real analysis
integration
calculus
function
real analysis unsolved

$ f(x)\equal{} e^{x^2}$ if $ x$ is a rational number. $ 0$ is x is an irrational number.

$ 1.$ Show that $ f$ is Lebesgue measurable.

$ 2.$ Compute $ \int_\Re f(x) dx.$

[hide="hint"] Each subset of $ f\left(\mathbb{R}\right)$ is countable; in particular $ \lbrace x\in \mathbb{R}: f(x) > a\rbrace$ is countable for each $ a\in \mathbb{R}$, and hence measurable (why?).  What can you say about the integral of a measurable functions that is nonzero only on a set of measure $ 0$?[/hide]

@JoeBlow: Is $ \{x: f(x)>\minus{}1\}$ countable? Should we maybe consider $ a\geq 0$ and $ a<0$ as separate cases?

Good point, but the case $ a<0$ gives a rather trivially measurable set. :)


Let $S$ be the sum we want to calculate.

\[(2n)! \cdot S = \sum_{k=0}^n \binom{2n}{n-k} = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{n}\]
\[ \Longrightarrow 2(2n)! \cdot S = \left(\binom{2n}{0} + \cdots + \binom{2n}{2n}\right) + \binom{2n}{n} = 2^{2n} + \frac{(2n)!}{(n!)^2}. \] We can now divide by $2(2n)!$ to get \[ S = \frac{2^{2n-1}}{(2n)!} + \frac{1}{2(n!)^2}, \] exactly what we wanted to prove.


The chord $ AB$ splits a circle into two arcs with midpoints $ M$ and $ N$.
These points are then rotated around $ A$.
$ B$ is mapped to $ B'$, $ M$ is mapped to $ M'$
If the midpoint of $ BB'$ is $ P$ then show that $ M'P$ is perpendicular to $ NP$

Nice problem  :lol:

[hide="Incomplete solution"]
 [geogebra]9671de1536f02353bfaf35261e233db055dc9b69[/geogebra]

$ \triangle ABB'$ is isosceles by symmetry, and $ P$ is the midpoint of $ BB'$, so $ \angle APB=\angle APB'=90^\circ$.

Because $ M$ is the midpoint of $ %Error. "wideparen" is a bad command.
{AB}$, and $ O$ is the center of the circle, $ OM$ bisects $ AB$, and $ \angle OCB=\angle OCA=90^\circ$.

Let $ \angle PBD=\theta$. Notice that $ PBCF$ is cyclic because opposite angles are both $ 90^\circ$. Therefore, $ \angle PFC=180-\theta$. $ EF+FC=EC$, so $ \angle EFP=180-(180-\theta)=\theta$.

At this point, if I could get another angle to be equal between $ \triangle PEF$ and $ \triangle PDB$, I could show that they are similar (they're actually congruent; you can verify this with GeoGebra). After that, this follows:

Let $ \angle PDB=\gamma$. Then $ \angle BPD=180-\gamma-\theta$. Because $ \triangle PEF\sim\triangle PDB$, we have $ \angle EPF=\angle DPB=180-\gamma-\theta$. Because $ \angle BPF=90^\circ$, we have $ \angle DPF=90-(180-\gamma-\theta)=\gamma+\theta-90$. Then, $ \angle DPE=\angle DPF+\angle FPE=\gamma+\theta-90+180-\gamma-\theta=90^\circ$.
[/hide]

Anyone see a way to show either that $ \angle EPF=\angle DPB$ or that $ \angle FEP=\angle BDP$?


geometry AMC AIME USA(J)MO USAMO calculus trigonometry \/closed

At the end of AoPS's [i]Introduction to Geometry[/i], it was stated that [i]Intermediate Geometry[/i] should be published in 2007.  I have looked on the forums and have not seen any posts addressing this issue.  My assumption is that the book's writing and release has been postponed because Mr. Rusczyk decided to write [i]Introduction to Algebra[/i] and [i]Intermediate Algebra[/i].  I am really looking forward to the release of [i]Intermediate Geometry[/i], because I plan to participate in the AMC 10 and 12, AIME, and maybe even USAMO for the next several years, and the release of the book would be great for me as well as others preparing for these competitions.

Another few questions about Art of Problem Solving textbooks: Will an intermediate number theory book be written?  Will an [i]Advanced[/i] series of books be written for the subjects in which [i]Introduction[/i] books have already been written?

I [i]love[/i] the AoPS textbooks, and would like to keep learning a lot from them throughout high school.  Thanks for your responses!  :lol:

LowTopology

The Intermediate Algebra book will be out in 7-8 weeks if all goes as planned.  My next book is an Intermediate Trig/Complex Numbers/beginning Linear Algebra book (yeah, that's a pretty crummy title!), which I expect will be finished next year sometime.  After that, I'll probably turn my attention to another geometry book.  Sorry the schedule has been so rearranged!

DPatrick is working on a calculus book now, and will likely work on a collegiate-level linear algebra book after that.  To the extent we address number theory again in a later book, we will likely integrate it with an introduction to group theory.  We haven't yet decided on this, though.

[quote="rrusczyk"]My next book is an Intermediate Trig/Complex Numbers/beginning Linear Algebra book (yeah, that's a pretty crummy title!), which I expect will be finished next year sometime.[/quote]

Wow, I'm totally getting that book! I really need some work with trig  :oops: .

Hmm, just a suggestion for the title: "Complex Algebra" with the subheadings of Trig, Complex Numbers, Linear Algebra.

I think complex numbers should be the header, like "Intermediate Complex Numbers" with those subheadings, but that's just me.

so if  I understood corectly we can expect intermediate algebra in the following weeks and  intermediate trig (and the rest of the title) somewhere in 2009?

[quote]DPatrick is working on a calculus book now[/quote]

just a question : many people use your books to learn math but also to prepare themselves for competitions and sice there are not many high school competitions that would requiere calculus I wonder why you don't think about making intermediate number theory book sooner ....

just a thought though ....

good luck  :lol:

[quote="Wichking"]so if  I understood corectly we can expect intermediate algebra in the following weeks and  intermediate trig (and the rest of the title) somewhere in 2009?
[/quote]

That's correct.


I am pretty sure this will work:
The chance that Louis will fail in guessing Becca's number the first time is 4/5.  The chance that Becca will succeed in guessing Louis's number is 2/5.  So far we have (2/5)*(4/5)=8/25.  Now, the chance that Louis will fail, Becca will fail, Louis will fail again, and Becca will succeed is (4/5)*(3/5)*(3/4)*(1/2)=9/50.  Keep on doing this and you will get 8/25+9/50+2/25+1/50=60%.  It shoudn't take that long because you can use a calculator.

[quote="davidlizeng"]Louis and Becca each choose a number from the set {1,2,3,4,5}. then they each take turns guessing each other's number. The first one to guess the other's number correctly wins. Louis will randomly guess each time and rule out incorrect guesses from previous guesses. Becca does the same thing but she has twice the chance to guess the right number than Louis. If Louis guesses first, find the probablility that Becca will win. Express as %.

I brute-forced it by multiplying all these fractions that canceled out pretty well, but got it wrong by a stupid mistake.

Is there a better way to solve this? the way i tried leaves me little time to check, resulting in stupid mistakes....... :([/quote]

This problem had a wording error on the real test. They never specifically said that the incorrect guess is ruled out all they said was that she picks from the remaining numbers. I interpreted remaining as 1 through 5 so when I actually took the test I screwed the problem up big time.

I was really stupid and I calculated the probabilities wrong. And I still don't know how I got them wrong. On the second one I got 17/100 not 9/50...

The way I did it was calculate all probabilities for both people, then I added up the ones for Louis, took the opposite, and multiplied it by 2/5 or 2/4 or 2/3 or whatever turn they were on.

i did it QuatoMaster's way.
the problem is that there's lots of room for careless
mistakes. after the test i did it two more times before not making some sort of error.
i'm looking for a better way to do it.
that is, if it exists.

I don't think there is a simpler way of doing this problem.  It shouldn't be too bad with a calculator though.

[quote="davidlizeng"]now that i think about it, there is no better way.
you just need a clear mind to thik the solution out
quickly.
[quote]This problem had a wording error on the real test. They never specifically said that the incorrect guess is ruled out all they said was that she picks from the remaining numbers. I interpreted remaining as 1 through 5 so when I actually took the test I screwed the problem up big time.[/quote]

sonny, i don't think there was a wording error. the problem made sense to me after i stared at it for awhile.....[/quote]

Ok not an error but a fault. It never said an incorrect guess was ruled out. All it said was that Louis picks next from the remaining choices. I thought that remaining meant 1-5 so I messed it up. By my interpretation the game goes on forever but I realized that after turning my test in. The problem should have been clearer.

[quote="davidlizeng"]true. The problem was not that clear...........
anyway, what did you get on the test overall?[/quote]

I screwed up...I missed like 3 sprints by stupid mistakes


This question is not as easy as the previous one but it can still be solved by the 11th with the knowledge they have (i presume)

25 ml of a solution of a weak monoprotic acid HX was titrated with a 0.0640 M solution of NaOH requiring 18.22 ml. The pH of the solution varied as a function of the percentage of HX titrated:

% titrated                               pH
0                                           3.39
33.3                                       5.14
66.7                                       5.74

(1) Calculate initial concentration of weak acid in 25 ml of solution.
(2) Determine Ka for each of these 3 conditions
(3) Calculate pH at equivalence point.
(4) Calculate no. of moles of NaX (salt) that must be added to produce a pH of 6 in 150 ml of the original solution.

don't know about others,but your assumption about one 11ther(me) is wrong
I am pathetic in studies(esp chem)and really don't know how to solve other parts right now,will think about them later(have noted down question but am sure I can't solve as have not much idea about Ka as well as how pH changes(i have only that much idea about pH that is in NTSE books)

is answer of 1 $ 0.0466 M$???(i know theory,but calculations may be wrong :oops: )

thanks  :) (at-least one of my solutions were right)

i'll try about other problems later

i think these ques can be done by simple maths and basic equilibrium concepts :roll:
but i m in no mood of calculations rite now

just the way common sense is not common to all,basic concepts are also not so basic for all :(

beyond my reach :(


[img]http://www.artofproblemsolving.com/Images/Cal2.gif[/img]

What is the maximum number of cubes in the above picture?
What is the minimum number of cubes in the above picture?
What is the maximum surface area in the above picture?
What is the minimum surface area in the above picture?

I hope this is easy enough...

maybe my graphics card sucks, but the image doesn't show up for me

It's just the AoPS logo...it's at the top lefthand corner of your screen.

The original image no longer exists.  Tare was referring to the diagram in the upper left corner of this page.

Same with me; all I see is a big red X mark.

same.....try another picture that will suit the question, if that doesnt work.....try another question

otherwise my card doesnt work either.

As rrusczyk said, you can see the picture in the top left corner of your screen. Above "Books".

Does this work?
[img]http://artofproblemsolving.com/Images/TemCorner.jpg[/img]

Does this have something to do with turning the picture over?


Well, when? I'm taking it at seventh grade.

You are seriosly taking the SATs in 7th grade? Wow.
[hide=" :D "] You must be reallly smart.[/hide]

Probably because there's not enough poll options.

I took it in 7th grade too

[quote="José"]And which was your result???? Because if you can approve it in 7º it means that the test is not as difficult as everybody thinks (Don't missunderstand me, I'm not saying "you approve it because the test was easy", but it won't be normal for a 7º grader to approve it while loads of 11º or 12º graders fail it)[/quote]

I'm sorry--I don't quite understand what you mean by "approving it"

I did horrible in verbal and writing. :wallbash_red:  :wallbash:

i'm taking it in 9th or 10th, probably 9th

I declined to do in in 7th. I didn't see the point.

[quote="José"][quote="mathnerd314"]I'm sorry--I don't quite understand what you mean by "approving it"[/quote]

Have a good result at the test.

BTW, have anybody got some SAT's tests?? I would like to see them[/quote]

Oh, I did fairly well. (Details withheld)

I think http://www.saab.org has a SAT practice page.

I took it in 7th and again in 8th; I didn't know much about it at the time, so I was like ok, whatever. I believe I did manage to win some money in 8th grade from it though, so it wasn't completely pointless :wink:. I took it one last time last December and that's the history of my SATs.

[quote="akalra1"]i m taking the psat in seventh[/quote]

what's the point, you don't really gain anything.

and me? :o

So anyway, I got a zero on the math section.
[size=42]I m kidding[/size]

I took it in 7th grade. Got a perfect score in math, but only slightly above average in verbal. :(

And that earned me $\$$100 (gift card to Barnes and Noble)  :D

You got a perfect in math in [b]7th grade[/b]?

it's not that hard. i only missed one question due to a careless error. the difficulty of the math only reaches Alg II.

