Putnam
calculus
geometry
linear algebra
number theory
graph theory

Hi I am the Doctor and my goal, as those who already know from reading my other thread http://www.artofproblemsolving.com/Forum/viewtopic.php?t=249664, is to study mathematics and physics in college and hopefully make sure that my performance on the Putnam and other contests is not as abysmal as my experience with high school math contests has been, and I suppose I would also love to gain acceptance at a good University.

Since I have essentially run out of course selection options at my high school for my senior year I will be taking virtually all of my classes at the local University in accordance with my goal since I have completed virtually all of my high schools graduation requirements I have decided to focus very heavily on mathematics classes. As I am permitted to take basically any undergraduate course I hope to take most if not all of the following classes:

Multivariable Calculus
Differential Equations
Linear Algebra (I,II)
Partial Differential Equations and Fourier Analysis (I,II)
Introduction to Abstract Mathematics
Number Theory
Combinatorics
Graph Theory
Group Representation Theory
Computer Science (I,II)
Data Structures

Essentially I am just wondering what my fellow AOPS'ers think of this idea?

Well, those are all good courses to take.  But it's unlikely that you'll succeed at taking 14 courses in a year -- that's just a huge amount, even for a full-time undergraduate.  Something more realistic would be to try to take a multivariable calculus course, a linear algebra course, an analysis course, and maybe an abstract algebra course, as well as a computer science sequence.  This is much more reasonable (three courses/semester).

There is also some danger in specializing too heavily.  I browsed through your post history and you mentioned not doing very well on contests.  That is hardly the end of the world, of course -- lots of excellent research mathematicians didn't do well on contests either.  But there will be people applying to excellent colleges with stronger mathematics credentials than you (olympiad qualifiers and the like), so if other parts of your application are weak, then your application will probably not be very compelling.  In your case, I'd at least want to be taking something like AP english, for example, and making sure that you do well on the non-mathematical portions of the SAT.  Extracurricular activities would probably be helpful as well.

Of course, if you've already completed a bunch of non-science APs and are near the top of your class GPA-wise, then you can disregard what I just said.

I admit it's a lot of math classes but theoretically as long as I am willing I should be bale to do it there are like three trimester things in which I will be taking classes and some of these are only one trimester classes the way it breaks down I don't think I have more than 17 credit hours each trimester, like around 5 classes which shouldn't be too bad since there won't be much I have to take at my high school, and after talking to a friend who goes to the University I have discovered that this is fairly typical.

As far as specialization goes I feel I have taken a fare range of different classes up until now, AP US History, AP Economics, and AP English Language and Comp are my non science and math AP's and there are others non math science classes of course that just weren't AP's and I also plan on taking AP English Literature and Composition next year, and I guess now that I have gotten rid of most of those graduation requirements (basically all but one English credit and a little bit of art) I really think I wanna spend next year devoting myself to trying to improve my skill in the things I am really interested (not to mention ive been meaning to learn CS for like forever) in and secondarily try and become more mathematically qualified for college applications. I do a decent variety of extracurricular activities like Math team, Science Bowl and what not I also tutor kids and this year started doing athletics as well. My standardized test scores are not to great, not the best test taker I guess, but I am fairly confident that i should be able to improve these with a little practice.

Take MV calc, LA and some discrete math but anything else is a waste of time because you'll do poorly trying to learn that much. If you want a head-start on putnam, learn calculus and LA inside and out.

qntty what exactly are you suggesting I take out. I mean all those would not be taken at the exact same time they would be during different parts of the year usually no more than 4 at any given time and while some meet 5 times a week for most its only 3. Idk I thought it might be a good idea since I would have virtually no other school obligations i.e it can't conflict with my school work because it would become my school work :) . English is basically the only other class I would need to take.

P.S classes are picked one trimester at a time so I suppose I don't need to be super committed to any one course of action right away. But since you need some classes to take other classes it may be a good idea to know what I wanna do so that I can sorta plan accordingly.

It doesn't matter so much [i]what[/i] you take; it's that your original plan is a serious overload. You want to have a full-time college schedule and a partial high-school schedule at the same time? Bad idea. You want to take four or five math classes at once? Bad idea.

Talk to an adviser at your local university, and trim this down to a reasonable load. Take some things that aren't math.

When I was a high school senior, the only class I actually needed for high school requirements was English. I still took a nearly full schedule of six classes, along with one math class per semester at a local university. That worked well.

I took four math courses and a CS course during my first semester of undergrad, and four math courses and an econ course during my second semester of undergrad.  But I had a stronger math background than the OP and a lot of those courses were on material that I understood fairly well but had never taken formally, so they were pretty straightforward.  I still think that trying to take that many courses while juggling a partial high school workload and being involved with extracurricular activities is not reasonable.

"You want to have a full-time college schedule and a partial high-school schedule at the same time?" No not really just the former I have essentially no high school schedule. It's just one English class so that is not a huge  issue. Extracurricular activities might be.  And I do still plan on taking a full schedule of classes I just want them to be Math and CS classes that's all. blahblahblah said "...a lot of those courses were on material that I understood fairly well but had never taken formally, so they were pretty straightforward...", but the truth of the matter is if  I take a full load of classes even if they are not math classes (like idk AP Human Geo)  they will  likely be on material that I have never learned and is new to me. Also qntty mentioned getting a head start on putnam while that would be great I can't really do that until I catch up on all the high school contest stuff (ergo the combinatorics number theory and graph theory classes) it has to be done if I am to improve and not do as badly in college as I did in high school. As blahblahblah mentioned my math background could be stronger, the hope that doing this will make it stronger is why I want to go through with this the idea is that this way there is no interference from school duties so to speak. I mean if I am going to have a full schedule of 6 classes anyway why not have most of them be math classes? I mean I honestly think I would rather take a class in linear algebra than like AP Human Geo or Intro to Political Science or something. People go to places like Math Camp or w/e where academically at least all they do day to day is math is there a huge difference between that and what I wanna do other than that mine is throughout the whole school year?

P.S I have done some HW on this my friend who goes to the University as an Undergrad normally has around 17-18  credit hours a trimester and a part time job and he says its quite manageable. ( By comparison if I am not mistaken I should not have more than 15-17 at any given time and if I am not mistaken since alot of the classes are 3 credit hours during one trimester it might be as low as 12)

a 'full schedule of 6 classes' in high school is very different than taking 6 undergrad courses.  i mean you really don't have to listen, and it's possible that you are a unique snowflake and can take a ridiculous courseload with no real repercussions, but for what it's worth, jmerry and I have both been undergrads and taken most of the courses you mention, and we both think that it is a horrible idea.

Let me add a third vote to the proposition, "you don't seem to have a good sense of the relative workloads of college versus high school classes, and would probably be better off with a somewhat more realistic initial goals."

Let me say first that I edited my last post a little bit and I think to some extent it addresses what JBL mentioned about not being to aware in terms or work load.

Also as far as what I take regardless of whether it is math or not there is a good chance it will be at the university ( I mean originally I was gonna take a physics and organic chem class etc. I decided to focus on the math later). This is simply to avoid to and from trips multiple times a day and because there are quite simply not many classes left that I am actually interested in taking at my high school. It may also be worth noting that these for the most part introductory courses, it's just the volume, does the fact that they are taken during different semesters not mitigate this at all? And true high school is not the same as college but I will be attending college after next year anyway right, so what's the huge diff? I have a friend whose a senior this year and he did this only he took all of his classes at the University and had a total of like 5 including like Linear Alg, Eng, and Phys. Does the fact that all mine are in one department really make it that much worse?

PS. I am curious to know what those who advocate that I eliminate things, think that I should eliminate? Oh and also I just want to say that I hope I am not coming off as not listening in truth I am taking every bit of advice I am receiving into consideration I am just trying to be as thorough as possible in my analysis of the situation, trying to make sure I have taken all things pro and con into consideration.

Why don't you just wait until college to take these courses? Unless you are planning on going to this local university for your undergrad degree, you will probably get a better grasp on the material if you take the courses at your undergrad university. Take one or maybe two classes at the university and spend the rest of your time this year devoted to searching for colleges and preparing your resumé. If you feel like you have too much time left over then try developing the high school math contest skills that you are weak in. If you don't have a strong foundation now, you will be blown away at the college level.
Then again, I'm just a high school junior...

Well I mean what do you mean by the college level, the Number Theory course for instance is an introductory course that from what I know does not require too much prior knowledge of the subject although since it is for a whole semester they probably still manage to get through quite a but even though they start with the basics. I think the Intro to Abstract math class may be a prerequisite for it though. Anyway from what I have heard its not terribly hard.  I think whoever suggested having a chat with my adviser and some like math professors or something was definitely right though.

In reference to blahblahblah's first post: I don't really want to take an Abstract Algebra class because getting the prerequisite classes to do that would be tricky plus I would not be able to finish the sequence.

Here is what I am wondering how much of people's distaste of this idea stems not from the number of classes from all the classes being in one subject area? The reason I ask is that my friend who is doing his undergrad at the University said he took around 17 credit hours ( i don't remember the exact number) his freshman year and he did ok he also said that this type of load is not very atypical at all and like I said my other friend who is a senior this year takes 5 classes there and he is also doing ok although I think there were a few times throughout the year that he mentioned being sleep deprived lol. Anyway would your opinions be different if I replaced some of those math classes with say some classes from the Physics and Chem departments (which was originally my plan before I decided to focus on the math)

[quote="The Doctor"]Here is what I am wondering how much of people's distaste of this idea stems not from the number of classes from all the classes being in one subject area? [/quote]

very little.  if you are very interested in math then of course it makes sense to take primarily math courses (along with appropriate supplementary courses in other subjects like CS and possibly physics).  my 'distaste' lies solely in the fact that taking 5+ math/cs courses in a quarter or semester or whatever is simply not wise when you have not completed ANY credits at the undergraduate level.

I personally had a tough time adjusting to the high school-to-college shift in workload requirements and expectation.  In college, you do much more learning on your own, outside of class, than you are expected to do in high school.  When you're a high school student and you don't know anyone else in the class, this means among other things that you have no one else to work with on p-sets, ask questions to, etc.  This is the source of a lot of my skepticism about your plans.  However, I agree with blahblahblah that taking a bunch of math classes isn't a bad goal, and that it is possible that you would actually have no problem taking 17 math classes next year, just that it's unlikely and you should have some contingency plans available.  Of the classes you listed, here are some comments (and you should be aware that these are subjectively generated from the point of view of a graduate student in algebraic combinatorics; perhaps those with other perspectives will weigh in, too):

The following ones are the sort of thing that every math major must take or understand at some point:
Introduction to Abstract Mathematics (possibly redundant with your knowledge and skills, but worth at least visiting)
Linear Algebra (I,II) (but I don't know exactly what the "II" means)
Multivariable Calculus (although, in fact, I could have gone through college without such a course)

Then the others fall into sort of several categories that I would describe as "fun if that's what you're interested in" -- it may very well turn out that you aren't interested in any of them, or that you're interested in all of them!

Differential Equations and Partial Differential Equations and Fourier Analysis (I,II) form an obvious couple.  Very much not to my personal taste, but hopefully someone else will step forward to defend them.

Combinatorics, Graph Theory, Computer Science (I,II), Data Structures form a quadruple of discrete math (although CS in practice is probably more about programming than theory?).  I love the Comb & Graph Theory side of this.

