{} \minus{} \lambda_1$. On the other hand,

$ \left | \begin{array}{ccc} 7 & \minus{} 12 & 6 \\
10 & \minus{} 19 & 10 \\
12 & \minus{} 24 & 13 \end{array} \right| \equal{} \left | \begin{array}{ccc} 7 & 1 & 6 \\
10 & 1 & 10 \\
12 & 1 & 13 \end{array} \right| \equal{} \left | \begin{array}{ccc} 1 & 1 & 6 \\
0 & 1 & 10 \\
\minus{} 1 & 1 & 13 \end{array} \right| \equal{} \left | \begin{array}{ccc} 1 & 1 & 6 \\
0 & 1 & 10 \\
0 & 2 & 19 \end{array} \right|$

$ \equal{} \left | \begin{array}{ccc} 1 & 1 & 6 \\
0 & 1 & 10 \\
0 & 0 & \minus{} 1 \end{array} \right| \equal{} \minus{} 1$

and so $ \minus{} \lambda_1^2 \equal{} \minus{} 1 \equal{} > \lambda_1 \equal{} 1$ and $ \lambda_2 \equal{} \minus{} 1$.[/hide]

So when $ \lambda\equal{}\minus{}1,$ the rank of $ A\minus{}\lambda I$ is 2. The only remaining question is the rank of $ A\minus{}\lambda I$ for $ \lambda \equal{}1.$ Carcul showed that that eigenvalue has algebraic multiplicity $ 2.$ Its geometric multiplicity has yet to be determined.

We are to solve the system

$ \left [\begin{array}{ccc} 6 & \minus{}12 & 6 \\
10 & \minus{}20 & 10 \\
12 & \minus{}24 & 12 \end{array} \right] \left [\begin{array}{c} x \\ y \\ z \end{array} \right] \equal{} \left [\begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] <\equal{}> \left [\begin{array}{ccc} 1 & \minus{}2 & 1 \\
1 & \minus{}2 & 1 \\
1 & \minus{}2 & 1 \end{array} \right] \left [\begin{array}{c} x \\ y \\ z \end{array} \right] \equal{} \left [\begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right]$

$ <\equal{}> z \equal{} \minus{}x \plus{} 2y$.

Therefore, all eigenvectors associated with $ \lambda \equal{} 1$ have the form $ (x,y,\minus{}x\plus{}2y)^T$, i e., they belong to $ Span \{(1,0,\minus{}1)^T,(0,1,2)^T \}$.

... which is to say that the eigenvalue $ \lambda\equal{}1$ has geometric multiplicity $ 2,$ which is to say that $ \operatorname{rank}(A\minus{}I)\equal{}1.$


[hide="Hint"]The matrices are simultaneously diagonalizable (why?), hence you can assume WLOG that all matrices are diagonal. Now is should be easy.[/hide]


A merchant bought some goods at a discount of $ 20\%$ of the list price. He wants to mark them at such a price that he can give a discount of $ 20 \%$ of the marked price and still make a profit of $ 20\%$ of the selling price. The percent of the list price at which he should mark them is:

$ \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 100 \qquad\textbf{(C)}\ 125 \qquad\textbf{(D)}\ 80 \qquad\textbf{(E)}\ 120$

[hide="Click for solution"]
Let $ C$ be the cost, $ L$ the list price, $ S$ the selling price and $ M$ the marked price.  $ C\equal{}\frac{4}{5}L$, $ S\equal{}C\plus{}\frac{1}{5}S$, $ S\equal{}\frac{4}{5}M$.  Thus, $ \frac{4}{5} \times \frac{4}{5}M\equal{}\frac{4}{5}L \implies M\equal{}\frac{5}{4}L$, so $ \boxed{\textbf{(C)}}$ is correct.
[/hide]


The numbers p and q are primes and p^2 + 1≡0(mod q) and q^2-1≡0(mod p).
Prove that p+q+1 is a composite.

All i could show is that q is either 2 or of the form 4t+1. If q=2 it follows trivially that p+q+1 is a composite but what if q=4t+1 ???

$p|q^2-1\Rightarrow p|p^2+2pq+q^2-1=(p+q+1)(p+q-1)$. Assume $p+q+1$ is a prime. In this case, $p|p+q-1\Rightarrow p|q-1\Rightarrow q=mp+1,\ m\ge 1$. $q$ also divides $p^2+1$, so $\frac{p^2+1}q=np+1,\ n\ge 0$. If $n>0$, then it's clear that $(mp+1)(np+1)>p^2+1$, which is false, so $n=0\Rightarrow m=p$, so $q=p^2+1$, which means that $p+q+1=p^2+p+2$, which is even and $>2$, and thus a composite.


floor function
inequalities
number theory proposed
number theory

These three are my own problems, and actually i only know the solution of 2 and 3 but not 1. :oops: but i think its a nice problem and you can solve it... :D

$1$.Find all positive integers $n>1$ such that there exist $2n$ distinct positive integers $a_{1},a_{2},\cdots,a_{n},b_{1},b_{2},\cdots,b_{n}$ such that:
$\sum^{n}_{i=1}a_{i}= \sum^{n}_{i=1}b_{i}$
$\prod^{n}_{i=1}a_{i}= \prod^{n}_{i=1}b_{i}$

$2$.$n$ is a positive integer. Let $T(n)$ be the number of all positive integers which do not have any $0$ digit and the sum of their digits (in base $10$) is $n$. Prove that $T(n) \leq 2^{n-1}$

$3$.$n$ is a positive integer and for all reals ${x}$, we denote $\left\{x\right\}$ as $x-\left\lfloor x\right\rfloor$ . Assume that $A=\sum^{6n-1}_{m=1}( \left\{\frac{m}{6n}\right\}+\left\{\frac{m}{3n}\right\})$
Find the smallest prime number $p$ such that $\left\lfloor A \right\rfloor \equiv 1 ( mod \: p)$

#2: We have that $T(n)$ is the number of compositions of $n$ into positive integers from 1 to 9.  But the number of compositions into positive integers is $\binom{n-1}{0}+\binom{n-1}{1}+...+\binom{n-1}{n-1}=2^{n-1}$, and we are counting a subset of those compositions.  Then the desired inequality follows.  Was this your solution?

exactly k81o7, that was my solution too :D , don't you have a solution for 1 ? :maybe:

well , here is the solution of $\#3$ : $A = \sum^{6n-1}_{m=1}(\frac{m}{6n}+\frac{m}{3n})-\sum^{6n-1}_{m=1}\left( \left\lfloor \frac{m}{6n}\right\rfloor+\left\lfloor \frac{m}{3n}\right\rfloor \right)$
notice that: if $1 \leq m \leq 6n-1$ then $\left\lfloor \frac{m}{6n}\right\rfloor = 0$
and:         if $1 \leq m \leq 3n-1$ then $\left\lfloor \frac{m}{3n}\right\rfloor = 0$
             if $3n \leq m \leq 6n-1$  then $\left\lfloor \frac{m}{3n}\right\rfloor = 1$

As a consequence we have :  $\sum^{6n-1}_{m=1}\left( \left\lfloor \frac{m}{6n}\right\rfloor+\left\lfloor \frac{m}{3n}\right\rfloor \right) = 3n$
Also its easy to prove: $\sum^{6n-1}_{m=1}( \frac{m}{6n}+\frac{m}{3n}) = \frac{3}{2}\times ( 6n-1 )$
Thus $A = 12n-\frac{3}{2}\Rightarrow \left\lfloor A \right\rfloor = 12n-2$ , and the smallest possible value of $p$ is $3$ .  :lol:

i still cant find any solution for $ 1$, it seems easy but a problem...! :huh:

$ 1)$ it is imposible for $ n=2$
Assume it is possible:
$ b_{1}=a_{1}+k$ with $ k$ integer $ \ne 0$ then $ b_{2}=a_{2}-k$
$ a_{1}a_{2}= b_{1}b_{2}$ implies that $ 0=k (a_{2}-(a_{1}+k) )$ contradiction

Solution for $ n=3$

$ (2,12,7)$ and $ ( 3, 14, 4)$
Sums 21
Product 168

[hide=" How to build a n+3 solution"]Notice that if we multiply by a k a solution we get another one
then multiply by big number (for example the maximum plus one) and append previous solution[/hide]
I try to post a solution for $ n=4$ and $ n=5$ and then it is posible for $ n\ge 3$
Edited: Solution for $ n=4$
$ (2,7,180,13)$ and $ (3,5,182,12)$
Solution for $ n=5$
$ (2, 7, 11, 975, 28)$ and $ (3, 5, 13, 980, 22)$


wasn't sure where to post it, so I decided to post here.
I didn't know about the source of the problems soon enough last year, so had some problems with misunderstandings.
Anyone who sees these questions (counting similar format with only a few names changed) that are obviously seeking aid, which is not permitted, please report/lock them until due date.  These came out a few days ago from my knowledge.

http://www.nzamt.org.nz/sites/cms/index.php?option=com_content&task=view&id=68&Itemid=82

yep.  Selection for this problem has finished :D
I'll post my solutions soon


Prove that

$ x^{n} \plus{} x^{n\minus{}2} \plus{} x^{n\minus{}4} \plus{} ... \plus{} \frac{1}{x^{n\minus{}4}} \plus{} \frac {1}{x^{n\minus{}2}} \plus{} \frac{1}{x^{n}} \ge n\plus{}1$,

for all positive numbers $ x$ and for all $ n \in \mathbb{N}$.

[hide]For even $ n$

$ x^n \plus{} \frac{1}{x^n} > 2$ by AMGM

There are $ \frac{n}{2}$ such pairs plus $ 1$

Therefore $ L.H.S. > n\plus{}1$

For odd $ n$

There are $ \frac{n\plus{}1}{2}$ such pairs,

Therefore $ L.H.S. \ge n\plus{}1$[/hide]

For $ n\equal{}1$, we have $ x\plus{}\frac{1}{x} \ge 2$, which is true because $ (x\minus{}1)^{2} \ge 0$.

Replacing $ x$ by $ x^{2}$, we get

$ x^{2} \plus{} \frac{1}{x^{2}} \ge 2$.

Adding $ 1$ to both sides, we have $ x^{2} \plus{} 1 \plus{} \frac{1}{x^{2}} \ge 3$.

Therefore $ x^{2} \plus{} 2 \plus{} \frac{1}{x^{2}} \ge 4 > 3$.