Like SplashD said, there's not really any benefit to taking it at such a young age especially since the SAT is straight foward problems that test what you know and what you memorized, not how to apply them, so you don't learn anything from them. The ACT is much harder anyway.

[quote="Adunakhor"]There might be some people who took the SAT in middle school so that they could attend CTY, or something similar. Out of curiosity, how'd people get money from taking the SAT  :maybe:?[/quote]

From CTY. You need a perfect score or something. :maybe:

cty gives teh sat (psat), the sat doesnt give u cty

[quote="Adunakhor"]There might be some people who took the SAT in middle school so that they could attend CTY, or something similar. Out of curiosity, how'd people get money from taking the SAT  :maybe:?[/quote]
Yes, there are several programs that do this, including CTY (http://www.jhu.edu/cty/) and TIP (http://www.tip.duke.edu/). What they do is they let you take the SAT/ACT in 7th/8th grade, and if you score a perfect on either math or verbal (although I'm not sure how they do it with the ACT or now the new SAT), you receive a full scholarship to one of their summer camps (which is about $\$$3,000).
Also, taking the SAT/ACT early is good "practice" for the future; you'll be more prepared for the SAT/ACT by the time you take them in high school, and you also get to see what kind of material is ahead of you, coming in just a few years.

[quote="jclarkemathL314159"]I took it in 7th grade (and bombed critical reading, btw) with the intention of going to CTY. I won one of the National award thingys for math, but then decided that the CTY summer program was crap and applied to MathPath instead.[/quote]

u take the psat thru cty
________________________________

u can also take the PLUS thru cty and qualify for one of their programs
i took the plus and i qualifed for one of their writing courses and it really helped
there is also a cty award ceremony for people who get good score on the PLUS / psat

btw, the cty courses r a bit expensive, $600+

you can? I thought CTY only offered the SAT.

[quote="nebula42"]Yes, there are several programs that do this, including CTY (http://www.jhu.edu/cty/) and TIP (http://www.tip.duke.edu/). What they do is they let you take the SAT/ACT in 7th/8th grade, and if you score a perfect on either math or verbal (although I'm not sure how they do it with the ACT or now the new SAT), you receive a full scholarship to one of their summer camps (which is about $\$$3,000).
Also, taking the SAT/ACT early is good "practice" for the future; you'll be more prepared for the SAT/ACT by the time you take them in high school, and you also get to see what kind of material is ahead of you, coming in just a few years.[/quote]

Yeah, I took the SAT with TIP, made a few stupid mistakes, and ended up a few points short of a perfect score. I got a scholarship to some camp run by a local university, but eventually went to Mathpath instead.

[quote="mathnerd314"]deimos, SAT is the Scholastic Aptitude Test (or something like that), a standardized test that high schoolers take in the US that largely determines what kind of college they go to.

[quote="Ignite168"]Taking the SAT at such a young age is stupid.

I'm taking it in 11th.[/quote]

You calling me stupid? :wallbash_red: ;) :rotfl:[/quote]

yes

i'd take it in 11th if i had a choice but i'm planning to apply to exeter for 11th grade (possibly 10th, most likely 11th though) so i'm taking it in 9th


analytic geometry
geometry
number theory unsolved
number theory

how many points of integer coordinates are in a circle of radius r ?

I don't think you will receive any exact evaluation. All that you can hope is some equivalent.

If $f(r)$ is the desired number, then $f(r) = \pi \cdot r^2 + O(r^{\alpha})$ for some appropriate $\alpha$.
I think the best known is $\alpha = \frac {46 } {73} + \epsilon$, and an old conjecture is that $\alpha = \frac 1 2$ is possible.

Note that $f(r)$ is equivalent to the area of the disk, which is a general result for the convex bounded domains.

Pierre.

a friend of mine told me that these numbers
are also called the integers of Gauss.
is this true?
if yes he tells that also there is a square innscriptible in the circle and that
area of that square is the number of points wich the problem asks for.

to reduce complexity of the problem should we not consider points on the circle but
only the ones inside it,also consider the center of the circle in the origin of carthesian
coordinates(the point(0,0) ) and we have like in the figure for example a circle
of radius 3 and we see there are 25 points inside of integer coordinates.

could we generalize and see that $N=(2\cdot [r]-1)^2$ is the number of points asked for ?

The gaussian integers are that of the form $a+bi;a,b \in \mathbb{Z}$

For your problem you can look here: http://mathworld.wolfram.com/GausssCircleProblem.html

and you can't generalize since it's obviosly wrong for bigger $n$

i don't really know about you but i can't really understand but
15% of what i read from mathlinks rising up to as much as 30%
if i really try hard.
so...that link won't do too much for me.

Your conjecture with $N=(2\cdot [r]-1)^2$ is e.g. wrong for $n=4$, where you get $45$ points inside the circle


Let $ P$ be a point in the interior of $ \angle QOR$. Find the segment $ AB$ of minimum length which contains $ P$ with $ A$ on the ray $ OQ$ and $ B$ on the ray $ OR$.

Is this so difficult problem or just nobody is interesting in it...


I am not looking for an answer to a math problem per se (sp?) however, I am looking for an answer regarding the AMC 12. Recently, my friend showed me a site where they post people's names and their AMC 12 score. Now, I don't feel too comfortable with people knowing my AMC 12 score or other's scores for that matter, however do these people have a right to do this?

Thank you,
The Idiot without a village

BTW, here's the link [only for people's scores in California]

http://www.santarosa.edu/mathematics/AMCPages/CAHONTMA12.HTML

And you can find scores of people from all the AMC contests here [again only for California]

http://www.santarosa.edu/mathematics/AMCPages/[/url]

I think it's a good thing. It allows math people in high schools to figure out which people from local middle schools are good at math and which people they should train. As for invasion of privacy, why does it really matter if other people know your AMC12 score? Who is going to go through that long list to look for you? I doubt it will be very many people...

I agree with Simon - it's not like people are actually going to memorize that list...I don't know if they have a right per se to do so...

Also for competitions like ARML, the coaches may use the AMC scores as one of the criteria for team placement. The SFBA team does that, but I didn't know how our coach got the scores before. I guess he uses that website or another one like it.

This isn't just true for California.  I found this page listing the Nevada results from 2002: [url]http://unr.edu/homepage/lamberth/AMCresults02.pdf[/url] and 2001: [url]http://unr.edu/homepage/lamberth/AMCresults01.pdf[/url]
Like the California page, this page lists the top scorer in each school, as well as high scorers and AIME qualifiers.

I personally don't have a problem with this.  These pages are there [i]to give you recognition.[/i]  They're there so that people can look and see your name.  They aren't there so that someone can make fun of your score, and in fact, if you didn't do very well and didn't have a very good score, it wasn't posted there.  Besides, some colleges may even be interested in this sort of thing, which would help you when you're applying for college.

(Remember the "bubble" on the AMC answer form asking if they can add your name to a list of high scorers, which colleges/math programs can buy?  Incidentally, I think this was discussed at the mathcounts.org message board as well - and Eve may remember that one also!)


Can somebody explain to me why the extraction method works? The one where you estimate the square routes, because I don't get why you have to multiply by 20 and why you do what you do

could you describe it quickly on the forum, or give a link?


analytic geometry
geometry
geometric transformation
rotation
trigonometry

The points (0,0), (a, 11), and (b, 37) are the vertices of an equilateral triangle.  Find the value of $ab$.
[hide="hint"]think about rotations[/hide]

Actually all you need to do is use cosine and sine and substitue back in..and this has already been posted..


search
number theory unsolved
number theory

m/n=1+1/2+1/3+...+1/p-1 m and n are integers with greatest common devisor equal to 1and p is a prime number >5.Show that p^2 devides m

For odd p we can divide the fractions into pairs:
$\frac m n=\sum_{i=1}^{\frac{p-1} 2} \frac 1 a+\frac 1 {p-a}=\sum_{i=1}^{\frac{p-1} 2}
\frac p {a(p-a)} $

[quote="Kondr"]For odd p we can divide the fractions into pairs:
$\frac m n=\sum_{i=1}^{\frac{p-1} 2} \frac 1 a+\frac 1 {p-a} $[/quote]
But that is not enough it will only show that p devides m

[quote="Arne"]Search for... Wolstenholme, I think. Well-known.[/quote]
I tried,  no results

Strange, I put "like Wolstenholme" and I got a huge proof of Darij. Of course, it can be made much shorter but it is very well explained. Try to do like me with the search button.

[quote="Arne"]Search for... Wolstenholme, I think. Well-known.[/quote]
Sorry, man I don`t know what you may think of me but I tried a few times and haven`t found anything.

Look here

[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=wolstenholme&t=302[/url]

for example.

I just can't undersatnd that, but probably its easier to write a solution here. Let $S$ your sum. Observe that $2S=\sum_{k=1}^{p-1}{\frac{p}{k(p-k)}}$ so you need to prove that $\sum_{k=1}^{p-1}{\frac{1}{k(p-k)}}$ has numerator a multiple of $p$. Observe that using congruences with fractions, the last sum is congruent to $-\sum_{k=1}^{p-1}{\frac{1}{k^2}}$ and the last one has numerator a multiple of $p$ (use the fact that $Z_p$ is cyclic group).


Hi,

I didn't know where I could post my topic, so I choose to post here.

I've a project for my highschool. I've to deal on the subject "fourth dimension" with the problematic "How can we visualize fourth dimension's concept thanks to mathematic tools ?". And I've a question, a neurobiological question :

Why humans can't see physically objects in fourth dimension ? What's the thing in our brain (or the lack of this thing) which prevents us to seeing objects in fourth dimension, and more ?

Thanks you.

Don't hesitate to give me websites about it !

Short answer:  we don't [b]need to[/b]!

Longer answer:  nothing in our evolutionary history puts any evolutionary pressure on us to visualize four-dimensional objects when the geometry of the world we can experience is patently three-dimensional.  Nothing your brain can do is useless; every basic function of our neurological tissue stems from a survival need (such as the ability to remember where to find food or water, or the ability to use tools to get food or water).

Anyway, this should've been posted in Biology or Physics.


I am confused about this

On the one hand you have E= m c ^2 and w f = v (where w= wavelength and f = frequency)
so a wave can have momentum which equals m v = (E/c^2)*v = E/c (since v=c)

But a wave doesn't have any really mass so how can it have any momentum?
(i.e m = 0  because if m was not zero then a wave would have weight m*g and I don't think this can be true.
A wave cann't have a nonzero weight ? )

I haven't read into relativity indepth but E=m c^2 is rest mass .
I thought mass is the intuitive idea of how much matter there is in an object.
And Matter is anything that takes up space and has mass.

But now I am reading their are different types of mass.
I know their are different types of atoms so we have different types of matter. However mass should
just be an inherant property of the mattter.

Because know what does F= ma mean what mass???
Could I sub E/c into F=ma . Don't think so?
It seems like having different definitions for mass would make formulas like F= ma confusing.
Is their like a default mass where if you don't specify it , it is like implied to be the mass definition your talking about.

any clarity on this would be great because I just don't see How momentum when mass equals zero can be mathematically true. Unless we are talking about different momentums which would even complicate things even more.

Waves due to duality can also be seen as particles -- the particles have momentum so waves have momentum.

$ P \equal{} mv$ for momentum is only for nonrelativistic (sp) speeds of particles i think.
I think a broader equation is $ E^2 \equal{} p^2 c^2 \plus{} m^2 c^4$, and then you'll find that waves do have momentum.

I seem to have remembered that from some browsing around earlier,
correct me if i'm wrong.

Ok but if a wave has momentum and
E^2 = p^2 c^2 + m^2 c^4

What is E=mc^2 for?
It is when p=0 but why is this equation so important.
What is so special about p = 0?
I am assuming that p = momemtum in this equation.

And if a wave had  nonzero mass then it would have a nonzero
weight and would fall to the ground. So this would mean that
m = 0 for waves. But then we have E = p c but then p = mv
and so E = m v c. But then v = c because it is an electromagnetic wave so we get E=mc^2 again?

Which would still mean it would have a nonzero mass which would imply a nonzero weight?

I think the problem has to do with either the type of momentum we are talking about.
I was under the impression that momentum is p = m v?
But if we are talking about angular or some other momentum then please let me know.

I also think that E=m c^2 is only when an object is at rest v = 0
(in the p = mv) . Hence why the call it rest mass , rest energy.

Anyway I am still confused to say the least.