Number Theory and Group Representation Theory are a little bit out on their own here.  I assume NT means "elementary number theory," not algebraic or analytic (each of which requires more knowledge), in which case you should have no problem with it (and often elementary NT classes are good courses in which to work on your mastery of formal proof).  Group Representation Theory is a beautiful subject, though I'm somewhat surprised to see it out on its own instead of as part of an algebra sequence.

I think among these latter courses, there's a lot to be said for choosing a course based on the quality of the teaching -- since none of these are required for you in the near future, try to take the courses that will give you the most enjoyable experience.  Usually, it is very easy to switch, add or drop courses for the first few weeks of the semester (though you should find out about the details!), so it might be reasonable to start off taking 5 classes even if you're only planning to end up with 3.

I think that people are being a bit harsh on The Doctor.  There's nothing *necessarily* wrong with taking only math classes, assuming that you're "done" with your high school's curriculum.  Doctor, I think that what alarms people is that your proposed schedule is not a full college load, but rather much larger than a full college load.  You listed 14 courses, while a typical full year of college courses is more like 8 courses, or at *most* 10 (not including summers).

Others have acknowledged that your proposed schedule is not necessarily impossible but just impossible for most human beings.  Since no one here seems to know you, no one can say for certain if your plan is sheer lunacy or not.  You also haven't said *which* university you'll be attending.  This does matter.  For one thing, instead of taking a bajillion math courses, isn't there an honors sequence you can take?  Quality is usually more important than quantity.  IF you are good enough at math to do well in all of these classes at once, AND most of the classes are just computational, then you'd actually just be wasting a lot of your time.  Contrary to what others seem to be inferring, for a sufficiently bright student, some college courses really are very easy.

However, although I know almost nothing about you, I see that you do not already know multivariable calculus, which means that your proposed pace of learning mathematics greatly exceeds your historical pace of learning mathematics.

From your list, definitely take multivariable calculus and linear algebra.  Other than that, take what sounds most interesting to you.

Finally, the most important piece of advice is to talk to a faculty member at your local university and get some serious individualized advice.

This is a quarter school, so it's not much larger than a normal load. My objection is based mainly on the fact that this is all math. Even advanced undergraduates don't take that much math; two or three math classes at a time and some other stuff is a normal load for them.

Another point of view: what could you possibly gain by taking all of this at once? You could graduate from college early, in two or three years. That's it. I don't think you even want to do that, so it's not worth the substantial risk that you won't be able to handle it.
It's not "sheer lunacy". It's a very normal bad idea.

Quick side ? here to begin with, just so I know if my grades aren't quite as high as they normally are which is the bigger factor that or that I took a risk by taking these harder classes and pursuing my interests.

Anyway some  things to take into account are that my school requires a minimum of 5 academic classes I think, although I may be able to convince them to let me off with a little less since I am doing this. Here is the thing I wanna avoid taking any classes at my school to meet that requirement mainly because if it works out somehow that you know a class isn't offered at the University during one quarter so I do that but then in a later quarter to take all the classes that I want to I need the space that's occupied by the class at school (Of course if I take the other class at the University it will be over at the end of that quarter and the space will be open again). So I may be forced to take a certain number of classes (preferably math but what I am hearing is that ppl think it might be good to throw some other stuff in there as well).

Anyway what ive found out is that not only might the idea be bad ( I think you guys may be right)  but that it might simply not work because of scheduling i.e what quarters certain classes are offered and their prerequisites. For instance I am starting to think that Group .... might not be possible and though I wanted to take CS 1 during the autumn quarter it isn't going to be offered next year then. But anyway for my original idea to work every class had to be taken at sort of like a certain time to make sure all the prerequisites matched up properly and what not. And since I might just not be able to move a given class with a scheduling prob to another quarter since that would end up with me needing to take like 8 classes during one quarter which I can't , some things might not be possible. (btw side ? here how bad would it be if I ended up not being able take CS until I was actually an undergrad bearing in mind that programming and CS in general is something I want to at some point at least be good at)

Based on the info that I presented in my last post I have decided that while before I sort of had to have everything planned out based on prereq's and stuff for now at least for the Autumn quarter I am just gonna see what is available and then pick what I think is good. Plz tell me what you think of this idea in general and the potential schedule I have thought of for the Autumn quarter. In general does 3 to 4 per quarter seem reasonable?
(unless otherwise specified I plan on spacing these credit hours out throughout the week as much as possible)

MultV Calc- 5 Credit hours
Differential Equations - 5 Credit hours
Linear Algebra I - 3 Credit hours

As you can see this is only three classes (it's all the math that I was interested in that was available) plus Eng at school makes 4 academics (I will also be taking some semester long art class at school). So what I am wondering is do you think I should try and convince them to let me off with this or not (I don't want it to look like I am taking it easy to colleges though if I do that  will that be an issue?) if i can't do that or if you don't think I should, what  extra 1 or more classes would you recommend adding, I was thinking maybe the data structures class which is like 3 or 4 more credit hours but I think CS I and II might be a prereq for it.


[color=purple] When I studied in inequality I met the following :
 $ \frac {bc}{c^2 \plus{} a^2} \plus{} \frac {ca}{a^2 \plus{} b^2} \plus{} \frac {ab}{b^2 \plus{} c^2}\le\frac {3}{2}$
   I know mathlinkers are very good at inequality so please help me in order to prove or disprove it. Thanks alot  :) [/color]

[color=purple] Are you sure ?  :) Anyway thank you very much . Can any one give me the more brightly [b]counter-example[/b] ? Thank ![/color]

If $ b\rightarrow\plus{}\infty$ then $ \frac{bc}{a^2\plus{}c^2}\rightarrow\plus{}\infty.$ :wink:

[color=purple] Sorry I have a mistake :)[/color]

[quote="evarist"][color=purple] What will happend if$ a,b,c$ are the sides length of a triangle ? :) [/color][/quote]
$ b\equal{}c\equal{}2,a\equal{}3$


[quote="pvthuan"]Let $x,y$ be positive real numbers. Denote by $a,g,h,r$ be the arithmetic means, geometric means, harmonic means, and root means of the two numbers. That is,
\[ a=\frac{x+y}2,\;\; \; g=\sqrt{xy},\;\;\; h=\frac2{\frac1x+\frac1y},\;\;\; r=\sqrt{\frac{x^2+y^2}2}.  \]
It is known that $\min(x,y)\leq h\leq g\leq a\leq r\leq \max(x,y)$. Prove or disprove each of the following inequalities

$\Bigl(\frac{a+g}2\Bigl)^2\geq ar$,  $\dfrac1a+\dfrac1h\geq\dfrac4{a+g}$, $\sqrt{gh}\geq\dfrac2{\frac1a+\frac1h}$, and $\frac{a+g+h}3\geq\sqrt[3]{rgh}$.

The problem will be very hard when we repeat the procedure or when when extend to more than two numbers.[/quote]\[ \frac{\frac{1}{a}+\frac{1}{h}}{2}\frac{a+g}{2}\geq\sqrt{\frac{1}{ah}}\sqrt{ag}=\sqrt{\frac{g}{h}}\geq 1 \\\Rightarrow\dfrac1a+\dfrac1h\geq\dfrac4{a+g} \]
All other Inequalities are wrong, as you can see by calculating with $x=1,y=2$.

OK, we can replace by the means to have correct inequalities. It is just my ideas.

Let $x,y$ be positive real numbers. Denote by $a,g,h,r$ be the arithmetic means, geometric means, harmonic means, and root means of the two numbers. That is,
\[ a=\frac{x+y}2,\;\; \; g=\sqrt{xy},\;\;\; h=\frac2{\frac1x+\frac1y},\;\;\; r=\sqrt{\frac{x^2+y^2}2}.  \]
It is known that $\min(x,y)\leq h\leq g\leq a\leq r\leq \max(x,y)$. Prove or disprove each of the following inequalities

$\frac{a+h}2\geq \sqrt{ag}$,  $\dfrac1a+\dfrac1h\geq\dfrac4{a+g}$, $\sqrt{ah}\geq\dfrac2{\frac1a+\frac1g}$, and $\frac{a+r+h}3\geq\sqrt[3]{arg}$.

The problem will be very hard when we repeat the procedure or when when extend to more than two numbers


Compute the following limit: $\lim_{n\rightarrow\infty}{\frac{5n^2-5n+2}{2^{n}}}$.

It is 0 , because      ${\lim_{n\rightarrow\infty}{\frac{n^2}{2^n}}} = 0$


Let $ f: [0;1]   \rightarrow [0;1]$ such that  $ \lvert\ f(x)\minus{}f(y)\rvert \leq \frac{\lvert\ f(x)\minus{}x\rvert\plus{} \lvert\ f(y)\minus{}y \rvert  }{2}$ Prove that the equation $ f(x)\equal{}x$ has uniquae  solution .

___________________________
Azerbaijan Land of the Fire   :ninja:  :roll:  :idea:

isn't it trivial?

assume $ f(x)\equal{}x, f(y)\equal{}y, x\neq y$. then $ 0<|f(x)\minus{}f(y)|\leq \frac{|f(x)\minus{}x|\plus{}|f(y)\minus{}y|}{2}\equal{}\frac{0\plus{}0}{2}\equal{}0$, contradiction.

[quote="pco"][quote="pluristiq"]isn't it trivial?

assume $ f(x) \equal{} x, f(y) \equal{} y, x\neq y$. then $ 0 < |f(x) \minus{} f(y)|\leq \frac {|f(x) \minus{} x| \plus{} |f(y) \minus{} y|}{2} \equal{} \frac {0 \plus{} 0}{2} \equal{} 0$, contradiction.[/quote]

I think you showed that there is at most one solution. It remains to prove that there is at least one.[/quote]

Assume that $ f(x)\equal{}y,x \neq y \Rightarrow |y\minus{}f(y)| \leq \frac{|y\minus{}x|\plus{}|f(y)\minus{}y|}{2}$
$ \Rightarrow |x\minus{}y|>|f(x)\minus{}f(y)|>|f(f(x))\minus{}f(f(y))|>...>|f^{(n)}(x)\minus{}f^{(n)}(y)|>... \geq 0$
Let $ n \rightarrow \infty$. Hence $ f^{(n)}(x)\equal{}f^{(n)}(y) \Rightarrow f^{(n)}(x)\minus{}f^{(n)}(f(x))$
With $ f^{(n)}(x)\equal{}a \Rightarrow f(a)\equal{}a$, contradiction.
Therefore , $ f(x)\equal{}x$ has at least one solution.


Let f(a,b) be a function of two variables, defined by the recurrences

f(0,b)=b+1
f(a,0)=f(a-1,1)
f(a,b)=f(a-1,f(a,b-1)).

Find f(4,2).

If you have seen this problem before, don't post a solution please.

I would guess that the best way to approach this is:

[hide]Start at (0,0) and work up and over to (4,2).[/hide]

Try it. The result is very surprising. So, as a warning, if you get a result that looks unreasonable, it may not be.

I think a similar prb was given at an IMO or something like that. But they asked f(4,1981) there. It's not that hard to calculate f(1,x), f(2,x), f(3,x) f(4,x) for positive integers x.

As for the result, f(4,x)=[hide]2^2^2^..^2-3, where 2 appears x+3 times, sp it's not hard to calculate f(4,2)[/hide]

That's correct. This function is called the Ackermann Function.