Therefore the inequality is true for $ n\equal{}1$ (1)  and for $ n\equal{}2$.  (2)

For the inductive step, we show that

if it is true for $ n\equal{}k$ then it is true for $ n\equal{}k\plus{}2$:

If $ x^{k} \plus{} x^{k\minus{}2} \plus{} ... \plus{}\frac{1}{x^{k\minus{}2}} \plus{} \frac{1}{x^{k}} \ge k\plus{}1$

then adding $ x^{k\plus{}2} \plus{} \frac{1}{x^{k\plus{}2}} \ge 2$,

we get the required result which with (1) implies that the inequality is true for all odd natural numbers and with (2) implies that it is true for all even natural numbers.

[quote="JBL"][quote="chrischris"]$ x^n \plus{} \frac {1}{x^n} > 2$ by AMGM[/quote]  Not quite.[/quote]

 :?: I don't understand why not...?

(and in particular equality is obtained when $ x \equal{} 1$)


Find all real polynomials $p(x)$ such that $p(x) = a$ has  $\deg{p}$ real roots,  for any $a > 1995$. Multiple roots are counted according to their multiplicities.

[quote="olorin"]:huh: $p(x)=a$ has at most $2$ real roots for $a$ large enough.[/quote]
Why?  :maybe:

If $\deg p$ is even, then the line $f(x)=a$ would intersect $p(x)$ at exactly two places for $a$ sufficiently large that $f(x)$ clears the tallest "bump" (local maximum) in the graph of $p(x)$. If $\deg p$ is odd, then we will similarly have only one intersection.

Yes.
$p$ has only finitely many local extremes $(x_{1},p(x_{1})),\ldots,(x_{r},p(x_{r}))$.
For $a>\max\{|p(x_{1})|,\ldots,|p(x_{r})|\}$, then $p(x)=a$ has at most $2$ solutions.

Not sure what you mean [b]N.T.TUAN[/b]??

Ok, you are right! Thanks olorin and mathisfun1. I edited it

[quote="N.T.TUAN"]Find all real polynomials $p(x)$ such that $p(x) = a$ has  $\deg{p}$ real roots for any $a > 1995$. Multiple roots are counted according to their multiplicities.[/quote]

$\deg p\leq 2$, so
$p(x)=mx+n$ for any $m,n\in\mathbb{R},m\not =0$
or $p(x)=kx^{2}+mx+n$ for any $k,m,n\in\mathbb{R},k>0$ with $n-{m^{2}\over 4k}\leq 1995$.


Maryland is going to field its first home school ARML team this year.  It draws from all over the state, but with the requirement that students must be fully homeschooled according to the laws of Maryland.  The state doesn't allow home school students to participate in extracurricular activities with the public schools, so the ARML administration is allowing us to field our own team.  If you are interested in trying out for the team, contact

Catherine Asaro, Coach
Howard Area Homeschoolers
asaro@sff.net

If you have scores from the AMC-10, AMC-12, AIME, Sigma League, or Mandelbrot competitions, please have those ready.

We had originally planned to take one team and a few alternates.  However, we've had more people than expected try out.  We may possibly field two teams; that remains to be seen.  Selection of the team(s) and alternates will be made the last week in March, so if you are interested please contact me ASAP.

And my thanks to the Art of Problem Solving for making these Forums available.  :-)

Best regards
Catherine Asaro

You might also consider posting this in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=240]Maryland forum[/url].

Thanks. I didn't realize Maryland had a forum.  I will go post over there.

Best,
Catherine


So i started doing contest math problems last September, right now i barely get a 3 or 4 on the practice AIME and since i'm a sophomore i will be taking AMC 12 next year which obviously kills my chance for USAMO.  I just wished that i knew the competition couple years earlier, so that i can be like most of you to qualify USAMO.

you are NOT a loser, i was in the same case you are currently in last year
don't give up, just keep doing problems and keep a positive attitude  :D

USAMO qualifiers make up 400 of approximately 11,000 AIME qualifiers every year.  Even on AoPS, I seriously don't believe that USAMO qualifiers make up the majority of the posters in the high school forums.  Contest mathematics shouldn't be about being the best; it should be about being exposed to interesting mathematical ideas you wouldn't see any of in a classroom setting.  Relax and have fun with it :)

Don't have such a negative attitude! I realize this sounds clichéd. Taking the AMC 12 doesn't kill your chances for USAMO - you'll be surprised how much you improve just by doing lots of problems.

I agree with t0rajir0u.
This is my first time taking AMC12, and I have learnt a lot through studying AoPS Volume 1.  Like t0rajir0u said, I think you shouldn't limit yourself to doing well on the contest.  Just enjoy problem solving!

advice:
but AoPS V1 and V2. They really help!


trigonometry
function
inequalities
geometry unsolved
geometry

Let $a,b$ be the smaller sides of a right triangle. Let $c$ be the hypothenuse and $h$ be the altitude from the right angle. Fint the maximal value of $\frac{c+h}{a+b}$.

Set $ABC$ be a right traingle such that $BC=a,\ CA=b,\ AB=c.$ and $\angle{BCA=90^\circ}.$ Let $\angle {ABC} =\theta ,$ we have $a=c\cos \theta ,\ b=c\sin \theta , h=a\sin \theta =c\sin \theta \cos \theta .$ Thus $\frac{c+h}{a+b}=\frac{c(1+\sin \theta \cos \theta)}{c(\sin \theta +\cos \theta)}.$ Let $x=\sin \theta +\cos \theta,$ by $(\sin \theta +\cos \theta)^2=1+2\sin \theta \cos \theta ,$ we have $\sin \theta \cos \theta =\frac{x^2-1}{2}.$ so we can rewrite this equation as $\frac{c+h}{a+b}=\frac{1}{2}\left(x+\frac{1}{x}\right).$ Now the range of $x$ is $1< x\leq \sqrt{2},$
because we can write $x=\sqrt{2}\sin \left(\theta +\frac{\pi}{4}\right),$ from $0<\theta <\frac{\pi}{2}\Longrightarrow \frac{\pi}{4} <\theta +\frac{\pi}{4}<\frac{3}{4}\pi ,$ yielding $1<x \leq \sqrt{2}.$ Since the graph of $y=\frac{1}{2}\left(x+\frac{1}{x}\right)$ is an increasing function on the range of $x.$ Therefore $\frac{c+h}{a+b}$ is maximized when $x=\sqrt{2}\Longleftrightarrow \theta =\frac{\pi}{4},$ the desired maximum value is $\boxed{\frac{3\sqrt{2}}{4}}.$

Then the smaller sides be a,b then the expression becomes

c+h/a+b = (a^2+b^2+ab)/(a+b)(sqrt(a^2+b^2))).

Then, be a simple substitution, this reduces to:

(k^2+k+1)/(k+1)(sqrt(k^2+1)).

This does not exceed 3/sqrt8.

To prove this inequality, aquare both sides and apply AMGM in six variables or weighted AM-GM,
in which case we are done.

Bomb

Sailor, I didn't know. Valentin Vornicu asked me to post the problems.

We will prove that the maximum value is $\frac{3\sqrt{2}}{4}$.
The inequality is equivalent to: $\frac{a^2+b^2+ab}{\sqrt{a^2+b^2}(a+b)}\leqslant\frac{3\sqrt{2}}{4}$
After squaring both parts we have:
$(a^2+b^2+ab)^{2}\leqslant (a^{2}+b^2)(a+b)^{2}\cdot \frac{9}{8} \Longleftrightarrow $
$8(a^4+b^4)+24(ab)^{2}+16ab(a^2+b^2)\leqslant 9(a^2+b^2)(a+b)^{2}= \\ =9(a^4+b^4)+18(ab)^{2}+18a^3b+18b^3a$.
After simplifying we obtain:
$a^4+b^4+2(a^3b+b^3a)\geqslant 6(ab)^2$ which is true from [hide="AM-GM"]
$a^4+2b^3a\geqslant 3\sqrt[3]{a^6b^6}=3a^2b^2$
$b^4+2a^3b\geqslant 3\sqrt[3]{a^6b^6}=3a^2b^2$ [/hide]


summer program
Mathcamp
calculus
geometry
search
probability
MathPath

What grade are you in now, what math class are you in now, and what will you take next year?

I'm a junior, taking BC Calc, and then I'm going to do multivariable over the summer at UIC (they've got a free program there) and then probably take real analysis first semester and complex second semester next year, though I've heard neither is really analysis, since it's UIC.  But I figure it'll be good practice for college.  Either that or I'll take some combinatorics or number theory.

I'm taking BC Calculus as a 9th grader; next year, I'll probably take AP Statistics (if I don't go to boarding school) or (if I do go to boarding school) Exeter's Transition III Math (super pre-calc, good for math competition preparation) plus the last term of calculus (which is basically calculus topics not covered by the AP exam).

Adv. Analysis II at AAST, ninth grade... BC Calc next yr.. multivariate calc in junior yr, and linear algebra in senior yr.

*too lazy to take math course this year*...sorry

I'm taking BC Calculus in 9th grade. Next year I am going to AP Stats. And if the IB Program  (new) doesn't offer Further Math then I will take University Courses like Linear Algebra.

Wow yall peoples are really smart!!!!! Im going to take AP calculus at the sophmore level hopefully but I have no clue how Im gonna get there from Algebra 1. Next year Im taking pre AP geometry and then Im gonna take a couple of college courses over that summer. I dont know how Im gonna do it but my mom wants me to do it and I want to do it. Funfun eh?

I'm a junior. This year I'm taking Algebra II. Next year I'm taking Honors Precalculus and AP Statistics.

I'm in 7th grade and I'm taking pre-algebra.  Depending on where I go to school next year I will probally take algebra 1.

whats bc calc? im taking calculus as a freshman.  why dont we start a poll?  i dont know how

[quote="pinkbubblez90"]calculus in 9th grade?!?!?! =O Calculus BC is the highest math level they [i]offer[/i] at the high school im going to.

i'm in eighth grade taking geometry[/quote]
I agree, quite depressing...most of the people in my school didn't even believe me when I told them that and they're not stupid mind you :)

We also only go up to Calculus BC even in the Honors Curriculum.

Btw I take Calc AB next year, and I'm PROUD OF IT! :lol:

[quote="Tare"]I agree, quite depressing...most of the people in my school didn't even believe me when I told them that and they're not stupid mind you :)[/quote]

Hold, on, when you told them what?

[quote="justinian"]8th grade, Int. III[/quote]

What kind of topics are in your integrated math III course?

Would most eighth-graders be in a lower level of the integrated math program at your school?

(This question applies to everyone in the thread, please.) What specific textbook series are you using in your school?

[quote="pinkbubblez90"]calculus in 9th grade?!?!?! =O Calculus BC is the highest math level they [i]offer[/i] at the high school im going to.

i'm in eighth grade taking geometry[/quote]

I always get bored during the summer. So I started to take courses at the one provided by the state board of education where I live. I skipped Algebra 1 and 2 so I can do Calc in 9th grade and can devote all my study time to it. Which I think is a good idea because I am going to be loaded down next year with AP Biology.