I think we should make it explicit we are speaking of electromagnetic waves, since for mechanical waves, for instance, it’s an obvious concept (I hope…)

Even in classical electromagnetic wave theory we can find the idea of momentum, when  analyzing Poynting Vector, say, the electromagnetic energy flux.
Remember that momentum is the ability to apply a force, and its not sufficient the mass: there is a need of velocity too.
Remember, also, that in 1871  Maxwell deduced theoretically the pressure of an electromagnertic wave, confirmed experimentally by Lebedev in 1900 and Nichols in 1901,and we can ”see” it on a Nichols radiometer.

Electromagnetic fields are able to apply a force. Thus they must carry momentum to ensure momentum is conserved.

[quote="Mathematics!"]Ok but if a wave has momentum and
$ E^2 \equal{} p^2 c^2 \plus{} m^2 c^4$

What is $ E\equal{}mc^2$ for?
It is when $ p\equal{}0$ but why is this equation so important.
What is so special about $ p \equal{} 0$?
I am assuming that $ p \equal{} \text{momemtum}$ in this equation.[/quote]

$ E\equal{}mc^2$ is the case for a particle at rest (with zero momentum).  It gives
us the rest energy of a particle.

[quote="Mathematics!"]And if a wave had  nonzero mass then it would have a nonzero
weight and would fall to the ground. So this would mean that
$ m \equal{} 0$ for waves. [/quote]
Indeed, $ m\equal{}0$ for waves$ ^{[1]}$.
The relativistic justifications for what I am about to say are beyond my level of understanding,
but briefly, any particle with energy is affected by gravity.  Electromagnetic waves carry energy
(see below), and so are affected by gravity.  Indeed, one of the well-tested predictions of
general relativity is that light rays should be slightly deflected when passing by massive objects
(resulting in phenomena like [url=http://en.wikipedia.org/wiki/Gravitational_lensing]gravitational lensing[/url].
($ [1]$: Unless you are using the convention of relativistic mass; see end of post for details).

[quote="Mathematics!"]But then we have $ E \equal{} p c$...[/quote]
Exactly true.  [quote="Mathematics!"]but then $ p \equal{} mv$[/quote]
Not quite.  This equation is only valid when ignoring relativistic effects$ ^{[1]}$.  In fact, the correct
equation is $ p\equal{}\gamma m v$, where $ \gamma\equal{}\frac{1}{\sqrt{1\minus{}\frac{v^2}{c^2}}}$.
($ [1]$: Unless you are using the convention of relativistic mass; see end of post for details).

[quote="Mathematics!"]...and so $ E \equal{} m v c$. But then $ v \equal{} c$ because it is an electromagnetic wave so we get $ E\equal{}mc^2$ again?[/quote]
Well, you'll note that the constant $ \gamma$ is undefined for particles with $ v\equal{}c$, so this doesn't work (because $ \gamma m$ is not a well-behaved expression).  Suffice to say, as noted above, that $ E\equal{}pc$ is entirely correct (you can check this with some calculus by examining the limit as a particle's speed goes to $ c$).  Since we usually know the energy of a photon (using $ E\equal{}h\nu)$, this equation is often more useful in the form $ p\equal{}E/c$, or using the common convention in relativity of $ c\equal{}1$, $ E\equal{}p$.

[quote="Mathematics!"]Which would still mean it would have a nonzero mass which would imply a nonzero weight?[/quote]
No and yes (sort of); the rest mass of the wave is still zero, but it is influenced by gravity (but in a way, I suspect, that is more complicated than $ F\equal{}\frac{E}{c^2}a$, although I don't have the background to say for sure).  See also the note [1] about relativistic mass below.

[quote="Mathematics!"]I think the problem has to do with either the type of momentum we are talking about.
I was under the impression that momentum is $ p \equal{} m v$?
But if we are talking about angular or some other momentum then please let me know.[/quote]
We are talking about the relativistic analog of $ p\equal{}mv$, which is $ p\equal{}\gamma m v$.$ ^{[1]}$.  This preserves the proprety $ \frac{dp}{dt}\equal{}F$.

[quote="Mathematics!"]I also think that $ E\equal{}m c^2$ is only when an object is at rest $ v \equal{} 0$
...Hence why the call it rest mass , rest energy.[/quote]
Entirely true$ ^{[1]}$.

[quote="Mathematics!"]Anyway I am still confused to say the least.[/quote]
Hope that helped.

------
Note:
[1]: Relatvisitic mass:  Some conventions in relativity maintain that particles have a [i]rest mass[/i] $ m_0$ and a [i]relativistic mass[/i] $ m\equal{}\gamma m_0$.  In this case, many of the comments I made above do not apply, since we truly are reduced to $ E\equal{}mc^2$, even for moving particles.  Note that for particles traveling at $ c$, $ \gamma m_0$ is undefined, so we set $ m\equal{}E/c^2$.  See [url=http://en.wikipedia.org/wiki/Relativistic_mass]wikipedia:Relativistic mass[/url] for more.

Ok, I forgot about the whole relativistic thing.
But when you say p = \gamma m v .
Won't the p = 0 always since  m = 0 always for waves.
I know that the limit of \gamma m v as v --> c-  is mc/0+ which is 0 * +inf
But this is not truely an undetermined form because m = 0 always it isn't approaching zero. And when v --> c+ we have the square root undefined.
So the limit doesn't exist only the right side limit which must be zero unless
the mass is a function of v in some way?

And you brought up another question E = h v. Uh, I thought we had
E = f h and if we put this equation interms of v we get E = v h/ w
(where w = wavelength I am using the equation w * f = v )
Unless we are talking about different h's somebody is wrong!

If your formula is correct E = hv and we have also E = p c then
we have hv = pc which implies p = h if v = c ???
What are you talking about convention in relativity c =1  E = p ???
[quote]
I suspect, that is more complicated than
F = \frac{E}{c^2}a
[/quote]
I agree but wouldn't you just use the derivitive of momentum definition instead
to find out F? I think this is the general definition of momentum. But I could be wrong. Your formula for force is assuming m is rest mass. This is where I am sort of lost. I think their should be only one mass definition or then wtf does F = ma
mean?

Anyway, thanks for the help I forgot to factor in relativistic stuff that was the flaw in my logic.
Because  I forgot that the standard Newton physic's doesn't work for thing's
close to the speed of light. Your equations for F is not relativistic either
I think even F = ma doesn't hold if we are close to v = c .

$ \vec{F} \equal{} m \vec{a}$ is indeed not valid for varying mass (or $ v\rightarrow c$). With some algebra you can get:
\[ \vec{F} \equal{} \gamma m \vec{a} \plus{} \gamma^3 m \frac{\vec{v} \cdot \vec{a}}{c^2} \vec{v}\]

[quote="Mathematics!"]Ok, I forgot about the whole relativistic thing.
But when you say p = \gamma m v .
Won't the p = 0 always since  m = 0 always for waves.
I know that the limit of \gamma m v as v --> c-  is mc/0+ which is 0 * +inf
But this is not truely an undetermined form because m = 0 always it isn't approaching zero. And when v --> c+ we have the square root undefined.
So the limit doesn't exist only the right side limit which must be zero unless
the mass is a function of v in some way?[/quote]
The limit in question should be one where energy remains the same, not mass.  We see that if $ E\equal{}\sqrt{(mc^2)^2\plus{}(pc)^2}$ is to be kept constant and $ m$ is to go to zero, then $ p$ must go to $ E/c$.

[quote="Mathematics!"]And you brought up another question E = h v. Uh, I thought we had
E = f h and if we put this equation interms of v we get E = v h/ w
(where w = wavelength I am using the equation w * f = v )
Unless we are talking about different h's somebody is wrong![/quote]
In fact, we are talking about the difference between $ v$ and $ \nu$; I was using the latter, the greek letter nu, which is used to denote frequency, just as $ f$ is.  $ E\equal{}h\nu$ and $ E\equal{}hf$ are the same equation.

[quote="Mathematics!"]What are you talking about convention in relativity c =1  E = p ???[/quote]
It is sometimes tiresome to write out all of the $ c$s when working with relativistic equations.  Sometimes we omit the $ c$s, with the understanding that if I write an equation like $ E\equal{}m$, there is an omitted factor of $ c^2$ on the right to make the dimesons correct.

[quote="Mathematics!"]Your formula for force is assuming m is rest mass.[/quote]
Which formula is this?  The only formula I assert for force is $ F\equal{}\frac{dp}{dt}$, unless you're referring to my assertion that $ F\equal{}\frac{E}{c^2}a$ is probably [i]not[/i] how the influence of gravity on massless particles works.

[quote="Mathematics!"]Your equations for F is not relativistic either
I think even F = ma doesn't hold if we are close to v = c .[/quote]
The equation $ F\equal{}\frac{dp}{dt}$ is correct in a relativistic setting; one might take it as the definition of force.  But you are correct that $ F\equal{}ma$ does not hold in the relativistic regime.

Question
Can nothing go faster then the speed of light
Or is it nothing can accelerate faster then the speed of light?

I am still wondering if their is a bound on an objects velocity or acceleration.

$ \frac{1}{\sqrt{1 \minus{} \frac{v^2}{c^2}}}$ is only vaild if  -c < v < c

But I haven't found a way to determine what happens at c or greater.
I would presume that it would be done by considering what happens to energy / momentum at v >= c.
Don't know?

I don't know, but see [url=http://en.wikipedia.org/wiki/Tachyon]Wikipedia:Tachyons[/url].

by SR nothing with mass can exceed the speed of light (requires infinite energy so not possible). theres a highly interesting field called quantum tunneling that a friend of mine did a tad bit of research in by which physicists are able to transfer information faster than the speed of light. im not exactly sure how this works tho...if you are interested i would definitely check it out.

no. information transfer is never possible faster than the speed of light. In fact, it is precisely this principle that physicists refer to when forbidding faster-than-light travel: theoretically, anything else is fine, as long as it does not transfer information. For example (this is very likely what you reference, as quantum tunneling is something entirely different), a quantum tangled state may be created of two or more particles that are then separated by some nonzero distance; if any is then measured by an observer, it will immediately collapse according to the measurement principle of quantum mechanics, and hence so too will all the others to the corresponding eigenstates in an example faster-than-light causality. This process cannot be used to transfer information, however.

OK, you would have infinity energy at the speed of light for any particle that has mass.

But what about a particle that had zero mass.
Is their any other particle then a photon that has zero mass?

Neutrinos are theorized to have zero mass in the Standard Model, but recent experiments involving neutrino oscillation have disproven this. I believe some of the gauge bosons in the other field theories are theorized as massless (well I'm pretty sure the W and Z are in fact pretty heavy, so probably I'm thinking of the gluons here). You seem to be pretty interested in this whole branch -- if you ever have some free time, I'd recommend wikipedia as an excellent information source. Just start at the page for Standard Model or whatever interests you and read on related topics as you wish.

Yes I have study some particle physic's but I really would like to know if anybody know's what would happen if a particle was mass less. Would the energy still be infinity so that not even massless
particles could go faster then the speed of light.

The proof of infinty energy for mass particles is simple.
It is just taking the limit as v -- > c . The energy approaches infinity. But when we have zero mass the formulas don't even apply.

So what would be the proof of when mass = 0.

Either way I think it will be impossible to go faster then the speed of light.

But could you accelerate faster then the speed of light for a short
period of time. Enough to not reach the speed of light.


Hello,

I have a couple really simple (for you guys) physics problems that I just can't get. I'd really appreciate any help by today since my test is tomorrow.

1) A skateboarder and skateboard have a combined mass of 68.8 kg. She is on an incline of .436 rad that has a coefficient of rolling friction between the wheels and road of .2. If she starts from rest, how fast will she be going after 25 m?

2) A heat engine operating between 40C and 380 C has an efficiency 60% of that of a Carnot engine opearting between the same temperatures. If the engine absorbs heat at a rate of 60kW, at what rate does it exhaust heat?

thanks a million in advance!

2) eficciency of a Carnot enegine working between $ T_{1}$ and $ T_{2}$ is $ \nabla\equal{}1\minus{}\frac{T_{1}}{T_{2}}$
also the efficiency of any engine is the $ \frac{W}{Q}$ where W is the work it does divided by the heat it absorbs
so just put the values and get the answer

1) use Work energy theorem
the frictional force is $ f\equal{}mg\mu  \cos \alpha$ work done by it in covering distance $ x$ is $ \minus{}fx$ also the work done by weight is $ mg x\sin \alpha$ also if $ v$ is the speed then
$ \frac{1}{2} mv^{2}\equal{}\minus{}fx\plus{}mgx \sin \alpha$

For 2) you must also take into account the first law of Thermodynamics:

$ Q_1 \equal{} W \plus{} Q_2$

where Q1 is the heat absorved from the hot reservoir and Q2 is the heat transferred to the cold reservoir.

is the answer for the ist question =22.65m/sec


a man is hunting. if the deer is 50 yards away, he has a 1/3 chance of hitting it. if the deer is 100 yards away, he will miss 8 times out of 9. if the deer is 150 yards away, he has a 26/27 chance of missing. the pattern continues. a deer is standing 50 yards away from him. after his first shot, the deer starts running away from him. the deer runs 75 feet per second. the man takes 4 seconds to reload his rifle. what is the probability that he hits the deer?