Yeah, I remember this was one of the first examples we had to try to implement in computer science class when we learned recursion in Pascal.

That is truly evil. When I first heard this problem (it was being told to a group of people at once), some people tried to write computer programs, and they said they let it run for half an hour, and it was still running. I think it overflowed the stack.


inequalities
number theory unsolved
number theory

see http://www.mathlinks.ro/Forum/viewtopic.php?p=234089#p234089  :wink:


trigonometry
vector
inequalities
LaTeX
inequalities proposed

Find the min and the max of $A=8x+6y-5z$ if $x^2+y^2+z^2=5$

I use Lagrange's multipliers method.

Let $L=8x+6y-5z+\lambda (x^2+y^2+z^2)$.

$\frac{\partial L}{\partial x} = 8+2x\lambda$
$\frac{\partial L}{\partial y} = 6+2y\lambda$
$\frac{\partial L}{\partial z} = -5 +2z\lambda$

So $(x,y,z)=\left(\frac{-4}{\lambda},\frac{-3}{\lambda},\frac{2,5}{\lambda}\right)$, so with contition we have:

\[(x,y,z)=\left(\frac{-4}{2,5},\frac{-3}{2,5},1\right)=P_1 \vee (x,y,z)=\left(\frac{4}{2,5},\frac{3}{2,5},-1\right)=P_2.\]

You should only check $d^2L(P_1)(\vec{h})$ and $d^2L(P_2)(\vec{h})$ is positive (it's minimum) or negative (it's maximum).

Using the scalar product,
$A=(8,6,-5)\cdot (x,y,z)=\sqrt{8^2+6^2+(-5)^2}\sqrt{x^2+y^2+z^2}\cos \theta$, where $\theta$ is the angle between 2 vectors ${(8,6,-5)\ and\  (x,y,z)}$,thus $A=5\sqrt{5}\cdot \sqrt{5}\cos \theta =25\cos \theta$.Since $-1\leq \cos \theta \leq 1$, we have $-25\leq A\leq 25$.

The equality holds as follows.

$A=-25\Longleftrightarrow \theta =\pi \Longleftrightarrow (x,y,z)=-\frac{\sqrt{5}}{\sqrt{8^2+6^2+(-5)^2}}(8,6,-5)=\left(-\frac{1}{5},-\frac{6}{5},1\right)$.

$A=25\Longleftrightarrow \theta =0 \Longleftrightarrow  (x,y,z)=\frac{\sqrt{5}}{\sqrt{8^2+6^2+(-5)^2}}(8,6,-5)=\left(\frac{1}{5},\frac{6}{5},-1\right)$.

Thus desired  minimum value is $-25$, maximum value is $25$.

Hi  Silouan.

  Well, I think your puzzle is not much difficult.
I use Bunhacopxki (As it is the name of inequality  I studied in Viet Nam) and found out that :  max(f(x))= 25 and min (f(x))=-25
Max (f(x)) = 25 when  x=8/5  y=6/5  z= -1
Min ((f(x)) = -25 when x=-8/5  y=-6/5  z= 1

  Here is  mine:

abs(F(x))= abs(8x+6y-5z) <= sqrt [(8^2+6^2+ (-5)^2) (x^2+y^2+z^2]

Because x^2+y^2+z^2=5 so abs(F(x))<= 25
and we can have min and max

 Sorry for my very weak at English and also Latex but I will try my best to overcome that.
  Hope it is right

[b][u]From Sir Gibbenergy:[/u][/b]

$C.B.S.\Longrightarrow [8\cdot x+6\cdot y+(-5)\cdot z]^2\le [8^2+6^2+(-5)^2]\cdot (x^2+y^2+z^2)=125\cdot 5=625$ $\Longrightarrow |8x+6y-5z|\le 25$.

Thus, $-25\le 8x+6y-5z\le 25$. I think that you know when we have the equalities (left  or right - see the

Gibbenergy's  reply). [i]Sir Gibbenergy[/i], learn quickly and immediately [b]LaTeX[/b] and ... the [b]English language[/b]!

In case of using Cauchy-Schwalz's inequality, how did you obtain when the equality hold? :?

In inequality $|8x+6y-5z|\le 25$ we have equality iff $\frac x8=\frac y6=\frac {z}{-5}=k\in R$. From the relation $x^2+y^2+z^2=25$ we obtain $k\in \left\{ \pm \frac 15\right\}$. Thus, we have equality in the right part for $k=\frac 15$ and we have the equality in the left part for $k=- \frac 15$. Is it O.K. ?


algorithm
modular arithmetic
number theory
Euclidean algorithm
number theory open

http://www.mathlinks.ro/viewtopic.php?t=289701

I am sorry.

$ p | z \minus{} y$ doesn't imply $ x | z \minus{} y$.

Abel Lemma:http://www.mathlinks.ro/viewtopic.php?t=214396&sid=f5a7c2df70627ae6415a60f194999e6d

 Let $ (x, n)\equal{}1$, then $ x\equal{}ab, z\minus{}y\equal{}a^n$.

Hence $ p|z \minus{} y$ implies $ x|z\minus{}y.$

How about finally stopping to post near to unreadable "proofs" of a theorem that seems to be out of reach for you¿
From your "proofs", I see two things:
- You are lacking even some basic number theory skills.
- You fail to be able to check your proofs for really simple errors.

ZetaX, Thank you for your kindness! Mathematics requests hardworking training, but I need help and advice.
t0rajir0u, Thank you for your kindness! I think it is plain. But I forget Abel lemma is not so wellknown. Sorry.

Training is not to try a thousend different proofs on Fermat's conjecture. You got a lot of advice like recommended books, I think (and if not, you may still ask).
So just read some serious books, take a look at existing works, there is no harm in it.

I have to interrupt my study to get a job.

But I am going to be back somehow someday.

God bless me!

Theorem: Suppose $ (x, y, z)$ is one solution of Fermat
Equation $ x^n \plus{} y^n \equal{} z^n.$ Under condition of $ (x, y) \equal{} (y, z) \equal{}
(z, x) \equal{} 1, n$ is odd prime, if $ (x, n) \equal{} 1,$ then we get $ x \equal{} ab, z
\minus{} y \equal{} a^n, (a, b) \equal{} 1.$

Proof: Since $ x^n \equal{} z^n \minus{} y^n \equal{} (z \minus{} y)\frac{x^n}{z\minus{}y}, (x, y) \equal{} (y,
z) \equal{} (z, x) \equal{} (x, n) \equal{} 1,$ hence $ (z \minus{} y, nzy) \equal{} 1.$ According to
Euclidean algorithm, we get $ (z \minus{} y, \frac{x^n}{z\minus{}y}) \equal{} 1,$ then we
get $ x \equal{} ab, z \minus{} y \equal{} a^n, (a, b) \equal{} 1.$

Theorem: If $ n$ is odd prime, then $ x^n \plus{}
y^n \equal{} z^n$ have no solution $ x,y,z$ such that $ (x, y) \equal{} (y, z) \equal{} (z,
x) \equal{} 1.$

Proof: For any prime $ p|x, (x, n) \equal{} 1.$ Let $ y^{p \minus{} 1} \equiv z^{p \minus{}
1}\equiv 1, \pmod p.$ In case of $ n\nmid p \minus{} 1:$ Since we have $ |x|>
|z \minus{} y|, (z, y) \equal{} 1,$ according to Euclidean algorithm, if $ (m, n) \equal{}
1,$ we get $ (z^m \minus{} y^m,z^n \minus{} y^n) \equal{} z \minus{} y.$ Then $ p|(z^{p \minus{} 1} \minus{}
y^{p \minus{} 1},z^n \minus{} y^n)\equal{}z \minus{} y.$

In fact we get $ x \equal{} ab, z \minus{} y \equal{} a^n,
(a, b) \equal{} 1.$ If $ p|(x, z \minus{} y) \equal{} a,$ then $ p|a^n \equal{} z \minus{} y.$ which
leads to a contradiction of that $ x|z \minus{} y.$

Then $ n|p \minus{} 1.$ Hence
$ x\equiv1, \pmod n$. Let $ u^n \plus{} v^n \equal{} w^n, 2|v.$ In case of $ z \equal{} w,
y \equal{} v, x \equal{} u,$ we have $ u\equiv1, \pmod n.$ In case of $ z \equal{} \minus{} w, y \equal{}
\minus{} v, x \equal{} \minus{} u,$ we have $ \minus{} u\equiv1, \pmod n.$ Then $ 0\equiv2, \pmod
n,$ which is contradictory.


algebra
polynomial
Ring Theory
superior algebra
superior algebra solved

Let

$S=\big\{a+x^{3}f(x,y)+xyg(x,y)+y^{3}h(x,y), \ \ \text{where}\ \ a\in \mathbb{R}\ \ \text{and}\ \ f(x,y),g(x,y),h(x,y) \in\mathbb{R}[X,Y]\big\}$

Then prove that:
a) $S$ is a subring of $\mathbb{R}[X,Y]$
b) $x^{3},xy,y^{3}$ are irreducible but not prime elements of the ring $S$

Alexandros

Let $R$ be a $k$-algebra and let $I$ be an ideal in $R$. Then $k+I$ is a sub $k$-algebra of $R$.

In our case, $k=\mathbb R$, $R=\mathbb R[x,y]$ and $I=(x^{3},xy,y^{3})$.

$x^{3},xy,y^{3}$ are homogeneous and $\mathbb R[x,y]$ as well as $\mathbb R+(x^{3},xy,y^{3})$ are graded $\mathbb R$-algebras. If for example $x^{3}$ is  a product of some polynomials $f,g$ then by taking the homogeneous parts of maximal degrees, we can assume that $f$ and $g$ are homogeneous. It is easy to prove that $x^{3},xy,y^{3}$ don't split as products of homogeneous elements of $R'=\mathbb R+(x^{3},xy,y^{3})$. Therefore these three elements are irreducible.
$x^{3}|(xy)^{3}$ in $R'$ but does not divide $xy$. The same argument holds for $y^{3}$.
$(xy)|x^{3}\cdot y^{3}$ in $R'$ but does not divide any of the factors.
Thus the three elements are not prime.


das hier betrifft alle, die an den IMO-vorauswahlklausuren teilgenommen haben. es geht um die aufgabe 2 der 1. IMO-vorauswahlklausur 2005:

[b]Aufgabe 2[/b]

Sei E = {-1, 0, 1}, und sei M eine Menge von Gitterpunkten der Ebene. Die Punkte aus M sind so durch ein Streckennetz (S) miteinander verbunden, dass man von jedem Punkt aus M zu jedem anderen Punkt aus M gelangen kann ohne (S) zu verlassen. Dabei können die Strecken des Streckennetzes außer den Gitterpunkten auch andere Endpunkte enthalten.

Man ermittle die kürzeste Gesamtlänge des Streckennetzes, falls:

[b]a)[/b] M = { (i, j) | i, j Elemente von E und $i\cdot j=0$ }.
[b]b)[/b] M = { (i, j) | i, j Elemente von E }.

diese aufgabe wurde schon auf http://www.mathlinks.ro/Forum/viewtopic.php?t=21000 besprochen, allerdings ohne echte vollständige lösung. nun hab ich heut die musterlösungen gekriegt, und da war ne ziemlich kurze lösung von dieser aufgabe; allerdings zweifele ich an deren vollständigkeit. kann jemand die argumentationen der art "aus symmetriegründen...", die in der lösung zweimal vorkommen, formalisieren? danke sehr, würde mich nämlich wirklich interessieren.