I'm in sixth grade and I'm doing algebra II, through [url=http://www.itcep.umn.edu/umtymp/]UMTYMP[/url]. Earlier this year I did algebra I through UMTYMP and geometry through JHU CTY [url=http://cty.jhu.edu/math/courses/geometry.html]distance education[/url]. Now I am doing EPGY [url=http://epgy.stanford.edu/courses/math/M015/]Honors Geometry[/url] through distance learning. These are the books that I'm using: [url=http://www.brookscole.com/cgi-wadsworth/course_products_wp.pl?fid=M2b&product_isbn_issn=0534371639&discipline_number=1]Algebra for College Students, 6th ed.[/url] for the UMTYMP course, [url=http://www.mcdougallittell.com/default.sph/ProductCatalogStub.class/com.hmco.ecommerce.productcatalog.ProductCatalogController?cmd=Browse&subcmd=LoadDetail&ID=1000600000006343&frontOrBack=F&division=M01&sortProductsBy=SEQ_TITLE&sortEntriesBy=SEQ_NAME]Geometry[/url] by Jurgensen, Brown, and Jurgensen for the JHU CTY course, and course materials from the instructors for the EPGY course.

This summer I will take the [url=http://www.dcu.ie/ctyi/summer/academic/c_desc04.htm#Modern%20Mathematics]Modern Mathematics[/url] course at the Centre for Talented Youth Ireland. Then I will go to the [url=http://www.jhu.edu/gifted/summer/catalogs/oscatalog.html]JHU CTY summer program[/url] in Lancaster, Pennsylvania, either for the Probability and Game Theory course or another course that I'm wait-listed for.

Next year I will be twelve (seventh-grade age) and I will take the [url=http://www.itcep.umn.edu/umtymp/hschool.html]second-year UMTYMP course[/url] doing geometry and math analysis (precalculus). In the summer of 2005 I will probably go to [url=http://www.mathpath.org/]MathPath[/url]. The following year of UMTYMP is Calculus I.

My school uses the HRW series for math...Algebra II is the class that the average junior is in...my school isn't really a 'math' school...

[quote="Silverfalcon"]8th grade. Taking Algebra II. Doing independent study but not doing so well, though.  Taking Geo and Trig next year and probably or maybe, pre-cal or Calculus in Sophomore.. Maybe before that but not sure..[/quote]

um...geometry comes [b]before[/b] algebra II, at least at my school it does.

What's the difference between Calculus AB and Calculus BC? i was reading the high school course guide and they sound exactly the same.

[edit] WOW 100 posts! lol

Actually, some places have A. II before Geo.
I'm taking geometry 8th  and Pre-Calc next year.

Half the people here are in the calculus trap.

My school is same..
We take Algebra II then geo then trig but I can take geo and trig together...
but I think i can geo now too....

Algebra I...8th grade. I will be taking Algebra II honours next year as a freshman.

will someone answer my question, s'il te plait? what's the difference btwn Calculus AB and Calculus BC? =/

at my school you take trig with Alg II

[quote="pinkbubblez90"]will someone answer my question, s'il te plait? what's the difference btwn Calculus AB and Calculus BC? =/[/quote]

The calculus AB syllabus and the calculus BC syllabus are defined by the College Board for purposes of administering the AP calculus tests. As the names imply by the repetition of the "B," there is some overlap between the two syllabuses. Here are the links to the College Board Web site, so you can look at the topic lists yourself:

[url=http://www.collegeboard.com/student/testing/ap/calculus_ab/topic.html]calculus AB syllabus[/url]

[url=http://www.collegeboard.com/student/testing/ap/calculus_bc/topic.html]calculus BC syllabus[/url]

I am still trying to figure out how uniform colleges and universities are in designating courses as "calculus I" or "calculus II." My preliminary Google search suggests those terms are less consistently defined than the AB and BC terms, which are set by a single organization. Every college seems to have a slightly different idea what is first-semester calculus and what is second-semester calculus. My son's calculus courses will be special honors sections of calculus I and calculus II and calculus III, with the exact syllabus being revised by the time he takes the first of those courses two school years from now.

Hope this helps!

Sophomore, taking Precalc and an independent study on problem solving. Calc BC Junior year, hopefully Calc 3 over the summer and  then Calc 4 and Linear Algebra senior year

Would somebody tell me all the possible things to take above Pre-Calc?

Pick your favorite university (not a community college) and look at all the math classes they offer. That should give you a pretty good idea.

Same as lunmu - "Advanced Analysis 2" AKA pre-calc now as a freshman; some form of AP calc next year.

A+MATH - I met a girl this past summer at Mathcamp who was in the UMTYMP program.  Are there many people in it, or just a few?  It seems cool.. I wish I'd had that in elementary/middle school..

Taking Pre-Cal as a Jr. (normal is Alg II)
I'll probably take AB/BC Cal next year. (maybe I just won't do math)

I'm in 8th grade and i'm technically taking 4 courses, but only 3 at my school, and only one is official.  The simple answer is that I'll be in BC Calc as a 9th grader.

Im a senior and I am taking Adv. Precalc at my HS.  This summer I will be taking Calc I w/ Analytic Geometry and Calc II w/Analytic Geometry at URI.  Hopefully I will be taking a difficult calc class in college next year.

[quote="Caroline"]A+MATH - I met a girl this past summer at Mathcamp who was in the UMTYMP program.  Are there many people in it, or just a few?  It seems cool.[/quote]

This year there are five sections of Algebra I & II that each have about 30 students. By the second semester (Algebra II) a few students have dropped out, so I'm estimating there are about 140 students taking algebra II. In the second year there are four sections of Geometry & Math Analysis students.

It's a cool program; they have their own entrance test. Last April 848 students tested to get into the program, and 183 were invited to participate. The teachers are nice and your teaching assistants (TAs) are nice. They grade your work more by steps then by the answer and they really take time to look at your work. Another cool thing is that you get to take this course into calculus and become a college student through this program.

What else would you like to know about UMTYMP? My dad found [url=http://minhua.com/mhc_article.php?artid=360]an article by another UMTYMP student[/url] online last year.


inequalities
quadratics
algebra
number theory solved
number theory

Find all positive integers n for which the equation
a+b+c+d=n*sqrt(abcd)
as a solution in positive integers.

i thnk it comes down to proving n<=4 if the equality is to be satisfied and a,c,b,d>=1.and then we have infinitely many solutions for n=1,2,4
(i guess).ihavent worked the inequality out.ne ideas
cheers

I've a soln
Hint: Prove n<=4
Choose  (ao,bo,co,do ) satisfy equa such that (ao+bo+co+do) minimum
WLOG, suppose ao>=bo>=co>=do
Quadratic equation:
(x+b+c+d)^2=n^2.x.bcd has another root a1:
a1+ao=n^2bo.co.do-2(bo+co+do)(1)
a1*ao=(bo+co+do)^2(2)
Since a1>=ao, (1)=> a1>=1/2.n*n*bo.co.do-(bo+co+do)
(2)=> a1==(bo+co+do)^2/ao <= 3(bo+co+do)
then we get n*n*bo.co.do <= 8(bo+co+do)
then n<=4
good???


group theory
abstract algebra
number theory
superior algebra
superior algebra unsolved

Suppose $ G$ is a group such that $ |G|$ is coprime to 2. Further suppose that $ |Z(G)|$ is not divisible by 3. Show that $ G$ has no order 3 normal subgroups.

supppse that $ G$ has a normal subgroup of order $ 3$. this is cyclic generated by an element $ x$. because $ |Z(G)|$ is not divisible by $ 3$, $ x$ does not lie in the center. so take $ y$ such that $ y^{ \minus{} 1} x y \neq x$. since $ \langle x \rangle$ is normal, we find $ 0 \leq i \leq 2$ such that $ y^{ \minus{} 1} x y \equal{} x^i$. clearly we must have $ i \equal{} 2$. then $ y^{ \minus{} 2} x y^2 \equal{} x^{2^2} \equal{} x$, and inductively $ y^{ \minus{} i} x y^i \equal{} x$ for all even $ i$. since $ |G|$ is coprime to $ 2$, $ y$ has odd order, so $ y^i \equal{} y$ for some even $ i$. this yields the contradiction $ y^{ \minus{} 1} x y \equal{} x$.

Thanks for the reply. After reading this i tried to use the same approach for when $ |G|$ is coprime to 4 and 5 doesn't divide $ |Z(G)|$ but ran into problems.
I got to that $ yxy^{\minus{}1}\equal{}x^i$ where $ 0\le i\le 4$ but an having trouble seeing which value $ i$ must be.

my issue was that in this case you could choose $ 0\le i\le 4$ but 0 and 1 are not possible so you are left with 2,3,4.
however i think i have managed to get around this.
what i found was that $ y^nxy^{ \minus{} n} \equal{} x^{r^n}$ so then by Fermat's little theorem for all multiples of 4  $ y^nxy^{ \minus{} n} \equal{} x$. And then from here I basically use the same ideas as in your previous post.


An $ 8$-foot by $ 10$-foot floor is tiled with square tiles of size $ 1$ foot by $ 1$ foot. Each tile has a pattern consisting of four white quarter circles of radius $ 1/2$ foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the floor are shaded?
[asy]unitsize(2cm);
defaultpen(linewidth(.8pt));

fill(unitsquare,gray);
filldraw(Arc((0,0),.5,0,90)--(0,0)--cycle,white,black);
filldraw(Arc((1,0),.5,90,180)--(1,0)--cycle,white,black);
filldraw(Arc((1,1),.5,180,270)--(1,1)--cycle,white,black);
filldraw(Arc((0,1),.5,270,360)--(0,1)--cycle,white,black);[/asy]$ \textbf{(A)}\ 80\minus{}20\pi \qquad
\textbf{(B)}\ 60\minus{}10\pi \qquad
\textbf{(C)}\ 80\minus{}10\pi \qquad
\textbf{(D)}\ 60\plus{}10\pi \qquad
\textbf{(E)}\ 80\plus{}10\pi$

[hide="Click for solution"]
The four quarter circles have the same area as a whole circle of radius $ \frac{1}{2}$ or $ \frac{\pi}{4}$ square feet.  Thus the area of the shaded portion of each tile is $ 1\minus{}\frac{\pi}{4}$ square feet.  There are $ 8 \times 10\equal{}80$ tiles in the entire floor, so the area of the total shaded region is $ 80 \left (1\minus{}\frac{\pi}{4} \right )\equal{}80\minus{}20\pi$, or $ \boxed{\textbf{(A)}}$.
[/hide]


How can I prove that there is a possible 2n+1 multiplications (when the operations are performed as indicated) using mathematical induction?