[hide]
The deer runs 300 ft every 4 secs.
$ p \equal{} \frac{1}{3^1} \plus{} \frac{1}{3^7} \plus{} ...$
$ 3^6p \equal{} 3^5 \plus{} p$
$ p \equal{} \frac{3^5}{3^6\minus{}1} \equal{} \frac{243}{728}$

I can never remember the formula for geometric series, but it's simple to do out.  :P
[/hide]

[hide="incorrect"]you missread the problem everything is in yards. except for the rate. so the deer runs 25 yards per second [/hide]

Oh Oops...  :blush:

Well, using that,

[hide]
$ p \equal{} \frac{1}{3^1} \plus{} \frac{1}{3^3} \plus{} ...$
$ 3^2p \equal{} 3 \plus{} p$
$ p \equal{} \frac{3}{8}$[/hide]

[hide]
He has a $ \frac{1}{3}$ chance of hitting it on his first shot, then after 100 yards (4 seconds for reloading), he has a $ \frac{1}{27}$ chance.
So, the probability is $ \frac{\frac{1}{3}}{1\minus{}\frac{1}{9}} \equal{} \frac{\frac{1}{3}}{\frac{8}{9}} \equal{} \frac{9}{24} \equal{} \frac{3}{8}$.[/hide]


Let be $ x,y,z\in (0,\infty)$ and $ n\in \mathbb{N}$ . Show that :

i)$ \frac{x^n\plus{}y^n}{x^{n\plus{}2}\plus{}y^{n\plus{}2}}\plus{}\frac{y^n\plus{}z^n}{y^{n\plus{}2}\plus{}z^{n\plus{}2}}\plus{}\frac{z^n\plus{}x^n}{z^{n\plus{}2}\plus{}x^{n\plus{}2}}\le \frac{x\plus{}y\plus{}z}{xyz}$

ii)$ \frac{x^{n\plus{}2}\plus{}y^{n\plus{}2}}{x^n\plus{}y^n}\plus{}\frac{y^{n\plus{}2}\plus{}z^{n\plus{}2}}{y^n\plus{}z^n}\plus{}\frac{z^{n\plus{}2}\plus{}x^{n\plus{}2}}{z^n\plus{}x^n}\ge xy\plus{}yz\plus{}zx$

[quote="alex2008"]Let be $ x,y,z\in (0,\infty)$ and $ n\in \mathbb{N}$ . Show that :

i)$ \frac {x^n \plus{} y^n}{x^{n \plus{} 2} \plus{} y^{n \plus{} 2}} \plus{} \frac {y^n \plus{} z^n}{y^{n \plus{} 2} \plus{} z^{n \plus{} 2}} \plus{} \frac {z^n \plus{} x^n}{z^{n \plus{} 2} \plus{} x^{n \plus{} 2}}\le \frac {x \plus{} y \plus{} z}{xyz}$

ii)$ \frac {x^{n \plus{} 2} \plus{} y^{n \plus{} 2}}{x^n \plus{} y^n} \plus{} \frac {y^{n \plus{} 2} \plus{} z^{n \plus{} 2}}{y^n \plus{} z^n} \plus{} \frac {z^{n \plus{} 2} \plus{} x^{n \plus{} 2}}{z^n \plus{} x^n}\ge xy \plus{} yz \plus{} zx$[/quote]

$ (x^{n\plus{}2}\plus{}y^{n\plus{}2})\minus{}xy(x^n\plus{}y^n)\equal{}(x^{n\plus{}1}\minus{}y^{n\plus{}1})(x\minus{}y) \ge 0$
[b]i)[/b] So $ \frac{x^n \plus{} y^n}{x^{n \plus{} 2}\plus{}y^{n\plus{}2}} \le \frac{1}{xy}$....

$ \frac {x^n \plus{} y^n}{x^{n \plus{} 2} \plus{} y^{n \plus{} 2}} \plus{} \frac {y^n \plus{} z^n}{y^{n \plus{} 2} \plus{} z^{n \plus{} 2}} \plus{} \frac {z^n \plus{} x^n}{z^{n \plus{} 2} \plus{} x^{n \plus{} 2}}\le \frac{1}{xy}\plus{}\frac{1}{yz}\plus{}\frac{1}{zx}\equal{}\frac {x \plus{} y \plus{} z}{xyz}$

[b]ii)[/b] $ \frac {x^{n \plus{} 2} \plus{} y^{n \plus{} 2}}{x^n \plus{} y^n} \ge xy$......

$ \frac {x^{n \plus{} 2} \plus{} y^{n \plus{} 2}}{x^n \plus{} y^n} \plus{} \frac {y^{n \plus{} 2} \plus{} z^{n \plus{} 2}}{y^n \plus{} z^n} \plus{} \frac {z^{n \plus{} 2} \plus{} x^{n \plus{} 2}}{z^n \plus{} x^n}\ge xy \plus{} yz \plus{} zx$


The circle has center (-2,-4) and passes through point (6,14). How would I find the equation of the circle described and sketch a graph about it?

The distance between center & point which passes through the circle, gives the radius.

Plug the values in $(x-h)^{2}+(y-k)^{2}=r^{2}$ (principal ec. of circle) and that's all.

[hide="solution"]We find the distance between the two points(the radius), and that's $2\sqrt{97}$.

Since the center is (-2,-4), The equation would be

$(x+2)^{2}+(y+4)^{2}=388$

We plug in (6,14), and we see that it works.[/hide]


inequalities
trigonometry
geometry
perimeter
function
trig identities
Law of Sines

Write as
\[\displaystyle \frac{1}{1+\cos{A}}+\frac{1}{1+\cos{B}}+\frac{1}{1+\cos{C}}\leq\frac{2R}{2r}\]
Dividing by $2R$ and using the law of sines the inequality becomes
\[\frac{\sin{A}}{a(1+\cos{A})}+\frac{\sin{B}}{b(1+\cos{B})}+\frac{\sin{A}}{c(1+\cos{C})}\leq\frac{R}{r}\]
using the tan half angle identity
\[\frac{\tan{\frac{A}{2}}}{a}+\frac{\tan{\frac{B}{2}}}{b}+\frac{\tan{\frac{C}{2}}}{c}\leq \frac{1}{2r}\]
It has been shown many times that $\tan{\frac{A}{2}}=\frac{r}{s-a}$.  WLOG $a+b+c=1$, so
\[\displaystyle \frac{1}{a(\frac{1}{2}-a)}+\frac{1}{b(\frac{1}{2}-b)}+\frac{1}{c(\frac{1}{2}-c)}\leq \frac{1}{2r^2}\]
Applying Jensen's to the LHS we must prove
\[\displaystyle r\leq\frac{\sqrt{3}}{18}\]
However $r=\frac{K}{s}=2K$ so
\[\displaystyle K\leq \frac{\sqrt{3}}{36}\]
But for a triangle with a given perimeter its area is maximized when the triangle is equilateral.  Take $a=b=c=\frac{1}{3}$ to achive the desired result.

I  dont think  your  function is  concave  at the  interval  $(0,\frac{1}{2})$

[quote="cezar lupu"]Prove that in any triangle $ABC$ the following inequality holds:

     $\frac{1}{1+cos A}+\frac{1}{1+cos B}+\frac{1}{1+cos C}\leq\frac{R}{r}$ ;)[/quote]

So we got :

$\sum_{\text{cyclic}}\frac{1}{1+\cos A}=\sum_{\text{cyclic}}\frac{2bc}{(b+c)^2-a^2}$
Now use Ravi's substitution:
$\sum_{\text{cyclic}}\frac{(x+z)(x+y)}{(x+y+z)x}=\frac{1}{2(x+y+z)}\times\sum_{\text{cyclic}}\frac{(x+y)(x+z)}{x}$
and also
$\frac{R}{r}=\frac{(x+y)(y+z)(z+x)}{4xyz}$
So it's enought to prove
$x+y+z\geq 2\sum_{\text{cyclic}}\frac{yz}{y+z}$
which is  obvously true from the fact that $\frac{yz}{y+z}\leq \frac{y+z}{4}$

let $cosA=x, cosB=y, cosC=z$. using the fact that $\frac{r}{R}+1=x+y+z$ we find that
$\frac{R}{r}=\frac{1}{x+y+z-1}$
 then the original inequality is in fact equivalent to
$\bigl(\sum_{cyc}(1+x)(1+y)\bigl)(x+y+z-1)\leq (1+x)(1+y)(1+z) \\\Leftrightarrow (x+y+z)^2+(x^2+y^2+z^2+2xyz)+\sum_{cyc}x^2y+x^2z\leq 4 \\\Leftrightarrow (x+y+z)^2+\sum_{cyc}x^2y+x^2z\leq 3$

now, it's very well-known that $x+y+z\leq \frac{3}{2}$, so, we'll have to show that
$\sum_{cyc}x^2y+x^2z\leq \frac{3}{4} \\\Leftrightarrow \sum_{cyc}(4x^2)(y+z)\leq 3$,
but by am-gm and am-qm we have that
$\sum_{cyc}(4x^2)(y+z)\leq\frac{2x+\sqrt{y+z}}{2}\leq(x+y+z)+\frac{\sqrt{6(x+y+z)}}{2}\leq \frac{3}{2}+\frac{3}{2}=3$, and we're done!!!  :D


Find all pairs of positive integers (x,y) s.t. x+y2+z3=xyz, where z=(x,y).

grobber the actual convention is ISL for [b]I[/b]mo [b]S[/b]hort [b]L[/b]isted ((c) orl :D)

I knew there was something wrong! :D Can the title be edited?


Asymptote function algebra polynomial quadratics

we use long division to find the slant asymptote: y=3x/2-1+1/2x:^2:. Okay, now as x gets really big, the 1/2x:^2: term pretty much goes away, so we're pretty much left with y=3x/2-1, which is the slant asymptote.

The horizontal asymptote will be y=c, where c is the value that the function approaches as x goes to :pm::infinity:. Your example function doesn't have a horizontal asymptote, as when x goes to :pm::infinity:, so does the function, but a function like (1-2x:^2:)/2x:^2: would have the horizontal asymptote y=-1, since the function approaches that value as the magnitude of x gets really big.

the power of the numerator has to be one greater than that of the denominator to have horizontal asymptotes.

Yeah horizontal is only when the power is the same.

for quotients of polynomials:

slant is when power of num. > power of denominator

horizontal is when power of num. :le: power of denominator

if power of num. < power of denom., the horizontal asymptote will just be y=0.

earlier i ment slant, sorry.

mystic are you sure slants happen when top degree is greater than bottom? i always though it has to be one more.

I'm pretty sure that in order to have a slant asymptote we must have top degree=bottom degree+1. Just think about how the slant asymptotes are calculated for a function f: first of all, you have to calculate the limit of f(x)/x as x->00 and x->-00, and in order to have slant asymptotes one of these limits must be finite and non-zero.

Yeah, after the degree is more than one bigger, you would get some sort of "quadratic" asymptote, etc. Its only a straight line if its one more..

Well, there is a general method to find out whether a function has a slant asymptote, this method also works for exponential, logarithmic, irrational, ... functions, not only rational ones.

Let f(x) be a given function.
We will examine whether f has a slant asymptote for x going to infinity.
Calculate limx->+oof(x)/x.
If limx->+oof(x)/x is 0 or oo, then there is no slant asymptote.
But if limx->+oof(x)/x is a real number, call it m, then there will be a slant asymptote.
So suppose limx->+oof(x)/x = m is real.
Then calculate limx->+oof(x) - mx = q.
In that case, y = mx + q is a slant asymptote for the graph of f.

I'll give an example.
We look for slant asymptotes of f(x) = ln(ex - x). Then
limx->+oof(x)/x
= limx->+ooln(ex - x)/x
= limx->+oo(ex - 1)/(ex - x)
= 1
after some applications of l'Hospital. So m = 1. Then
limx->+oof(x) - mx
= limx->+ooln(ex - x) - ln(ex)
= ln[limx->+oo(ex - x)/ex)]
= ln 1 = 0
So q = 0. Hence y = mx + q = x is a slant asymptote.


Hi all

It's great that we Africans now have our own forum! So come on, Mathlinkers, participate in this forum and encourage more people from your country to become members!

Let's all introduce ourselves. I'm Melissa Munnik, from Cape Town, South Africa. I'm 16 years old, in 10th grade. I like doing maths (obviously), playing piano, playing chess, riding bicycle and swimming.

