calculus
integration
geometry
rotation
function
3D geometry
sphere

I've been able to find the "washer method" of calculating a volume independently, but this alternative method of shells really baffles me. Although I understand how to derive the shell method's formula from the washer method's, deriving the formula for the shell method from scratch has really been a burden. It is only now that I realize my personal conjecture that integrating a lateral area gives a volume is true (sorry if I'm being vague with details). Now, being so accustomed to 2D Euclidean geometry, this concept seems slightly abstract to interpret geometrically. The way I see it, we can have a straight line y=k in 2 dimensions rotated about the x-axis to give a dimensionless cylinder (at least in it's thickness). Integrating this basically shades the entire region within the cylinder. But mathematically, what does this mean in terms of functions? I'm pretty sure a function in 3 dimensions must be a function of at least 2 variables, which would complicate things beyond my current knowledge of calculus. [b]Am I right in saying that proving the integral of an area to be a volume in some rigorous manner requires multivariable calculus?[/b]

Volume in single-variable calculus is pretty limited to revolving around an axes.  To truly calculate volumes, you need multivariable calculus because you are right, a function in 3 dimensions requires more variables.  This is because a single-variable function ($ y\equal{}x^{2}$ for example) is a single plane while $ z\equal{}x^{2}\plus{}y^{2}$ is a surface.  By definition, $ \text{Vol(S)}\equal{}\iiint_{S}\ dV$ where $ S$ is some surface.  If you want to know more, just post again and I will try to answer your questions.

I've been thinking about this lately, and have come up with the following conjecture:

The volume of any solid shape can be defined as $ V(x) \equal{} \int_{a}^{x} f(y) \; \Bigg( \int_{b}^{c} \; g(z) \; dz \Bigg) \; dy$

For instance, let's suppose you want a shape such that its cross section is given by $ f(z) \equal{} z^{2}$ (where $ b \leq z \leq c$) and whose height varies according to $ g(y) \equal{} \sin(y)$ (where $ a \leq y \leq x$ ) then the resulting volume would be:

$ V(x) \equal{} \int_{a}^{x} \frac {1}{3}\sin(y)(c^{3} \minus{} b^{3}) \; dy$

Is there any validity to this conjecture?

Other then the clumsy notation of re-using the letter $ f$, you're inching toward the development of double integrals. The full development of that includes what to do if the region you're integrating over isn't a rectangle.

[quote="Kent Merryfield"]The full development of that includes what to do [b]if the region you're integrating over isn't a rectangle.[/b][/quote]

I don't understand what you mean by this. Do you mean the formula I gave only applies to volumes defined on cartesian planes, and doesn't apply to volumes on hyperbolic planes, for instance...?

I also forgot to mention I don't quite get the notation used in defining volume as $ \text{Vol(S)} \equal{} \iiint_{S} dV$ For one, I'm not sure why there's 2 improper integrals followed by an integral with no upper bound but a lower bound $ S$. Moreover, if V represents volume, what purpose does it serve in using V as your differential? Maybe someone could give a simple example of such a formula being applied...?

I should probably start independently learning multivariable calculus rather than just asking questions about things that are currently beyond my level of understanding.

That is basically notation for you.  $ S$ is the surface you're integrating over.  Once you know what $ S$ is, then you fill in your boundaries.  For example, if $ S$ is the unit sphere, then it's pretty easy to determine your region of integration.  You'd have

$ \int_{\minus{}1}^{1}\int_{\minus{}\sqrt{1\minus{}x^{2}}}^{\sqrt{1\minus{}x^{2}}}\int_{\minus{}\sqrt{1\minus{}x^{2}\minus{}y^{2}}}^{\sqrt{1\minus{}x^{2}\minus{}y^{2}}}\ dz\ dy\ dx$

Of course, it's a lot easier to just put it into spherical coordinates.

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

I still haven't given up on this, and I think I figured out what was making my little conjecture incomplete.

$ V(x) \equal{} \int_{a}^{x} f(y) \Bigg( \int_{b(y)}^{c(y)} g(z) \; dz \Bigg) \minus{} h(y) \Bigg( \int_{b(y)}^{c(y)} i(z) \; dz \Bigg) \; dy$

One of the problems with the original statement $ V(x) \equal{} \int_{a}^{x} f(y) \; \Bigg( \int_{b}^{c} \; g(z) \; dz \Bigg) \; dy$ is that it doesn't apply the notion of "areas between curves" to volumes, so it can only really find the volume of shapes with a flat bottom surface. Moreover, having $ b$ and $ c$ as constants also guarantees the sides of the shape will be flat, so the potential for finding volumes of various shapes is actually quite limited in this formula.

Still, I have to wonder, what do you do if you want to integrate such that you don't cut off a function with two vertical lines, but rather have the cut-off be its own unique function. Does this require computing three separate integrals, or is there another way to do this?

I also applied the same idea for volumes in trying to formulate a formula for the lateral surface area and base area of a solid.

$ LSA(x) \equal{} \int_{a}^{x} \Bigg( \int_{b(y)}^{c(y)} \sqrt {1 \plus{} (f'(z))^{2} } \plus{} \sqrt {1 \plus{} (g'(z))^{2} } \; dz \Bigg)$

$ \plus{} \; h(y)( f(b) \plus{} f(c) ) \minus{} i(y)( g(b) \plus{} g(c) ) \; dy$

*The area of the base is found by removing the first integral sign and plugging in $ y \equal{} a$

A more convenient expression would be to say something like $ LSA \equal{} \int_{a}^{b} p(x) \; dx$ where $ p(x)$ is the perimiter of the cross section of the solid in question.

[quote="Prime factorization"][quote="Kent Merryfield"]The full development of that includes what to do [b]if the region you're integrating over isn't a rectangle.[/b][/quote]

I don't understand what you mean by this. Do you mean the formula I gave only applies to volumes defined on cartesian planes, and doesn't apply to volumes on hyperbolic planes, for instance...?[/quote]
Just as a definite integral on the line is defined with respect to an interval, a definite (double) integral on a plane is defined with respect to a region.  The things about regions (even convex ones) is that, unlike intervals, they can be specified by nearly arbitrary functions, and the integrals you've written so far only work for certain very specific types of regions.

You are correct that you should probably learn multivariable calculus proper.  The definitions and development of the double and triple integrals are exactly what you're looking for.


geometry
3D geometry
pyramid
trigonometry
analytic geometry
geometry proposed

Prove that the volume of a triangular pyramid, in general case, is given by

$V={abc\over 3}\sqrt{\sin\phi\sin(\phi-\alpha)\sin(\phi-\beta)\sin(\phi-\gamma)}$

where $a,b,c$ are the side edges; $\alpha, \beta, \gamma$ are the angles between them; $\phi=\frac{\alpha+\beta+\gamma}{2}$

Sorry, but a,b,c are which edges? Also which angles are they?...

If the base is $\triangle ABC$ and the tip is $T$, then $a=TA, b=TB, c=TC, \alpha=\angle BTC,\beta=\angle ATC,\gamma=\angle ATB$.

Source: A friend (from a book, don't know which one).
Solution: Available

Seeing as no one is posting, why don't you post the derivation. I'm bad at geometry myself, and would like to see it.

Yes, could we find the solution? I'm in need of it as well  :lol:

[hide="Two hints"]1. Prove:
\begin{eqnarray*}4\sin\phi\sin(\phi-\alpha)\sin(\phi-\beta)\sin(\phi-\gamma)=1-\cos^{2}\alpha-\cos^{2}\beta-\cos^{2}\gamma+2\cos\alpha\cos\beta\cos\gamma\end{eqnarray*}

2. Show that the proof can be reduced to the case $a=b=c=1$ and then use coordinate geometry.[/hide]


Circles $O$ and $P$ are tangent externally at point $T$. Chords $TA$ and $TB$ of circle $O$ are extended through $T$ to meet circle $P$ again at points $D$ and $C$ respectively.  If $TA=3$, $TB=5$, $TD=7$, compute $TC$.

i got the answer
[hide]$TC= \frac{35}{3}$[/hide]

Could you explain how you got that?[/list]


I can't update my profile (the page with the password, email, AIM, signature, etc.)
I click on it and it shows a page with only the Art of Problem Solving bar at the top and the [u]Profile/Private Messages/Faq/Memberlist/Search/AoPS Blogs/Contest/My Blog/Usergroups/Logout/Forum Index[/u] BAR underneith it. There is no links on the left side.
My Windows Explorer is version 6.0.2900.2180.xpsp_sp2_gdr.050301-1519.

I'm using AoPS skin, not Mathlinks.
I can't switch to Mathlinks because my profile edit doesn't work.

Make a print screen and attach it here (make sure you make it into a reasonable size before that, compress it into a gif or something :) ).

I'm using AoPS style right now and I see no problems :|

Yeah, I asked for a .gif and you give me a .zip with a .html inside  :| Anyway the file is ok. Why can't you click on profile?!

What browser are you using?

He said in his first post:

[quote]My Windows Explorer is version 6.0.2900.2180.xpsp_sp2_gdr.050301-1519.[/quote]

The problem was that you had an username with 2 letter, which triggered errors in the displaying functions. You will have to tell me another username you would like, for now tz1 is the one you can use.

Problem fixed. Any user who has this problem (profile link not working, displaying an half empty page) please pm or email me.


trigonometry
inequalities
geometry
circumcircle
function
inequalities unsolved

[quote="mahbub"]Well, I am facing some problems with the Section 1.1 of kiran kedlaya’s note. Here are those problems:

6.3 $\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}\leq \frac{1}{8}$

8. Let $x_1, x_2, \cdots$ be a sequence of positive real numbers. If $a_k$ and $g_k$ are the AM and GM respectively of $x_1, x_2, \cdots x_k$ prove that

$\frac{a_n ^n}{g_k ^n} \geq \frac{a_{n-1} ^{n-1}}{g_k ^ {n-1}}$

I am a bit suspicious about the correctness of problem 8.

Anyhow, I would be glad if some one give the solutions for these two problems![/quote]
x,y,z are reals
you can factorise:
cosx+cosy+cosz-cos(x+y+z)
then:  use x+y+z=pi and then you can find

$\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}= \frac{cosA+cosB+cosC-1}{4} \leq \frac{1}{8}$
$ \frac{cosA+cosB+cosC-1}{4} \leq \frac{1}{8}$

$ cosA+cosB+cosC \leq \frac{3}{2}$
try to prove this

We have:
$a_n=\frac{x_1+\cdots+x_n}{n}=(n-1)\frac{x_1+\cdots+x_{n-1}}{n(n-1)}+\frac{x_n}{n}$
Using weighted AM-GM we have:
$(n-1)\frac{x_1+\cdots+x_{n-1}}{n(n-1)}+\frac{x_n}{n}\geq n\sqrt[n]{\frac{(x_1+\cdots+x_{n-1})^{n-1}x_n}{n^{n-1}(n-1)^{n-1}n}}=\sqrt[n]{a_{n-1}^{n-1}x_n}$
Together:
$a_n\geq \sqrt[n]{a_{n-1}^{n-1}x_n}$
It implies:
$a_n^n\geq a_{n-1}^{n-1}x_n$
But we can easily see that $x_n=\frac{g_n^n}{g_{n-1}^{n-1}}$
So we get:
$a_n^n\geq a_{n-1}^{n-1}\cdots\dfrac{g_n^n}{g_{n-1}^{n-1}}$
And it's equivalent to:
$\frac{a_n ^n}{g_k ^n} \geq \frac{a_{n-1} ^{n-1}}{g_k ^ {n-1}} $

Thanks! Eh those were easy! Actually after seeing the solutions it always seems ez :P

Another nice proof:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cos&t=21188

The second inequality is Rados??? The first one nicely solved by Jensen's or Euler's result .

Bomb

It's easy but i wonna send a geometric solution.
Let r and R be the radii of inscribed-circle and circumcircle of triangle ABC. Let K is the foot of altitude from A and
a=BK+KC.Thus,
BK/r =ctgB/2 and KC/r=ctgC/2  then a=BK+KC=r.[ctgB/2 + ctgC/2]=[r.cosA/2]/[sinB/2.sinC/2] and
using the law of sine : A=4R.cosA/2.sinA/2= [r.cosA/2]/[sinB/2.sinC/2] so
........................sinA/2. sinB/2. SinC/2 =r/4R

An easy method for proving  cosA+cosB+cosC<=3/2

cosA+cosB=2cos[(A+B)/2] cos[(A-B)/2]=2sin[C/2]cos[(A-B)/2] and cosC=1-2sin^2[C/2]

So cosA+cosB+cosC=2sinC/2 .[ cos(A-B)/2 -sinC/2]+1<2sinC/2 .[ 1-sinC/2]+1<3/2

Equality holds for A=B=C=60

well my idea to solved the cosine inequality is just to use Cosine law :P

The second one can also be derived by Heron's in combination with other stuffs.