$\displaystyle\sum_{i=0}^n a\sub_i x^i = a\sub_0 +a\sub_1 x+a\sub_2 x^2 +...+a\sub_n x^n$

Thank you.

I'm not sure what you mean.  Do you mean the total number of multiplications needed to evaluate such a sum?  Because the total number of multiplications would be $\frac{n(n+1)}{2}$... so I'm not sure what you mean.

Or do you mean the minimum number of multiplications needed?  Because this number is $n$... yet again, I'm not sure what you mean.

Can you prove, using mathematical induction that the total number of multiplications is [n(n+1)]/2?

[quote="shyce"][quote="scorpius119"]Actually, i see a way to have $2n-1$ multiplications (if $n>0$) without converting to synthetic-division-ish form:

[hide]Base case: $n=1$: 1 multiplication here. That is $2(1)-1$.
Inductive step: The polynomial $\sum_{k=0}^n a_kx^k$ takes $2n-1$ multiplications according to inductive hypothesis. During these multiplications, the value of $x^n$ is determined (as shown below for $n>1$: for $n=1$ it is given).

Then add on $a_{n+1}x^{n+1}$. First multiply $x^n$ by $x$ (now the value of $x^{n+1}$ has been determined). Then multiply by $a_{n+1}$. Two multiplications were required here, so the total is $(2n-1)+2=2(n+1)-1$. Done.[/hide]

unless, of course, you wanted the multiplications to be done in left-to-right order: then it would be $\frac{n(n+1)}{2}$: this is similar to above, except

[hide]$n+1$ multiplications are needed for $a_{n+1}x^{n+1}$.[/hide][/quote]

Are you proving there are 2n+1 or 2n-1 multiplications.. your last step confused me.  Can you please prove via induction how there are $\frac{n(n+1)}{2}$ multiplications?[/quote]

I proved that there were $2n-1$ (the induction assumes true for $m$, needs to be proven for $m+1$: when $n=m+1$ then $2m+1=2n-1$).

[hide="solution for n(n+1)/2"]Base case ($n=0$): no multiplication needed. $0=\frac{0(0+1)}{2}$.

Inductive step: (assuming true for $m$) The polynomial $\sum_{k=0}^m a_kx^k$ will require $\frac{m(m+1)}{2}$ multiplications.

To find the number of multiplications for $\sum_{k=0}^{m+1} a_kx^k$, add on the number of multiplications from the $x^{m+1}$ term. This is $a_{m+1}*\underbrace{x*x*x*\cdots*x}_{m+1}$. There is exactly one $*$ before each $x$ so $m+1$ multiplications.

The total for $m+1$ is then $\frac{m(m+1)}{2}+(m+1)=\frac{m(m+1)+2(m+1)}{2}=\frac{(m+1)(m+2)}{2}=\frac{([m+1])([m+1]+1)}{2}$. Done.[/hide]


[quote="13375P34K43V312"]How many ways are there to spell out the way MTHOMBA going from top to bottom below?
[code]
 M
TT
HHH
OOOO
MMM
BB
A[/code]

You can only move to adjacent letters directly below.[/quote]
You stole my question :P
[hide]16[/hide]

[quote="davidli"]can you post a solution?[/quote]
Ok.
[hide]Notice that from the M on top, you have two choices. The same is for the Ts, and the same is for the Os. That is 8. Now it gets tricky, but notice that there are 8 ways from the bottom a to any o as well, so our answer is 8*8=64[/hide]

No, that's incorrect.

There was this dude named Pascal long ago...

[quote="ProtestanT"]hm wut about him?[/quote]
He liked triangles.

[hide]At each letter I wrote how many ways there were to get to that letter by adding the 1 or 2 above it. The answer is 20. You can also draw a Pascal's triangle over it and the number that coincides with the A is a 20.[/hide]

"You can only move to adjacent letters directly below."

This makes no sense.


I have never done chemistry before, and I was wondering a few things about atoms.

*What do atoms have to do with cells? Do the atoms make up the cells?
*If quarks make up atoms, then is there anything more basic than quarks?

Quarks r the most basic products known till now......

cells r made up of atoms.....quarks never exist independently they always combine to make one or the other  atom or structure

so if we see the complete hierachy of life,atoms will be found at bottom--below cellular level

Did scientists just stop at quarks, because they did not have good enough technology to see anything smaller?

As far as I'm aware even the existence of quarks is debatable. They started off as useful theoretical tools but they make such good predictions that people generally believe they exist. It's not so much that we cannot see them (after all there are a lot of things we can't see) but that we have no need to postulate something more elementary yet.

[quote="BanishedTraitor"]As far as I'm aware even the existence of quarks is debatable. They started off as useful theoretical tools but they make such good predictions that people generally believe they exist. It's not so much that we cannot see them (after there are a lot of things we can't see) but that we have no need to postulate something more elementary yet.[/quote]

Okay, I get it. Thanks you two for your help.

[quote="rituraj007"]and well i think the LHC proposes to see quarks(bosons)

more specifically, i think the scientists who 1st thought abt quarks just gave the name quark --to the smallest possible particle---and in future wat we call today as quarks may not be called so[/quote]

Yeah, because we don't have the technology, so they just kind of guess that there is something that makes up the atoms. Right?

Why do you say (bosons)?

they r trying to use the tech to prove themselves correct... :P

well LHC proposes to see bosons which r a type of elementary particle :)

LHC proposes to support the existence of Higgs Bosons (not just any bosons because we know they exist e.g. proton, neutron etc.).


[b][size=100][color=DarkBlue]Δίνεται τρίγωνο $ \bigtriangleup ABC$ και έστω $ A',\ B',\ C',$ τα σημεία επαφής του εγγεγραμμένου κύκλου $ (I),$ στις πλευρές $ BC,\ AC,\ AB,$ αντιστοίχως. Εάν $ I_{a},\ I_{b},\ I_{c},$ είναι τα κέντρα των παρεγεγραμμένων κύκλων που αντιστοιχούν στις ως άνω πλευρές, αποδείξτε ότι οι ευθείες $ I_{a}A',\ I_{b}B',\ I_{c}C',$ τέμνονται στο ίδιο σημείο έστω $ P,$ και ότι το σημείο αυτό ανήκει στην ευθεία $ OI,$ όπου $ O,$ είναι το περίκεντρο του $ \bigtriangleup ABC$[/color][/size][/b].

Κώστας Βήττας.

Πολύ ωραία άσκηση ... Μοιάζει το σκεπτικό της με αυτή του Εσωτερικού διαγωνισμού.
Λοιπόν έχουμε $ A'C'//I_aI_c$ ,$ B'C'//I_bI_c$, $ A'B'//I_aI_b$ επομένως τα τρίγωνα $ A'B'C'$ και $ I_aI_bI_c$ είναι ομοιόθετα
επομένως οι ευθείες $ I_aA',\ I_bB',\ I_cC'$ συντρέχουν στο $ P$ που είναι και το κέντρο ομοιοθεσίας των δύο αυτών τριγώνων.
Αν τώρα θεωρήσουμε την ομοιοθεσία κέντρου $ P$ που στέλνει το τρίγωνο $ A'B'C'$ στο $ I_aI_bI_c$ αυτή η ομοιοθεσία θα στείλει το έγκεντρο $ I$ (που είναι περίκεντρο του $ A'B'C'$ ) στο $ I'$ που είναι το περίκεντρο του $ I_aI_bI_c$ επομένως
τα $ I,P,I'$ είναι συνευθειακά . Όμως καθώς το $ I$ είναι ορθόκεντρο του $ I_aI_bI_c$ η ευθεία που ορίζεται από τα $ I,I',P$ είναι η ευθεία Euler tou $ I_aI_bI_c$ . Όμως το $ O$ είναι το κέντρο του κύκλου του Euler sto $ I_aI_bI_c$ επομένως $ I,I',P,O$ συνευθειακά and we are done . :)


calculus
integration
calculus computations

how to do the following integration:

t is a constant,
Integral from 0 to t of (y^2)(e^[(t^2)/(y^2)]) dy  ?????????

I'm fairly sure that requires complex analysis. I don't think that's going to have a nice solution over the reals.

ok...then I think I approach this problem wrongly...

the original question is:

lim (t->0) of  [ (1/t) A]

where A= double integral of e^[(tx)/(y^2)] dxdy over the region [0,t] x [0,t]

Is there a theorem or a trick that I missed????

That question [i]again[/i]? What's the source?

And of course, it simply diverges. The value is $ \plus{}\infty.$ Rather than linking to where I said that, how about you show some initiative and find my post.

I've actually done something to it in the past few minutes..

what about applying green's theorem to the double integral??
So integral of Pdx+Qdy over the boundry (square in our case)= double integral of dQ/dx-dP/dy over [0,t]x[0,t]
set P=0 and dQ/dx=e^[(tx)/(y^2)]
so Q=(y^2)[e^[(tx)/(y^2)] t^(-1)

now by setting C1 and C3 be the two horizontal boundry (where dy=0), and C2 and C4 be the two vertical boundry (where y is ranging from 0 to t (when x=t) and t to 0(when x=0))

so the integral over the boundry over C1 and C3 vanishes.
and the whole thing becomes:

lim, as t->0, [ (integral of Q dy from 0 to t)- (t^2)/3 ] / t
by L-hopital's rule and FTC,
lim, as t->0, te-(2/3)t = 0

Is it wrong?? And if yes, where is my mistake?


[quote="Daniel Gutekunst"]This may be one of the senseless posts ... but I just looked at your picture, Valentin, and wondered, if you make gel in your hair or not. It looks a little bit wet, but decent (very nice). I for myself always use a little bit gel to style my hair in a similar way.

I thought Romania was a poor country, so no one should use gel there  :D[/quote]
about the last phrase, Romania is far away from a poor country. Just that a few people are rich, but those who are, they are really rich. Many people are poor, but for example, on my street I can see 3 mercedez, 2 BMWs, and a lot of other very expensive cars. :D

Anyway, usually most young people in romania (and I might say Japan, because I know :D) use a looot of gel. I personally never use gel, so that's just what my hair looks like :D.

btw,how old are u in this pic of urs,valentin? :D (in india,it's considered  bad manners if u ask sum1 elder than u his/her(more importantly) the
age)
but i'll take the liberty coz i'm asking ur age in the pic,not ur present age :D  :D

this pic is like 8 months old. so i'm 19 and 10 months old. :)

The $4$ oldest posts are locked, so this is the oldest unlocked post!