Please reply!

Hello, I am Arne, 18 years old, and although I have the Belgian nationality, I am "partially South African :D". I am starting my first year of university now (mathematics and physics), I went to IMO twice, I am IMO coach in Belgium (and last year I was coach in South Africa). I enjoy swimming, cricket, the piano, classical music, friends, etcetera :)
And, by the way, I will be the moderator of this forum.

Well I'm Ixeus, 25 years old, also from South Africa - Live in Cape Town!  Nice to see something for South Africa here, hope we can get some more people here.  I remember math olympiads when I went to school :D

i'm fermat -euler !!!!! i'm morrocon i like maths and i want that every africain participate in this forum  :lol:

[quote="Arne"]I think the "official languages" in this forum should be English and French...

Any comments?[/quote]
As I know, there are thousands of languages spoken in Africa. But I think those who can get to know the Mathlinks will certainly master English (and French).

[quote="Melissa"]Well, Fermat-Euler, you could start a thread in the African forum that is in Arabic (or whatever language you speak) :)  Maybe this will encourage other Morocceans to participate.[/quote]

i prefer speak french or  english in this forum because there are a members can't speak arabic

 :gathering:

Oh obviously I don't care, it was just a suggestion.

not probléme miss melissa it's just my opinion . and i can write in arabic sometimes

hi i'm maria.
i'm currently not in Africa, but i lived there most of my life. i'm originally egyptian and moved to NY 5 years ago. right now i'm a frosh in MIT :). hopefully this community will blossom ;)

well, i can write in arabic, just not on my computer...

Hi everybody
I am senouy (younes) from Algeria.
I am very old comparing to you .But comparing to your math level i am by far wearker than you for the moment beeing.Since i am married (four kids) i dont have to much time for leisure.Actually i am teacher in premium school,and i used to be a business man before.I am what we call a BAC+5 math major.But i stopped doing math for more than 20 years(you get an idea about my age).I came back to math when one day i was given a problem which is Collatz conjecture, and i tried to think about solution for more than four month day in day out......
But when i discovered mathlink.ro website it's becoming a passion now.
Here in Algeria we dont have any  organisation to attrack talented youth to math so as to participate in olympiads.I am thinking about it.And waiting for your advise to create something for  algerien math talented and how to do it.
Many thanks.
Younes.

bienvenue au forum african senouy (younes)  :welcome:

Hi,  I'm Brad I'm a postgrad student in Cape Town in South Africa.  I've been helping out with our IMO and PAMO training for the past while,  I was deputy leader at the last two PAMOs.  I went through the program when I was at school and went to IMO 99.

I hope to go to imo 2007 too  :lol:
Maybe I'll meet you there

Hey! I'm Kylie.  I live in Cape Town but originally lived in East London (both in SA).  I'm going to PAMO this year and would love to go to IMO next year but i really really doubt i will  :(  It's really cool there are so many Africans on here.... how do African countries do in IMO?
I know SA usually comes in the 30's or 40's.

[quote="Ashegh"]hi...

im ashegh, im 17 years old, i study in atomic energy high school.

i live in tehran , the capital of iran,

my intrest is inversion,

i love math...

i have $1$, girl friend which is $16$ , her name is $jigartala$.

and she is also here, in math links. :D[/quote]

USA have problem with you! Hahahaha!

I'm Fitim Dika (18 years), from Macedonia! I love maths and literature.

[quote="Fitim"][quote="Inspired By Nature"][quote="Fitim"]USA have problem with you! Hahahaha![/quote]
 :?: What is that supposed to mean?[/quote]

What are you doing?

USA have problems with Iran for Nuclear Atomic. Big problems!!!

 :ninja:[/quote]
oh, i didn't know you were talking in that sense.

[quote="Inspired By Nature"][/quote][quote="Fitim"][quote="Inspired By Nature"][quote="Fitim"]USA have problem with you! Hahahaha![/quote]
 :?: What is that supposed to mean?[/quote]

What are you doing?

USA have problems with Iran for Nuclear Atomic. Big problems!!!

 :ninja:[/quote][quote="Inspired By Nature"]
oh, i didn't know you were talking in that sense.[/quote]

Sorry!

dears i think it is not a good place to talk about politics, :wink:

i only say that nuclear energy is our definite right,

and USA cant privent us from it. :D

[quote]USA have problems with Iran for Nuclear Atomic. Big problems!!! [/quote]

Why big problems?

[quote="Inspired By Nature"]no need to apologize, happy to meet you fitim. :)[/quote]

why no?

fitim shouldnt say those words,

why doesnt he came and answer amir.s question?

(im sorry to changing the sense of this topic, i had to do it.) :wink:

[i]my name is okeke arthur ugochukwu iam from nigeria.its quite fun knowing all of you.i love maths and i love you all.to express my love this a question for you all.if d(n) is the laregest odd of n that is d(2)=1,d(6)=3 find d(1)+...+d(2^99)[quote] i am falling in love at just 16 years[/quote][/i]

my name is okeke arthur i am a 16 yera old nigerian.i love math.in our place the girls are usually weak in math but seeing mellisa wow! thumbs up.how can i know if there are other nigerians around.i love you all. :lol:

hello, i am the a 18 yr old  nigerian....ı participated in IMO 2007 and PAMO in 2008....ıts nice to meet u all here


Determine all ordered pairs $(m,n)$ of integers which satisfies the following equation.

\[3^{2^m}+3^n=2^{m+1}+2^{3^n}\]

I presume that you mean in non-negative integers?
In that case, the solutions are $m=n=0$ and $m=n=1$.

Assume that $m,n \geq 2$ satisfy the equation.
Let $m=3p+r$, with $r=0,1,2$.
Then the LHS is 0 modulo 9, since the RHS is $-1+(-1)^p2^{r+1}$, which forces $r=2$.
It follows that the RHS is now 0 modulo 8, since the LHS is $1+3^n$, and the latter is never 0 modulo 8, a contradiction.

Therefore, $m < 2$ or $n < 2$. Direct checking leads to the above solutions.

Pierre.

Merci beaucoup pour votre reponse, Pierre.

You  are right.

kunny

Wow! Are you a friend of Harazi?  :D  :D  :D  :D  (sorry, private joke).

Pierre.

Come on, Pierre, admit that I amazed you with my magnific :D  :D  knowledge of french last Sunday!  ;)  ;)


Let a checkerboard have the size $n\times n$ (for $n>1$). And distinct integers from $\{1,2,...,n^{2}\}$ are placed on this checkerboard. We say that two squares are "adjancent" iff they are side by side (horizontal or vertically). Show that no matter how these integers are placed there are two adjacent squares whose difference is at least $n+1$.

My work so far was to select the biggest difference possible among these and claim it mist be $\geq n+1$. Because it not assume that $\leq n$ and arrive at a contradiction, but I cant get passed that step.  :(

It doesn't seem to be true... try the simple $2 \times 2$ square:

---------
| 1 | 2 |
---------
| 3 | 4 |
---------

No adjacent squares have difference at least $3$. In fact, placing the $n^{2}$ simply in order is always a counterexample since for any square, the left square is one less, the right square is one more, the square above is $n$ less, and the square below is $n$ more (in the cases that those squares exist).

EDIT: me@home, your hint should work when the problem includes diagonals as adjacent, I think.

Oh I misread and thought adjacent includes diagonals. :oops:

This is so embarrasing guys  :blush: , so sorry, adjacent means diagnolly also. No wonder I coundnt prove it, cuz it false! I think now I can do this problem.

[quote="me@home"][hide="Hint:"]consider paths between largest and smallest[/hide][/quote]

[hide="Further Hint"]
$n \cdot n < n^{2}+1$

[hide="Even Further Hint"]
Pigeonhole Principle.  :wink:  [/hide] [/hide]

[quote="me@home"]This is really lame :oops: the only reason we can sound smart and give hints is because this was like the first problem they went over in woot :P[/quote]
At least you admit that you done these problems before. I think I am beginning to realize IMO (or close to it)  level problems are a lot about experience not just creativity.

In fact, it is true that there are always two adjacent squares differing  by at least $n$, not counting diagonals as adjacent. This is considerably harder to prove.

We write $1$~$n^2$ one by one.If at some step the number of blank squares which are adjacent to the written squares is over $n$,the conclusion holds.We consider the time at which some row(or some column) is completed(all squares on some row are written numbers).

[u][b]Case1[/b]:only some row is completed[/u]
Because any column isn't completed,there is blank square on each column which is adjacent to the written squares.Then the conclusion holds.

[u][b]Case2[/b]:only some column is completed[/u]
This case is the same as case1.

[u][b]Case3[/b]:some row and some column are simultaneously completed[/u]
Just before completing some row and some column,there is blank square on each column which is adjacent to the written squares.Then the conclusion holds.

From case1,2,3,the proof is completed$\blacksquare$ :D

[hide=Solution to Diagonals Problem]
Assume for sake of contradiction that each difference is $\leq n$.
Take the square where $1$ is placed.
All other squares are within $n-1$ adjacencies of this square, thus all squares are $\leq n^2-n+1<n^2$, a contradiction.
[/hide]
This problem is one of the example problems for the extreme principle found here:
http://www.artofproblemsolving.com/wiki/index.php?title=Extreme_principle

Hello,gradysocool. :D

We don't count diagonal as adjacent.So we cannot say that all squares are $\le n^2-n+1$.

Thanks


....and it was quite easy. A few of the sentence completions had some vocab that I was unfamiliar with, and I had a little trouble thinking of examples for the essay, but other than that, the test was easy. Don't even get me started on the math  ;)

Overall... some messy parts, but easier than expected!

Any thoughts on today's test?

Eh, I don't know...

It was very, very LOOOOOONG, plus I had horrible neck pain from sleeping 12 hours last night...very tedious. Not to mention the pressure to beat my 1530 from October..

But I knew all the vocab exc 1 (but I got it after eliminating the 4 others :) I really couldn't afford to mess up on those sentence completions), and I didn't get lost on any reading passage like I did last time. But I HATED those reading analogies!! Those "The relationship between this and that mentioned in the passage is most similar to:" were mind-wracking. I'd've preferred those old analogies, but oh well.

The essay was okay... My intro was, as usual, very flat and boring (2-3 sentences basically restating the prompt) but at least I didn't blank out and thought of some examples (I used a movie I'd seen on Wednesday :D and parts of the world history material I'd studied for yesterday's final). I wrote about a page (25 lines) but managed to fit in 10+ words per line, so length was good.

Math was, as expected, boring. My average time for each section was less than 10 minutes, so I had to spend the rest of the time checking, re-checking, and checking again.

Anyway, I'm glad it's over.. (and I get to see Star Wars III today!! w00t!)

the sat was long and boring.. i also have a really bad neck pain..

te english part was a bit hrd for me because english is not my first language but i think i did ok.

I only wish I had a bit more time for the essay. I had to stop midsentence and in the middle of explaining an idea.....I only my grader can infer from my introduction and my previous sentences.

If I do not get an 800 on this math..... :mad:

I hated some of the "choose the best answer...." For some of those, I could have given a good argument for at least 2 choices. That is what frustrates me the most, because in some cases there is no one correct answer. For math (at least this math ;) ) there is only one.

Still, I had pretty good stamina for this test. I only mind-blocked on the 10th section, and then I could just persever the rest of the questions. This test is just as much endurance as knowledge, perhaps more so.

mcalderbank, nice avatar.

i just saw Diaries of a Motocyclist.. the movie..
great great movie..

did you see it?

[url=http://rogerebert.suntimes.com/apps/pbcs.dll/article?AID=/20041001/REVIEWS/40920008/1023]Ebert's review[/url] of The Motorcycle Diaries is entertaining.

As for the SAT, it was all right, but I had to use the bathroom very badly and couldn't concentrate on one of the sections (math with the grid-in); I blame this on the test's  new attrocious length.  Except for that extraterrestrial passage, I thought it was all right.

[quote]But I HATED those reading analogies!![/quote]

I liked them  :lol:

Oh, and there were supposedly many versions of the test this time, crazy ....

what did u guys get for:

the stab in the dark and what it meant:... it was either vast task or hopeless (i put vast)

the writing that had: "the gardners will use fertilizers more if they can be shown to work" (i put "as it is" (A), although all the answers seemed wrong...)

also, does anyone know what a 10 on the essay and one wrong equates to (scaled)?

"A stab in the dark" means a guess.  As in, "I took a stab in the dark to choose the answer to that question."

[quote="tokenadult"]I didn't see the exact wording of the question, but "stab in the dark" implies an effort with very little chance of success, so "hopeless" makes sense as an answer choice.[/quote]

The article was about the search for extraterrestrial contact, and the "stab in the dark" was one of the methods the scientists were using to try and accomplish this.