Bomb

another easy way to prove $cosA+cosB+cosC \le \frac{3}{2}$ is that it is equivalent to $(cosA+cosB-1)^2+(sinA-sinB)^2 \ge 0$

u know there are a lot of proves of that inequality in Angel :P

an easy way to prove $\cos{A}+\cos{B}+\cos{C}\le \frac{3}{2}$ is to use Jensens.

unfortunately jensen will not work so well. because cos function is not entirely convex or concave in [0, pi]. So if u wanna use jensen u need to take care of special case ;)


Ok matheads, here's a question for you.

I'm doing a math project on music. Specifically, I'd like to end up with a core formula containing a variable into which I would plug different values. The resulting formula would spit out a sequence of notes which I would then attribute to different tracks - melody, bass, percussion. You can foresee the problem.. how can I translate that sequence of numbers into actual notes? MIDI, right? but how? I have access to Finale 2007 (thanks John) and Ableton Live. Any thoughts on this?  Maybe a formula isn't necessary, maybe I can just decide to feed in different number sequences into Finale or Ableton and see what happens...

I'll be popping in and out for the next few days so feel free to give this deep thought :)

I am not clear about your wording, but there are some applications that can analyze music from one composer and produce a series of numbers that correspond to various properties (EMI).  See http://www.wnyc.org/shows/radiolab/episodes/2006/04/21 at the bottom.  The information can then be used to make another similar piece.

wow that sounds cool. i'll def check it out. not looking to copy notes though. it's hard to explain.. essentially i'm just trying to convert a sequence of numbers into musical notes. one sequence of numbers would correspond to one track, and so on..

In my humble opinion, I think that the beauty of music is its incompatibility with mathematics. I'm sure you can create a decently sounding track with some type of series or function but masterpieces are created with an uncomprehensible randomness and elegance that makes music so much more special. I just think that your project needs to mention this  :lol:

I encourage you to keep searching for a function but it might be harder than you think.  :wink:

[quote=xdl615x]In my humble opinion, I think that the beauty of music is its incompatibility with mathematics.[/quote]
I agree with him, but the same goes for paintings where a good painting will beat any fractal but fractals are still kinda cool.

I thought about doing this at one point, but it never materialized.  I was going to try to use a series and different series woudl offer different sequence of notes, I think to have a few number you would want a computer to generate the notes based off an algorithm, b/c it would probably be a lot of add/subrtacting/multipling/dividing just to do a hand ful of notes.  I would decide on a scale first, I'd strongly reccomend a diatonic scale at first.  You'll need a way to play it too, a midi player would work if you can create your own midi files.

Good Luck


algebra
polynomial
factorial
number theory
number theory proposed

Find all complex polynomials $f$ with the property that the equation $ f(x)=n!$ has at least a rational root for all natural numbers $n$ greater than 2005.

Let $ f(x) \equal{} \sum_{i \equal{} 0}^sa_ix^i$ where $ a_i\in\mathbb{C}$, and let $ \{1\}\cup\{c_{\lambda}: \lambda\in\Lambda\}$ be Hamel basis of $ \mathbb{C}$ over $ \mathbb{Q}$. Let $ a_i \equal{} r_{i,0} \plus{} \sum_{\lambda\in\Lambda}r_{i,\lambda}c_{\lambda}$, where $ r_{i,0},r_{i,\lambda}\in\mathbb{Q}$, and all but finitely many of $ r_{i,\lambda}$ are equal to $ 0$. Then $ f(x) \equal{} \sum_{i \equal{} 0}^s\left(r_{i,0} \plus{} \sum_{\lambda\in\Lambda}r_{i,\lambda}c_{\lambda}\right)x^i \equal{} \sum_{i \equal{} 0}^sr_{i,0}x^i \plus{} \sum_{\lambda\in\Lambda}c_{\lambda}\left(\sum_{i \equal{} 0}^sr_{i,\lambda}x^i\right)$. The equation $ f(x) \equal{} n!$ is now equivalent to $ \sum_{i \equal{} 0}^sr_{i,0}x^i \equal{} n!$ and $ (\forall\lambda\in\Lambda)\left(\sum_{i \equal{} 0}^sr_{i,\lambda}x^i \equal{} 0\right)$. If there were $ \overline{\lambda}\in\Lambda$ such that $ \sum_{i \equal{} 0}^sr_{i,\overline{\lambda}}x^i\not\equiv 0$, the equation $ \sum_{i \equal{} 0}^sr_{i,\overline{\lambda}}x^i \equal{} 0$ would have only finitely many solutions in $ x$; therefore, the expression $ \sum_{i \equal{} 0}^sr_{i,0}x^i$ would assume only finitely many different values, and therefore there would be no way for $ \sum_{i \equal{} 0}^sr_{i,0}x^i \equal{} n!$ to have a solution in $ x$ for each $ n > 2005$. Therefore, $ (\forall i)(\forall\lambda\in\Lambda)(r_{i,\lambda} \equal{} 0)$.

In the other words, all the coefficients $ a_i$ must be rationals. If $ s \equal{} 1$, then for each $ n > 2005$ the equation $ a_1x \plus{} a_0 \equal{} n!$ has a rational solution $ x \equal{} \frac {n! \minus{} a_0}{a_1}$, and therefore all rational polynomials of degree $ 1$ satisfy our condition. Assume, therefore, $ s\geqslant 2$. Multiplying the polynomial with their common denominator, say $ b$, leaves us with the equation $ \sum_{i \equal{} 0}^s\tilde{a}_ix^i \equal{} bn!$, which, by assumption, has a solution in rationals for every $ n > 2005$. This is the same as stating for each $ n > 2005$ exist coprime $ p,q\in\mathbb{Z}$ such that $ \sum_{i \equal{} 0}^s\tilde{a}_i\frac {p^i}{q^i} \equal{} bn!$, or equivalently, $ \sum_{i \equal{} 0}^s\tilde{a}_ip^iq^{s \minus{} i} \equal{} q^sbn!$. From this follows that $ q\mid\tilde{a}_s$. Since $ \tilde{a}_s$ has only finitely many divisors, it follows that there is only finitely many possibilities for $ q$. Divide all integers $ n > 2005$ into classes accordingly to which of these possibilities corresponds to $ n$. Since there is only finitely many classes, at least one of them must have positive density. From now on, we should fix $ q$ to the one corresponding to this class. If we denote $ A_i \equal{} \tilde{a}_iq^{s \minus{} i}$ and $ B \equal{} q^sb$, the condition is reduced to the statement that the set of $ n\in\mathbb{N}$ such that the equation $ \sum_{i \equal{} 0}^sA_ip^i \equal{} Bn!$ has an integer solution in $ p$ is of positive density. However, we now cite the paper of D. Berend and C. F. Osgood, On the equation $ P(x) \equal{} n!$ and a question of Erdős, [i]J. Number Theory[/i] [b]42[/b] (1992), 189–193:
[quote][b]THEOREM 1.[/b] Let $ P$ be a polynomial of degree $ \geqslant 2$ over $ \mathbb{Z}$. Then the set of positive integers $ n$, for which the equation
\[ P(x) \equal{} n!
\]has an integer solution $ x$, is of density zero.

...

[b]THEOREM 1’.[/b] In the setup of Theorem 1, the same conclusion is true if (1.1) is replaced by
\[ P(x) \equal{} s\cdot n!,
\]$ s$ being a fixed non-zero integer.[/quote]
Therefore, our assumption contradicts [i]Theorem 1'[/i] of this paper. In the other words, the polynomials we are asked for are precisely all rational polynomials of degree $ 1$. The problem is solved.


.........when you click on alcumus.

Alcumus is currently undergoing maintenance as we upgrade to a new system.
We do not have an estimated time for Alcumus to be back online. In the meantime, you can view all the Alcumus related videos.

details any1?

I'm not exactly sure...but I sure hope they are adding in new problems, lessons from different topics, like Number theory, Algebra, Geometry, etc  :D

i hope that they are making it less complicated to see, it is complicated to understand some stuff.


On the index, we all know that if there are new posts, the icon changes color. When I click on a forum that has new topics, new topics are obviously shown. If I view them all (or click mark all posts viewed) and then click on the Forum link at the top of the site, I still get the new posts icon. Similarily, I sometimes log in, and the site displays me as not logged on. If I click on a forum however, it automatically "logs" me back on. Or in another similar case, I click on a forum, just to find out then I am not logged on.

Opera 9.5 I think
Windows (XP)

[quote="Dojo"]This error is still occuring. I tried to click on the login button, but it redirects me to http://www.artofproblemsolving.com/Forum/index.php.php, which I believe is a 404.[/quote]It's definitely an issue on your side. Try clearing the cookies and/or using another browser.

If I'm logged in, and go back (in my history) to the login page, the same thing happens.

[quote="Dojo"] If I view them all (or click mark all posts viewed) and then click on the Forum link at the top of the site, I still get the new posts icon. [/quote]

This happens to me sometimes also.

Sometimes, an error like this is still happening. I log in, go to the index, and it looks like I'm not logged in... there's no My blog icon and there still is a Login icon at the top. However, when I go into any forum I get logged back on.

I have a issue like Dojo's. Whenever I log on and go to any different page (Community, FTW, Alcumus), then it acts like I haven't logged in. (Not able to chat, login button still there, not able to post). The strange thing is, this problem is only on my mom's laptop, but not in my computer.

hmmm, this has happened to me once before too... and louiethechewy123, make a  new topic for that

Louie - Make sure cookies are enabled on your mom's computer. Internet security software can interfere with cookies as well, so disable that stuff. Contact your computer manufacturer or internet security software developer for information on how to disable security software.

Dojo - Probably just your cache making it look like you're not logged in (browser not reloading page)

Well I cleared my cache and went into a forum, and clicked back on the forum tab. It doesn't seem to happen regularly, but occasionally, it does.

If you can't constantly recreate the "bug" then it's hard to impossible for us to fix it (if there is anything to fix here.)


Suppose $n \in R$ such that $(\frac{sinx}{x})^{n}>cos x$ $\forall 0<x<\frac{\pi}{2}$
Show that: $n \leq 3$

[quote="Rust"]If x smal and n>3 ${1-\frac{x^{2}}{2}+O(x^{4}}=cosx.$[/quote]
why? I don't understand it. :blush:

If n>3  $1-\frac{nx^{2}}{6}<1-\frac{x^{2}}{2}$, therefore for small x $(\frac{sinx}{x})^{n}<cosx$.


Posted by: Valentin Vornicu @ Sun Feb 08, 2004 2:55 pm:
[quote]
Hi folks!

Just a remainder that the MathLinks Contest 3rd Edition, Spring 2004 is due to begin on the 22th of February 2004. More about it, in HERE
[/quote]

Do you mean remainder or reminder?  :)


[b]I was just wondering if AMSP provides transportation to campus from San

Francisco int'l airport on the first day, and to the airport from campus on the

last day.[/b]

I'm an international student, so I have to book flights in advance...and I have

to match the flight schedule to the transportation schedules.....

If there is no transportation avaliable, well....guess there arises a serious problem :(

What was the case in the past??

Yes, transportation is provided.  There were people arriving from 11 AM to 6 PM last year.  Earlier and later as well, but mainly by 6.  Don't worry too much about it if you already booked.  AFAIK, I went last year, if you send in your flight information, then you shouldn't be left behind.
Edit: ... AIME15 is right.  Sorry for not being helpful.

I think he is asking this because it says that you are supposed to go to San Jose International Airport, not San Francisco.

15 days past from his post so I assume he already solve the situation.

Toronto Airport Limo is a good taxi service which help the US people a lot while they are in very hurry to reach their destination..Or sometimes to catch their plan quick.Airport Limos caters to all your special requests.Toronto Airport Limo mainly concentrates on the airport pick up and drop and they are considered to be experts as they are the masters of airport travel.I liked this site as this tells about the transportation fact of world..And if you guys want well qualities transportation to the airport you can have some sites like this one. "http://www.airportslimo.com".
Plan your trip anytime with Airport Limo.So,ready to let the good times roll and make sure that your journey is a safe one.


For every natural number $ n$, find all polynomials $ x^2\plus{}ax\plus{}b$, where $ a^2 \ge 4b$, that divide $ x^{2n}\plus{}ax^n\plus{}b$.