(sorry daß ich die musterlösung hier nicht posten kann - mein scanner geht net...)

  grüße,
  darij

[quote="darij grinberg"]allerdings zweifele ich an deren vollständigkeit. kann jemand die argumentationen der art "aus symmetriegründen...", die in der lösung zweimal vorkommen, formalisieren? danke sehr, würde mich nämlich wirklich interessieren.
[/quote]

nach genauerer betrachtung verstehe ich die musterlösung auch nicht.  meiner meinung nach beweist die symmetrie nur, daß das zweite teilstreckensystem genauso lang ist wie das erste.  allerdings vermisse ich den beweis, daß das kürzeste streckensystem aus diesen beiden teilen besteht.
 :|  :huh:  :wacko:

ich stimmt christopher absolut zu...werd vielleicht noch mal drüber nachdenken, aber irgendwie sieht die lösung verdammt unvollständig aus,

[size=117]Guten Tag,

eure Fragen / Zweifel sind zwar schon 9 Tage alt, aber falls noch jemand eine Antwort vermisst (und sei es ein zufällig vorbeisurfender Leser):

Zur Vervollständigung des Beweises zu a) genügt eigentlich folgende Vorüberlegung. Betrachten wir zunächst den Teilgraphen, der nur aus O(0|0) besteht. Für keinen der vier übrigen Punkte gibt es einen kürzeren Anschluss an diesen Teilgraphen als einen der Länge 1. Falls der gesuchte Gesamtgraph kürzer als 4 sein soll, muss es folglich einen Hilfspunkt P in irgendeinem Quadranten (o.B.d.A. im ersten) geben, über den einer der zu diesem Quadranten gehörenden Punkte, z.B. A(1|0) mit O verbunden wird. Der Gesamtgraph kann sich aber trivialerweise nur dann verkürzt haben, wenn der andere zu diesem Q. geh. Punkt, also C(0|1), ebenfalls an die Strecke OP angeschlossen wird. Für jeden der übrigen beiden Punkte beträgt aber die kürzeste Entfernung zu irgendeinem Punkt des "Dreier-Teilgraphen" immer noch Eins. Außerhalb des 3. Quadranten gibt es keinen Hilfspunkt, über den der Anschluss an diesen Teilgraphen verkürzt (insgesamt kleiner als 2) werden kann, also muss man ihn innerhalb suchen, und es ist eine symmetrische Situation zu der im 1. Quadranten gegeben.

Bei b) kann man ähnlich argumentieren. Wenn es einen Graphen geben soll, der kürzer als 8 ist (Länge z.B. eines Hakenkreuzes -- pardon!), muss dazu mindestens ein Hilfspunkt abseits der Achsen, o.B.d.A. innerhalb des 1. Quadranten bemüht werden. Nur für die drei zu diesem Quadranten gehörenden Punkte A, B und C kann sich aber dadurch eine Verkürzung der Netzstrecke ergeben, die zu ihrer Anbindung an O(0|0) insgesamt nötig ist. Also ist ein kürzester Verbindungsgraph Tg für die vier Punkte O, A, B und C zu wählen, wie er von der Form her aus a) bekannt ist.
Jeder weitere Hilfspunkt, der die Gesamtlänge des Graphen verringern könnte, kann nun nur noch in einem der drei übrigen Quadranten liegen. Nimmt man je einen Punkt im 2. und im 4. Quadranten an, legt man damit den unter diesen Umständen kürzesten Gesamtgraphen fest, erhält aber kein optimales Ergebnis [sondern 2 + sqrt(2) + sqrt(3) + sqrt(6)]. Nimmt man ihn im 3. Quadranten an, erzwingt man dort einen weiteren Teilgraphen mit der Länge von Tg, und es bleiben nur die zwei Punkte (1|-1) und (-1|1) übrig, die mit der minimalen Verbindung der Länge 1 angeschlossen werden, eine weitere Verkürzung ist also nicht möglich.

Mit dieser Vorüberlegung hätte man also wieder die Punktsymmetrie zum Ursprung, die man schon bei a) genutzt hat. Allerdings darf man sich wohl darüber streiten, ob das alles so trivial ist, dass man es in zwei Worten („aus Symmetriegründen“) zusammenfassen kann. Zumal in b) der berechnete halbe Graph nun gerade nicht symmetrisch ist. Jedoch einfach mal postuliert wird, der minimale Fünfergraph sei ein minimaler Vierer mit Anhängsel.

Aber vielleicht lassen sich ja die Symmetriegründe viel einfacher begründen? ;) [/size]

Pardon, hab mich noch nicht vorgestellt: [size=117]Als am Wettbewerbsgeschehen sehr interessierter Lehrer bin ich vor einer Woche rein zufällig auf eine ähnliche Aufgabe* gestoßen, deshalb hab ich mich spaßeshalber angemeldet und gepostet. Verzeiht mir, falls ich umständlich oder gar fehlerhaft argumentiert haben sollte, aber Lehrer sind naturgemäß dümmer als Deutschlands klügste Schüler...

*Nämlich die, das kürzeste Netz zwischen vier im Quadrat angeordneten Punkten zu finden. Bin genau so vorgegangen wie in a): Annahme, der gesuchte Graph sei kürzer als die Summe der Diagonalen --> 1. Hilfspunkt P --> Symmetrie --> 2. Hilfspunkt Q --> Sei x die Strecke PQ und S(x) die Gesamtlänge des Graphen --> Minimum der Funktion S(x) ermittelt.[/size]

zugegebenermaßen kann ich zu a) einen echten beweis geben, jedoch verstehe ich in der obigen lösung nicht, warum es z.b. bei a) keinen weiteren hilfspunkt geben kann, also in etwa so etwas(unterstriche dienen nur als platzhalter):

A
_\
_ \
_  P
_/ \
O  Q---C
_  /
_/
/
B

wieso muss C DIREKT mit P verbunden sein? durch einen umweg zu Q kann das argument, dass die anbindung von B an das restliche streckennetz mindestens länge 1 hat zerstört werden.

ich denke, dass ein vollständiger beweis da noch einige fälle mehr diskutieren muss...aber eigentlich will ich gar keinen beweis kennenlernen - die aufgabe is einfach viel zu eklig :D

Zu deinem Einwand: Mir ist zwar klar, dass die Darstellung oben nur eine prinzipielle Überlegung veranschaulichen soll, aber man sieht ihr doch trotzdem sofort an, dass der Graph kürzer wäre, wenn C direkt mit P und B direkt mit O verbunden wäre.

Läge Q im ersten Quadranten, wäre die Strecke BQ keine Verkürzung gegenüber BO, also müsste sich zum Ausgleich das Netz der anderen Punkte verkürzen. Vorausgesetzt, das ginge. Da O sowieso nicht "abgekoppelt" werden darf, wäre es dann immer noch günstiger, B wieder direkt an O anzuschließen, egal, welche Rolle der Punkt Q spielt.

Und im zweiten (oder vierten) Quadranten gilt wieder: Ein Hilfspunkt Q macht nur dann Sinn, wenn mehr als ein Punkt über ihn angeschlossen wird. Da ist aber außer B (bzw. D) keiner mehr, der noch nicht angeschlossen wäre, also ist durch Hilfspunkte außerhalb des dritten und ersten Quadranten keine weitere Verkürzung möglich, den Rest besorgt die in der Musterlösung gezeigte Optimierung.

Kann es sein, dass Ihr Problemlöseprofis manchmal zu kompliziert denkt? Schmunzeln musste ich z.B. über die irgendwo im AoPS-Forum gepostete Aufgabe mit dem israelischen Parlament, wo mehrere Löser mit "Matrizen auf Spatzen" schossen, einer auch noch in die falsche Richtung. Dabei ging´s nur darum, dass man bei zwei zufällig anwesenden Parlamentariern nicht garantieren kann, dass sie unterschiedlichen Ausschüssen angehören, wenn es 120 Abgeordnete, aber nur 40 Ausschüsse gibt.

Da kommt mir der Klaviermeisterschüler in den Sinn, der mit virtuosem Liszt brilliert, aber plötzlich vor lauter technischer Perfektion keine langsamen Stücke mehr spielen kann.

Doch das ändert nichts an meinem Riesenrespekt vor euren Leistungen!! Unter Klausurbedingungen fänd ich wahrscheinlich jede Aufgabe eklig ...


function
calculus
derivative
integration
logarithms
rate problems
calculus computations

Let f be the function defined by f(x)=x+lnx. What's the value of c for which the instantaneous rate of change of f at x=c is the same as the average rate of change of f over [1,4]?
A) 0.456  B)  1.244  C) 2.164  D)2.342  E)2.452.
I think the average rate of change is f(4)-f(1)/4-1, and instantaneous one is f'(x), but they never equal?? the previous one is negative. the latter one is positive?
thanks for help.

This is the mean value theorem. Or rather, the intermediate value theorem for the derivative. If the derivative of $ f$ is continuous (which implies $ f$ is differentiable) on the interval then

$ \frac{f(b)\minus{}f(a)}{b\minus{}a}\equal{}f^{\prime}(c)$

Or stated from the derivative's point of view, $ \frac{1}{b\minus{}a}\int^b_a f^{\prime}(x)dx \equal{} f^{\prime}(c)$
The former is most definitely faster here. $ \frac{f(4)\minus{}f(1)}{3}\equal{}\frac{\ln{4}\plus{}3}{3}\equal{}\frac{1\plus{}c}{c}$
$ 1\plus{}\frac{\ln{4}}{3}\equal{}1\plus{}\frac{1}{c}$
$ \Leftrightarrow c\equal{}\frac{3}{\ln{4}}$

Or, $ c\equal{}2.164$ after rounding. The answer is C. :) I also might take a moment to state that A can automatically be thrown out because [i]it is not in the interval[/i].

[quote="igiul"]This is the mean value theorem. Or rather, the intermediate value theorem for the derivative.
..
I also might take a moment to state that A can automatically be thrown out because [i]it is not in the interval[/i].[/quote]  This is not accurate: the theorems you've quoted guarantee a point in the interval with the desired property -- they don't tell you that the [i]only[/i] such points are in that interval (and it's certainly not true in general).  As a matter of "finding the answer in a multiple choice exam as quickly as possible," though, you have an excellent point.

That's actually a really REALLY good point. Because the function did not say 'in the interval [1,4]' this does mean it can be applied to it's entire domain. Perhaps if one answer had a multiplicity of values and one just had one in the interval this problem would be tricky because you automatically think MVT which then leads you to the value in the interval, but nothing outside of it.


function
logarithms
ceiling function
floor function
number theory unsolved
number theory

Is Bertrand's Postulate simple???
Bertrand's Postulate says : "Between $n$ and $2\times n$ there is always a prime number $p$".

You can do an elementary counting argument (= But its not simple.

[quote="zengmao"]Is there a simple elementary proof of $\pi\left(2^n\right)>n$?[/quote]

Yes, there is  ;) . The following one-line argument of Erdos is really strikingly simple :D . Let $p_1,\ldots,p_m$ be all primes up to $x$. Because every positive integer may be (uniquely) expressed as the product of a square-free number and a perfect square, every number up to $x$ may be written as $p_1^{e_1} \cdots p_m^{e_m}q^2$, where $e_i \in \{0,1\}$. There are $2^m$ choices for the $e_i$ and $\sqrt{x}$ choices for $q$, so it must be that $2^m\sqrt{x} \geq x$, i.e. $\pi(x) = m \geq \frac{\log{x}}{2\log{2}}$. Although this is a bit weaker than what you asked for by a factor of $1/2$, the argument may be refined.