The question was one of those "what does in mean in this context" questions.

The author appeared to be on the side of the scientists, so to say they are dabbling in a hopeless venture didn't seem right.  The only other reading question I was ambivalent over was "what did the scientist mean by 'It's only fair game.'"

[quote="albertfermat"]the writing that had: "the gardners will use fertilizers more if they can be shown to work" (i put "as it is" (A), although all the answers seemed wrong...)

also, does anyone know what a 10 on the essay and one wrong equates to (scaled)?[/quote]

I thought that this was a pronoun error, since we can't be sure which "they" the sentence is refering to, but it looked like all the other choices had errors, so I chose A.

sonar...i think the fair game was was ez...tell me what u put and i will tell if it is correct (i am 100% sure of my answer...I just dont remember what it was...)

the vast i think is correct because it mean that it is difficult, but not hopeless (or impossible) because stab in the dark means there is room for success...

i agree with the A for the question...i think ALL of the questions had THEY or THEM...so the only logical response would have to be A...anyone else who is a grammar expert want to say what the answer is (or what they think it is)

I can't remember, but I was deciding between one and something about obligations.

sorry...i dont remember...if anyone can remind me what the choices were...i could respond..

[quote="tokenadult"][quote="juicybooty911"]how do you study for the verbal part? i just can't remember all those words :?[/quote]

The way to study for the verbal part is always the same: READ, READ, READ, and READ. Read about subjects you like. Turn off the TV and read. Put away the video game and read. Read on the bus. Read while hanging around outside. Just be sure to read.

Oh, by the way, did I mention it's important to read to do better on the SAT verbal section?[/quote]

This really is true.  My freshman SAT verbal and junior SAT verbal are almost identical (690 and 720 respectively), and I honestly only read a few books in that period (sad, yes, I know).  However, I read quite a bit in elementary and even more in middle school.

Read a newspaper, especially since it will help prevent you from becoming ignorant of the times.

This may sound a little bit crazy, but umm besides reading, which is a must, i think rapping, yes i said rapping, might help especially if u like rap. Check out the flocabulary song on SparkNote. Its hilarious.

what do u guys think i would get in each section if i got:

one wrong math
two wrong verbal
two wrong writing, 8-12 essay (give a range for this)

can anyone tell me my scores based on previous curves??

[quote="themonster"][quote="sonar"]
Read a newspaper, especially since it will help prevent you from becoming ignorant of the times.
[/quote]
Newspapers typically don't have a ton of obscure vocabularly, but it's still good to know what is going on in the world.[/quote]

But the SAT's vocabulary isn't obscure--I knew all of the vocab on this SAT (I think I misbubbled one though (a friend and I disagreed on answer "imperious" was)).  Granted this is probably a result of my English class.

Nevertheless, if you read publications like The New York Times, The Washington Post, The New Yorker, Atlantic Monthly, et cetera, you will pick up "SAT vocab."  You can memorize vocabulary lists too; I bought Word Smart I and II, but I didn't even get through C in the first one (memorizing lists is very boring stuff).  In any case, even if you do memorize all the words, you still need to be able to understand what you read, since the critical reading section is mostly just that--reading.  You will not get through that part if all you know is words.

One wrong math - 760 (the curve is harsh now)
Two wrong verbal - 780-800
Writing, I have no idea

What did u guys put for the writing question about two kangaroos and water and had the phrase "the other kangaroo drinks [u]the greater amount[/u] of water..."? I think I put E, but many are saying the answer is [u]the greater amount[/u]. It just seems weird to me because if that weren't E, then the only other E for the Sentence ID's would be the question that was: "Shana had had..." One E for 18 questions seems wrong...

albert... You're not the only one who thinks it is E, but I really hope it is the greater amount :lol: .  It just sounds wrong.  If it was like "a greater amount" it would be better.... I'm not sure though.

Ragingg (or anyone else who wants to answer...),

      Who else is arguing it's E...?...also, how many Es did u get for the sentence IDs...i am pretty sure that ONLY shana had had was E (besides the greater amt, which might be wrong...)...doesnt that seem weird (one (or two) E out of 18...)??

Also, what did u put for the gardners question?... the one that had fertilizers and how using them can be shown to increase crop production....they all seemed wrong to me...so i just put A (as it is)...

btw...we get our scores on the 20th...

You know, the answers are not completely random (I think).

bump...
u
m
p
.
.
.

can someone explain what they put for the "gardeners and fertilizers" question and what they put for the "kangaroo and the greater amount" questions... and how many Es they got in the sentence IDs section...


algebra
polynomial
group theory
abstract algebra
number theory unsolved
number theory

[quote="MeKnowsNothing"]Determine all solutions of $ a^{3}+b^{3}+c^{3}=abc$ as $ a,b,c\in\mathbb{Z}$.[/quote]

$ a^{3}+b^{3}+c^{3}\geq 3abc \geq abc$ by AM-GM.

So $ a=b=c=0$ is the only possible solution.

I noted that too. What about if, for example, $ a<0$? At least $ (a,b,c)=(0,t,-t)$ and its permutations are solutions for every $ t\in\mathbb{Z}$.

Let $ x =\frac{a+b}{3a+3b+c}, y = \frac{a-b}{3a+3b+c}$.  Verify that the equation is equivalent to

$ y^{2}= 4(3x-1)^{3}-4x^{3}+x^{2}$

Which is an elliptic curve in (nearly) Weierstrass form.  Later I will find the rational points on this curve.

[hide="Is it a hint?"]If $ (a,b,c)$ is a solution then $ (k a, k b, k c)$ is also a solution[/hide]

Doesn't help much; that's true of any homogeneous polynomial, which is part of the motivation for the transformation I gave.  (The rest of the motivation comes from the theory of algebraic geometry.)

With a little help from SAGE, it turns out that the elliptic curve has rank zero and a torsion subgroup of order $ 3$, meaning the only nonzero solutions to the original equation are $ (1:-1 : 0), (1 : 0 :-1), (0 : 1 :-1)$ (projectively).  (Rank computations are possible by hand, but unfortunately a general method doesn't exist and the details are dense.  Perhaps a more elementary method would be in order for this particular curve; this method is, however, relevant to my current subject of research, so I took it up as an exercise  :) )


induction
combinatorics unsolved
combinatorics

Let $M=\{1,2,\ldots,3 \cdot n\}$. Partition $M$ into three sets $A,B,C$ which $card$ $A$ $=$ $card$ $B$ $=$ $card$ $C$ $=$ $n .$
Prove that there exists $a$ in $A,b$ in $B, c$ in $C$ such that or $a=b+c,$ or $b=c+a,$ or $c=a+b$

[i]Edited by orl.[/i]

[hide]Suppose $n$ is odd, then we have $\frac{3n-1}{2}$ pairs  of numbers which sum to $3n$. To avoid making $3n$ each pair is in the same set. BUT, since $n$ is odd the partition sets are odd size,and we split up a pair to cover the 3 sets.

Say $n$ is even, then we have $\frac{3n-2}{2}$ pairs which make $3n$. To preserve parity, both $3n$ and $\frac{3n}{2}$ must be in the same set. Now we see it is possible [b]not[/b] to make $3n$. However consider $3n-1$, it also has $\frac{3n-2}{2}$ pairs which sum to $3n-1$, a [i]different[/i] $\frac{3n-2}{2}$ pairs. So, for each number in the set with $3n$, and $\frac{3n}{2}$, (except $3n$), there are two different numbers which also must be in the same set. So we need $\frac{3n}{2}$, and $\frac{3n}{2}-1$, and $\frac{3n}{2}+1$, but then we need $\frac{3n}{2}-2$,....etc. we generate all of $\{1,2,3....,3n\}-\{3n\}$ in this manner.[/hide]

[quote="TheJerk"][hide]
To avoid making $3n$ each pair is in the same set. [/hide][/quote]

Wrong.

It's a really nice problem, but I won't yet post a solution, to give others the pleasure to find one.  :)

actually more is true: if the integers $1,2,...,n$ are colored with 3 colors with each color set receiving more than $n/4$ elements, then the equation $x =y+z$ has a solution with all $x,y,z$ receiving different colors.

the version i stated is actually a problem in one of the Schweitzer competitions(i don't remember which).i had a solution (sometime ago) but it is a rather long and nasty one. maybe vess can give something shorter and smarter.

I'm quite sure to have post a solution of the Schweitzer one here, but I can't find it anymore  :?
Idem for the original one  :?  :?  ;)

Pierre.

here is a solution for the problem in question(not the Schweitzer version):suppose not.then clearly $n \geq 2$.wlog $1 \in A$.consider $X=[3n] \setminus A$.since there is no solution to the equation $a=b+c$ with a,b,c in different sets, it follows that the elements of $X$ do not have consecutive elements in $B,C$ respectively.so X is partitioned(into B,C) and further B,C consist of unions of intervals(singleton sets are also regarded as intervals).suppose $[b_1,b_2] \subset B,[c_1,c_2] \subset C$ then there exists no element of $A$ of the form b-c.so if $b_2 < c_1$ then the interval $[c_1-b_2,c_2-b_1]$ is contained either in B or in C.starting with maximal such intervals of length $k,l$ respectively the length of the new interval is $c_2-c_1+b_2-b_1+1=(k-1)+(l-1)+1=k+l-1$.so if both $k,l \geq 2$ then this conradicts the choice of maximal intervals for both B and  C.
but if say $k=1$,i.e.B consists of isolated points only then $|A| \geq 2|B|-1 \Rightarrow |B| \leq 1$.contradiction.

[quote="Myth"]Please, excuse me my stupid questions:
1. why do you say that $[c_1-b_2,c_2-b_1]$ is contained either in B or in C. maybe some parts is contained in B and some parts in C :?
2.  your final contradiction "if B consists of isolated points then $|A|\geq 2|B|-1$" is unclear for me :blush:[/quote]
sorry for the lack of clarity.
1)firstly, this interval cannot contain any element of A,so it is contained in $B\cup C$. now since $1 \in A$ we cannot have 2 consecutive elements  of $X$ with one of them in B and the other in C.if the interval in question had points from both B and C then there would be consecutive elements with one of them in B and the other in C.(hope i am saying it clearly enough)
2)yah what i thought initially is not correct,since there can be strings of B's separated by the A's but i think it can be modified to still get the contradiction.the neighbors of the points of B cannot be points of C and so have to be points of A.so whenever the points of B occur they must occur in the form ABABA..BA,except if $3n \in B$ in which case that particular string contains equal number of A's and B's.now consider the strings ABAB..A as above.clearly each such string contains more elements of A than of B.Also,since there are elements of C present there have to be atleast 2 such strings and so $|A| >|B|$.(hopefullly this answers your questions)

Sorry, I was very tired last evening, and I was too lazy to think about your proof, so I wrote these stupid questions.

Yes!!!  :)
I found back the Schweitzer one :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=contest&t=3781

Pierre.

I remember seeing the following generalization:

Partition mn numbers (1,2,...,mn) to m sets such that each set has n elements, where m,n are arbitrary positive integers greater than 2. Show that it is possible to pick one element from every set such that sum of [m/2] of these chosen numbers equal to the sum of the rest m-[m/2].

Here is my solution :
Supose prroblem is false.and 1,2,...,k-1 in A,k in B.k>=2.
Let c in A => c-1 not in B.
if c-1 in C we have c-ik,c-ik-1 in C .Exist such that c-ik <= k contraction.
Since then c-1 in A.Let C={c_1,....,c_n} then A={c_i-1,i=1,2,...,n} .Wee have k-1 inA =>k in C (contraction

I think he means more or less the following, which is the solution we gave when we used the problem in our training :

We will say that $(a,b,c) \in A \times B \times C$ is good if one of $a,b,c$ is the sum of the two others.

Assume that there is no good triples.
Wlog, we may assume that $1 \in A$ and that the least number which is not in $A$, say $k$, belongs to $B$.
Let $x \in C$. We prove that $x-1 \in A$ :
Clearly, since there is no good triple, $x-1 \notin B$.
Suppose that $x-1 \in C$, and from the minimality of $k$, it follows that $x-1 > k$.
Since none of the triples $(x-k,k,x)$ and $(k-1,x-k,x-1)$ is good, it follows that $x-k \notin A \cup B$. Similarly, using the triples $(x-k-1,k,x-1)$ and $(1,x-k-1,x-k)$ we deduce that $x-k-1 \notin A \cup B$. Thus both of them belong to $C$.
By induction, it is now easy to prove that all the positive integers of the form $x-ik$ or $x-ik-1$ belong to $C$.
But, for some $i$, one of the numbers $x-ik$ or $x-ik-1$ is positive and not greater than $k$, so that it belongs to $A \cup B$. A contradiction.