If n=1, then a,b any numbers,
If $ b\equal{}0$, then $ a^{n\plus{}1}(a^{n\minus{}1}\plus{}(\minus{}1)^n)\equal{}0$, therefore $ a\equal{}0$ or , $ a\equal{}\minus{}1$ or $ a\equal{}1, n\minus{}odd$.
If $ b\not \equal{}0$, then from $ (x_1^n,x_2^n)\in (x_1,x_2)$ roots we get $ |x_1|\equal{}|x_2|\equal{}1$ we get $ b\equal{}\pm 1$.
If $ x_1\equal{}x_2\equal{}c\equal{}\pm 1$, then $ (x\minus{}c)^2|(x^n\minus{}c)^2\to$ $ a\equal{}\minus{}2,b\equal{}1$ or $ a\equal{}2,b\equal{}1, n\minus{}odd$.
If $ x_1\equal{}1,x_2\equal{}\minus{}1, a\equal{}0,b\equal{}\minus{}1$.


articles
geometry
geometric transformation

Dear chinese people i 've got some questions i have been wondering for quite a long time:
- How do you type chinese characters? Do you just have really huge keyboards or do you type in pinyin (or however chinese in latin characters is called) en let the computer pick out the right character? And doesn't that goes way more slowly then just typing in english?
- Is it true that it is difficult to find a spot in china where you have acces to the internet without having a government-sent watchdog at your side watching every move you make? Those are the stories told abot china here in europe and i have always wondered if they could be true.

grtz
(please reply in english :P )

(1) We type pinyin

(2) Actually, people do not care if the government is watching or not. There are mountains of articals criticizing the government in the forums of nearly every Chinese site.

[quote="K.W.M.A.N"] Do you just have really huge keyboards?[/quote]
You can never expect the invention of a keyboard with more than 10000 buttoms...

[quote="lccs"][quote="K.W.M.A.N"]
- Is it true that it is difficult to find a spot in china where you have acces to the internet without having a government-sent watchdog at your side watching every move you make? Those are the stories told abot china here in europe and i have always wondered if they could be true.
[/quote]
西方真是把中国"妖魔化"了……[/quote]
Of course it's just exaggeration. However, there are really people sentenced imprisonment because of posting articles criticizing the Chinese government in some forums.

[quote="mecrazywong"]10000 buttoms...[/quote]Thank You :harhar:

[quote="mecrazywong"]It was reported in Hong Kong several months ago, and the excuse for this is something like "危害國家安全，洩露國家秘密，顛覆國家政權，破壞國家統一"...[/quote]

I remember this piece of news is related to yahoo as yahoo revealed the content of the emailbox of an user to help the investigation of the case?

[quote="K.W.M.A.N"]Dear chinese people i 've got some questions i have been wondering for quite a long time:
- How do you type chinese characters? Do you just have really huge keyboards or do you type in pinyin (or however chinese in latin characters is called) en let the computer pick out the right character? And doesn't that goes way more slowly then just typing in english? [/quote]
There are many different methods to enter Chinese characters. In Hong Kong, the most popular way is called Cang Jie, and a simplified version of it. The main idea is to break a character into several parts, and each English alphabet represents a list of the different parts.

[quote="lccs"][quote="shobber"]Really!!?? I don't know it....[/quote]
The same to me.... :roll:[/quote]
當然了，這些報道全都給壓下來...

[quote="shobber"](2) Actually, people do not care if the government is watching or not. There are mountains of articals criticizing the government in the forums of nearly every Chinese site.[/quote]
The worst thing that I've found so far is that the government block wikipedia.

[quote="shobber"]Yep. I've heard about that... What does the Wikipedia contain to make the government do that?[/quote]
There are lots of stuffs on Wikipedia like this:
六四事件
維基百科，自由的百科全書

由北京中央美術學院製作的「民主女神」塑像，成了1989年中國民主運動後期的象徵。原件被毀，在溫哥華、舊金山等城市有複製品立在公共場所。六四事件，也稱六四民運、八九學運、六四風波、六四運動、六四屠城、八九學潮、天安門事件，簡稱六四，是1989年4月15日至6月4日間及其後一場發生在中國大陸的政治事件，以大規模的學生、民眾的較為平和的遊行、示威運動開始，但是學生團體和政府之間在交涉中未能達成共識和政治妥協，最後以軍隊武力鎮壓告終。一般認為事件的中心是北京的天安門廣場，此外上海等城市也在這一期間進行了各種政治主張的示威遊行。在學生遊行中一些城市發生了打砸搶等暴力事件，許多鐵路被阻斷以致交通中斷。一般政治評論家認為此次事件導致了自1978年以後中華人民共和國的政治改革步伐的停步甚至倒退。而事件中的眾多爭論也尚未解決。

事实上当时全国都在乱。学生都嚷着要上街游行，但其实其中有多少是被利用的就不得而知了。

[quote="leepakhin
"]當然了，這些報道全都給壓下來...[/quote]
可是我需要这些报道干什么呢?我周围从来没发生这些所谓的"报道"一类事
不知居心何在?
意识形态的事难道真要分出个你死我活?

[quote="ifai"]怎麼由"how to type chinese"扯到政治來?[/quote]
This is his second question.

Well, as far as what i know, http://www.voanews.com , plus all kinds of other American websites and many Taiwan's are blocked. And since my cousin(actually, he's a funny hacker) hate them, he once revealed the methods of how to un-blocking those websites, but I don't have time to read his articles. And since I'm canada now, no real need for me to do those stuffs within a few years, haha.
Chinese government choose to make everybody silence to what happened on the day of 1989.6.4, but, remember this sentence:
[b]
You can fool all of the people some of the time, and some of the time all of the time, but you CANNOT fool ALL of the people all of the time --
by Abraham Lincoln in 1863, Washington D.C.[/b]
Anyways, I hope that Prime Minister Hu Jintao can recognize this point and make it into real process.
And we don't really need to talk about this kinda topic at this time of the year.
Merry Christmas.
Happy New Year! New year, new wishes, new Hope.
Frank

[quote="Kalimdor"]Well, as far as what i know, http://www.voanews.com , plus all kinds of other American websites and many Taiwan's are blocked. And since my cousin(actually, he's a funny hacker) hate them, he once revealed the methods of how to un-blocking those websites, but I don't have time to read his articles. And since I'm canada now, no real need for me to do those stuffs within a few years, haha.
Chinese government choose to make everybody silence to what happened on the day of 1989.6.4, but, remember this sentence:
[b]
You can fool all of the people some of the time, and some of the time all of the time, but you CANNOT fool of the people all of the time --
by Abraham Lincoln in 1863, Washington D.C.[/b]
Anyways, I hope that Prime Minister Hu Jintao can recognize this point and make it into real process.
And we don't really need to talk about this kinda topic at this time of the year.
Merry Christmas.
Happy New Year! New year, new wishes, new Hope.
Frank[/quote]
I think you have made a little typo.  ;) Don't you mean
[b]
You can fool all of the people some of the time, and some of the people all of the time, but you CANNOT fool of the people all of the time --
by Abraham Lincoln in 1863, Washington D.C.
[/b]

Ooooooooops! I'm really sorry
Missing the most important word in the whole post sounds really bad.
I'll correct it.
Merry Christmas Again!
Frank

哎,不想再多说什么了,人就是这样，斗争为乐
 ;) I don't want to talk about it any more.

[b]EDIT:[/b] Good idea, Kalimdor. I have inserted "...!x!..." in a number of words which could be of danger, particularly into the link - remove the "...!x!..." from the link in order to make it work!

For everyone interested in that topic, I would recommend http://cyber.law.harva...!x!...rd.edu/filtering/China , where you can find, in particular, a long list of bloc...!x!...ked sites. I am copying part of the article here since I can guess the above mentioned site is bl...!x!...ocked in China as well (I am too lazy to check):

[quote][u][b]Analysis & Summary Statistics[/b][/u]

Among the specific bl...!x!...ocked pages are the following categories of content:

[b]Dissident/democracy sites.[/b] bl...!x!...ocked sites includes sites about democracy and human rights generally and sites specific to China. Of the top 100 sites returned by Google in response to a search for "democracy China," 40 were found to be bl...!x!...ocked, while 37 "dissident China" sites were bl...!x!...ocked, 32 were bl...!x!...ocked for "freedom China," and 30 for "justice China." Specific bl...!x!...ocked sites included Amnesty International, Human Rights Watch, the Hong Kong Voice of Democracy, the Direct Democracy Center, and dozens of Fal...!x!...un Gong and Fal...!x!...un Dafa sites.
[b]Health.[/b] bl...!x!...ocked sites included sites about health generally and about health in China specifically. Of the top 100 Google results for "hunger China," 24 were bl...!x!...ocked; for "famine China" 23; for "AIDS China" 21; for "sex China" 19; for "disease China" 14. Specific bl...!x!...ocked sites included the AIDS Healthcare Foundation, the Internet Mental Health reference, and the Health in China research project. We found bl...!x!...ocking of a total of 139 sites listed in Yahoo's Health directory categories and subcategories.
[b]Education.[/b] bl...!x!...ocked sites included a number of well-known institutions of higher education, including the primary web servers operated by Caltech, Columbia, MIT, and the University of Virginia. bl...!x!...ocked non-university sites included the Learning Channel, the Islamic Virtual School, the Music Academy of Zh...!x!...eng, and the web sites of dozens of public and private primary and secondary schools. We further found evidence of bl...!x!...ocking of 696 sites listed in Yahoo's Education directory categories and subcategories.
[b]News.[/b] The BBC News was consistently unreachable, while CNN, Time Magazine, PBS, the Miami Herald, and the Philadelphia Inquirer were also often unavailable. Of Google's top 100 results for news, 42 were bl...!x!...ocked. We further found evidence of bl...!x!...ocking of 923 sites listed in Yahoo's News and Media directory categories and subcategories. Nonetheless, some news sites that were previously bl...!x!...ocked became accessible during the course of our testing; for example, Reuters was bl...!x!...ocked through April 29, but was subsequently accessible, while the Washington Post was bl...!x!...ocked through May 6 and was subsequently accessible. This reduction in bl...!x!...ocking of entire news sites may reflect that certain new filtering technologies (discussed in greater detail in the appendix) allow bl...!x!...ocking only of the particular sections and articles that are particularly controversial in China. As a result, our results should not be taken to suggest that every Washington Post article is now accessible in China.
[b]Government sites.[/b] bl...!x!...ocked sites included a variety of sites operated by governments in Asia and beyond. As discussed below, government sites of Taiwan and Tibet were targeted specifically. Also bl...!x!...ocked was the entirety of uscourts.gov, including the many federal district and appellate courts in the United States, as well as the United Kingdom's Court Service and Israel's Judicial Authority. The communication sites of various governments were bl...!x!...ocked, including the United States' Voice of America, as well as travel sites from Australia, Israel, Korea, Switzerland, and Wales. Government military department sites were also bl...!x!...ocked, including the US Department of Defense, though others remained reachable (the CIA). A variety of additional government sites were bl...!x!...ocked, without manifest pattern, both in the United States and beyond; examples include the site of Seattle's King County, the main Australian Federal Government index site, the Philippines Bureau of Customs, the British Insolvency Service, the Office of the Governor of Makkah in Saudi Arabia, and the Legislative Assembly of British Columbia. bl...!x!...ocked sites included 516 sites in Yahoo's categories and subcategories pertaining to governments.
[b]Taiwanese and Tibetan sites generally.[/b] bl...!x!...ocked sites included business sites (like the A&D Company of Taiwan), non-commercial sites (the Taiwan Health Clinic and a total of 709 .edu.tw sites, as well as the Voice of Tibet), and government sites (the Office of the President of Taiwan and the Taiwanese Parlia...!x!...mentary Library among 936 other Taiwanese government sites, and the Official Website of the Tibetan Government in Exile). More than 60% of Google's top 100 "Tibet" sites were found to be bl...!x!...ocked, and more than 47% of the top "Taiwan" sites were bl...!x!...ocked. Taiwanese content was also bl...!x!...ocked disproportionately, relative to its representation in our testing sample; fully 3,284 .TW sites (13.4% of .TW sites tested) were bl...!x!...ocked, while our overall bl...!x!...ock rate was approximately 9.3%. (Of course, comparisons of bl...!x!...ock rates must be performed with care given the subjective formation of the list of sites tested. For lack of a domain name specifically associated with Tibetan sites, it is more difficult to perform such a comparison on the bl...!x!...ock rate of Tibetan content.)
[b]Entertainment.[/b] bl...!x!...ocked sites included the movie Deep Impact, the Canadian Music Centre, the Taiwanese site of MTV (mtv.com.tw) and multiple sites providing off-color jokes. We also found bl...!x!...ocking of a total of 451 sites in Yahoo's categories and subcategories pertaining to Entertainment.
[b]Religion.[/b] bl...!x!...ocked sites included the Asian American Baptist Church, the Atheist Network, the Catholic Civil Rights League, Feng Shui at Geomancy.net, the Canberra Islamic Centre, the Jewish Federation of Winnipeg, and the Denver Zen Center. We found bl...!x!...ocking of a total of 1,763 sites in Yahoo's categories and subcategories pertaining to religion.[/quote]

I think these categories pretty much reveal the reasons behind the bl...!x!...ocking policy. However, you don't need to have anything to do with politics in order to get on the "black list": for instance, the cen...!x!...soring system completely bl...!x!...ocks the webspace provider Yahoo GeoCities and all sites on this provider, among them, e. g., my geometry website. I find it really strange that MathLinks is still allowed.

By the way, you can always find out whether some certain website is bloc...!x!...ked in Ch...!x!...ina by trying to access it from a Chinese pro...!x!...xy server. This, however, involes finding a functioning pro...!x!...xy server, what could take half an hour, so I usually don't bother.