I also showed a simple, elementary & 1.5-line proof somewhere on the forum that $[1,2,\ldots,n]>2^n$ for $n \geq 9$; it follows that $n^{\pi(n)} > \prod_{p \leq n}p^{\lceil \log_p{n} \rceil} = [1,2,\ldots,n] > 2^n$, i.e. $\pi(n) > \frac{n \log{2}}{\log{n}}$ --- which differs from the 'truth' only by (roughly) a factor of $\ln{2}$!  :)

[quote="vess"]it follows that $n^{\pi(n)} > \prod_{p \leq n}p^{\lceil \log_p{n} \rceil} = [1,2,\ldots,n]$...[/quote]
Just a small correction, instead of $\prod_{p \leq n}p^{\lceil \log_p{n} \rceil}$ there should be $\prod_{p \leq n}p^{\lfloor \log_p{n} \rfloor}$.


geometry
parallelogram
geometry proposed

Let $ABCD$ be a convex quadrilateral, $M$ the midpoint of $AB$, $N$ the midpoint of $BC$, $E$ the intersection of the segments $AN$ and $BD$, $F$ the intersection of the segments $DM$ and $AC$. Prove that if $BE = \frac 13 BD$ and $AF = \frac 13 AC$, then $ABCD$ is a parallelogram.

Let the vertices A, B, C be fixed. The midpoints M, N of AB, BC and the point $F \in AC$ such that AC = 3AF are then also fixed. The locus of the point E is the segment AN. The locus of the remaining vertex D is a ray of the fixed line MF opposite to the ray FM and a segment $M'N' \parallel MN$, such that M'A = AM = MB and BN = NC = CN', i.e., M'B = 3MB and BN' = 3BN. The ray of the line MF opposite to the ray FM intersects the segment M'N' at a unique point D, which means that a unique convex quadrilateral ABCD satisfying the given condition exists. But since the given conditions are trivially satisfied for a parallelogram obtained from the triangle $\triangle ABC$ by reflecting the vertex B in the midpoint of the segment AC (when E is the centroid of the triangle $\triangle ABC$), this unique quadrilateral ABCD is a parallelogram.

[hide="Here is the first method and a little generalization."]
[color=darkblue][b]A proof using the vectorial calculus.[/b] Choose the origin $I\in AC\cap BD$ of the vectorial system and denote $\frac{IC}{IA}=a>0$, $\frac{ID}{IB}=b>0$, i.e. $\boxed {C=-aA\ ,\ D=-bB}$. Thus, $2M=A+B$, $2N=B-aA$ and appling the Menelaus' theorem to the transversals $\overline s {AEN}$, $\overline {DFM}$ in the triangles $BCI$, $ABI$ obtain $\boxed {E=\frac{1}{a+2}B}$, $\boxed {F=\frac{b}{2b+1}A}$. Therefore,
$3\cdot BE=BD$, $3\cdot AF=AC$ $\Longleftrightarrow$ $\boxed {E=\frac{2-b}{3}B}$, $\boxed {F=\frac{2-a}{3}A}$ $\Longleftrightarrow$ $\frac{1}{a+2}=\frac{2-b}{3}$, $\frac{b}{2b+1}=\frac{2-a}{3}$ $\Longleftrightarrow$ $ab=2(a-b)+1$, $2ab=b-a+2$, $a>0$, $b>0$ $\Longleftrightarrow$ $a=b=1$ $\Longleftrightarrow$
$IC=IA$, $ID=IB$ $\Longleftrightarrow$ the quadrilateral $ABCD$ is a parallelogram.[/color]

[color=darkred][b]A little generalization.[/b] Let $ABCD$ be a convex quadrilateral. Define: the middlepoints $M$, $N$ of the sides $[AB]$, $[BC]$ respectively; the intersections $E\in BC\cap AN$, $F\in AC\cap DM$. Prove that the quadrilateral $ABCD$ is a parallelogram if and only if $2\cdot \frac{ED}{EB}=2+\frac{FC}{FA}$.[/color][/hide]
[hide="Here is the second method."][/hide]


Let be $ x,y,z\in \mathbb{R}$ such that $ (xy\plus{}yz\plus{}zx)^3\equal{}yz(x\plus{}y\plus{}z)^3$ . Prove that $ 2x\equal{}y\plus{}z$ or $ 2y\equal{}z\plus{}x$ or $ 2z\equal{}x\plus{}y$ ($ x,y,z$ are in arithmetic progression) .

Two things wrong with the statement:

1. It should read $ (xy\plus{}yz\plus{}zx)^{3}\equal{}xyz(x\plus{}y\plus{}z)^{3}$

2. $ x,y,z$ are forming a geometric, not an arithmetic sequence

[hide="Proof of the corrected problem"]Since the equation is homogenous in $ x,y,z$, we can WLOG assume $ xyz\equal{}1$, which yields

$ xy\plus{}yz\plus{}zx\equal{}x\plus{}y\plus{}z$

$ xy\plus{}{1\over x}\plus{}{1\over y}\equal{}x\plus{}y\plus{}{1\over xy}$

$ x^2y^2\plus{}y\plus{}x\equal{}x^2y\plus{}xy^2\plus{}1$

$ x^2y^2\minus{}xy^2\minus{}x^2y\plus{}y\plus{}x\minus{}1\equal{}0$

$ xy^2(x\minus{}1)\minus{}y(x^2\minus{}1)\plus{}x\minus{}1\equal{}0$

$ (x\minus{}1)(xy^2\minus{}xy\minus{}y\plus{}1)\equal{}0$

$ (x\minus{}1)\left[xy(y\minus{}1)\minus{}(y\minus{}1)\right]\equal{}0$

$ (x\minus{}1)(y\minus{}1)(xy\minus{}1)\equal{}0$

But $ xy\equal{}1\iff z\equal{}1$, hence in any case one of the numbers is equal to $ 1$, therefore the cubic root of their product is equal to one of the numbers, which means they form a geometric progression. QED[/hide]

An easier way would be
$ (xy\plus{}yz\plus{}zx)^3 \minus{} xyz(x\plus{}y\plus{}z)^3 \equal{} (x^2\minus{}yz)(y^2\minus{}zx)(z^2\minus{}xy)\equal{}0$  giving the required result.


Let $ r(1) \equal{} 1$ and for $ k > 1$ let $ r(k)$ equal the product of the prime divisors of $ k$. A sequence of natural numbers $ a_1,a_2,...$ with arbitrary first term $ a_1$ is defined recursively by the relation $ a_{n\plus{}1} \equal{} a_n \plus{} r(a_n)$. Show that for any positive integer $ m$, the sequence $ (a_n)$ contains $ m$ consecutive terms in arithmetic progression.

[hide="Hint"]
$ Lemma \ 1.$
In the sequence r(a_1), r(a_2), ..., each term divides the next.

$ Lemma \ 2.$
For any $ m$, there exists an integer $ n$ s.t. $ z_n \equal{} z_{n\plus{}1} \equal{} z_{n\plus{}2} \equal{} \cdots \equal{} z_{n\plus{}m\minus{}2} \equal{} 1$
$ \left(z_n \equal{} \frac{r(a_{n\plus{}1})}{r(a_n)}\right)$
[/hide]


If $mn|n^2+m^2+m$ then prove that m is a perfect square.($m,n$ are positive integers)

Is it that boring? Dont you have any idea?

Well, in fact, it was already here ;) :
http://www.mathlinks.ro/Forum/viewtopic.php?t=24081


For a given triangle $ABC$ , Let $X$ be a variable point on the line $BC$ such that $C$ lies between $B$ and $X$ , and the incirlces of the triangles $ABX$ and $ACX$ intersect at two different points $P$ and $Q$ . prove that the line $PQ$ passes through a point independent of $X$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=41033
http://www.mathlinks.ro/Forum/viewtopic.php?t=40772

  darij


can anyone help me in getting eboks on algebra geometry and number theory online. :)

Sure man...try http://www.esnips.com or http://www.4shared.com.....if u are unable to do that then u can contact me...i have a large collection of ebooks of NT

TX BOTH.
sorry, manzil .those were social websites. couldnt find anything there.

http://fileshunt.com
http://www.ebdb.net
http://www.addebook.com
I love  Indian.
I like Ramanujan  so  much.

these are all the sites where we have to pay, do u know anything which are free?

Try these [url=http://www.btscene.com/details/879315/]Geometric Algebra for puter Science: An Object-Oriented Approach to Geometry[/url]  and [url=http://www.btscene.com/details/232872/]All Math Learning Books (Total 120+ Books, Geometry, Algebra, Calculus...)[/url]

Most e-books can be downloaded free from the site http://www.gigapedia.org

You just need to sign up with your email adress and nothing else. I use this site for years now to see what books to order from amazon. :)

Google search is always the solution, if you want specific results try [url=http://www.mininova.org/sub/50]http://www.mininova.org/sub/50[/url] or [url=http://www.realtorrentz.com/cat.php?id=7]http://www.realtorrentz.com/cat.php?id=7[/url]


Let m, n, k be natural number such that  5^n - 2, 2^k - 5 are both divided to 5^m- 2^m. Show that GCD(m, n) = 1.

Easy but nice. Suppose d is a common divisor of m and n. Then $ 5^d-2^d| 5^m-2^m$ and $ 5^d-2^d| 5^{kn}-2^{kn}$. But $ 5^{kn}-2^{kn}=(5^n)^k-(2^k)^n (mod 5^m-2^m)=3(mod 5^m-2^m)$. From these three relations we find that$ 5^d-2^d|3$ from where d=1.

harazi's solution is nice .Oh I have this problem if (tex) 2^n -2 divided 5^m-2^m then n = km .Is this problem true? thank


[quote="MCrawford"]After one-fifth of the girls left a school dance, the ratio of girls to boys was 2:3.  Then, 44 boys left and the ratio of boys to girls was 2:5.  How many students remained at the dance?[/quote]
This was a Nationals problem...

[hide="solution"]
(4g/5) / b = 2 / 3
4g / 5b = 2 / 3
10b = 12g
5b = 6g

(b - 44) / (4g/5) = 2 / 5
(5b - 220)/4g = 2 / 5
(6g - 220)/4g = 2 / 5
(30g - 1100)/4g = 2
30g - 1100 = 8g
22g - 1100 = 0
22g = 1100
g = 50

Plugging it back in:

(4g/5) / b = 2 / 3
40 / b = 2 / 3
2b = 120
b = 60

Since there were 50 girls and 60 boys to start with, after one-fifth of the girls left, there were 40 girls and 60 boys (ratio = 2:3 - check). After 44 boys left, there were 16 boys and 40 girls left (ratio = 2:5 - check). That means [b]56 students remained[/b].

*dies*
[/hide]

[hide][color=indigo]the number of girls to boys is:
at first 50:60
then 40:60
then 40:16
So 56 kids are still there.[/color][/hide]

[quote]isn't this a problem from AMC8 year 2004(~2005) ???