It follows that, for all $x \in C$, we have $x-1 \in A$.
Let $C = \{c_1, \cdots, c_n \}$. Thus each of the numbers $c_1-1, \cdots , c_n -1$ belongs to $A$. But these numbers are pairwise distincts and all are greater than $1$ (since they all the $c_i$'s are greater than $k$). Thus $|A| \geq n+1$, a contradiction. And we are done.

Pierre.

Oh, it is so great to see nice solutions in the forum!


inequalities
calculus
derivative
function
inequalities unsolved

Master, you should delete the other duplicate post:

1) It is a duplicate
2) It is not correct

x\lim0 (sin x)/x = 1. How is ((sin x)/x)3>cos x, I'll leave it for someone else to do it, maybe Peter :P

[quote="darkmaster"]This inequality is correct . Even with 0<x<3 \pi/2, it 's still correct . I'll very happy if you can solve it.[/quote]
There is a root in this \pi<x<3\pi/2 interval and so it is not correct.
It is only correct as it is stated right now.
[quote="Master"]not really difficult but i have thought a lot
let 0 < x < \pi /2
(sinx/x)^3>cosx[/quote]

so can you give me the right problem and solution?
try!!!!

This one can be proved violently by using derivative of this function
 f(x) = (sinx)^3-x^3*cosx.
   I will give you my solution later because I have no time to type now. It may take a little bit long to do that. Sorry! :?


limit
combinatorics unsolved
combinatorics

There is a bug that jumps in $ [0;1]$. From the point $ x$ $ (x\in [0;1])$, it can jump to point $ \large\frac{x}{\sqrt{3}}$ or point $ \large\frac{x}{\sqrt{3}}\plus{}1\minus{}\frac{1}{\sqrt{3}}$. With this way of moving, can the bug jump to a point that the distance of this point to the point it started jumping is not greater than $ \frac{1}{100}$?

Every number can be written in the form \[ \left(1\minus{}\frac{1}{\sqrt{3}}\right)\sum_{i\equal{}1}^{\infty}\sqrt{3}^{\minus{}a_{i}}\] for some sequence $ a_{1}, a_{2}, ...$ where $ 0\le a_{1}<a_{2}<...$ and $ a_{i}\in\mathbb{Z}$ for each $ i\in\mathbb{N}$.  We first put the bug very close to zero, and then we generate, digit by digit, the above representation of the starting number.  We get arbitrarily close to the starting point.

I'm sorry but I don't understand your proof, K81o7. How can you "generate digit by digit"? Do I have to calculate $ \large\lim_{i\to\plus{}\infty}\sum_{i \equal{} 1}^{\plus{}\infty}\sqrt{3}^{\minus{}a_{i}}$ ?

Let's say you represent the number you're at by a sequence of $ 1$'s and $ 0's$ multiplied by $ 1\minus{}\frac{1}{\sqrt{3}}$, that is, in a sense, in "base $ \sqrt{3}$".  So, for example, the number $ 1.10011011$ would represent
\[ \left(1\minus{}\frac{1}{\sqrt{3}}\right)(\sqrt{3}^{0}\plus{}\sqrt{3}^{\minus{}1}\plus{}\sqrt{3}^{\minus{}4}\plus{}\sqrt{3}^{\minus{}5}\plus{}\sqrt{3}^{\minus{}7}\plus{}\sqrt{3}^{\minus{}8})\]
Then notice the following.  In this representation, the first kind of jump defined is euivalent to shifting the whole representation to the right by one space, while the second kind of jump is equivalent to shifting the whole representation to the right by one space and adding $ 1$ to the units place.  Clearly, any number has a unique base $ \sqrt{3}$ representation (even if it's infinite).  So what one can do is, from the point $ x$, first go arbitrarily close to $ 0$.  Then one chooses a starting point in the representation of $ x$ and gradually builds it up backwards.  You can thus approximate $ x$ to as close a level of accuracy as you want.

[quote="JBL"][quote="K81o7"]Clearly, any number has a unique base $ \sqrt{3}$ representation (even if it's infinite).[/quote]

"Clearly" is a dangerous word.  Every number has a representation of this sort because $ 1 <\sqrt{3}\leq 2$, but the representation won't be unique because $ \sqrt{3}< 2$.[/quote]

This is true.

Well I should say then that for any number, there exists a representation. (by a partial sum argument.)


geometry
3D geometry
sphere
algebra
polynomial
function
modular arithmetic

There is a discussion about the China TST Quizzes in this forum. After reading the posts I checked my computer and I find out that I translated the 2002 quizzes in the summer of 2006. Here is the PDF of problems (Ignore the date of tests, I messed up with it......) :

Problem 1: http://www.mathlinks.ro/viewtopic.php?p=1389490#1389490
Given $ n \geq 3$, $ n$ is a integer. Prove that:
\[ (2^n \minus{} 2) \cdot \sqrt{2i\minus{}1} \geq \left( \sum_{j\equal{}0}^{i\minus{}1}C_n^j \plus{} C_{n\minus{}1}^{i\minus{}1} \right) \cdot \sqrt{n}\]
where if $ n$ is even, then $ \displaystyle 1 \leq i \leq \frac{n}{2}$; if $ n$ is odd, then $ \displaystyle 1 \leq i \leq \frac{n\minus{}1}{2}$.

Problem 2: http://www.mathlinks.ro/viewtopic.php?p=1389494#1389494
There are $ n$ points ($ n \geq 4$) on a sphere with radius $ R$, and not all of them lie on the same semi-sphere. Prove that among all the angles formed by any two of the $ n$ points and the sphere centre $ O$ ($ O$ is the vertex of the angle), there is at least one that is not less than $ \displaystyle 2 \arcsin{\frac{\sqrt{6}}{3}}$.

Problem 3: http://www.mathlinks.ro/viewtopic.php?p=1389497#1389497
The positive integers $ \alpha, \beta, \gamma$ are the roots of a polynomial $ f(x)$ with degree $ 4$ and the coefficient of the first term is $ 1$. If there exists an integer such that $ f(\minus{}1)\equal{}f^2(s)$.

Prove that $ \alpha\beta$ is not a perfect square.

Problem 1: http://www.mathlinks.ro/viewtopic.php?p=1389505#1389505
If $ P_n(x)\equal{}a_0 \plus{} a_1x \plus{} \cdots \plus{} a_nx^n$, $ n \geq 2$ is a real-coefficient polynomial. Prove that if there exists $ a > 0$, making $ \displaystyle P_n(x) \equal{} (x \plus{} a)^2 \left( \sum_{i\equal{}0}^{n\minus{}2} b_i x^i \right)$, $ b_i \in R^\plus{}$, $ i\equal{}0,1,2, \cdots, n\minus{}2$, and there is some $ i$, $ 1 \leq i \leq n\minus{}1$, making $ a_i^2 \minus{} 4a_{i\minus{}1}a_{i\plus{}1} \leq 0$.

Problem 2: http://www.mathlinks.ro/viewtopic.php?p=1389507#1389507
$ A_1$, $ B_1$ and $ C_1$ are the projections of the vertices $ A$, $ B$ and $ C$ of triangle $ ABC$ on the respective sides. If $ AB\equal{}c$, $ AC\equal{}b$, $ BC\equal{}a$ and $ AC_1\equal{}2t AB$, $ BA_1\equal{}2rBC$, $ CB_1\equal{}2 \mu AC$. Prove that:
\[ \frac{a^2}{b^2} \cdot \left( \frac{t}{1\minus{}2t} \right)^2 \plus{} \frac{b^2}{c^2} \cdot \left( \frac{r}{1\minus{}2r} \right)^2 \plus{} \frac{c^2}{a^2} \cdot \left( \frac{\mu}{1\minus{}2\mu} \right)^2 \plus{} 16tr \mu \geq 1\]

Problem 3: http://www.mathlinks.ro/viewtopic.php?p=1389515#1389515
Let $ p_i \geq 2$, $ i\equal{}1,2, \cdots n$ be $ n$ integers such that any two of them are relatively prime. Let:
\[ P\equal{} \{ x \equal{} \sum_{i\equal{}1}^{n} x_i \prod_{j\equal{}1, j \neq i}^{n} p_j \mid x_i \text{is a non\minus{}negative integer}, i\equal{}1,2, \cdots n \}\]
Prove that the biggest integer $ M$ such that $ M \not\in P$ is greater than $ \displaystyle \frac{n\minus{}2}{2} \cdot \prod_{i\equal{}1}^{n} p_i$, and also find $ M$.

Problem 1: http://www.mathlinks.ro/viewtopic.php?p=1389521#1389521
Given triangle $ ABC$ and $ AB\equal{}c$, $ AC\equal{}b$ and $ BC\equal{}a$ satisfying $ a \geq b \geq c$, $ BE$ and $ CF$ are two interior angle bisectors. $ P$ is a point inside triangle $ AEF$. $ R$ and $ Q$ are the projections of $ P$ on sides $ AB$ and $ AC$.

Prove that $ PR \plus{} PQ \plus{} RQ < b$.

Problem 2: http://www.mathlinks.ro/viewtopic.php?p=1389523#1389523
For any two rational numbers $ p$ and $ q$ in the interval $ (0,1)$ and function $ f$, there is always $ \displaystyle f \left( \frac{p\plus{}q}{2} \right) \leq \frac{f(p) \plus{} f(q)}{2}$. Then prove that for any rational numbers $ \lambda, x_1, x_2 \in (0,1)$, there is always:

\[ f( \lambda x_1 \plus{} (1\minus{}\lambda) x_2 ) \leq \lambda f(x_i) \plus{} (1\minus{}\lambda) f(x_2)\]

Problem 3: http://www.mathlinks.ro/viewtopic.php?p=1389528#1389528
Sequence $ \{ f_n(a) \}$ satisfies $ \displaystyle f_{n\plus{}1}(a) \equal{} 2 \minus{} \frac{a}{f_n(a)}$, $ f_1(a) \equal{} 2$, $ n\equal{}1,2, \cdots$. If there exists a natural number $ n$, such that $ f_{n\plus{}k}(a) \equal{} f_{k}(a), k\equal{}1,2, \cdots$, then we call the non-zero real $ a$ a $ \textbf{periodic point}$ of $ f_n(a)$.

Prove that the sufficient and necessary condition for $ a$ being a $ \textbf{periodic point}$ of $ f_n(a)$ is $ p_n(a\minus{}1)\equal{}0$, where $ \displaystyle p_n(x)\equal{}\sum_{k\equal{}0}^{\left[ \frac{n\minus{}1}{2} \right]} (\minus{}1)^k C_n^{2k\plus{}1}x^k$, here we define $ \displaystyle \frac{a}{0}\equal{} \infty$ and $ \displaystyle \frac{a}{\infty} \equal{} 0$.

Problem 1: http://www.mathlinks.ro/viewtopic.php?p=1389533#1389533
Given a positive integer $ n$, for all positive integers $ a_1, a_2, \cdots, a_n$ that satisfy $ a_1\equal{}1$, $ a_{i\plus{}1} \leq a_i \plus{} 1$, find the sum $ \displaystyle \sum_{i\equal{}1}^{n} a_1a_2 \cdots a_i$.

Problem 2: http://www.mathlinks.ro/viewtopic.php?p=1389537#1389537
Two circles meet at points $ A$ and $ B$. Through point $ A$ construct tangents to the circles and meet the other circle at $ C$ and $ D$. The incentres of triangle $ ABC$ and $ ABD$ are $ I_1$ and $ I_2$. Segment $ I_1I_2$ meets $ AB$ at $ E$.

Prove that $ \displaystyle \frac{1}{AE} \equal{} \frac{1}{AC} \plus{} \frac{1}{AD}$.

Problem 3: http://www.mathlinks.ro/viewtopic.php?p=1389538#1389538
Find all groups of positive integers $ (a,x,y,n,m)$ that satisfy $ a(x^n \minus{} x^m) \equal{} (ax^m \minus{} 4) y^2$ and $ m \equiv n \pmod{2}$ and $ ax$ is odd.

Problem 1: http://www.mathlinks.ro/viewtopic.php?p=1389548#1389548
In acute triangle $ ABC$, show that:
\[ \sin^3{A}\cos^2{(B\minus{}C)} \plus{} \sin^3{B}\cos^2{(C\minus{}A)} \plus{} \sin^3{C}\cos^2{(A\minus{}B)} \leq 3\sin{A} \sin{B} \sin{C}\]
and find out when the equality holds.