As for circ...!x!...umventing the bl...!x!...ocking, this is surely ille...!x!...gal (in Ch...!x!...ina), but I heard it is not too difficult. Just today I posted [url=http://anonymouse.org/anonwww_de.html]a link to an anony...!x!...mization service[/url] which is currently not ba...!x!...nned. However, I am not quite sure it will remain so in future... There should be better ways in any case. I am a bad hacker (no, I'm not a hacker at all).

  Darij

Well, Darij, the link you gave us http://anonymouse.org/
is actually what my cousin thought about at first. The other way to do that (also my coz's thought) is to put some sort of "virus" to block the "cats" proposed by governments, disable them (fully or partially). I guess we should be able to do that.(for example, whatever a user is browsing, the cats can only see that they are browsing their homepage, and blahblahblah...)
And We can try to make a Chinese translation of the anonymouse.org, since most chinese can't read neither english nor deutch, unlike mathlinkers. In order to make this website useful for chinese people who want to know the outside world via internet, that's necessary. :roll:
Here's my translation: 世上有许多"老鼠"认为,它们在浏览www时 是安全的,隐蔽的, 可惜的是,事实并非如此. 每当你为了一片奶酪而冒险时, 你都会留下一张名片, 告知别人你的一切,当有许多只猫斗了解了你的时候,它们就会抓住你!
这项服务,允许了全世界广大网络"鼠"民在不被"坏猫"的跟踪下,从容的吃奶酪. That's it. :ninja:
Well,I think we have to avoid some important keywords in order notto be found by the chinese government through webpage searching. and to publish ads for this website, we can try google. a commercial link on google is good enough. :D
Frank

[quote="K.W.M.A.N"]- Is it true that it is difficult to find a spot in china where you have acces to the internet without having a government-sent watchdog at your side watching every move you make? Those are the stories told abot china here in europe and i have always wondered if they could be true.[/quote]
Actually i don't think u have to worry about the watchdog. It is just some automatic devices that can block visits to the websites not favored by the government.... if you aren't visiting those websites no one would know what you are doing online, and when u r trying to visit those webs, the worst outcome is that u can't access the the information u want, they won't accuse u for "attempting to hear".

Problems only occur on the people who try to post on forum and spread the information. That is, things only happen to "writers", not "readers".

the rumor is not true obviously...
cuz they cant watch 1.3 million ppl at the same time ^ ^
but it's partly true.
some sites are banned bcz they have sumthing to do with 六四JUNE FOURTH MOVEMENT...and now in SHENZHEN(my born place) there is a group called 警警&察察(online police ) who shield these sites.
for the typing part. i use 智能ABC which can type chinese characters through their latinized pronounciation 拼音

[quote="nirvanatear"]...and now in SHENZHEN(my born place) there is a group called 警警&察察(online police ) who shield these sites.
[/quote]
Well, yes.They came to being on Jan.2nd.

[quote="shobber"]Really!!?? I don't know it....[/quote]

That's because of the excellent Chinese censorship.  You should see American newspapers!


inequalities
inequalities proposed
algebra

[quote="nguoivn"]Given $ a, b, c$ are the real numbers satisfy $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Prove that:
\[ \frac {3 \plus{} a^2b^2}{(a \minus{} b)^2} \plus{} \frac {3 \plus{} b^2c^2}{(b \minus{} c)^2} \plus{} \frac {3 \plus{} c^2a^2}{(c \minus{} a)^2} \ge \frac {21}{2}\]
PS: Hope that you'll like it  :)[/quote]
The inequality follows from adding the two known inequalities
$ 2(a^2 \plus{} b^2 \plus{} c^2)\left[ \frac {1}{(b \minus{} c)^2} \plus{} \frac {1}{(c \minus{} a)^2} \plus{} \frac {1}{(a \minus{} b)^2}\right] \ge 9,$
and
$ \frac {1 \plus{} a^2b^2}{(a \minus{} b)^2} \plus{} \frac {1 \plus{} b^2c^2}{(b \minus{} c)^2} \plus{} \frac {1 \plus{} c^2a^2}{(c \minus{} a)^2} \ge \frac{3}{2}.$

Remark. The former was created by Dao Hai Long and the latter was created by Nguyen Van Thach.

Yes, you are right. It's a weak ineq  :oops:
Maybe the following is harder:
Given $ a, b, c$ are the real numbers. Prove that:
\[ \frac {1 \plus{} a^2}{(b^2 \minus{} c^2)^2} \plus{} \frac {1 \plus{} b^2}{(c^2 \minus{} a^2)^2} \plus{} \frac {1 \plus{} c^2}{(a^2 \minus{} b^2)^2} \ge \frac {2(a \plus{} b \plus{} c)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\]
PS: [hide]Anh ko dc lam` bai` nay` nua~ dau nha' :-w [/hide]:)

[quote=can_hang2007]
$$ \frac {1 \plus{} a^2b^2}{(a \minus{} b)^2} \plus{} \frac {1 \plus{} b^2c^2}{(b \minus{} c)^2} \plus{} \frac {1 \plus{} c^2a^2}{(c \minus{} a)^2} \ge \frac{3}{2}.$$[/quote]
[url=https://artofproblemsolving.com/community/c6h2254533p17393729]h[/url]


[quote="kaijin"]First of all let me state that I am not a studying Mathematics. But after reading Fermat's Last Theorem by Simon Singh, i have a new passion for numbers. i do have one questiion though. hope someone can answer it.

To count we use the number 1 then move to 2.

But between 1 and 2 there's 1.1, which (obviously) is bigger than 1.0 but smaller than 1.1 is 1.01 which is still greater than 1 but smaller than 2. This can go on infinitely as 1.000000000000000000000000000001 is smaller than 2 but still greater than 1 by a small (significant) margin.

this is more of a philosophical question. how do we move from 1 to 2 if we can infinitely go down decimal places that are greater rthan 1.0 but smaller than 2.
 does counting ignore the infinite gap between 1.0 and 2.0? or am i missing something simple?[/quote]
Okay, you can walk from point a to point b. But no matter how small of a step you take on the way, there is an even smaller possible step. You can walk one milimeter, micrometer, nanometer, picometer, femtometer or even one attometer.
But regardless, you can still walk from A to B.

Yeah, you skip a lot of stuff. There's nothing wrong with it. It's because when you count, you are only counting certain kinds of numbers (that is, natural numbers, 0, 1, 2, 3, ...). There are no natural numbers between 1 and 2, so you're not actually skipping anything when you're counting. You're skipping a bunch of other rational numbers (like 5/4) and real numbers (like $\pi / 3$), but those "don't count." ;)


Does any one know how the smiley limit works?
I think I have a smiley limit of two.  :mad:  :mad:

Yea me too. But I have seen some people put like 10 smilies in one post....Whats up with that?

Hi.  :alien:  :huh:  :starwars:  :stink:
[hide][hide][hide]Get me outta here!!!:rotfl:[/hide][/hide][/hide]
 :rotfl: [b]I love smileys!!![/b]

And yes. Dogs eat cheese. ([hide]Or at least my dog does.[/hide])

how does mbabbitt2003 do so much?

it can t be based on post ranks! gobraves would also have been able to do it.

Nor can it be based on join date...
 :mad:  :mad:

Hmm, I wonder if I have a limit...

Yes, my limit is 2 as well.  What's the deal?

 :rotfl:  :ninja:

:P  :P

EDIT: Dang, I can't edit them in either. The best way to get a bunch of smilies woud be to link to a large .gif of smiles.
EDIT AGAIN: Wait, you can. Just link directly to the image!
[img]http://www.artofproblemsolving.com/Forum/images/smiles/tongue.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smiles/tongue.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smiles/tongue.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smiles/tongue.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smiles/tongue.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smiles/tongue.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smiles/tongue.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smiles/tongue.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smiles/tongue.gif[/img]

MORE EDIT:
OBSERVE!
[img]http://logicfury.com/public/smilies/smilies.png[/img]

i think this topic should be locked because it is an easy topic to spam in.

When in the MATHCOUNTS forum, I get about 8 or 9.

I think it's because of the moderatorship.
 :rotfl:  :rotfl:

Yeah but Penguin, if you quote mbabbitt2003, you see his smileys were actual smileys, not IMGs

I can only have 2 too :(

The smiley number should primary depend on the number of letters in your post.
Thus just take more letters in it ;)
 :)  :lol:  :D  :roll:  :rotfl:

I really dont get this its kinda weird. I wonder what happens if i ramble on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on. will i have a bigger smiley limit?  :?  :?  :?  :?  :?  :?  :?

:ninja:  :ninja: blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah  :ninja: :ninja:

First I tried to post 80 smilies:[quote]Too many smilies in your post: 80.
You can have at most 5 smilies in this message. Please remove some smilies before submitting this message.[/quote]
After I added that message to my 80 smilie post, I got:[quote]Too many smilies in your post: 80.
You can have at most 7 smilies in this message. Please remove some smilies before submitting this message.[/quote]
I repeated and got higher smilie limits, so I think it does depend on the number of characters.
@ math 92: Games and Fun is the forum, so this is pretty much on topic. :D  :D  :D  :D  :D  :D  :D  :D  :D

[quote="Klebian"]How about this?

The test:
[hide][hide][hide]

WHAT a waste of time.

[/hide][/hide][/hide][/quote]

Wow.  I'm really glad I just hit "quote".

 :ninja:  :idea:  :oops:  :mad:  ;)  :(  :)  :lol:  :o  :!:

And now I get 10 smilies!

Wow. I, too, am very concerned about this potential hazard.

[quote="Klebian"]How about this?

The test:
[hide][hide][hide]

WHAT a waste of time.

[/hide][/hide][/hide]
[/quote]

[quote]
Too many smilies in your post: 32.
You can have at most 10 smilies in this message. Please remove some smilies before submitting this message.
[/quote]
[quote]
Too many smilies in your post: 32.
You can have at most 10 smilies in this message. Please remove some smilies before submitting this message.
[/quote]
[quote]
Too many smilies in your post: 32.
You can have at most 10 smilies in this message. Please remove some smilies before submitting this message.
[/quote]
[quote]
Too many smilies in your post: 32.
You can have at most 10 smilies in this message. Please remove some smilies before submitting this message.
[/quote]
[quote]
Too many smilies in your post: 32.
You can have at most 10 smilies in this message. Please remove some smilies before submitting this message.
[/quote]
 :)  :(  :?  :D  :D  :lol:  ;)  :blush:  :P  :mad:
Hooray!
But it does not cross 10. Why?

:(  that is kinda funky, but obviously its to remove spammers!!!!!!!! : because last year u could post as many as u want but these stupid spammers are just too spammy!!!!!!!!!!:spam:

[quote="PenguinIntegral"]:P  :P

EDIT: Dang, I can't edit them in either. The best way to get a bunch of smilies woud be to link to a large .gif of smiles.
EDIT AGAIN: Wait, you can. Just link directly to the image!

[img]http://www.artofproblemsolving.com/Forum/images/smiles/tongue.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smiles/tongue.gif[/img]

MORE EDIT:
OBSERVE!
[img]http://logicfury.com/public/smilies/smilies.png[/img][/quote]
 :rotfl:
 :rotfl:
 :rotfl:
 :rotfl:
 :rotfl:
[hide] :D [/hide]
[hide] :D [/hide]
[hide="more"] ;) [/hide]

I THINK that the smiley limit has been removed.

EDIT:  Never mind.  It's just because I'm a mod of the SC Forum that I can post a buch of smilies.

We need one of the Admins to answer this question.

I started going through it but soon thought better of it.
It was actually to increase the size!

[quote="easyas3.14159..."][quote="Klebian"]How about this?

The test:
[hide][hide][hide]

WHAT a waste of time.

[/hide][/hide][/hide]

And the result:
[hide] :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  A total of 10 smilies. -_- [/hide][/quote]

Geez Klebian did you have to do that? I went through who knows how many hide tags?[/quote]

Yea...I went through all of them too.....gosh....it took like....20 minutes....(laptop)...not very touch sensitive.....


Putnam
pigeonhole principle
arithmetic sequence

Let $ A$ be any set of $ 20$ distinct integers chosen from the arithmetic progression $ 1,4,7,........,100$. Prove that there must be two distinct integers in $ A$ whoe sum is $ 104$.

[hide="Hint"] Pigeon Hole Principle and in fact the $ 20$ can be replaced with $ 19$ (unless I made a mistake).[/hide]

[hide] It is true for 19 integers. Partition into the pairs

{4,100}
{7,97}
...
{49,55}

and {1,52}.

It follows from Pigeonhole that there exists at least two pairs in which we pick both integers, as there are only 17 pairs. Then, of those two, at least one must be of the first 16. But all of the first 16 pairs sum to 104, so we're done.[/hide]

[quote="BanishedTraitor"]Let $ A$ be any set of $ 20$ distinct integers chosen from the arithmetic progression $ 1,4,7,........,100$. Prove that there must be two distinct integers in $ A$ whoe sum is $ 104$.
[/quote]

By the way we can even replace the $ 19$ with $ 18$  :!:  But then we need to change the problem into: "... prove that there must exists two integers (possibly the same) in $ A$ such that the sum is $ 104$".