Let $f_0=f_1=1$ and $f_{i+2}=f_{i+1}+f_i$ for all $n\ge 0$. Find all real solutions to the equation
\[x^{2010}=f_{2009}\cdot x+f_{2008}\]

[hide="reduction"]
$ x^{2010} \minus{} f_{2009} \cdot x \minus{} f_{2008} \equal{} (x^2 \minus{} x \minus{} 1)(\sum_{i \equal{} 0}^{2008} f_{2008 \minus{} i} x^i )$ by $ ij$
Two roots are given by the solutions to $ x^2 \minus{} x \minus{} 1 \equal{} 0$; they are $ \frac {1 \pm \sqrt {5}}{2}$.
[/hide]

by calculating $(x^{2010}-f_{2009}x-f_{2008})''=2010*2009x^{2008}$
we know that $x^{2010}-f_{2009}x-f_{2008}$ is a convex function
hence it has no more than $2$ zeroes on $R$
we can confirm that $x=\frac{1+\sqrt5}{2}$ and $\frac{1-\sqrt5}{2}$ are solutions,so they are the only roots.


hey folks can you tell me the TOP 5 rock gutarists you would choose as the best?

here's my list....

1 SLASH(gnr)

2 Mark knophler(dire straits)

3 Blackmoor(deep purple)

4 Kirk(metallica)

5 jimi hendrix

I'm going to move this thread to Fun and Games based on a previous statement by AoPS management of what the topic scope of the Music and Art Forum is.

um, i don't know many rock gutarists but i think jimmy hendrix was good.

i would also have to say santana. his music has a great blend of classic rock and latin. jimi hendrix was a great guitarist as well.

need to add tom morello (rage against the machine/audioslave) to the top 5 list.

My two favorites: Jimmy Page and Edward Van Halen.

Here's mine...

1# Carlos Santana

2# Matt B------ (forgot his name, it's weird) From Muse

3# Jack White... (white stripes)

But see, those are only the ones I could name, I have heard better. I just don't know their names. I can't remember Jimmy Hendrix playing guitar even though I know he does, I just remember his voice... I'm more of a voice person when it comes to rock! Thus my liking for White Stripes and MUSE

I dont have favorites, but all of the above mentioned are good, though I think Slash is a little bit overrated. He's good... hes not legend-type stuff but everyone seems to think hes a guitar god...He is good though.


calculus
integration
function
trigonometry
analytic geometry
parameterization
algebra

Let $ C$ be the circle with radius 1 centered on the origin. Fix the endpoint of the string with length $ 2\pi$ on the point $ A(1,\ 0)$ and put the other end point $ P$ on the point $ P_0(1,\ 2\pi)$. From this situation, when we twist the string around $ C$ by moving the point $ P$ in anti clockwise with the string streched tightly, find the length of the curve that the point $ P$ draws from sarting point $ P_0$ to reaching point $ A$.

Parametrize the movement of P by $ P_t: \equal{}(x(t),y(t))$ where $ t$ is the polar angle of the point of tangency of the sting to C (in Radians). Then the functions $ x\equal{}x(t)$ and $ y\equal{}y(t)$ satisfy the system of equations:

\[ \begin{array}{rr} (x\minus{}\cos t)^2\plus{}(y\minus{}\sin t)^2 \equal{} (2\pi \minus{} t)^2 \\ \frac{y\minus{}\sin t}{x\minus{}\cos t}\equal{}\minus{}\cot t \end{array}\]

where the first equation was deduced from the length of the string and the second equation requires the sting be tangent to C when it "leaves it".

The required value is then

\[ S\equal{}\int_{0}^{2\pi}\sqrt{\left(\frac{dx}{dt}\right) ^2\plus{}\left(\frac{dy}{dt}\right) ^2}\, dt\]

Solving the system I get

\[ \begin{array}{ll} x(t)\equal{}\cos t \plus{} (2\pi \minus{}t) \, |\sin t| \\ y(t) \equal{} \minus{}\cos t \left( \mbox{sgn} (\sin t) (2\pi \minus{}t) \plus{} \cot t\right) \end{array}\]

And I'm going to smoke while mull over the nagging feeling that the is a clever way to formulate that integrand without solving the system explicitly...

[quote="kunny"]Regrettably that is incorrect.[/quote]
yeah sorry made a blunder.. will fix it and post asap!

We have to evaluate $ \int_{0}^{2\pi\minus{}\arctan{2\pi}}\sqrt{\frac{(1\plus{}\varphi^2)^2\plus{}\varphi^2}{1\plus{}\varphi^2}}d\varphi$.

Is the answer $ 2 \pi ^2$ by any chance?

I used the plane polar coordinate system.  I let the origin of the polar coordinate system be the center of the circle $ C$, and the half-line $ \theta\equal{}0$ be the $ x$ axis. Then, a point in the curve can be denoted by $ \bold{\hat{r}}\plus{}(2\pi\minus{}\theta)\bold{\hat{\theta}}$. Applying the formula for calculating length of a curve which is $ \int_{\theta_i}^{\theta_f}\sqrt{r^2\plus{}(dr/d\theta)^2}d\theta$, we get the integral($ r\equal{}\sqrt{1^2\plus{}(2\pi\minus{}\theta)^2}$, $ \theta_i\equal{}\arctan{2\pi}$, $ \theta_f\equal{}2\pi$).

 Something is wrong in my approach, if the answer is surely $ 2\pi^2$. Mathematica says the value of my integral is  $ 13.883$ approaximately.

The error was the meaning of $ \theta$. $ \theta$ in $ \bold{\hat{r}}\plus{}(2\pi\minus{}\theta)\bold{\hat{\theta}}$ and in $ \int_{\theta_i}^{\theta_f}\sqrt{r^2\plus{}(dr/d\theta)^2}d\theta$ is different. Let the latter $ \theta$ be $ \Theta$ and the former be $ \theta$, the relation $ \theta\equal{}\Theta\minus{}\arctan (2\pi\minus{}\theta)$ holds because $ \theta$ is the angle of the tangent point, not a point in the curve.

 Using polar coordinate system is useless. We can simply find an expression of $ P(x,y)$ in Cartesian coordinate system enough, using parameter $ \theta$ which is the angle of the tangent point.  $ x\equal{}\cos\theta\minus{}(2\pi\minus{}\theta)\sin\theta$ and $ y\equal{}\sin\theta\plus{}(2\pi\minus{}\theta)\cos\theta$ holds. Then, we can apply the formula $ L\equal{}\int_{\theta_i}^{\theta_f}\sqrt{(dx/d\theta)^2\plus{}(dy/d\theta)^2}d\theta$($ L$ is, of course, the length of the curve between $ \theta\equal{}\theta_i$ to $ \theta_f$) based on the Phytagorian theorem.

It gives $ \int_{0}^{2\pi}\sqrt{(\minus{}\sin\theta\minus{}2\pi\cos\theta\plus{}\sin\theta\plus{}\theta\cos\theta)^2\plus{}(\cos\theta\minus{}2\pi\sin\theta\minus{}\cos\theta\plus{}\theta\sin\theta)^2}d\theta$
      $ \equal{}\int_{0}^{2\pi}\sqrt{4\pi^2\minus{}4\pi\theta\plus{}\theta^2}d\theta\equal{}\int_{0}^{2\pi}(2\pi\minus{}\theta) d\theta\equal{}\boxed{2\pi^2}$.

[quote="kunny"]Yes.[/quote]
Sadly I was rotating it in clock wise direction :(

[quote="J.Y.Choi"]
 Using polar coordinate system is useless. We can simply find an expression of $ P(x,y)$ in Cartesian coordinate system enough, using parameter $ \theta$ which is the angle of the tangent point.  $ x \equal{} \cos\theta \minus{} (2\pi \minus{} \theta)\sin\theta$ and $ y \equal{} \sin\theta \plus{} (2\pi \minus{} \theta)\cos\theta$ holds. Then, we can apply the formula $ L \equal{} \int_{\theta_i}^{\theta_f}\sqrt {(dx/d\theta)^2 \plus{} (dy/d\theta)^2}d\theta$($ L$ is, of course, the length of the curve between $ \theta \equal{} \theta_i$ to $ \theta_f$) based on the Phytagorian theorem.

It gives $ \int_{0}^{2\pi}\sqrt {( \minus{} \sin\theta \minus{} 2\pi\cos\theta \plus{} \sin\theta \plus{} \theta\cos\theta)^2 \plus{} (\cos\theta \minus{} 2\pi\sin\theta \minus{} \cos\theta \plus{} \theta\sin\theta)^2}d\theta$
      $ \equal{} \int_{0}^{2\pi}\sqrt {4\pi^2 \minus{} 4\pi\theta \plus{} \theta^2}d\theta \equal{} \int_{0}^{2\pi}(2\pi \minus{} \theta) d\theta \equal{} \boxed{2\pi^2}$.[/quote]

Yes, so did I. :lol:


http://en.wikipedia.org/wiki/Rebus

Rebus marathon? Solve the previous rebus and post a new one. Or, just post a new one, and start a queue of unsolved rebuses, but you don't get any points for doing this.

There are two classes of rebuses: Those that are tricky and those that are just unbearably weird.

For example.
[img]http://www.webmikey.com/images/wallpaper/starfire.jpg[/img] + [img]http://www.x-mencharacters.com/images/Scott.gif[/img] + [img]http://media.gamersroundtable.com/images/naruto/naruto.gif[/img]  + [img]http://www.mad-cow.org/00/turtle.gif[/img] = ??

Or

[code]

CCCSAILINGCCCC

[/code] = ?

[quote="Koda"]I solved the second one:
sailing in the seas, right?[/quote]

Nope, sailing in the seven seas :D

The first one is:

Teenage Mutant Ninja Turtle

[quote="jackdillon"][quote="jhredsox"]try this one

[b]sta4nce[/b][/quote]
That is easy, [b][u]for instance [/u][/b]I can think of a harder one!

[u]ONHOLEE[/u][/quote]
hole in one

YOUJUSTME

[code]

1D  5U
2R  6L
3A  7A
4C

[/code]

15S68A20F9E41T28Y975

[code] D R
S   I
 K N
HOUSE[/code]

just between you and me.

the others i can't get the exact things:
dracula's letters?
safety among the numbers? (as in crowd)

no idea about the last one - how are drinks adn house related?

Just between you and me

Safety in numbers

Drinks all around on the house

Count Dracula.

HERE'S A REALLY GOOD ONE

GEGS

$\text{H}_{2}\text{O}$

Count Dracula was great!