Problem 2: http://www.mathlinks.ro/viewtopic.php?p=1389550#1389550
$ m$ and $ n$ are positive integers. In a $ 8 \times 8$ chessboard, $ (m,n)$ denotes the number of grids a Horse can jump in a chessboard ($ m$ horizontal $ n$ vertical or $ n$ horizontal $ m$ vertical ). If a $ (m,n) \textbf{Horse}$ starts from one grid, passes every grid once and only once, then we call this kind of Horse jump route a $ \textbf{H Route}$. For example, the $ (1,2) \textbf{Horse}$ has its $ \textbf{H Route}$. Find the smallest positive integer $ t$, such that from any grid of the chessboard, the $ (t,t\plus{}1) \textbf{Horse}$ does not has any $ \textbf{H Route}$.

Problem 3: http://www.mathlinks.ro/viewtopic.php?p=1389555#1389555
$ n$ sets $ S_1$, $ S_2$ $ \cdots$, $ S_n$ consists of non-negative numbers. $ x_i$ is the sum of all elements of $ S_i$, prove that there is a natural number $ k$, $ 1<k<n$, and:
\[ \sum_{i\equal{}1}^n x_i < \frac{1}{k\plus{}1} \left[ k \cdot \frac{n(n\plus{}1)(2n\plus{}1)}{6} \minus{} (k\plus{}1)^2 \cdot \frac{n(n\plus{}1)}{2} \right]\]
and there exists subscripts $ i$, $ j$, $ t$, and $ l$ (at least $ 3$ of them are distinct) such that $ x_i \plus{} x_j \equal{} x_t \plus{} x_l$.

Problem 1: http://www.mathlinks.ro/viewtopic.php?p=1389560#1389560
Given that $ a_1\equal{}1$, $ a_2\equal{}5$, $ \displaystyle a_{n\plus{}1} \equal{} \frac{a_n \cdot a_{n\minus{}1}}{\sqrt{a_n^2 \plus{} a_{n\minus{}1}^2 \plus{} 1}}$. Find a expression of the general term of $ \{ a_n \}$.

Problem 2: http://www.mathlinks.ro/viewtopic.php?p=1389564#1389564
$ \odot O_1$ and $ \odot O_2$ meet at points $ P$ and $ Q$. The circle through $ P$, $ O_1$ and $ O_2$ meets $ \odot O_1$ and $ \odot O_2$ at points $ A$ and $ B$. Prove that the distance from $ Q$ to the lines $ PA$, $ PB$ and $ AB$ are equal.

(Prove the following three cases: $ O_1$ and $ O_2$ are in the common space of $ \odot O_1$ and $ \odot O_2$; $ O_1$ and $ O_2$ are out of the common space of $ \odot O_1$ and $ \odot O_2$; $ O_1$ is in the common space of $ \odot O_1$ and $ \odot O_2$, $ O_2$ is out of the common space of $ \odot O_1$ and $ \odot O_2$.

Problem 3: http://www.mathlinks.ro/viewtopic.php?p=1389570#1389570
There is a game. The magician let the participant think up a positive integer (at least two digits). For example, an integer $ \displaystyle\overline{a_1a_2 \cdots a_n}$ is rearranged as $ \overline{a_{i_1}a_{i_2} \cdots a_{i_n}}$, that is, $ i_1, i_2, \cdots, i_n$ is a permutation of $ 1,2, \cdots, n$. Then we get $ n!\minus{}1$ integers. The participant is asked to calculate the sum of the $ n!\minus{}1$ numbers, then tell the magician the sum $ S$. The magician claims to be able to know the original number when he is told the sum $ S$. Try to decide whether the magician can be successful or not.

( For instance, if we take number $ 210$, then $ S\equal{}201\plus{}120\plus{}102\plus{}021\plus{}012\equal{}456$, where $ 012\equal{}12$ and $ 021\equal{}21$.)

Problem 1: http://www.mathlinks.ro/viewtopic.php?p=1389583#1389583
Circle $ O$ is inscribed in a trapzoid $ ABCD$, $ \angle{A}$ and $ \angle{B}$ are all acute angles. A line through $ O$ intersects $ AD$ at $ E$ and $ BC$ at $ F$, and satisfies the following conditions:

(1) $ \angle{DEF}$ and $ \angle{CFE}$ are acute angles.

(2) $ AE\plus{}BF\equal{}DE\plus{}CF$.

Let $ AB\equal{}a$, $ BC\equal{}b$, $ CD\equal{}c$, then use $ a,b,c$ to express $ AE$.

Problem 2: http://www.mathlinks.ro/viewtopic.php?p=1389586#1389586
Find all non-negative integers $ m$ and $ n$, such that $ (2^n\minus{}1) \cdot (3^n\minus{}1)\equal{}m^2$.

Problem 3: http://www.mathlinks.ro/viewtopic.php?p=1389588#1389588
Given positive integer $ m \geq 17$, $ 2m$ contestants participate in a circular competition. In each round, we devide the $ 2m$ contestants into $ m$ groups, and the two contestants in one group play against each other. The groups are re-divided in the next round. The contestants compete for $ 2m\minus{}1$ rounds so that each contestant has played a game with all the $ 2m\minus{}1$ players. Find the least possible positive integer $ n$, so that there exists a valid competition and after $ n$ rounds, for any $ 4$ contestants, non of them has played with the others or there have been at least $ 2$ games played within those $ 4$.

Problem 1: http://www.mathlinks.ro/viewtopic.php?p=1389597#1389597
$ A$ is a set of points on the plane, $ L$ is a line on the same plane. If $ L$ passes through one of the points in $ A$, then we call that $ L$ passes through $ A$.

(1) Prove that we can divide all the rational points into $ 100$ pairwisely non-intersecting point sets with infinity elements. If for any line on the plane, there are two rational points on it, then it passes through all the $ 100$ sets.

(2) Find the biggest integer $ r$, so that if we divide all the rational points on the plane into $ 100$ pairwisely non-intersecting point sets with infinity elements with any method, then there is at least one line that passes through $ r$ sets of the $ 100$ point sets.

Problem 2: http://www.mathlinks.ro/viewtopic.php?p=1389600#1389600
Does there exist $ 2002$ distinct positive integers $ k_1, k_2, \cdots k_{2002}$ such that for any positive integer $ n \geq 2001$, one of $ k_12^n + 1, k_22^n + 1, \cdots, k_{2002}2^n + 1$ is prime?

Problem 3: http://www.mathlinks.ro/viewtopic.php?p=1389607#1389607
$ c$ is a positive integer, $ \alpha, \beta, \gamma$ are pairwise distinct positive integers, and \[  \{ \begin{array}{c} \displaystyle a= \alpha + \beta + \gamma \\ \\ b= \alpha \cdot \beta + \beta \cdot \gamma + \gamma \cdot \alpha \\ \\ c^2 = \alpha\beta\gamma \end{array} \]
A number $ \lambda$ satisfies the condition $ \lambda^4 -2a\lambda^2 + 8c\lambda + a^2 - 4b = 0$.

Prove that $ \lambda$ is an integer if and only if $ \alpha, \beta, \gamma$ are all perfect squares.


Prove X^N + Y^N does not equal Z^N, for N greater than or equal to Z , if x,y,z,n are positive integers.[/code]

Wiles' proof of Fermat's Last Theorem is sufficient since this is a special case  :D

On a related note, prove for the special case $N = 4$ (the proof is much simpler).

[quote="t0rajir0u"]Wiles' proof[/quote]
Sure, especially if you can give the whole proof in competition conditions  :D

But here it is a very special and much easier case.

If $n > z$ then since $x < y$ and $y < z$ we have $2 \leq z^{n-z} = (x/z)^z + (y/z)^z < 1 + 1$ and we have a contradiction.

Edited : sorry so many typos ...

[quote="tµtµ"]If $n > z$ then since $x < y$ and $y < z$ we have $2 \geq z^{n-z} = (x/z)^n + (y/z)^n < 1 + 1$ and we have a contradiction.[/quote]

How did that last step go? I was able to prove it for primes bigger than z, namely

$x^n+y^n=z^n \rightarrow x+y \equiv z \mod{n}$

But since there all less than n, this implies $x+y=z$

Which is a contradiction for non-trivial solutions.

Should your $z^{z-n}$ in fact be $z^{n-n}$  ???


4. A solid, uniform cylinder with mass M and radius R is initially at rest. It is gently placed onto a conveyor belt moving at constant velocity v as shown above. The cylinder’s axis is perpendicular to the belt’s velocity. The coefficients of static and kinetic friction are nonzero constants. (Ignore other energy losses such as rolling resistance.)

a) What is the final speed of the cylinder’s center of mass?
b) If a cylinder with twice the radius and four times the mass had been used in place of the original cylinder, what would have been its final speed?
c) If a ring with the same mass and radius had been used in place of the cylinder, what would have been its final speed? Assume the ring has negligible thickness.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=381[/img]

Thanks! Post your solutions :)

Is this posted as a challenge, or as something you want help for?

[hide] $\Sigma F=ma$ and $\tau=I*\alpha$

second hint:
The cylinder stops being accelerated when the speed of its surface matches the speed of the conveyer belt
[/hide]

this was posted as a help problem.  :? I'm still not too sure I get this....

It means that:

The force that the conveyor belt applies on the cylinder is constant, $F=Mgf$ ($f$ is the friction coefficient). Therefore both the acceleration and the angular acceleration of the cylinder are constant:

$a=F/M$

$\alpha=FR/I$.

The force is applied until what daveed said happens - that is:

$v_\text{cylinder}+\omega R=v_\text{conv. belt}$

and the rest is simple. (The cylinder is moving with uniform acceleration so you should be able to find the exact $v$ and $\omega$ in any given time.)

To give a more detailed answer...

Have you learned about rotational dynamics?
Anyway, the Moment of Inertia of that big cylinder is $\frac{1}{2}MR^2$

Letting $F$ be the frictional force of the conveyer...

Punch that into your angular acceleration- $FR=I\alpha$

So $\frac{F}{M}=\frac{1}{2}\alpha R$

So, in terms of linear velocity, the acceleration of the surface of the cylinder is $2\frac{F}{M}$

We also know that $F=Ma$ for the center of mass of the cylinder
So we know that the Center of Mass of the solid cylinder accelerates at half the rate as the outside.

So the final velocity of the Center of Mass would be $\frac{v}{3}$.

This velocity only depends on the moment of inertia of the object- in the case of part B, the same answer would have been reached.  However, because a Ring has a different $I$, the final velocity of the center of mass would be different.


Find the probability of rolling a 12 with three 6-sided die.

I'll assume the sides numbered 1-6,

[hide]is the answer $\frac{25}{216}$? I made a list (I love lists, even though I might've missed one). [/hide]


algebra
polynomial
number theory
number theory open

Hello,Mathlinkers.What is your favorite number theory problem?Can you introduce it?

Mine:If $ p\equal{}4k\plus{}1$ is a prime,then $ k^k \equiv 1 (mod p)$

[quote="KDS"]Hello,Mathlinkers.What is your favorite number theory problem?Can you introduce it?

Mine:If $ p \equal{} 4k \plus{} 1$ is a prime,then $ k^k \equiv 1 (mod p)$[/quote]

Mine:Prove Lagrange's 4-square theorem.

Mine is quite standard:

[i]Prove that the equation $ x^2 \plus{} y^2 \equal{} 3z^2$ has no positive integer solutions.[/i]

Or this one, although I don't know if it is number theory or combinatorics:

[i]Prove that among $ n$ integers there are always some of them whose sum is divisible by $ n$.[/i]

About the ones I posted on the forum. This includes:

-If $ p \equiv 1 \mod 4$ is prime, then $ \frac{p^p\minus{}1}{p\minus{}1}$ is not prime ( http://www.mathlinks.ro/viewtopic.php?t=169135 ). And the closely related Gauß&Lucas' formulas ( http://www.mathlinks.ro/viewtopic.php?t=146020 ).

- Clausen van Staudt, that is: describing Bernoulli numbers $ \mod 1$ ( http://www.mathlinks.ro/viewtopic.php?t=91510 ).

- Euclidean polynomials to prove Dirichlet's theorem on primes in aritmetic progressions in some cases ( http://www.mathlinks.ro/viewtopic.php?t=100567 )

- Erdös-Ginzburg-Zif (or however they are spelled): from $ 2n\minus{}1$ integers, one can pick $ n$ such that their sum is divisible by $ n$ ( http://www.mathlinks.ro/viewtopic.php?p=389030 ).

There would be some more if we would speak about theorems instead of problems.

For every integer $ n$ the interval $ [n^2,(n \plus{} 1)^2]$ contains a prime

The problem was proposed 97 years ago and is unsolved.

Mi favorite problem is: The Goldbach´s conjeture.  :)

The giuga's conjecture is very interesting and can be considered as one of my favorites .
Is it still a conjecture or is it proved ?