[hide="Solution"]
Consider $ 1,4,7,...,100$ there are $ 34$ terms in this progression. Let $ \{a_1,...,a_{18}\}$ be a set consisting of $ 18$ distinct elements from the progession. Now consider: $ a_1,a_2,...,a_{18}, 104 \minus{}a_1,104 \minus{} a_2, ... 104 \minus{} a_{18}$ now  $ a_i \not \equal{} a_j$ and $ 104 \minus{} a_i \not \equal{} 104 \minus{} a_j$ we have $ 36$ integers in total these are our pigeons. The holes are their values, now $ a_i$ is among $ 1,3,...,100$ but $ 100 \minus{} a_j$ is among $ 3,...,103$ so the possible ways these $ 36$ integers take are: $ 1,3,...,100,103$ in total $ 35$ possibilities. By pigeonhole we have that one of $ a_i$ is equal to $ 104 \minus{} a_j$ so $ a_i \plus{} a_j \equal{} 104$.  :wink:
[/hide]


trigonometry
inequalities proposed
inequalities

[color=darkblue]Consider any four numbers in the interval $ \left[\frac{\sqrt{2}\minus{}\sqrt{6}}{2},\frac{\sqrt{2}\plus{}\sqrt{6}}{2}\right]$. Prove that there are two of them, say $ a$ and $ b$ such that:[/color]
$ \left|a\sqrt{4\minus{}b^{2}}\minus{}b\sqrt{4\minus{}a^{2}}\right|\leq2.$

[u][b]Solution[/b][/u]
let the no.s chosen be $ a,b,c,d$
first note that the interval $ \left[\frac{\sqrt{2}\minus{}\sqrt{6}}{2},\frac{\sqrt{2}\plus{}\sqrt{6}}{2}\right]\equiv [2\sin (\minus{}\frac{\pi}{12}) , 2\sin\frac{5\pi}{12}]$
hence since $ 2\sin x$ is bijective in this interval we have every real no.$ a$ in this interval can be written as $ 2\sin x_{a}$ for some real $ x_{a}$ belonging to $ [\frac{\minus{}\pi}{12},\frac{5\pi}{12}]$
hence what we actually have to prove that there exists $ x_{a}$ and $ x_{b}$ belonging to $ [\minus{}\frac{\pi}{12},\frac{5\pi}{12}]$
 such that among a collection of $ x_{a},x_{b},x_{c},x_{d}$ coresponding to choice of $ a,b,c,d$
$ |\sin ( x_{a}\minus{}x_{b})|\leq\frac{1}{2}$
divide this interval into $ [\minus{}\frac{\pi}{12},\frac{\pi}{12}]$ , $ [\frac{\pi}{12},\frac{\pi}{4}]$ and $ [\frac{\pi}{4},\frac{5\pi}{12}]$ of length $ \frac{\pi}{6}$ .
hence by $ PHP$ two of them belong to the same interval say $ x_{a}$ and $ x_{b}$
thus $ \frac{\minus{}\pi}{6}\leq (x_{a}\minus{}x_{b})\leq\frac{\pi}{6}$ thus
$ |\sin ( x_{a}\minus{}x_{b})|\leq\frac{1}{2}$


trigonometry
inequalities
inequalities unsolved

$ a\plus{}b\plus{}c\equal{}1$ and $ a,b,c$  positive number . prove :

$ \frac{a}{a\plus{}bc}\plus{}\frac{b}{b\plus{}ac}\plus{}\frac{\sqrt{abc}}{c\plus{}ab}\leq 1\plus{}\frac{3\sqrt{3}}{4}$

Let

$ \color{white}.\quad.$ $ \cos^2{\frac{\alpha}{2}}\equal{}\frac{a}{a\plus{}bc}$

$ \color{white}.\quad.$ $ \cos^2{\frac{\beta}{2}}\equal{}\frac{b}{b\plus{}ca}$

$ \color{white}.\quad.$ $ \cos^2{\frac{\gamma}{2}}\equal{}\frac{c}{c\plus{}ab}$

where $ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi$. Then it follows that

$ \color{white}.\quad.$ $ \sin^2{\frac{\gamma}{2}}\equal{}\frac{ab}{c\plus{}ab}$

and we have

$ \color{white}.\quad.$ $ \cos^2{\frac{\gamma}{2}}\sin^2{\frac{\gamma}{2}}\equal{}\frac{abc}{(c\plus{}ab)^2}$

Hence the inequality becomes

$ \color{white}.\quad.$ $ \cos^2{\frac{\alpha}{2}}\plus{}\cos^2{\frac{\beta}{2}}\plus{}\cos{\frac{\gamma}{2}}\sin{\frac{\gamma}{2}}\leq1\plus{}\frac{3\sqrt{3}}{4}$

which after some juggling becomes

$ \color{white}.\quad.$ $ \cos{\alpha}\plus{}\cos{\beta}\plus{}\sin{\gamma}\leq\frac{3\sqrt{3}}{2}$

Well, that looks like a reasonable inequality to me (equality holding at $ \alpha\equal{}\beta\equal{}\frac{\pi}{6},\gamma\equal{}\frac{2\pi}{3}$), anyway. :roll:

in fact from [url=http://www.mathlinks.ro/viewtopic.php?t=181810]here[/url] it follows that
$ \sum _{cyc} \cos \alpha \leq \frac{3}{2} < \frac{3\sqrt{3}}{2}$


The problem can be solved similarly to the case n=3  ;)


hi,

can anyone solve this problem .

[u]Problem : [/u]

find the sum of the divisors of $104060401$ .

thanks in advance.

The number is $101^{4}$ as one sees by the binopmial theorem. Now it's easy.

sorry ,what is [quote]binopmial theorem [/quote] ?
 :huh:

sorry but i don't know neither binomial theorem ? could you explain more ,please ?

It's $(a+b)^{n}= \sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}$, here for $n=4, a=100, b=1$.
For any given special case, it's trivial to check. In general, it can be shown by induction.

[quote="Rust"]$x=\sum_{k=0}^{4}101^{k}=105101005.$[/quote]
So what¿ Read my post again!

thanks so much ZetaX but the problem now is that i don't

understand how you got n=4,a=100,b = 1 .i know They realize

the relation but how did you get them ,please ?

Look at the numbers: 1, 0, 4, 0, 6, 0, 4, 0, 1. Now omit all zeros: 1, 4, 6, 4, 1. Compare those numbers with Pascal's triangle.

waw ! i didn't pay attention ! thank you puuhikki !


I did horrible on the first 6 problems...

anyways. from number 5.

5. 500
6.51 i know it is wrong
7. 203
8. 27
9.973
10.289
11.625
12.134
13.144
14.460

number 5 is 766, i put 500 too though. it was a really bad question, too easy to mis interpret. getting 500 as your answer meant you interpreted the question as having the workers working for the allotted time and not finishing the work. in actuality, for each 'quarter,' the workers work overtime and finish the work. it sucks.

number 7 is 293

your number 9 is right

everyone i know, except for one person, put 500 for #5. the wording on that one was tricky.


topology
real analysis
real analysis theorems

Let S be an orientable smooth k-dimensional manifold in $ R^n$. Tell me if the following assertions are correct:

1) The connected components of S are open in S
2) The connected components of S are themselves k-dimensional manifolds
3) (Assuming 2) The connected components of S are orientable
4) (Assuming 2) The union of (orienting) atlases of the connected components of S provides an (orienting) atlas for S

1) is true. More generally, if $ X$ is a locally connected topological space, then any connected component of $ X$ is open in $ X$.
2) follows from 1). An open subset of a manifold is a manifold, of the same dimension
3) is true. If $ \omega$ is a nonvanishing $ k$-form on $ S$ (which is a common way to define orientation), then the restriction of $ \omega$ to any open subset of $ S$ is also a nonvanishing $ k$-form.
4) is true, because one can assemble $ k$-forms that are defined on disjoint open sets into one $ k$-form.

nonvanishing means nonzero on frames ('independent sets'), right?

And it's true that S has at most countably many connected components?

Thanks. I haven't learned topology so I have to reinvent the wheel when studying manifolds.

[quote="Soarer"]"And it's true that S has at most countably many connected components?"
Yes, since S is second countable (inherited from R^n)

What is "frame" by the way?[/quote]

Frame is the terminology my book uses to denote a basis for the tangent space of a point of a surface.

[quote="§outh§tar"]nonvanishing means nonzero on frames ('independent sets'), right?[/quote]
Yes, that's correct. It's a multilinear form that acts on tangent space, and nonvanishing means that it assumes a nonzero value on any basis of the space (and thus distinguishes the bases into "positive" and "negative").


calculus
integration
logarithms
calculus computations

$ \frac{\pi^{2}}{12}$
because for $ |x|\leq 1$ we can do the next step :
$ \ln (1\plus{}x) \equal{} x\minus{}\frac{x^{2}}{2}\plus{}\frac{x^{3}}{3}\minus{}...$
therefore
$ \int_{0}^{1}\frac{\ln (1\plus{}x)}{x}dx \equal{}\int_{0}^{1}x\minus{}\frac{x}{2}\plus{}\frac{x}{3}\minus{}... \equal{}\frac{\pi^{2}}{12}$
last step you  can calculate yourself
edit

hmm, I know this solution. I'm interrested if there is another solution, without Taylor expansion of $ \log(1\plus{}x)$  :wink:

yes we can by the next substitution $ t \equal{} x\plus{}1$
 then we have
$ \int_{0}^{1}\frac{\ln x}{1\minus{}x}dx \equal{}\int_{0}^{1}\sum_{i \equal{} 0}^{\plus{}oo}x^{i}\ln xdx$
last we can integration term by term because :
$ |\int_{0}^{1}(\sum_{n\geq N}x^{n}\ln x)dx| \equal{}\int_{0}^{1}\frac{x^{N}|\ln x|}{1\minus{}x}dx\leq\int_{0}^{1}x^{N\minus{}1}dx \equal{}\frac{1}{N}$

We also have to verify that integral pass through series.

Mmmm. I hate using infinite series. I was thinking of using a substitution like $ \sqrt{u^{2}\minus{}1}$ or something similar to make the numerator an inverse hyperbolic function. But I can't be bothered trying it out.

[quote="Carcul"]Extremal, your substitution doesn't seem to be right.[/quote]
Why do yiou think so ?  :)
After substitution I use
$ (1\minus{}x)^{\minus{}1}\equal{}1\plus{}x\plus{}x^{2}\plus{}...$

The substitution $ t\equal{}x\plus{}1$ is wrong.

I think the purpose of Cezar Lupu is not to use any power series expansion.

If not already known, an interesting reference is

M. Lutzky,
The American Mathematical Monthly, Vol.77, No.10 (Dec.,1970),pp.1080--1082

[quote="Cezar Lupu"]hmm, I know this solution.....[/quote]


geometry
geometric transformation
reflection
3D geometry
sphere
induction

Lastly Iranian result

Nima Ahmadipour           7   7   1   7   3   7             total               32
Aliakbar Daemi               7   7   7   7   7   7            total              42
Mostafa Eynollahzadeh    7   7   0   7   6   7            total                34
Omid Hatami                  1  7    6   7   7   1            total                29
Ali Khezeli                     7  7    1  7    7   0             total               29
Morteza Saghafian          7  7    7  7    7   0            total                 35

Total score is 201

I don't know. I don't know the results of the other teams. We must wait and see the results of other teams

But I'm partly sure Mostafa, Morteza and Aliakbar are Gold

Congrats, especially to Daemi for a perfect score :first:

I checked the total scores of all the teams

Iran is totally 4th.( 2 Golds and 4 silvers)

Well Done Every one. :coolspeak:

#1 China
#2 USA
#3 Russia
#4 Iran

Realy good . khaste nabashid , chiefly Ali Daemi  :D  ;)

Daste hamegan dar nakonad,
  :coolspeak:
Team IRAN as parsal kheli behtar natije gerefte(kheli bishtar az kheli, emsal 4th jahan shodim  :first:  yani team khafan shode :lol:

Natayeje parsal :
IRA1 7 7 2 7 7 1 ===>31
IRA2 7 7 0 7 7 2===>30
IRA3 7 7 2 7 6 2===>31
IRA4 6 3 3 7 6 2===>27
IRA5 6 1 7 7 7 5===>33
IRA6 7 6 1 7 3 2===>26
sum =178
(By sam_n)

natayeje emsal:
IRAN, ISLAMIC REPUBLIC OF
Score P1 P2 P3 P4 P5 P6 Final Medal
IRN1 7  7  1  7  3  7  32  Silver (Nima ahmadipor)
IRN2 7  7  7  7  7  7  42  Gold   (adaemi)
IRN3 7  7  0  7  6  7  34  Silver (eanollahzade)
IRN4 1  7  6  7  7  1  29  Silver (hatami)
IRN5 7  7  1  7  7  0  29  Silver (khezeli)
IRN6 7  7  7  7  7  0  35  Gold (saghafiyan)
TOTAL 201

[quote="fatehi"]afareen beh hame bachehaye team iran.
khasteh nabasheen

one question.. may I ask the schools of team members?
medoonam ke az mashhad emsal kesee ro nadashteem, valee az esfahan chand danesh amooz boodand?

this is certainly out of place but can any one tell me the results of other international olympiads?  (physics and Chemistry Biology and computer)[/quote]
4 members was from Tehran,1 from Isfehan and 1 from Kermanshah.