[hide]
$x^2 + y^2 + z^2 = (x+y+z)(x+y+z)$
Expanding right hand side, then simplifying, then multiplying by (x+y+z)
$(x+y+z)(xy + yz + xz) = 0$
$3xyz + x^2(y+z) + y^2(x+z) + z^2(y+x) = 0$
x+y+z = 7, so we sub to simplify..
$3xyz + x^2(7 - x) + y^2(7 - y) + z^2(7 - z) = 0$
simplifying, then subbing back in
$3xyz + 7*49 - 7 = 0$
therefore, xyz = -112
[/hide]

[quote="shadysaysurspammed"][hide="short and sweet"]$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$

$\implies xy+yz+zx=0$

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

Substitute values and get the answer[/hide][/quote]

Jajaja, short and sweet....... nice one!


See here : http://www.mathlinks.ro/viewtopic.php?t=56071

[quote="duc 95"]Given a,b,c>0
prove[/quote]

$ \frac {a}{\sqrt {a \plus{} b}} \plus{} \frac {b}{\sqrt {b \plus{} c}} \plus{} \frac {c}{\sqrt {c \plus{} a}}\geq \frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}{\sqrt{2}}$


superior algebra
superior algebra unsolved

[size=134]Find a splitting field $ K \subseteq \mathbb{C}$ for $ x^5\minus{}1 \in \mathbb{Q}[x],$ and determine $ [K: Q]$[/size]

$ x^n \minus{} 1 \equal{} \prod_{k\equal{}0}^{n} (x \minus{} e^{ \frac{2 \pi i k}{n} } )$ over $ \mathbb{C}$, so the splitting field of $ x^n \minus{} 1$ over the rationals is the [url=http://mathworld.wolfram.com/CyclotomicField.html]cyclotomic field[/url] $ \mathbb{Q}[ e^{\frac{2 \pi i}{n} } ]$.


function
real analysis
real analysis unsolved

Give example of $f \in L^2(\mathbb{R}) - L^1(\mathbb{R})$ such that $\widehat{f} \in L^1(\mathbb{R})$. Under what circumstances can this happen ?

A function that has very fast decay but is not very smooth will have a Fourier transform that is quite smooth but very slow decay.

Suppose $\widehat{f} = \chi_{[-1, 1]}$. Then $f$ is $\frac{\sin(2\pi \theta)}{\pi \theta}$, I believe. Does this work?


projective geometry
geometry proposed
geometry

Triangle $ABC$ is acute-angled.$M$ is a midpoint of bisector $AA_{1}$.$P$ and $Q$ are points on $MB$ and $MC$ respectively.$\measuredangle{APC}=\measuredangle{AQB}=\frac{\pi}{2}$.Prove that $A_{1}PMQ$ is cyclic.

[hide] This problem is hard. But if you make inversion in point A  it will become an easy exersise for using symmetry. :wink:
There is other two provings.They didn't use inversion.
One is hard and long.
Second is long but not hard.Consider the line $l \parallel AM, B \in l$
Consider  the altitude from A to l and circle AQB.We get some eqeations between angles.And then a lot of calcules...
[/hide]

[quote="Huyền Vũ"]let $l$ be a line through $B$ and parallel with $AM$. $CM$ meets $l$ at $D$
We have $MC/CD=MA_{1}/BD=AM/BD=EM/ED$ so $(C,E,M,D)=-1$
But $AM$ is the bisector of $\angle EAC$ so $DA\bot AM$
Otherwise $BD//AM$ so $BD\bot AD$
This means $A,D,B,Q$ are cyclic
So $\angle MCA+\angle MAC=\angle A_{1}MC=\angle BDC=\angle BAQ=\angle BAM+\angle MAQ$
so $\angle MCA=\angle MAQ$
So triangle $MAQ$ and $MCA$ are similar
thus $MA_{1}^{2}=MA^{2}=MQ*MC$
so  triangle $MQA_{1}$ and $MA_{1}C$ are similar
So $\angle MQA_{1}=\angle MA_{1}C$
Similarly $\angle MPA_{1}=\angle MA_{1}B$ . But $\angle MA_{1}B+\angle MA_{1}C=180$
so $\angle MQA_{1}+\angle MPA_{1}=180$
This means $M,Q,A_{1},P$ are cyclic $q.e.d$[/quote]

[b] NOTE[/b]
 Please ,let me ''improve'' (make easier?) this [color=red]nice[/color] solution in the first step , avoinding the harmonic division:
 If line $BD$ meets line $CA$ at $Z$ , then since $BZ // AA_{1}$ we take that triangle $ABZ$ is isosceles and since $M$ is the mid point of  $AA_{1}$, point $D$ will be the mid point of $BE$.Thus $AD \perp BZ$.

[u] Babis[/u]

OH! Your note is very nice stergiu. Thanks

[Hide=solutions]My solutions
[url]https://artofproblemsolving.com/community/c6h1832889p12277985[url]

$(MA_{1})^2=MA^2=MP.MB$ $\rightarrow$ $\angle{MA_{1}P}=\angle{MBC}$
$\angle{APM}=\angle{AQM}$$\rightarrow$ $BPQC$ is concyclic $\angle{MBC}=\angle{PQM}$
$\angle{MA_{1}P}=\angle{PQM}$ $\rightarrow$ $MPA_{1}Q$ is cyclic


[quote="Harry Potter"][quote="Sunjee"]Let $ a,b,c,d$ non-negative numbers such that $ a^{6}\plus{}b^{6}\plus{}c^{6}\plus{}d^{6}\equal{} 1$, then
\[ \sqrt{a^{4}\plus{}b^{4}\plus{}c^{4}\plus{}d^{4}}\geq\sqrt[3]{a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}d^{3}}\]
[/quote]
I don't think it is true[/quote]

uhm, for $ 3$-variable, Sunjee 'inequality is true ( follow my proof ) .

And for $ 4$ - variable , i think, the inequality must be :

$ \sqrt{a^{4}\plus{}b^{4}\plus{}c^{4}\plus{}d^{4}}\le\sqrt[3]{a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}d^{3}}$

I am true ??  :blush:

[quote="Sunjee"][quote="fjwxcsl"]The inequality inversely holds!

see:http://www.irgoc.org/bbs/dispbbs.asp?boardid=12&id=2696&star=2#8228[/quote]
Sorry I dont understand this language...If its inversely holds,show me please example.[/quote]

I have root this solution from  fjwxcsl'link .

$ (x^{3}\plus{}y^{3}\plus{}z^{3})^{2}(x^{6}\plus{}y^{6}\plus{}z^{6})\minus{}(x^{4}\plus{}y^{4}\plus{}z^{4})^{3}\equal{}\sum f(x,y,z)(x\minus{}y)(y\minus{}z)$

with $ f$ is a 3-variable function so that if $ x\ge y\ge z$ then: $ \sqrt{f(x,y,z)}\minus{}\sqrt{f(z,x,y)}\plus{}\sqrt{f(y,x,z)}\ge 0$

then after some argument, we have $ \sum f(x,y,z)(x\minus{}y)(y\minus{}z)\ge 0$.

I have seen a same argument :

Let $ f:\mathbb{R^{\plus{}}}\to\mathbb{R^{\plus{}}}$ so that :

$ \sqrt{f(x)}\plus{}\sqrt{f(y)}\ge\sqrt{f(z)}$ with$ x\ge y\ge z > 0$ .
Then :
$ \sum_{\text{a,b,c}}f(a)(a\minus{}b)(b\minus{}c)\ge 0$

But it is easier than own inequality . :blush:

hi Sunjee $ a\equal{}b\equal{}0$ $ c\equal{}\frac{1}{2}$ $ d\equal{}\frac{\sqrt[6]{63}}{2}$ not true

Who can you solve the inequality by the way from the link :

http://www.irgoc.org/bbs/dispbbs.asp?boardid=12&id=2696&star=2#8228

 :)

[quote="Sunjee"]Let $ a,b,c,d$ non-negative numbers such that $ a^{6}\plus{}b^{6}\plus{}c^{6}\plus{}d^{6}\equal{} 1$, then
\[ \sqrt{a^{4}\plus{}b^{4}\plus{}c^{4}\plus{}d^{4}}\geq\sqrt[3]{a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}d^{3}}\]
[/quote] Sorry everybody,in my proof i mistyped,so
\[ \sqrt{a^{4}\plus{}b^{4}\plus{}c^{4}\plus{}d^{4}}\leq\sqrt[3]{a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}d^{3}}\]
in general case: if $ \sum x^{6}_{i}\equal{}1$ then
\[ \sqrt{\sum x^{4}_{i}}\leq\sqrt[3]{\sum x^{3}_{i}}\]
thank you.

[quote="quangpbc"]The inequality is equivalent to :

$ (x^{4}\plus{}y^{4}\plus{}z^{4})^{3}\le (x^{3}\plus{}y^{3}\plus{}z^{3})^{2}(x^{6}\plus{}y^{6}\plus{}z^{6})$.

By Holder ,

$ (x^{3}\plus{}y^{3}\plus{}z^{3})(x^{3}\plus{}y^{3}\plus{}z^{3})(x^{6}\plus{}y^{6}\plus{}z^{6})\ge (\sum\sqrt [3]{x^{3}x^{3}x^{6}})^{3}\equal{} (x^{4}\plus{}y^{4}\plus{}z^{4})^{3}$. :wink:[/quote]
Ok.If you used HOlder's inequality,equality occurs when $ (a,b,c)\equal{}(t,t,t)$ here $ 3t^{6}\equal{}1$...
In my inequality equality occurs if $ n\equal{}3$ then $ (a,b,c)\equal{}(t,t,t)$ and $ (a,b,c)\equal{}(m,m,0)$ here $ 2m^{6}\equal{}1$ and $ (a,b,c)\equal{}(1,0,0)$
so i think holder not work here. I want general solution...

[quote="Sunjee"][quote="quangpbc"]The inequality is equivalent to :

$ (x^{4}\plus{}y^{4}\plus{}z^{4})^{3}\le (x^{3}\plus{}y^{3}\plus{}z^{3})^{2}(x^{6}\plus{}y^{6}\plus{}z^{6})$.