[/quote]
i thought so too

[hide]This is from a MathCounts nationals--I'm sure.
The answer is 56--I got it the same way that Treething got, so I won't explain.[/hide]


Hey I was thinking that it'd be really cool if there could be an automatically generated "solutions" pdf where we could export a PDF with latex solutions from the posts; perhaps have these dynamically generated by the user clicking on which post to designate as the solution to a given problem

Yeah that's what I was implying
and for those which didn't have the solution we could hand pick

[quote="worthawholebean"]We already have the ability to designate replies in a thread as solutions to problems, so such a feqture could simply use those.[/quote]That is used only for the resources pages.

But I'm saying we could just automatically use those as the solutions.


6. If $ x \plus{} 1/x \equal{} 3$ , what is the value of $ x^4 \plus{} 1/x^4$??

1. Solve $ 8!\minus{}10  / 8!\plus{}10$. Express your answer as a common fraction.

And if you don't mind, post the solutions for me...

I think #1 was the problem in the sprint round that we took last time.. (for jenks people..)

:) [b] Happy Holidays!![/b]

[quote="kunny"]$ x^4 \plus{} \frac {1}{x^4} \equal{} \left(x^2 \plus{} \frac {1}{x^2}\right)^2 \minus{} 2.$[/quote]

No.... that's not the right answer (Nice try, though). I don't know if you misunderstood me or not, but the answer is wrong...   :(

[quote="kunny"]I just gave a hint to you, jogaman323 :lol:[/quote]

I knew that... :oops:

Ok.. I'll try to use your clue... thanks


geometry
calculus
rotation
3D geometry
sphere
integration
probability

Show the derivation of the formulas for the suface area and the volume of a sphere.

Proofs are too hard for Classroom Math level.

lol, we need calculus to do that  :rotfl:

[quote="mewto55555"]I know for volume you wold rotate a semi circle around the x-axis.... You would do the same thing for surface area but with a different formula, I'll look it up later.... I remember I forgot formula fro volume of a sphere at MC state and had to rederive it lol.[/quote]

Also one could note that to find the volume you simply integrate the surface are formula (and remove constant of integration) and to find surface are you simply differentiate the volume, so once you know one you know the other....

[quote="FantasyLover"][quote="Bachukas"][quote="FantasyLover"]well, i dont think theres other way to prove that.[/quote]

Of course there are... but Stereometry proofs are too hard for Classroom level - it's for high school  :wink:[/quote]

and calculus is not too hard for middleschool level?

excuse me then, i didnt know there was one.[/quote]

Calculus is for checking formula not for proving, there is a probability that calculus shows a right digits just by coincidence.


geometry
3D geometry
pyramid
geometry open

Consider pyramid S-ABC. M is the midpoint of BC, O is the midpoint of AM.
A', B', C' lie on SA, SB, SC respectively, so that O is the centroid of triangle A'B'C'.
Question: How to construct plane (A'B'C') ?

Let $ N\in AM$ so that $ \overrightarrow{ON} \equal{} \frac {1}{2}\overrightarrow{AO}$
Construct a line through $ N$, parallel with $ SA$ cut $ SM$ at $ M'$.
Construct a line through $ M'$, parallel with $ BC$ cut $ SB,SC$ at $ B',C'$.
Let $ OM'\cap SA \equal{} A'$.
It is very easy to check that: $ A',B',C'$ satisfy condition of problem!

:P Thnz lotz ^ ^ your solution's really easy to understand !


hello, since $ 75^2\equal{}5625$ we have $ \sqrt{5625}\equal{}75$.
Sonnhard.

it also helps if you know the trick for squaring a number that ends in a $ 5$.

suppose we have to do $ 1,325^2$, we could pretend that the $ 5$ at the end doesn't exist. if we did that, we get $ 132$. now take that number $ n$ and multiply it by $ (n \plus{} 1)$. $ (132)(133) \equal{} 17,556$. now just "plop" a 25 at the end. so $ 1,325^2 \equal{} 1,755,625$.

similarly, if we are given a number that ends in a $ 25$, take the other digits before it (say it is $ x$), and solve the quadratic $ n(n \plus{} 1) \equal{} x$


Topic moved to Games and Fun factory.

Could some moderator delete this topic now.

Mastermind is awesome. But how would such a tournament work? Also, you could probably get more responses in the Games and Fun Factory instead of here.


Lots of team trophies went to the New York City schools yesterday--they swept five of the six top teams

1) Hunter College High School (Manhattan)
2) Mark Twain Intermediate School 239 (Brooklyn)
3) Middle School 74 Nathaniel Hawthorne (Queens)
4) Horace Mann (Bronx)
5) Twelve Corners Middle School (Rochester)
6) JHS190 Russell Sage (Queens)

Other top schools from New York City included:  #9 NYC Lab School (Manhattan), #12 Marie Curie Intermediate (Queens), and #13 Dalton (Manhattan)

NYC also had the three top individuals:

#1 Andre Arslan (Hunter)
#2 Hon Wei Khor (Mark Twain)
#3 David Lu (Hunter)

The fourth member of the NYS team this year is Ben Lei from Arlington (near Poughkeepsie), only a little over an hour's commuter train ride from the city, so it should be easy to schedule lots of practices this year.

Best of luck to the New York team at Nationals this year!

Thanks, Mary, for the information.  Congratulations to all those listed!


[b]Let [/b]$p\in N\,p\ge 2.$ [b]Ascertain[/b] ([b]without L'Hospital[/b])[b]:[/b]

$L_1=\lim_{x\searrow 0}\ \frac{1}{x^2}\cdot \left(\sqrt [p] {1+ax+bx^2+d_3 x^3\ldots+d_p x^p}+\sqrt [p] {1-ax+cx^2+e_3 x^3\ldots+e_p x^p}-2\right)$[b];[/b]

[b]Answer:[/b]  $L_1=\frac{b+c}{p}-a^2\cdot\frac{p-1}{p^2}.$

[b]Generalization[/b][b]:[/b] $L_n=\lim_{x\searrow 0}\frac{1}{x^2}\cdot\left(\sum_{k=1}^n \sqrt [p] {1+a_kx+b_kx^2+c_{k,3}x^3+\ldots+c_{k,p}x^p}-n\right)$[b], where[/b] $\sum_{k=1}^n a_k =0.$

[b]Answer:[/b]  $L_n=\frac{1}{p}\cdot\sum_{k=1}^n b_k-\frac{p-1}{2p^2}\cdot\sum_{k=1}^n a^2_k.$

Could you check for the generalisation, maybe $1/x^2$ missing

For $u$ near $0$
 $\sqrt{1+u}=1+(1/2)u-(1/8)u^2+o(u^2)$ for the generalisation


If $a,b>0$ then prove that $\displaystyle \frac{2}{\frac{1}{a}+\frac{1}{b}} + \frac{a+b}{2} \geq 2\sqrt{ab}.$

cheers! :D :D

What a trivial inequality. With AM-GM we see:

$\frac{2ab}{a+b}+\frac{a+b}{2}\geq 2\sqrt{\frac{2ab(a+b)}{(a+b)2}}=2\sqrt{ab}$

Misha


In a class of twenty students, five like to go camping and ten like to ride bikes. What is the greatest possible number of students that like to do neither?

Assume that all the students who like to go camping also like to ride bikes.  Then the greatest possible number of students that like to do neither is $ 20\minus{}10\equal{}\boxed{10}$

We have to find the maximum nuber that like to do both. So assume all five that like camping like biking.
So 20-10 (the five campers included) =10
So 10 is the greatest possible number of students that like to do neither.
 :P

OOOHHH! I was so stupid...thinking not out of the box. XD
Well, thanks for taking time on this problem...I finally understand!


analytic geometry
trigonometry
conics
ellipse
calculus
calculus computations

Let $ L_1,\ L_2$ be the ray with the origin as end point in a plane . $ L_1$ makes angle of $ \minus{}\alpha$ with the $ x$ axis and $ L_2$ makes angle of $ \alpha$ with the $ x$ axis, where $ 0<\alpha <\frac{\pi}{3}$. Take points $ P,\ Q$ on $ L_1,\ L_2$ respectively such that $ PQ\equal{}1$ and $ R$ in the opposite side  to $ O$ with respect to the line $ PQ$ such that $ \triangle{PQR}$ is regular triangle.

(1) If the segment $ PQ$ is perpendicular to the $ x$ axis, then find the coordinate of $ R$.

(2) When two points $ P,\ Q$ move on $ L_1,\ L_2$ with keeping $ PQ\equal{}1$, prove that the locus of $ R$ is a part of some ellipse.

How can $ {PQ}$ be perpendicular to $ {x}$ axis?

[hide="(1)"]

If $ N$ is the mid-point of $ PQ$, then $ PN \equal{} NQ \equal{} \frac {1}{2}$.

From triangle $ ONQ$, $ \tan \alpha \equal{} \frac {QN}{ON}$

Therefore $ ON \equal{} \frac {1}{2 \tan \alpha}$

From triangle $ RNQ$, $ RN \equal{} \frac {\sqrt {3}}{2}$.

Therefore $ R \equal{} (\frac {1}{2}( \cot \alpha \plus{} \sqrt {3}), 0)$.[/hide]

For problem 2:first , call the angle formed by line $ {L_1}$ and $ {PQ}$    $ {\angle \theta}$.Our propose is to use $ {\theta}$to express the  coordinate of point $ {R}$ .
Obviously ,$ {\angle OQP \equal{} \frac {2\pi}{3} \minus{} \theta}$, by the law of sine , we get the lenth of $ {OP}$ and $ {OQ}$, $ {OP \equal{} \frac {2}{\sqrt {3}}\sin(\alpha \plus{} \frac {\pi}{3})}$; $ {OQ \equal{} \frac {2}{\sqrt {3}}\sin \alpha}$ .
we draw a line through point  $ {R}$ perpendicular to the $ {x \minus{} axis}$ insects L1,L2 at points $ {S,T}$ ,easy to know $ {\triangle OPQ,\triangle QTR,\triangle SRP}$ are congruent, so $ {OS \equal{} OP \plus{} OQ}$ , with the length of $ {OS}$, we can get $ {R_x}$, notice that $ {R_y \equal{} \sin (\frac {\pi}{6})(OP \plus{} OQ) \minus{} OP}$, we can get $ {R_y}$ with trigongometry.SO the coordinate of ponit $ {R: (\sqrt {3}\sin (\theta \plus{} \frac {\pi}{6}),\frac {1}{\sqrt {3}}\cos (\theta \plus{} \frac {\pi}{6}))}$, $ {\theta \in [0,\frac {2\pi}{3}]}$.
we can easily get the equation of the locus of point $ {R}$: $ {\frac {x^2}{3} \plus{} 3y^2 \equal{} 1}$which is a part of an ellipse .


geometry
angle bisector
Menelaus
Germany
TST
Team Selection Test
2004

In a triangle $ABC$, let $D$ be the midpoint of the side $BC$, and let $E$ be a point on the side $AC$. The lines $BE$ and $AD$ meet at a point $F$.

Prove: If $\frac{BF}{FE}=\frac{BC}{AB}+1$, then the line $BE$ bisects the angle $ABC$.

An old easy problem that remains unsolved...
 :D

Use Menelaus' theorem in triangle $BEC$ with $\overline{AFD}$ as tranversal.....