The Computer intl Olympiad has nat begun yet... it's a long time till then!!
But in Physics Iran was 6th( 2 golds -2 silvers- 1 bronze)
And in Biology they were 25th between 40 countries( :oops_sign: )
Nothing about the ChMO yet...

Excuse me every one for replying to an out of place question!! :lol:

[quote="payman_pm"]4 members was from Tehran,1 from Isfehan and 1 from Kermanshah.[/quote]
Are there any from Tehran that attended the allameh helli highschool (tizhooshan)?

[quote="Aidin Kashigar"]
Are there any from Tehran that attended the allameh helli highschool (tizhooshan)?[/quote]
I want to say that all of them were in Allameh-helli(not only one) :10:  :first:

be hame tabrik migam !! vaghean haghe iran ham hamin bood !! joze 4 ta time avval bodan !! :)

[quote="lomos_lupin"]are , dar bareye aghaye[b]Nima ahmadi pore [/b], eshon
toye dabirestane tizhoshan dars mikhonan.(dar hale hazaer az sale 2vom be payeye 3 vom miran)[/quote]

ok,hamon tori ke man midonam Nima ahmadi poor as payeye avale dabirestan be dovome dabirestan mire :lol:

Arne=as you wish arne but a little persian wont hurt anyone in the IRANIAN results.
but as you wish
Ehsan=I am afraid you are wrong about the grade of Nima.and here is a simple proof
Every member of team ,won a gold medal in the end of last summer(1383),Therefor has participate in the second round exam in the last spring (ordibehest 1383)
But only the high school students are allowd to paticipate.
Nima ahmadipor was accepted in Iran math olympiad in the first year of high school so after a year and half of trainig courses he is going to go to 3th grade.

lomos_lupin ,I'm sure about the grade of Nima ahmadi pore!
 Do you know Mr. Davoode vakili (the greatest Iranain in goemetric). Last year he said to me Nima studyed in the first grade of high school.That is very very interesting ,is'n it?

lomos_lupin, in fact it is a long-established rule that we should use English instead of our own languages in the international forums. We can use our own languages in the forums of our country though.  :)

siuhochung:Ok ,As you wish Mr.Inequality  :D

Ehsan:This Topic is going out of Its title("Iranian results"),but to clear this :
Yes I know Mr.Vakili(He has a gold medal in iran national olympiad in 2000 & he is Pretty good at geometry).
Basiclly ,Mr.ahmadipor Is the first one between then Iranian delegations which won a gold medal  when he was in the first year of high school.
Then He attend a year of Training courses  for imo in his second year of high school.
And now after Imo, he is going to 3th garde.
These arguments are according to the Rules of selecting Iran delegations ,regardless of any individual's opinion.

[quote="morteza-k"]beravo bacheha[/quote]

[quote="arne"]PLEASE use English in the international forums.[/quote]

Same remark! :mad:

ehsan2004,it doesnt mather wich grade he is.but he is in grade 3 now... :)

:clap2: hey guys i wont to applause your team...

[quote="shobber"][quote="Fitim"][color=blue]Good Luck!!![/color][/quote]
The competition has already ended. Why are you still saying good luck?[/quote]
You are right Shobber, but i like to say 'Good Luck' for another IMO. Its good now?! :lol:

[quote="Amir.S"][quote="Fitim"][color=blue]Good Luck!!![/color]
[color=black]You knows who is Al Farisi???[/color][/quote]

no, who is he/she?

maybe he/she is Arab ,no?[/quote]

He is Arab! But, if you like to knows really for Al Farisi read:
Al-Farisi is also known as Kamal al-din. His full name is Kamal al-din Abu'l Hasan Muhammad ibn al-Hasan al-Farisi. He made two major contributions to mathematics, one on light, the other on number theory. His work on light, colour and the rainbow is discussed in [1] but no mention of his work on number theory (nor mention of any other work at all by al-Farisi) occurs in that article written by R Rashed. On the other hand his contributions to number theory are discussed the references [2], [3], [4], [7], [8], and [9], most of which are also written by R Rashed but relate to discoveries made after the article [1] was written. (http://www-history.mcs.st-and.ac.uk/history//Biographies/Al-Farisi.html)

Al-Farisi was a pupil of the astronomer and mathematician Qutb al-Din al-Shirazi (1236 - 1311), who in turn was a pupil of Nasir al-Din al-Tusi. His work on light was prompted by a question put to him concerning the refraction of light. Al-Shirazi advised him to consult the Optics of ibn al-Haytham and al-Farisi made such a deep study of this treatise that al-Shirazi suggested that he write what is essentially a revision of that major work. Al-Shirazi himself was writing a commentary on works of Avicenna at the time.

Now al-Farisi went much further, for he undertook a project to study all the optical work of ibn al-Haytham. His major work the Tanqih (which means revision) was far more than a commentary on ibn ibn al-Haytham's optical writings. Al-Farisi does not seek merely to explain the works of a master in a more elementary form, rather he is quite prepared to suggest that some of ibn al-Haytham's theories are incorrect and to propose alternative theories himself.

The most important part of this work by al-Farisi is his theory of the rainbow. Ibn al-Haytham had indeed proposed a theory, but al-Farisi considered both this theory and another proposed by Avicenna before giving his own. The theory proposed by al-Farisi was the first mathematically satisfactory explanation of the rainbow.

Ibn al-Haytham had proposed that light from the sun is reflected by a cloud before reaching the eye. It was a theory which did not allow for a possible experimental verification. Al-Farisi, on the other hand, proposed a model where the ray of light from the sun was refracted twice by a water droplet, one or more reflections occurring between the two refractions. This model did allow an experiment to be conducted with a transparent sphere filled with water. Of course this introduced two additional sources for refraction, namely at the surface between the glass container and the water. Al-Farisi was able to show that the approximation obtained by his model was good enough to allow him to ignore the effects of the glass container.

In order to explain the colours in the rainbow, however, al-Farisi had to produce some new ideas about how colours were formed. The view before al-Farisi was that colours were produced a mixing darkness with light. This could not explain the rainbow so, based on the experimental evidence of the colours that he had observed with his transparent sphere experiment, al-Farisi proposed that the colours occurred because of the superimposition of different forms of the image on a dark background. He wrote (see for example [1]):-

... If the images then interpenetrate, the light is again intensified and produces a bright yellow. Next, the blended image diminishes and becomes a darker and darker red until it disappears when the sun is outside the cone of rays refracted after one reflection.

There have been arguments between modern scholars as to whether al-Farisi's theory of the rainbow was due to him or whether it was a theory proposed by his teacher al-Shirazi. Boyer writes in [5]:-

... the discovery of the theory should presumably be ascribed to [al-Shirazi], its elaboration to [al-Farisi].

Rashed discusses the claims of Boyer and others that the innovation in the theory of the rainbow was from al-Shirazi, but gives sound arguments for his claim that ascribing the theory to al-Shirazi is unconvincing.

Al-Farisi made a number of important contributions to number theory. He noted the impossibility of giving an integer solution to the equation

[hide="Index"]x^{4} + y^{4} = z^{4}[/hide]

but he attempted no proof of this case of Fermat's Last Theorem. Al-Farisi's most impressive work in number theory is on amicable numbers. Suppose that, in modern notation, S(n) denotes the sum of the aliquot parts of n, that is the sum of its proper quotients. The numbers m and n are called amicable if S(n) = m, and S(m) = n.

In Tadhkira al-ahbab fi bayan al-tahabb (Memorandum for friends on the proof of amicability) al-Farisi gave a new proof of the following theorem by Thabit ibn Qurra on amicable numbers:

For n > 1, let pn = 3.2^{n} - 1 and qn = 9.2^{2n-1} - 1. If pn-1, pn, and qn are prime numbers, then a = 2^{n}pn-1pn and b = 2^{n}qn are amicable numbers.

It was not a simple modification that al-Farisi made. Rather he produced a major new approach to a whole area of number theory, introducing ideas concerning factorisation and combinatorial methods. In fact al-Farisi's approach is based on the unique factorisation of an integer into powers of prime numbers, and, according to Rashed, he states and attempts to prove this, the so-called fundamental theorem of arithmetic, in this work. Whether al-Farisi proved or attempted to prove the fundamental theorem of arithmetic is also discussed in [4].

At the end of his treatise al-Farisi gives the pairs of amicable numbers 220, 284 and 17296, 18416, obtained from using Thabit's rule with n = 2 and n = 4 respectively. To check that Thabit's theorem gives amicable numbers with n = 4, al-Farisi has to show that p3, p4, and q4 are prime numbers. Now p3 = 23, p4 = 47 and q4 =1151 and, to show that 1151 is prime al-Farisi uses a number of lemmas including an application of the sieve of Eratosthenes.

The pair of amicable number 17296, 18416 are known as Euler's amicable pair. There is no doubt that al-Farisi proved these to be amicable numbers long before Euler. However, al-Farisi was probably not the first to discover these amicable numbers. In [6] Hogendijk argues that they were known to Thabit ibn Qurra himself.

Al-Farisi saw the relation between polygonal numbers and the binomial coefficients and he presented arguments, using an early type of mathematical induction, which showed a relation between triangular numbers, the sums of triangular numbers, the sums of the sums of triangular number, etc., and the combinations of n objects taken k at a time.

Article by: J J O'Connor and E F Robertson


A man, not knowing any better, has put three 100 W bulbs in series in a regular household circuit and is wondering at the result. If we assume that the resistance of the bulbs is a constant, what is the energy dissipated in each bulb?
A) 120 W
B) 77 W
C) 60 W
D) 11 W
E) 30 W

D 11 watts

Intermediate answers:  R = 144 ohms/bulb
                                    R (3 series) = 432 ohms
                                    I (3 series) = 0.277 amps
                                    V (3 series) = 40 v/bulb

easier way, possibly - note that $P = \frac{V^2}{R}$, where $V$ is the typical voltage, but since there are $3$ bulbs in a series, only $\frac{1}{3}$ of the voltage goes through each bulb; as the resistance doesn't change (assuming these are ohmic), we easily see that $P$ must be $\frac{1}{9}$ of what it was supposed to be.

Easier mathmatically, but if one must have more understanding of electical circuits and their equations to use the easier math.


Points $ M$ and $ K$ are marked on the sides $ BC$ and $ CD$ of a square $ ABCD$,
respectively. The segments $ MD$ and $ BK$ intersect at $ P.$ Prove that
$ AP \perp MK$ if and only if $ MC \equal{} KD.$

It is not true at all because $ AP\perp MK$ when $ MC\equal{}KC<>KD$.


By how much does the expression with the larger value exceed the expression with the smaller value?

\begin{tabular}{cc}
$ \frac{6!}{4!}$ & $ \frac{7!}{5!2!}$
\end{tabular}

I think it means what is $ \frac{6!}{4!} \minus{} \frac{7!}{5!2!} \equal{} 30\minus{}21 \equal{}\boxed{9}$


Nice proof.

When I made this problem, I thought a substituition: $x=sin^{2}A, y=sin^{2}B, z=sin^{2}C$.

So there can be made more tight inequalities like:
If $x^{2}+y^{2}+z^{2}+4xyz=2xy+2yz+2zx$ and $x,y,z \in [0,1]$, prove that $\sqrt{x}+\sqrt{y}+\sqrt{z}\leq \frac{3\sqrt{3}}{2}$.


Apparently for a simple random walk which starts at the origin and has a step length of $1$, the expected distance of the walk from the origin after $n$ steps is asymptotic to
\[\sqrt{\frac{2n}{\pi}}\]
(which presumably involves some calculation with Stirling's formula).  How do I carry out this calculation?  I have that the answer is
\[2\sum_{1\leq j\leq \left\lfloor \frac{n}{2}\right\rfloor}\binom{n}{j}\left(\frac{1}{2}\right)^{-n}(n-2j) \]
which I can't evaluate - maybe there's a better way?

e: I made a mistake the first time I posted; I'll see if I can solve it now.