By Holder ,

$ (x^{3}\plus{}y^{3}\plus{}z^{3})(x^{3}\plus{}y^{3}\plus{}z^{3})(x^{6}\plus{}y^{6}\plus{}z^{6})\ge (\sum\sqrt [3]{x^{3}x^{3}x^{6}})^{3}\equal{} (x^{4}\plus{}y^{4}\plus{}z^{4})^{3}$. :wink:[/quote]
Ok.If you used HOlder's inequality,equality occurs when $ (a,b,c) \equal{} (t,t,t)$ here $ 3t^{6}\equal{} 1$...
In my inequality equality occurs if $ n \equal{} 3$ then $ (a,b,c) \equal{} (t,t,t)$ and $ (a,b,c) \equal{} (m,m,0)$ here $ 2m^{6}\equal{} 1$ and $ (a,b,c) \equal{} (1,0,0)$
so i think holder not work here. I want general solution...[/quote]

 :D For $ n\equal{}3$ i have posted a solution above by Muirhead , but the inequality occurs if $ a\equal{}b\equal{}c$ or or $ a\equal{}b,c\equal{}0$or $ b\equal{}c\equal{}0$ and cyclic . ( i have typed only $ a\equal{}b\equal{}c$. Sorry  :blush:  .Do you have a solution for $ n\equal{}4$. I think it can be solve by Muirhead, ( simirlary for $ n\equal{}3$ )

Holder works for this one, as well as the $ n$-variable case
\[ \left(\sum_{k\equal{}1}^{n}a_{k}^{6}\right)\left(\sum_{k\equal{}1}^{n}a_{k}^{3}\right)^{2}\geq\left(\sum_{k\equal{}1}^{n}a_{k}^{4}\right)^{3}\]
as 6+2*3=3*4.

There's nothing wrong with the equality case, either. There will be equality as long as there exist nonzero constants $ x,y,z$ such that
\[ xa_{k}^{6}\equal{}ya_{k}^{3}\equal{}za_{k}^{4}\]
for all $ k$, which not only occurs when all $ a_{k}$ are equal, but when all [i]nonzero[/i] $ a_{k}$ are equal. (For a simpler example, Cauchy, a special case of Holder, proves $ (a^{3}\plus{}b^{3})(a\plus{}b)\geq (a^{2}\plus{}b^{2})^{2}$ for nonnegative $ a,b$ even though there's equality when $ b \equal{} 0$.)


Let's $a,$ $b$ and $c$ are non-negative numbers and $f$ is  increasing positive function.
Prove that $\frac{ab+ac-2bc}{f(b)+f(c)}+\frac{ab+bc-2ac}{f(a)+f(c)}+\frac{ac+bc-2ab}{f(a)+f(b)}\geq0.$

[quote="hydro"][hide]
since f is positive and increasing, let $f(a) \ge f(b) \ge f(c)$
so
$f(a)+f(b) \ge f(b)+f(c)$
$f(a)+f(b) \ge f(a)+f(c)$

and $\frac{ab+ac-2bc}{f(b)+f(c)}+\frac{ab+bc-2ac}{f(a)+f(c)}+\frac{ac+bc-2ab}{f(a)+f(b)}\ge$$\frac{ab+ac-2bc}{f(a)+f(b)}+\frac{ab+bc-2ac}{f(a)+f(b)}+\frac{ac+bc-2ab}{f(a)+f(b)}$
and $\frac{1}{f(a)+f(b)}(2ab+2ac+2bc-2ab-2ac-2bc) \ge 0 \Longleftrightarrow 0 \ge 0$[/hide][/quote]
Why $\frac{ab+ac-2bc}{f(b)+f(c)}+\frac{ab+bc-2ac}{f(a)+f(c)}\ge\frac{ab+ac-2bc}{f(a)+f(b)}+\frac{ab+bc-2ac}{f(a)+f(b)}$ $?$ :wink:
Thank you.


Four tangent circles are centered on the x-axis. Circle O is centered at the origin. The radius of Circle A is twice the radius of Circle O. The radius of Circle B is three times the radius of Circle O. The radius of Circle C is four times the radius of Circle O. All circles have integer radii and the point (63,16) is on Circle C. What is the equation of Circle A?

[quote="7h3.D3m0n.117"]
This should go into Intermediate  :?:[/quote]

Ok, sorry.

Im just going to assume that they are in the order on the x axis, O, A, B ,C.

let the radius for O be $r$, this implies that the radius for A is $2r$, B is $3r$ and C is $4r$.

This means that the center of C is $r+2*2r+2*3r+4r = 15r$ from the origin

So we can set up an equation for circle C
$(x-15r)^{2}+y^{2}= (4r)^{2}= 16r^{2}$

Since C is on the point (63,16), we can plug in these x and y values into our equation and solve for r.

$(63-15r)^{2}+16^{2}= 16r^{2}$
$\Rightarrow 63^{2}-2(63)(15r)+225r^{2}+256 = 16r^{2}$
$\Rightarrow 3969+256-1890r+225r^{2}= 16r^{2}$
$\Rightarrow 209r^{2}-1890r+4225 = 0$
$\Rightarrow r_{1,2}= \frac{1890 \pm 200}{418}$
$r_{1}= 5$, $r_{2}= 4.04306....$
$r_{2}$ is inadmissable since the radii are of integer lengths.

Since we now know the radius, we can begin to construct an equation for A.

Circle O is centered on the Origin with radius 5, thus its max point on x is 5. Circle A follows, but its radius is two times that of circle O, it follows that the center of circle A is on $x = 15$
So our equation for circle A is:
$\boxed{(x-15)^{2}+y^{2}= 100}$

[quote="sharkboy"]nope. i was told to move it to intermediate.[/quote]

psh by "move" i was talking to the mods. i didn't mean for u to make a new topic.


Find all $ f : \mathbb R \rightarrow \mathbb R$ functions such that for all reals $ x$ and $ y$ we have:

$ f(x^{3})\minus{}f(y^{3})\equal{}(x^{2}\plus{}xy\plus{}y^{2})(f(x)\minus{}f(y))$

[color=green]This problem is easy. And the same problem is the problem below

Find all functions $ f : \mathbb R^ \plus{} \longrightarrow \mathbb R^ \plus{}$ such that for all positive reals $ x$ and $ y$ we have:

$ f(x^{3}) \minus{} f(y^{3}) \equal{} (x^{2} \plus{} xy \plus{} y^{2})(f(x) \minus{} f(y))$.[/color]

[quote="delegat"]Find all $ f : \mathbb R \rightarrow \mathbb R$ functions such that for all reals $ x$ and $ y$ we have:

$ f(x^{3}) \minus{} f(y^{3}) \equal{} (x^{2} \plus{} xy \plus{} y^{2})(f(x) \minus{} f(y))$[/quote]
Here is my solution
$ g(x) \equal{} f(x) \minus{} f(0)$ then
$ g(x^3) \minus{} g(y^3) \equal{} (x^2 \plus{} xy \plus{} y^2)(g(x) \minus{} g(y))$
Take $ y \equal{} 0$ then
$ g(x^3) \equal{} x^2g(x)$
So $ x^2g(x) \minus{} y^2g(y) \equal{} (x^2 \plus{} xy \plus{} y^2)(g(x) \minus{} g(y))$
$ \Leftrightarrow (x \plus{} y)(xg(y) \minus{} yg(x)) \equal{} 0$
Take $ y \equal{} 1$ then
$ g(x) \equal{} g(1)x$ for all x is different from $ \minus{} 1$ (1)
Replace to the equation we have
$ g(1)x^3 \minus{} g( \minus{} 1) \equal{} (x^2 \minus{} x \plus{} 1)(g(1)x \minus{} g( \minus{} 1))$
So $ g( \minus{} 1) \plus{} g(1) \equal{} 0$ (2)
From  (1)+(2) then
$ g(x) \equal{} ax$
It follow that $ f(x) \equal{} ax \plus{} b$

[color=green]For the second problem in this topic, who can solve it?  :) [/color]

Set $ g: \mathbf{R}^ \plus{} \to \mathbf{R}$, $ g(x) \equal{} f(x) \minus{} f(1)$. Hence $ g(1) \equal{} 0$ and $ g$ satisfy same equation. Setting $ y \equal{} 1$ gives $ g(x^3) \equal{} g(x)(x^2 \plus{} x \plus{} 1)$. This implies $ g(x)(x \plus{} 1 \minus{} xy \minus{} y^2) \equal{} g(y)(y \plus{} 1 \minus{} xy \minus{} x^2)$, which gives $ g(x)(1 \minus{} y) \equal{} g(y)(1 \minus{} x)$ . Therefore $ g$ is linear function and so it is $ f$.


algebra
polynomial
modular arithmetic
quadratics
superior algebra
superior algebra unsolved

Let $ p$ be a primer number. Determine the number of points in the quadric of $ \mathbb{Z}_{p}^{2}$ given by: $ x^{2}\plus{}y^{2} \equal{} a$, $ a\in \mathbb{Z}_{p}$.

Assume $ p > 2, a \neq 0$. Then the associated projective variety is isomorphic to $ \mathbf{P}^1_{\mathbf{F}_p}$ as it has a rational point (counting argument). Now count the points at infinity.

something more elementary? this problem is desinged to be in a undergraduate contest

why is this useful? can you write down a complete proof pleasE?

$ \ldots \equal{} p \plus{} \sum_{x\equal{}0}^{p\minus{}1}\left(\left(\frac{x}{p}\right) \plus{} \left(\frac{a\minus{}x}{p} \plus{} \left(\frac{x(a\minus{}x)}{p}\right)\right)\right) \equal{} \sum_{x\equal{}1}^{p\minus{}1}\left(\frac{x^{\minus{}1}}{p}\right)^2\left(\frac{x(a\minus{}x)}{p}\right) \equal{} \ldots$

Note that the number of solutions in $ \mathbf{F}_p$ to $ z^2 \equal{} x$ is equal to $ 1\plus{}\left(\frac{x}{p}\right)$.

[quote="Lousin Garckz"]I don't know what is [b]$ F_{p}$[/b] please, can you write down all details? it would be great ...  :lol:[/quote]

$ F_p$ is the field on $ p$ elements. Its another way to write $ \mathbb{Z}_p$ if you don't want to confuse it with the $ p$-adics and if you want to emphasize that it's a field.

There's not many details to write down there: he just wrote that there are two solutions if $ x$ is a square mod $ p$, and zero otherwise.

EDIT: Or 1, if the RHS is zero.

If $ F$ is a field of characteristic different from 2, so that $ 2a \equal{} 0 \implies a \equal{} 0$, then $ z^2 \equal{} x$ has either 0 or 2 roots -- 1 iff $ x \equal{} 0$. It can't have more than two roots, since a polynomial of degree $ n$ has as most $ n$ roots over a field, and if $ r^2 \equal{} x$ then $ ( \minus{} r)^2 \equal{} x$, so a root $ r$ implies the existence of a second roots $ \minus{} r$, provided $ r$ is nonzero.

Does that make sense?

yes, I understand it, but I don't understand how is related it with the topic, I mean, how can I use it to find the number of pairs $ (x,y)$ such that $ x^{2}\plus{}y^{2} \equal{} a$.