By Menelaus
$\frac{BF}{FE}\frac{EA}{AC}\frac{CD}{DB}=1$
But $CD=DB$ and $AC=AE+EC$
Then
$\frac{BF}{FE}=\frac{AE+EC}{EA}=1+\frac{EC}{EA}$
Finally
$\frac{EC}{EA}=\frac{BC}{AB}$ and then the line $BE$ bisects the angle $ABC$ :lol:

You can also use the Van Aubel´s theorem.

Extend $ CF$ to meet $ AB$ at $ G$.
Given : $ \frac{BF}{FE}\equal{} \frac{BC}{AB}\plus{}1$
By Van Aubel Theorem
$ \frac{BF}{FE}\equal{} 1 \plus{} \frac{BG}{GA}$ (since $ BD\equal{} DC$)
Plugging  this in the given relation

We get : $ 1\plus{} \frac{BG}{GA} \equal{} 1 \plus{} \frac{BC}{AB}$

Which implies $ \frac{BG}{GA}\equal{}\frac{BC}{AB}$.

Now, $ AD, BE, CG$ concur at $ F$. So by Ceva's Theorem, $ \frac{CE}{EA}\frac{GA}{GB}\equal{} 1$

which implies $ \frac{GB}{GA}\equal{}\frac{CE}{AE}$

Therefore, $ \frac{BC}{AB}\equal{}\frac{CE}{EA}$ which implies $ BE$ is the angle bisector of angle $ ABC$.

$ \text{QED}$.

The given relation $\implies \frac{BF}{FE}=\frac{AB+BC}{AB}\ (\ *\ ) $.
Applying [b]Menelaos[/b] to $\triangle BCE$ with the transversal $\overline{AFD}$ we get $\frac{BF}{FE}=\frac{AC}{AE}\ (\ 1\ )$.
Extend $(AB$ with $BK=BC$, so $(*)$ becomes $\frac{BF}{FE}=\frac{AK}{AB}\ (\ **\ )$; with $(1)$ we get $CK\parallel BE$, i.e. $BE$ angle bisector of $\angle ABC$.

Best regards,
sunken rock

join $C,F$ and extend it to meet $AB$ at $X$.
then It's well known that $BF/FE=BD/DC+BX/XA=1+BX/XA$. so , we have $BX/XA=BC/AB$
by Ceva's theorem , $CE/EA.AX/XB=1$ . so , $CE/EA=BC/AB$ , which implies that $BE$ is the bisector . :D


If $ x^2\plus{}xy\plus{}y^2\equal{}40$ and $ x\plus{}y\equal{}10$, let s=number of ordered pairs (x,y) that satisfy the two equations, and m=maximum value of $ xy$ such that x and y satisfy the two equations. Find m+s.

[hide="Hmm?"]
$ x^2 \plus{} xy \plus{} y^2 \equal{} (x\plus{}y)^2 \minus{} xy \equal{} 100 \minus{} xy \equal{} 40 \Longrightarrow xy \equal{} 60$; then $ x$ and $ y$ are the roots of the quadratic $ z^2 \minus{} 10z \plus{} 60 \equal{} 0$, and the answer is $ 2 \plus{} 60 \equal{} 62$.
[/hide]

What do you mean " [b]s= number of pairs (x,y)[/b] " ?


function
real analysis
topology
real analysis unsolved

Please someone answer this.
Suppose that $ f$ is an uniformly continuous [b]real[/b] function on a dense subset $ E$ of a metric space $ X$. Then we have to prove that $ f$ can be uniquely extended to a continuous function on whole of $ X$.

I tried it in the following way:

We know that a uniformly continuous function carries cauchy sequences into cauchy sequences.
So if $ x\in X\setminus E$, there exists a sequence $ x_n$ in $ E$ such that $ x_n\to x$.
Therefore $ x_n$ is cauchy and hence $ f(x_n)$ is cauchy, and so $ f(x_n)$ is an cauchy sequence in $ \mathbb{R}$ and hence converges to some point in $ \mathbb{R}$ which I define to be $ f(x)$.
And I have proved that this extension is well defined and unique.

I have tried a lot to prove the  continuity $ f$ in $ X\setminus E$. Can anyone please help me? Thanks in advance.

please someone answer this. I desperately need this.

Suppose $ x,y\in X$ with $ d(x,y)<\delta,$ and suppose $ x_n,y_n\in E$ with $ x_n\to x,y_n\to y.$ What can you say about $ d(x_n,y_n)?$ What can you say about $ |f(x_n)\minus{}f(y_n)|?$

[quote="Kent Merryfield"]Suppose $ x,y\in X$ with $ d(x,y) < \delta,$ and suppose $ x_n,y_n\in E$ with $ x_n\to x,y_n\to y.$ What can you say about $ d(x_n,y_n)?$ What can you say about $ |f(x_n) \minus{} f(y_n)|?$[/quote]

Ohh.. yes I think we can show that $ d(x_n,y_n)\le d(x,y)$ . So using uniform continuity on $ E$ we can then show  that  $ |f(x_n) \minus{} f(y_n)|< \epsilon$.
Now since  $ |f(x) \minus{} f(y)|\le |f(x_n) \minus{} f(x)| \plus{} |f(x_n) \minus{} f(y_n)| \plus{} |f(y_n) \minus{} f(y)|$, we are done. Am I right sir?

Thank you very much Professor Kent Merryfield.


function
logarithms
number theory
Diophantine equation
number theory open

When reading about Matiyasevich's theorem on Wikipedia (the famous theorem which showed that the answer to Hilbert's tenth problem is negative), I stumbled across the following:

"Other applications concern what logicians refer to as pi-0-1 propositions, sometimes also called propositions of Goldbach type. These are like the Goldbach Conjecture, in stating that all natural numbers possess a certain property that is algorithmically checkable for each particular number. The Matiyasevich/MRDP Theorem implies that each such proposition is equivalent to a statement that asserts that some particular Diophantine equation has no solutions in natural numbers. [b]A number of important and celebrated problems are of this form: in particular, Fermat's Last Theorem, the Riemann Hypothesis, and the Four Color Theorem. [/b] In addition the assertion that particular formal systems such as Peano Arithmetic or ZFC are consistent can be expressed as pi-0-1  sentences. The idea is to follow Kurt Gödel in coding proofs by natural numbers in such a way that the property of being the number representing a proof is algorithmically checkable."

I used emphasis for the bit that particularly caught my attention. Now, what it says about Fermat is obvious, but is this true (it is wikipedia after all)? Is it true that the Riemann hypothesis and Four-Colour theorem can be reduced to statements amount a particular diophantine equation having no solution in the natural numbers? The former is about a complex valued functions, and the latter about graphs, so how can they both be about diophantine equations?

Is it the case that the emphasised statement above merely guarantees that the problems could be so formulated, and that the actual diophantine equation in question might be extremely complicated? Or is it simply a mistake? If it's true, then it's a pretty incredible thing, to me at least.

[quote="ZetaX"]I'm pretty sure that the statement is correct.
A even more impressing result is: there is a diophantine equation that has a solution in integers iff the axiom of choice is true.[/quote]

OK, so in principle, you could solve the Riemann hypothesis by doing diophantine analysis. What would such a diophantine formulation of the Riemann hypothesis look like, I wonder?

Now, about the axiom of choice thing you mentioned, that really is unbelievable! I thought, from my limited knowledge of the "forcing" method by which independence of AC from ZF was proved, that truth or falsity of AC does not relate to any statements in number theory, because Cohen's method of forcing does not affect the arithmetical statements true in the model.

I know this goes for CH, so CH is kind of "independent from arithmetic" in as much as whether you believe it to be true or false, does not change what you believe to be true or false in arithmetic. I assumed the same went for AC because it was proved using a similar method. But you are saying it does not go like this for AC. Are you sure? Do you have any reference I could take a look at?

[quote="mathmanman"][quote="Tanequil"]
OK, so in principle, you could solve the Riemann hypothesis by doing diophantine analysis. What would such a diophantine formulation of the Riemann hypothesis look like, I wonder?[/quote]
I guess it might be this one :
[i]There exists a constant $C$ such that
\[|\pi(x)-\text{Li}(x)| \leq C\sqrt{x}\log x \]
for $x \geq 2.$[/i]
If a number theorist were to prove this relation using the hypothesis that the Peano arithmetic is non-contradictory, then, ok, we'd have a proof of RH.
Yet, it is not clear that such a proof exists.[/quote]

Well, I was thinking in terms of giving a diophantine equation, and saying that it has no natural number solutions. That kind of formulation was what I was talking about.

I don't think, though I could be wrong, that AC has such a formulation (or any other kind of number theoretic formulation), which would go some way to explaining why it is such a "controversial" axiom of set theory. Surely if AC had a "this equation has no natural number solutions" formulation it would be true, because if it were false you could come up with a counterexample (and by independence from ZF we know such a counterexample does not exist)? So it couldn't have such a formulation, otherwise people wouldn't debate about it.

Update: http://math.berkeley.edu/~poonen/nt_is_hard.pdf
here I found confirmation of ZetaX's confirmation of what was said on wikipedia. The presentation is called "number theory is hard" and talks about diophantine formulation of the Riemann hypothesis, and also gives a diophantine equation in 26 variables which spits out the primes, and only the primes, as it's positive values. Very interesting.


calculus
integration
number theory unsolved
number theory

Classify, if possible, integral solutions (X,Y,Z) such that

$ X^2/(Z\plus{}1) \minus{} Y^2/Z \equal{} 1$.

let a,t integer both positiv or negativ or a=0, t integer

$ z\equal{}t, y^2\equal{}at, x^2\equal{}(1\plus{}a)(1\plus{}t)$ are desired solutions...

Seems like wauwau's response begs the question; i.e. for which values of $ a,t$ are $ (at)$ and $ [(1\plus{}a)(1\plus{}t)]$ perfect squares?


ratio geometry perimeter inradius trapezoid AMC AIME

In right $ \triangle ABC$ with hypotenuse $ \overline{AB}$, $ AC \equal{} 12$, $ BC \equal{} 35$, and $ \overline{CD}$ is the altitude to $ \overline{AB}$. Let $ \omega$ be the circle having $ \overline{CD}$ as a diameter. Let $ I$ be a point outside $ \triangle ABC$ such that $ \overline{AI}$ and $ \overline{BI}$ are both tangent to circle $ \omega$. The ratio of the perimeter of $ \triangle ABI$ to the length $ AB$ can be expressed in the form $ \displaystyle\frac{m}{n}$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m\plus{}n$.

[hide="responding to my own post"]
First we see $ AB\equal{}37$ by the pythagorean theorem. We find $ CD \equal{} \frac {35*12} {37}$ using the area of the triangle. It follows by the Pythagorean theorem that $ BD \equal{} \frac {35^3} {37}, AD \equal{} \frac {12^2} {37}$. In triangle ABI, let the tangents from I have length a. We have $ s \equal{} 37 \plus{} a$.
By herons formula, $ [ABI] \equal{} \frac {35*12} {37}\sqrt {(37 \plus{} a)(a)}$. Since $ s \equal{} A/r$,
We have $ 37 \plus{} a \equal{} \frac {\frac {35*12} {37}\sqrt {(37 \plus{} a)(a)}} {\frac {35*12} {2*37}} \equal{} 2\sqrt {(37 \plus{} a)(a)}$.
So, $ \sqrt {37 \plus{} a} \equal{} 2\sqrt {a}$, and $ a \equal{} \frac {37} {3}$. We want $ \frac {2s} {37}$ which is $ 2\frac {\frac {37} {3} \plus{} 37} {37} \equal{} \frac {8} {3} \Rightarrow \boxed{011}$

[/hide]

[hide]
First, we note that $ 12^2\plus{}35^2\equal{}37^2$, so $ AB\equal{}37$.