The better way is to employ the Central Limit Theorem. The distribution of positions at time $n$ is approximately a multivariate normal distribution. (I'll let you work out the variances.) Use that.

as usual Dr. Merryfield gives a very good tip which leads to a quick solution of the problem, but for those of a masochistic or a combinatorial bent (does one imply the other?), we have the following:

If $n$ is even, then
\[\sum_{1\leq j\leq \lfloor n/2\rfloor}\binom{n}{j}j = \frac{n2^{n}}{4}\]
and
\[\sum_{1\leq j\leq \lfloor n/2\rfloor}\binom{n}{j}n = \frac{n2^{n}}{2}+\frac{n}{2}\binom{n}{n/2}\]
and if $n$ is odd, then
\[\sum_{1\leq j\leq \lfloor n/2\rfloor}\binom{n}{j}j = \frac{n2^{n}}{4}-\frac{n}{2}\binom{n-1}{(n-1)/2}\]
and
\[\sum_{1\leq j\leq \lfloor n/2\rfloor}\binom{n}{j}n = \frac{n2^{n}}{2}\]
and these, combined with a simple application of Stirling's formula, suffice to prove the result.


algebra
polynomial
inequalities
trigonometry
calculus
derivative
complex analysis

Find all polynomials with complec coefficients f such that we have the equivalence: for all complex numbers z, $ z\in[-1,1]$ if and only if $ f(z)\in [-1,1]$. Everyone who wants a long struggle should try this one.

I see you read "Questions & Rponses" of the RMS
You can also send your answer of the "Questions" to  RMS

Well, Moubi, I just read. As for the answer, if it were, I would send it. But since the problems published there are fare above my small brain, I just play the role of the lecturer.  :(   Nice problems, but damn hard!  Pssst for this problem, please!

The obvious solutions are the Chebyshev polynomials; $T_n(x)=\cos(n\cdot \cos^{-1}x)$.

These and their negatives are probably the only solutions, but I don't have a proof yet.

[quote="dickchimney"][quote="harazi"]Find all polynomials with complec coefficients f such that we have the equivalence: for all complex numbers z, $ z\in[-1,1]$ if and only if $ f(z)\in [-1,1]$. [/quote]

 If $z$ is complex. why can we have $ z\in[-1,1]$ :?:
 And if that is according to "real" meaning, then the Chebysev polynomials are not the only solutions. Consider $x^n$ for example!![/quote]
Equation $x^n=1$ has complex roots!

It is obvious that $f$ must be a real polynomial.
 We can show that ( using arguments on roots, local min, max points )

    \[f^2(x)-1 = C(x^2-1)f'^2(x)\]
 To deduce that $f$ must be Chebysev polynomials or their negatives from the above equation is not difficult, I think :)

[quote="dickchimney"]
 We can show that ( using arguments on roots, local min, max points )

    \[f^2(x)-1 = C(x^2-1)f'^2(x)\]
[/quote]

 The proof for that fact is :

$1$. $f'$ has [b]enough[/b] and all [b]single [/b]real roots in $(-1,1)$. Reason:
 -$f=k$ must have[b] enough [/b]real roots for all $k$ in $[-1,1]$ and any multiple root must be root of $f'$....
 - so we can choose $k$ ( which is quite arbitrarily ) so that the root of $f=k$ are diffrerent than those of $f'$, and hence must be distinct, then comes $1$.
$2$. $f(1)$ and $f(-1)$ must be either $1$ or $-1$ and those cannot be maximum or minimum point, and so not a root of $f'$!!
$3$. Every root of $f=1$ inside $(-1,1)$ must be local maximum point and therefore a [b]single[/b] root of $f'$, and hence double root of $f=1$ it self.
$4$. Same argument for $f=-1$
$5$. Since $f=1$ and $f=-1$ have [b]enough[/b] real roots, with $1. 2. 3. 4.$ it shows

      \[f^2(x)-1 = C(x^2-1)f'^2(x)\]

$6$. If we take the derivative we get

     \[ f= 2C(xf' + (x^2-1)f'') \]

  This is solved uniquely for a polynomial of $n$ degree up to the highest
coefficent!! So $f$ must be $cT_n$ and $c$ must be either $1$ or $-1$

 Can anyone check this, please!! :!:

Indeed, a superb solution. The solution published in the magazine is the same and it is absolutely beautiful. They also say that there exists a solution using the theory of holomorphic funtions. Does anyone know such a solution?

[quote="harazi"]The solution published in the magazine is the same[/quote]

Wich year,  issue, page ?

Uppps, I do not know this, I will have to look for. In fact it is the issue you saw yesterday. I will wite you a message soon and tell you where you can find it.

Now I know RMS 2004 n°2 page 181
They gave a reference

Iterations Rational Functions
Springer GTM 132


These questions are from florida Mu Alpha Theta tests in Algebra 2 division..
First:

500
:Sigma:  k 2
k=27

The answer is[hide] 41,785, 549[/hide]

The second one is more of a geometry question, but our school does alg.2 before geometry so I don't know how to solve it..

A unicycle travels 6km.  The diameter of the wheel is 1m.  How many FULL revolutions did the wheel make during the trip?

The answer is:  [hide]1909[/hide]

Please help~~ :)

I'm not far enough to help you with the first one  :blush:  :?  :( .

But the second is easy.

The radius of the wheel is 1/2m, and the wheel travels 6000m.

1: Find the circumfrence of the wheel.

2rpi

1/2*2pi=pi=3.14

2.Divide the lenth by the circumfrence.

6000/3.14=1910.83

So the wheel makes 1910 full revolutions.

Side question: what grade is this algebra 2 for?

EDIT:HAHAHAHA, I got it wrong, rounding pi was a baaaad move.

6000/pi=(approx)1909.86

 :blush:  :blush:  :blush:  :blush:  :blush:

I still don't really understand the first one, can some one use the summation equation to find it.. or maybe you did, and I just can't understand...

I understand the second one though, thank you...~
I take Alg2 in 9th grade....

Thanks

Unless I'm mistaken, the question is asking k=27 to in which it is 500.

Using the squares of sum formula, which is $\frac {n(n+1)(2n+1)}{6}$, we get the additions from 1 through 500 as 41791750. But we don't want the numbers from 1 through 26, so we subtract them, which is 6201.

Thus, answer is 41785549, which is what exactly he put.

See if you can prove this, which is formula I used:

$1^2+2^2+3^2+...n^2 = \frac {n(n+1)(2n+1)}{6}$

If you're interested in a proof of the formula Silverfalcon gives, do a search -- there have been several recent threads discussing it.

What city are you in?
CAuse in my grade 9 acedemic, we're doing realllllyyy comperatively easy math.

[quote="Silverfalcon"]I'm in 9th grade also and I take geometry and trig.

Not a surprise that there are students in this forum, who are even younger than me, taking calculus.

Again, skipping school curriculum is not the best thing in the world to do.

Best is to develop problem solving.[/quote]

I think this is highly misleading, in a sense.  Most of the advanced problem solving techniques use university level math at some point and understanding these techniques is hugely advantageous.  At a point, it becomes artificial for students to not start studying things such as analysis, or linear algebra, or (university level) number theory.

What is important is that you understand the school curriculum at a reasonably deep level.

I'm in 9th grade......
But I'm in an advanced class...
Only for people doing Mu Alpha Theta (math competition)..
We get through the book in half a year, spend the rest of the year
working on Mu Alpha Theta practice tests, etc...
So, I'm fairly lost, not everything has sunk in yet....
Geometry will be very easy compared to Algebra 2~
Where in Canada are you??

[quote="mna851"]How does one become a member of Mu Alpha Theta?[/quote]

It's a high school club.  You must be a high schooler to join.  I am in 6th grade, taking pre-calc.  So, my school made me an associate member.  I can take the test for practice only.

You have to be 10th grade or above, as well as have completed or be enrolled in certain math courses.

People under 10th grade are assosciate members.

the requirement for a member is 4 (or perhaps 6) completed semesters of highschool level of algebra and higher, completed at the highschool (classes taken in Middle school don't count)  with a 3.5 math gpa and 3.2 gpa for all classes.

Here is the official link for Full Member and Associate Member:

[url]http://www.mualphatheta.org/How_to_Join/MAO_Eligibility_Individual.htm[/url]

[b]Full Membership, High School Chapter[/b]: High school students in grades 9 through 12, at the school where their permanent record resides, who have completed the equivalent of four semesters of college preparatory mathematics and in addition have completed or are enrolled in a still more advanced course, are eligible for full membership providing their mathematical work was done with distinction. On the 4-point grading scale, this shall mean at least a 3.0 grade point average.

[b]Associate Membership[/b]: High school students in grades 9 through 12, at the school where their permanent record resides, who have completed two semesters of algebra or the equivalent with distinction and are enrolled in, or have completed, a semester of geometry or of second year algebra, are eligible for associate membership. Associate members do not pay the initiation fee, but should be registered with the national office so that they may attend the National Convention. They are not entitled to vote on national policy, however. They are entitled to attend and be heard at meetings and presumably are likely candidates for full membership.


limit
real analysis
real analysis unsolved

Hi,

$ \sum\limits_{n\equal{}1}^\infty \frac{1}{n^n} \notin \mathbb{Q}$ ? How can I prove it?

Thanks in advance. Any ideas are welcome.

I'd try to adapt the proof of the irrationality of $ e$.

That was my first idea too, but it doesn't work. I think it's not as easy as it looks.  :)

Ok, I was able to prove $ \sum\limits_{n\equal{}1}^\infty \prod\limits_{k\equal{}1}^n k^{\minus{}k} \notin \mathbb{Q}$

try to consider :
$ \frac {\prod _{all primes < n^{n}} p_{n} }{n^{n}}$
prove that it wiil be tends to 0.
after try to consider :
$ S: \equal{} your sum$
and
$ |S \minus{} \sum^{N}| \equal{} \sum _{N}^{\infty}.. < \frac {C}{N^{N}}$
after use the fact that:
if there exist such sequance $ \frac {p_{n}}{q_{n}}$ such as
$ |\alpha \minus{} \frac {p_{n}}{q_{n}}|<\frac {\epsilon_{n}}{q_{n}}$ and $ \lim _{n \minus{} > \infty}\epsilon_{n} \minus{} > 0$ then $ \alpha$ is irrational


In five years, Tom will be two thirds as old as his friend Mac. Three years ago he was half as old as Mac is now. How many years old is Mac now?

Bugi's solution is incorrect.
The system of equations we get is:
$t+5=\frac{2}{3}(m+5)$
$t-3=\frac{1}{2}m$
The second equation gives us $t=\frac{1}{2}m+3$. Substitute into the first equation:
$\frac{1}{2}m+3+5=\frac{2}{3}(m+5)$
$\frac{1}{2}m+8=\frac{2}{3}m+\frac{10}{3}$
$\frac{14}{3}=\frac{1}{6}m$
$m=28$


For nonnegative a,b,c,d such that $ a\plus{}b\plus{}c\plus{}d\equal{}2$, prove that

$ \frac{a\plus{}1}{a^2\plus{}1}\plus{}\frac{b\plus{}1}{b^2\plus{}1}\plus{}\frac{c\plus{}1}{c^2\plus{}1}\plus{}\frac{d\plus{}1}{d^2\plus{}1} \leq \frac{24}{5}$

[quote="NikolayKaz"]For nonnegative a,b,c,d such that $ a \plus{} b \plus{} c \plus{} d \equal{} 2$, prove that

$ \frac {a \plus{} 1}{a^2 \plus{} 1} \plus{} \frac {b \plus{} 1}{b^2 \plus{} 1} \plus{} \frac {c \plus{} 1}{c^2 \plus{} 1} \plus{} \frac {d \plus{} 1}{d^2 \plus{} 1} \leq \frac {24}{5}$[/quote]

$ 0 \le x \le2$

we have

$ \frac {x \plus{} 1}{x^2 \plus{} 1} \le\frac {32 \minus{} 4x}{25}$

$ \frac {x \plus{} 1}{x^2 \plus{} 1} \le\frac {32 \minus{} 4x}{25} \Longleftrightarrow \frac {(x \minus{} 7)( \minus{} 1 \plus{} 2x)^2}{25(x^2 \plus{} 1)} \le 0$

Therefore
$ \frac {a \plus{} 1}{a^2 \plus{} 1} \plus{} \frac {b \plus{} 1}{b^2 \plus{} 1} \plus{} \frac {c \plus{} 1}{c^2 \plus{} 1} \plus{} \frac {d \plus{} 1}{d^2 \plus{} 1}$

$ \leq \frac {32 \minus{} 4a}{25} \plus{} \frac {32 \minus{} 4b}{25} \plus{} \frac {32 \minus{} 4c}{25} \plus{} \frac {32 \minus{} 4d}{25} \equal{} \frac {24}{5}$

[quote="NikolayKaz"]For nonnegative a,b,c,d such that $ a \plus{} b \plus{} c \plus{} d \equal{} 2$, prove that

$ \frac {a \plus{} 1}{a^2 \plus{} 1} \plus{} \frac {b \plus{} 1}{b^2 \plus{} 1} \plus{} \frac {c \plus{} 1}{c^2 \plus{} 1} \plus{} \frac {d \plus{} 1}{d^2 \plus{} 1} \leq \frac {24}{5}$[/quote]

Let $ f(a,b,c,d)\equal{}\frac {a \plus{} 1}{a^2 \plus{} 1} \plus{} \frac {b \plus{} 1}{b^2 \plus{} 1} \plus{} \frac {c \plus{} 1}{c^2 \plus{} 1} \plus{} \frac {d \plus{} 1}{d^2 \plus{} 1}$,

we can prove that $ f(a,b,c,d)\le f\left(\frac{a\plus{}b}{2},\frac{a\plus{}b}{2},c,d\right)$,

so ...