[quote="myan"]Compute $ \sum_{x \plus{} y \equal{} a}\left(1 \plus{} \left(\frac {x}{p}\right)\right)\left(1 \plus{} \left(\frac {y}{p}\right)\right)$[/quote]

This computes the number of ways to write $ a \equal{} x \plus{} y$, and to furthermore find $ z$ and $ w$ such that $ z^2 \equal{} x$ and $ w^2 \equal{} y$. So, it counts all solutions to $ z^2 \plus{} w^2 \equal{} a$.

I got that part, but, actually what is $ (\dfrac{x}{p})$ is this the Jacobi symbol? can you compute the sum please?

It's the Legendre symbol (since $ p$ is prime). myan started expanding the sum. You should give it a try first :).

ok I'll try it, i'm really sorry about my poor knowledge in number theory and been so insistent, I will try!..

Nukular, i can't get the result  :( .... help  :blush:

This is just an outline... if you need more, I can flesh out the details.

You should be able to reduce it to a calculation of $ \sum_{x \equal{} 1}^{p \minus{} 1} \left( \frac {x(a \minus{} x)}{p} \right)$. Then you can replace that with $ \sum_{x\equal{}1}^{p\minus{}1} \left( \frac{ax^{\minus{}1} \minus{} 1}{p} \right)$, factoring out the $ x^2$.

You get two cases: either $ a \equiv 0 \pmod{p}$ or not. The first case should be relatively easy. I'll assume we're not there for now.

Then $ ax^{\minus{}1}$ runs over all nonzero residues mod $ p$; therefore $ ax^{\minus{}1} \minus{} 1$ runs from $ 0,1,\dots,p\minus{}2$. So then you compute $ \sum_{y \equal{} 0}^{p\minus{}2} \left( \frac{y}{p} \right) \equal{} \minus{}\left(\frac{\minus{}1}{p}\right) \equal{} \minus{}(\minus{}1)^{(p\minus{}1)/2}$.

I think the answer then ends up being something like $ p \minus{} (\minus{}1)^{(p\minus{}1)/2}$, but I might be off by one or something.

(Sorry, written in haste and organized for crap. If you need more help I'll clean it up a bit and take care of the constants that fall out.)

nop, i have problem manipulating the Legendre symbol, i don't know why i can't do nothing  :(

Re-check my previous post, I edited it while you were posting to have some more helpful details.

Useful facts:

$ \left(\frac{x}{p}\right)\left(\frac{y}{p}\right) \equal{} \left(\frac{xy}{p}\right)$
$ \sum_{i\equal{}0}^p \left(\frac{i}{p}\right) \equal{} 0.$
$ \left(\frac{\minus{}1}{p}\right) \equal{} (\minus{}1)^{(p\minus{}1)/2}.$

And the ever-useful (though not useful here) Quadratic Reciprocity Law, which you should learn at some point: for odd primes $ p$ and $ q$:
$ \left(\frac{p}{q}\right)\left(\frac{q}{p}\right) \equal{} (\minus{}1)^{(p\minus{}1)(q\minus{}1)/4}.$

Those should get you started.

OK, let's see: Using Nukular arguments and recopilating to write more clear:

$ \sum_{x\plus{}y\equal{}a}(1\plus{}(\dfrac{x}{p}))(1\plus{}(\dfrac{y}{p})) \equal{} \sum_{x\plus{}y\equal{}a}1\plus{}(\dfrac{y}{p})\plus{}(\dfrac{x}{p})\plus{}(\dfrac{xy}{p}) \equal{} \sum_{x\equal{}0}^{p\minus{}1}1\plus{}(\dfrac{a\minus{}x}{p})\plus{}(\dfrac{x}{p})\plus{}(\dfrac{x(a\minus{}x)}{p}) \equal{} \sum_{x\equal{}0}^{p\minus{}1}1\plus{}(\dfrac{x(a\minus{}x)}{p}) \equal{} p \plus{} \sum_{x\equal{}0}^{p\minus{}1}(\dfrac{x(a\minus{}x)}{p}) \equal{} 1 \plus{} p \plus{}\sum_{x\equal{}1}^{p\minus{}1}(\dfrac{x(a\minus{}x)}{p}) \equal{} 1 \plus{} p \plus{} \sum_{x\equal{}1}^{p\minus{}1}(\dfrac{xa^{\minus{}1}\minus{}1}{p})$. As nukular said, let's think about this by part. If $ a \equiv 0 (p) \Rightarrow ax^{\minus{}1}\minus{}1 \equiv \minus{}1 (p) \Rightarrow (\dfrac{ax^{\minus{}1}\minus{}1}{p}) \equal{} (\minus{}1)^{\dfrac{(p\minus{}1)}{2}}$ and the final result is $ 1 \plus{} p \plus{} p (\minus{}1)^{\dfrac{(p\minus{}1)}{2}}$. By the other hand, if $ a \neq 0 (p)$ then $ ax^{\minus{}1}$ runs over all non-zero residues mod $ p$ then $ \sum_{x\equal{}1}^{p\minus{}1} (\dfrac{ax^{\minus{}1}\minus{}1}{p}) \equal{} \sum_{y\equal{}0}^{p\minus{}2} (\dfrac{y}{p}) \equal{} \minus{}(\dfrac{\minus{}1}{p}) \equal{} \minus{}(\minus{}1)^{\dfrac{(p\minus{}1)}{2}}$ and the final result is $ p\plus{}1\minus{}(\minus{}1)^{\dfrac{(p\minus{}1)}{2}}$. Is all this right?

P/D: Nukular, about what you wrote in other post of mine, believe me that I'm working so hard on all that problems, but I have pretty much work to do and is for tomorrow, and i Can't, i mean, I know that this is not of your interest but at least i want to give an explanation.

It looks fine, except that $ \left(\frac{0}{p}\right) \equal{} 0$, not $ 1$.

[quote="Lousin Garckz"]
P/D: Nukular, about what you wrote in other post of mine, believe me that I'm working so hard on all that problems, but I have pretty much work to do and is for tomorrow, and i Can't, i mean, I know that this is not of your interest but at least i want to give an explanation.[/quote]

Oh, if it's for homework, then, good luck with it! I know the stress of a problem set due the next day firsthand :).

all right, and thank you very very much Nukular!!!! you are great!


analytic geometry
function
algebra solved
algebra

We have a curve in space (you may consider it to be a polygonal line). Prove that there are 4 coplanar on this curve which divide it in 4 parts of equal lengths.

First of all, I only solved the problem for a CLOSED CURVE of FINITE LENGTH. If it wasn't closed then the four points would divide the curve into five parts. If the curve didn't have a finite length, then there is no point in defining 'equal lengths' for the pieces. Assume that the curve has length L>0. We fix a point O on the curve and we say that a point X has coordinate 0<=u<=L and we write X(u) if the length of the arc of the curve from O to X when going in the clockwise direction is u.

Also by convention O(0) and the point with coordinate u+L has coordinate u, for all real u.

Now, take an x>0 , 0<=x<=L and consider the points A(x) , B(x+L/4) , C(x+2L/4) , D(x+3L/4). If they lie on the same plane, we are done. Let us define the function f(x) as follows : if A(xa,ya,za) , B(xb,yb,zb) , C(xc,yc,zc) , D(xd,yd,zd), then

|xa ya za 1|
|xb yb zb 1| = f(x)
|xc yc zc 1|
|xd yd zd 1|

It's easy to see that f(x) = 0 <=> A,B,C,D lie on the same plane. Now we have :

|xd yd zd 1|
|xa ya za 1| = f(x+3L/4) = - f(x)
|xb yb zb 1|
|xc yc zc 1|

f is obviously continuous and f(x) + f(x+3L/4) = 0 and f(x) is not 0 => f(x)f(x+3L/4) < 0 => there is an x0 in (x,x+3L/4) such that f(x0)=0 , QED.

Of course! :D Sorry about the incomplete text: the curve must indeed be closed and it must have finite length.


Hi,

I am kind of confused about the derivations of the equations: $ y\equal{}a(x\minus{}h)^2\plus{}k; x\equal{}a(y\minus{}k)^2\plus{}h.$ I am finding myself basically memorizing these 'formulas' and I can't really understand them. Any explanation would be highly appreciated.

I am also confused as to why $ y\equal{}ax^2\plus{}bx\plus{}c, x\equal{}ay^2\plus{}by\plus{}c$ are useful and why they work. Maybe it is just that I am misunderstanding some basic concept.

Thank you very much for the help :)

Not sure if this is gonna help but:
A most primitive parabolic equation is: $ y\equal{}x^2$

Now, if I alter the equation by adding any number (constant) to the end, to form $ y\equal{}x^2 \plus{} k$, all the values for [i]y[/i] in this equation will be [i]k[/i] more than the values in the first equation for the respective [i]x[/i] value. Basically a shift upward of [i]k[/i].
Example: $ y\equal{}x^2$ vs $ y\equal{}x^2\plus{}2$
If we plug in $ x\equal{}1$ then $ y\equal{}1$ for the first, but $ y\equal{}1\plus{}2\equal{}3$ in the second.

Now, about the [i]h[/i]. Let's look back at the example of $ y\equal{}x^2$. Using an example, say $ y\equal{}(x\plus{}2)^2$.
If plugging in values, notice that for the second equation, if $ x\equal{}1$, then the result is $ y\equal{}(1\plus{}2)^2$ which is $ y\equal{}3^2$. (Basically the same result as $ y\equal{}x^2$ if we had plugged in "3" for x).
Thus, the [i]y[/i] value we obtain from the second equation by plugging in [i]x[/i] is actually the same as the y value from the first equation if we plug in $ x\plus{}2$.
Then entire graph of $ y\equal{}x^2$ will be reproduced two units to the left because each y value of $ y\equal{}(x\plus{}2)^2$ is the one from $ y\equal{}x^2$ that is 2 to the right (or, I mean, +2 units on the x-axis). So the result is a shift of the graph to the left. Which is why the form is $ y\equal{}(x\minus{}k)^2\plus{}h$ instead of $ y\equal{}(x\plus{}k)^2\plus{}h$. We want positive [i]h[/i] to correspond with a shift of the graph to the right, for some reason.

I hope it was at least a bit helpful.

Thank you for the reply, dChampionLance4. That basically cleared up my conception about the $ y\equal{}(x\minus{}k)^{2}\plus{}h$ forms. That was really helpful. That was my main point of confusion. I kept confusing $ ax^2\plus{}bx\plus{}c \equal{} 0$ with something like $ y\equal{}ax^{2}\plus{}bx\plus{}c, x\equal{}ay^{2}\plus{}by\plus{}c ,$ and that was after all a pretty big misconception. Your example cleared that up for me :) (I probably wasn't paying that much attention when the book was explaining that gulping down a pound and a half of Nachoz ;))

Thanks again,
limac