[quote="kuing"][quote="NikolayKaz"]For nonnegative a,b,c,d such that $ a \plus{} b \plus{} c \plus{} d \equal{} 2$, prove that

$ \frac {a \plus{} 1}{a^2 \plus{} 1} \plus{} \frac {b \plus{} 1}{b^2 \plus{} 1} \plus{} \frac {c \plus{} 1}{c^2 \plus{} 1} \plus{} \frac {d \plus{} 1}{d^2 \plus{} 1} \leq \frac {24}{5}$[/quote]

Let $ f(a,b,c,d) \equal{} \frac {a \plus{} 1}{a^2 \plus{} 1} \plus{} \frac {b \plus{} 1}{b^2 \plus{} 1} \plus{} \frac {c \plus{} 1}{c^2 \plus{} 1} \plus{} \frac {d \plus{} 1}{d^2 \plus{} 1}$,

we can prove that $ f(a,b,c,d)\le f\left(\frac {a \plus{} b}{2},\frac {a \plus{} b}{2},c,d\right)$,

so ...[/quote]

Why

$ f(a,b,c,d)\le f\left(\frac {a \plus{} b}{2},\frac {a \plus{} b}{2},c,d\right)$

?


Hi!

I've started a xanga that covers a few cool topics in number theory. It'll be updated once or twice a week, check it out!

http://www.xanga.com/learn_number_theory

Thanks!

since it might be spamming if us two chat in that topic, i'm pming u...

sure, i'll tell my friends. Neal already knows, and i'll put it onto my xanga and AIM profile.

It has some good stuff in it. Will it increase in difficulty?

edit: sigh... i guess i didn't pm it then... wrong button... hehehe

iw.oav doesnt frequent this forum anymore so pming would be futile

[quote="Idiot_without_a_village"]I've started a xanga that covers a few cool topics in number theory. It'll be updated once or twice a week, check it out!

http://www.xanga.com/learn_number_theory[/quote]

(After editing:) I love to read more about number theory, and I especially love to encourage math learners to write more about it online, so I found the first choice of colors on the site regrettable, as they were hard for me to read.

The standard advice on how to make a Web site usable can be found at

http://www.useit.com/

a site with an abundance of free resources. The short answer is to NEVER specify a background color or a text color: leave that to each browser-user's defaults. Best wishes for much success in sharing your enthusiasm for number theory.

I'm horrible with colors, any suggestions?

Also, it should increase in difficulty, but I'm trying to make it easier to understand as well.

Thanks!

Well my suggestions with colours is not to use bright colours, the easiest to read is just simple dark writing on a lightish background...


Is given a triangle ABC outscribed the circle . Let $ m_{a},m_{b},m_{c}$ be  the segments of tangents that pass respectively through A ,B and C. Prove that :

$ \frac {m_{a}}{a} \plus{} \frac {m_{b}}{b} \plus{} \frac {m_{c}}{c} \ge \frac {3}{2}$

From your problem we get :

   $ m_{a}$ + $ m_{b}$ = $ c$   ,    $ m_{a}$ + $ m_{c}$ = $ b$   ,    $ m_{b}$ + $ m_{c}$ = $ a$

   if we add first two equations and descent the third equation  ,  then we get :

   $ 2m_{a}$ = $ b \plus{} c \minus{} a$    or   $ m_{a}$ = $ \frac {1} {2} (b \plus{} c \minus{} a)$

    in the same way we find :

   $ m_{b}$ = $ \frac {1} {2} (a \plus{} c \minus{} b)$      and    $ m_{c}$ =$ \frac {1} {2} (a \plus{} b \minus{} c)$    .

    Now we have :

    $ \frac {m_{a}}{a}$ + $ \frac {m_{b}}{b}$ + $ \frac {m_{c}}{c}$ = $ \frac {1} {2}(\frac {b \plus{} c \minus{} a}{a} \plus{} \frac {a \plus{} c \minus{} b}{b} \plus{} \frac {a \plus{} b \minus{} c}{c})$ =

          =  $ \frac {1} {2}((\frac {a}{b} \plus{} \frac {b}{a}) \plus{} (\frac {a}{c} \plus{} \frac {c}{a}) \plus{} (\frac {b}{c} \plus{} \frac {c}{b}) \minus{} 3) \geq \frac {3} {2}$

    because  $ (\frac {a}{b} \plus{} \frac {b}{a}) \geq 2$

    We have equation just if $ a \equal{} b \equal{} c$

     Done  :)

What is your problem asking ,because I'm having some troubles understanding the problem [english problems]


function
real analysis
real analysis unsolved

Build a bijective function $f: [0,1]\to[0,1]$ such that $f$ is not monotonic on any specific interval $J\subseteq[0,1]$

$f(x)=x$ if $x$ is irrational and $f(x)=1-x$ otherwise ?


geometry
combinatorics proposed
combinatorics

We have 100 points in the space such that every 4 are not in the same plane.We consider all the planes determined by three points.We know these are not parralel each two.Find out the number of lines resulted from the intersection of these planes.

(Sorry about the initial missing in the problem text).

It seems strange that the number of lines does not depend of the position of the points. Are you sure that it is not asked for the maximum number of lines? Or that no two planes are parallels? or whatever....

Pierre.

[quote="xxxxtt"]Fin lines dout the number of lines[/quote]

what does that mean??? :?

you mean "Find out the number of lines" ?

well in whatever way I interpret this, you either got 0 points or all 100

so in the first case that means 0 lines, in the other case 100*99/2 = 4950 lines :?


calculus
geometry
real analysis
real analysis solved

hi, help me with this problem :

the graph of y=f(x) is concaved down over the interval (A,B)
ie f''(x)<0 for A=<x=<B
the tangent of the curve at point P(k,f(k)) meets lines x=A and x=B at P and Q respectively.
the value of k is such that the area bounded by the curve, the tangent and lines x=A and x=B is minimized.
Prove that k is independent of f(x).

thanks

Indeed, described figure is a tropezium, so its area is $(B-A)\cdot \frac{y_1+y_2}{2}$, where $y_1$ and $y_2$ are tropezium's sides on lines $x=A$ and $x=B$. But, on the other hand,  $\frac{y_1+y_2}{2}$ is a length of a middle line of the tropezium. Draw this line. It lies on $x=\frac{A+B}{2}$ and has two endpoints: one is a middle of $\big((A,f(A))(B,f(b))\big)$ and second is a middle of a tangent segment. Since the tangent line lies below the graph of $f$, this "middle" segment intersects the graph of $f$, so its length is at least length of a segment between a middle of $\big(((A,f(A))(B,f(b))\big)$ and a point $\left(\frac{A+B}{2},f\left(\frac{A+B}{2}\right)\right)$. It follows that the optimum point $k$ is $frac{A+B}{2}$.


search
superior algebra
superior algebra unsolved

find no. of irreducible poly of degree n in Fp[x],p a prime no.


45. A car located 150 m from a street corner moves away from the corner at the uniform rate of 30. m/s. After maintaining this velocity for 10.s, it decelerates to a stop in 5.0 s. It then rests for another 5.0 s, then accelerates away from the corner at the rate of 2.0 m/s^2 for 10. s. Construct a celocity-time graph representing the motion of the car for the entire trip.

[It may to difficult to construct a graph on CC, but please try your best to construct a table so I can use the data to make the graph.]

51. What is the magnitude of the vector sum of the two concurrent corces represented in the diagram? (1) 2.5 N (2) 3.5 N (3) 3.0 N (4) 4.0 N

[Please explain answer to 51.]

/l\
l
l
l 2.0 N
l
l___________\
1.5 N /

54. Two concurrent forces act at right angles to each other. If one of the forces is 40 newtons and the resultant of the two forces is 50 newtons the magnitude of the other force must be (1) 10 N (2) 20 N (3) 30 N (4) 40 N

[Please explain answer to 54.]

55. If two 10.-newton concurrent forces have a resultant of zero, the angle between the forces must be (1) 0 degrees (2) 45 degrees (3) 90 degrees (4) 180 degrees

[Please explain answer to 55.]

45. You'd be better off if you could do this yourself but... oh well.

Time / Velocity:
0 s ..... 30 m/s
10 s ..... 30 m/s
15 s ..... 0 m/s
20 s ..... 0 m/s
30 s ..... 20 m/s

If you put these five points on your graph and connect them with straight lines, you get what you need!

You should know how to do 51. It is a very elementary problem, using vector addition.

Ans. 2.5
When you add vectors its always head to tail
so when you place them head to tail it looks like this
------------------------->
^
|
|
|
|
|
|
|

The  two forces have magnitude 1.5 and 2 so $\sqrt{2^2 +1.5^2}$ will give you the hypotenuse of the triangle, which is the resultant vector.

And the same for 54. and 55. If you have two forces at right angles, the resultant force is a hypotenuse in a right triangle with the two forces as the other two sides. So from Pythagorean theorem:

$40^2+x^2=50^2$
$x=30$

55. is 180°, at any other angle the sum you got as qweretyq described will be obviously nonzero.


If 4800 gallons of gasoline occupy $ 60\%$ of a storage tank's capacity, how many more gallons are needed to fill the tank?

I'll just pick up where yaofan left off. $8000\cdot 0.40=3200$ or $8000-4800=3200$.

Also:
$\frac{3}{5}x=4800$ We're trying to find $\frac{2}{5}x$ so we can just multiply $4800$ with $\frac{2}{3}$ giving us $\boxed{3200}$


[quote="mhuarancca"]donde puedo conseguir los examenes anteriores de la
 o. asia pacifico???? :?  :?[/quote]

Entra aquí http://www.kalva.demon.co.uk/apmo.html

Por ahora he encontrado en la pagina de olimpiadas colombias los examenes del 1999 y 2000 en español

[url=http://olimpia.uanarino.edu.co/apmo/apmo99pe.htm]http://olimpia.uanarino.edu.co/apmo/apmo99pe.htm[/url]

[url=http://olimpia.uanarino.edu.co/apmo/apmo00pe.htm]olimpia.uanarino.edu.co/apmo/apmo00pe.htm[/url]

Espero  ayudarte con eso...

$Tipe$

[url=http://www.oma.org.ar/enunciados/]Aquí[/url] hay algunos examenes de la [i]Asian-Pacific[/i] en español.


Let $1= d_{1}<d_{2}<...<d_{k}=n$ be the positive divisors of n. Find the smallest positive integer n such that
$d_{1}^2 + d_{2}^2+...+d_{k}^2 = (n+3)^2$


integration
trigonometry
logarithms
calculus
calculus computations

Let $ t\in\mathbb{R}$.

Find the value of following:

$ g(t) \equal{} \int_{ \minus{} \infty}^{\infty}e^{ \minus{} \frac {x^2}{2}}\cos (tx) dx$,

and find the vales at $ t \equal{} 0$,$ t \equal{} 1$,$ t \equal{} \sqrt {2\log\pi}$ and other insteresting values.

edit : $ t \equal{} 1$.

Just wondering why $ \sqrt{2\log\pi}$ is interesting :O

$ \sqrt{2\log{\pi}}$ makes g(t) algebraic and contains $ e$ and  $ \pi$ .

\begin{eqnarray*}g'(t)
& = & - \int_{-\infty}^{\infty}x e^{-\frac {x^2}{2}}\sin (tx) dx \\
& = & \left[ e^{-\frac {x^2}{2}}\sin (tx) \right]_{-\infty}^{\infty} - t \int_{-\infty}^{\infty} e^{-\frac {x^2}{2}}\cos (tx) \, dx \\
& = & - t g(t)
\end{eqnarray*}

This, together with the initial condition $ g(0) = \sqrt{2\pi}$, shows that $ g(t) = \sqrt{2\pi}e^{-\frac{t^2}{2}}$.

Hi, sos440.  Thanks a lot.

So if $ t \equal{} 0$ , $ \int_{ \minus{} \infty}^{\infty}e^{ \minus{} \frac {x^2}{2}} dx \equal{} \sqrt{2 \pi}$ is well-known , but

$ t \equal{} 1$ , $ \int_{ \minus{} \infty}^{\infty}e^{ \minus{} \frac {x^2}{2}} \cos{x} dx \equal{} \sqrt{\frac{2 \pi}{e}}$ is so interesting for me.

My [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=249955]course notes[/url] devote a whole chapter to computing the Fourier transform of the Gaussian - which this is. There are multiple methods.