Now, the area of $ ABC$ is $ \frac{1}{2}(12)(35)\equal{}210$.  Thus,

$ \frac{1}{2}(CD)(37)\equal{}210\Rightarrow CD\equal{}\frac{420}{7}$.

Now, note that triangles $ CBD$ and $ ABC$ are similar.  Thus

$ \frac{BD}{35}\equal{}\frac{35}{37}\Rightarrow BD\equal{}\frac{1225}{37}$.

This means that $ AD\equal{}\frac{144}{37}$.  Also, if we let $ BI$ be tangent to the circle at $ E$ and $ AI$ be tangent to the circle at $ F$, $ BE\equal{}\frac{1225}{37}$ and $ AF\equal{}\frac{144}{37}$ by equal tangents.

Now, we have $ IE\equal{}IF$.  Let this length be $ x$.  We find the area of triangle $ ABI$ in two different ways.  First, we see that the inradius of the circle is $ \frac{210}{37}$.  Also, the semiperimeter is clearly $ x\plus{}37$.  Now, we use the $ K\equal{}rs$ formula and Heron's formula, and set them equal:

$ \frac{210}{37}(x\plus{}37)\equal{}\sqrt{(x\plus{}37)(x)(\frac{144}{37})(\frac{1225}{37})}$.

Pulling squares out of the right side:

$ \frac{210}{37}(x\plus{}37)\equal{}\frac{12(35)}{37}\sqrt{(x\plus{}37)(x)}$

$ x\plus{}37\equal{}2\sqrt{(x\plus{}37)(x)}$.

Squaring both sides:

$ (x\plus{}37)^2\equal{}4(x\plus{}37)(x)\Rightarrow x\plus{}37\equal{}4x$.

Thus, $ x\equal{}\frac{37}{3}$.  So the desired ratio is

$ \frac{2\frac{37}{3}\plus{}2(37)}{37}\equal{}\frac{8}{3}$.

The answer is $ \boxed{011}$.
[/hide]

Hm, I found a solution not using the given side lengths at all...

[hide="outline"]Construct the tangent at $ C$ with intersection $ E$ on $ IA$ and $ F$ on $ IB$. Then $ EF//AB$. Using pythagorean theorem in right trapezoid $ ADCE$, we get $ 4AD\cdot CE \equal{} CD^2$. Since $ CD/AD \equal{} BD/CD$ by $ ACD \sim ABD$, this gives $ CE \equal{} BD/4$. Similarly, $ CF \equal{} AD/4$. Now just use $ IEF\sim IAB$ to set up some ratio and calculate that $ IJ \equal{} IK \equal{} AB/3$ and the answer equals $ 8/3$.
[/hide]

Blargh, I wasted much time trying to convince myself of this solution...(I can't figure out what $ \sqrt {12^2 \plus{} 35^2}$ is...)

hmm

[hide]
Lemma. Let triangle $ ABC$ have inradius $ r$, and let the incircle (with center $ I$) intersect $ AB$, $ BC$, $ CA$ at $ F, D, E$ respectively. Let $ AF\equal{}AE \equal{} x$, $ BF \equal{} BD \equal{} y$, $ CD \equal{} CE \equal{} z$. Then $ z \equal{} \frac{x\plus{}y}{\frac{xy}{r^2}\minus{}1}$.

Proof. Let $ 2\alpha \equal{} \angle FIE, 2\beta \equal{} \angle DIF, 2\gamma \equal{} \angle EID$. Then $ \alpha \plus{} \beta \plus{} \gamma \equal{} 180$, so $ \alpha \plus{} \beta \equal{} 180 \minus{} \gamma$. Hence,

\[ \tan(\alpha \plus{} \beta) \equal{} \frac{\tan{\alpha} \plus{} \tan{\beta}}{1\minus{}\tan{\alpha}\tan{\beta}}\equal{} \minus{}\tan \gamma\]

Note that $ \tan \alpha \equal{} x/r, \tan \beta \equal{} y/r, \tan \gamma \equal{} z/r$. Hence,

\[ \frac{(x\plus{}y)/r}{1\minus{}(xy/r^2)} \equal{} \minus{}z/r \Leftrightarrow z \equal{} \frac{x\plus{}y}{(xy/r^2)\minus{}1}\] as desired.

so now we apply it to our diagram... we want to find the ratio (letting $ \omega$ hit $ AI, BI, AB$ at $ X,Y,Z$ respectively, and letting $ IX \equal{} z, BY \equal{} y, AX \equal{} x$) $ 2(x\plus{}y\plus{}z)/(x\plus{}y)$. Since $ r$ is half the altitude of a right triangle to the hypotenuse, $ r^2 \equal{} xy/4$; i.e., $ z \equal{} (x\plus{}y)/3$. So, the desired ratio is $ 2(x\plus{}y \plus{} (x\plus{}y)/3)/(x\plus{}y) \equal{} \frac{8}{3}$.
[/hide]

[quote="1=2"][[1999 AIME Problems/Problem 12]][/quote]

Yeah, I was lucky enough to to choose the 1999 AIME as the last AIME to practice on and when I couldn't get #12 I was also lucky enough to thoroughly understand the solution.

Let $ AD\equal{}x$, $ BD\equal{}y$, $ IA\equal{}x\plus{}z$ (so $ z$ is the length of the third tangent in $ IAB$) and $ r$ be the inradius of $ IAB$. Since $ ABC$ is right and $ CD$ is an altitude, $ CD^2\equal{}AD\cdot BD\implies r^2\equal{}\frac{yz}{4}$. By a well-known formula, $ r^2\equal{}\frac{xyz}{x\plus{}y\plus{}z}\equal{}\frac{yz}{4}\implies x\equal{}\frac{y\plus{}z}{3}$. The perimeter of $ IAB$ is $ (2)(x\plus{}y\plus{}z)\equal{}\frac{8}{3}(y\plus{}z)\equal{}\frac{8}{3}\cdot AB$. Therefore the answer is $ 8\plus{}3\equal{}\boxed{011}$.

[quote="Altheman"]By a well-known formula, $ r^2 \equal{} \frac {xyz}{x \plus{} y \plus{} z} \equal{} \frac {yz}{4}\implies x \equal{} \frac {y \plus{} z}{3}$. quote]

How is this proven?  Using that formula reduces the problem greatly (there would then be a direct way to find the 3rd tangent without using trigo or whatnot)...  Would also mean that Area can be expressed as $ \frac{xyz}{r}$, so u have 2 equations linking Area, r, x, y, z...... useful.

Well, note that from Heron's: $ A \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)( s \minus{} c)}$, that $ x \equal{} s \minus{} a, y \equal{} s \minus{} b, z \equal{} s \minus{} c$ from equal tangents. Also, from equal tangents, $ x \plus{} y \plus{} z \equal{} s$. So
\[ A^2 \equal{} r^2s^2 \\
\Longleftrightarrow r^2s \equal{} \frac {A^2}{s} \equal{} \frac {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}{s} \equal{} xyz \\
\Longleftrightarrow r^2 \equal{} \frac {xyz}{s} \equal{} \frac {xyz}{x \plus{} y \plus{} z}
\]
.

You're right, no trig is needed.

edit:

[quote="timwu"] Using pythagorean theorem in right trapezoid $ ADCE$, we get $ 4AD\cdot CE \equal{} CD^2$.
[/hide]

[/quote]

how did you get that?

Can someone provide a picture? It is difficult for me to get a good diagram
Thanks

[hide="Solution"]First, we see that $O$ is the incenter of triangle $ABI$. Thus, $\angle IOB=90^\circ+\frac{1}{2}\angle IAB=90^\circ+\angle OAD$. Also, $\angle QOB=90^\circ-\angle OBQ=90^\circ-\angle OBD$, so $\angle IOQ=\angle OAD+\angle OBD$. Now, we can use trigonometry to solve for $IQ$, which we also know is going to equal $IP$: $$\tan \angle IOQ=\tan(\angle OAD+\angle OBD)$$$$=\frac{\tan \angle OAD+\tan \angle OBD}{1-\tan \angle OAD\tan \angle OBD}$$$$=\frac{\frac{210}{144}+\frac{210}{1225}}{1-\left(\frac{210}{144}\right)\left(\frac{210}{1225}\right)}$$$$=\frac{\frac{35}{24}+\frac{6}{35}}{1-\left(\frac{35}{24}\right)\left(\frac{6}{35}\right)}$$$$=\frac{4}{3}\left(\frac{35}{24}+\frac{6}{35}\right)$$$$=\frac{35}{18}+\frac{8}{35}.$$ This equals $\frac{IQ}{\frac{210}{37}}$, so we find that $IQ=\frac{210}{37}\left(\frac{35}{18}+\frac{8}{35}\right)=\frac{1225}{111}+\frac{48}{37}=\frac{1369}{111}=\frac{37}{3}$. Thus, the perimeter of $\triangle ABI$ is $2\left(\frac{37}{3}\right)+2\times37$ because $AP+BQ=AD+BD=37$, and the desired ratio is $$\frac{2\left(\frac{37}{3}\right)+2\times37}{37}=\frac{8}{3}.$$ Therefore, $m+n=8+3=\boxed{011}$.[/hide]

I bashed for my life on this problem and kept on getting $m+n>1000$  :stretcher:

[hide=No similar triangles]
Let $BI = a, AI = b, AB = c$. Let $r, s$ be the semiperimeter, inradius of $AIB$.
Let $K$ be the reflection of $C$ over $AB$. Since $ABC$ is right, $ABCK$ is cyclic. So, by PoP,
$$CD * DK = (2r)^2 = AB * DB = (s-a)(s-b),$$
The latter equality coming from the fact that D is the contact point of incircle $\omega$ with $AB$, since $CD \perp AB$.
Hence,
$$r = \frac{1}{s} * [AIB] = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}= 2r \sqrt{\frac{s-c}{s}} \implies \frac{s-c}{s} = 1/4$$
So, $$\frac{a+b+c}{c} = \frac{2s}{c} = 2 * \frac{1}{1 - \frac{s-c}{s}} = \frac{8}{3} \implies \boxed{011}$$ [/hide]

Bro the Official MAA Solution didn't even use the lengths of that right triangle - in fact $\triangle ABC$ itself is irrelevant  :dry:

This might've been the trolliest #12 I've ever seen lol

I am really, truly, a dumb person.

I spent like 15-30 minutes (ok this isn't that much but still) thinking that the circle didn't have to be tangent to AB. With this method, I got an answer of $2$, scrolled the thread, saw the diagram (#13) and started laughing at myself.

:wallbash_red:

Firstly, [b]WE NOTE THAT SINCE $CD$ IS A DIAMETER AND $\angle ADC = 90$, LINE $AB$ IS TANGENT TO THE CIRCLE![/b]

We can easily find that $AD = \frac{144}{37}$ and $DB = \frac{1225}{37}$. Let the tangency point on $AI$ be $P$ and let the tangency point on $BI$ be $Q$. Then $AP = \frac{144}{37}, BQ = \frac{1225}{37}$ from equal tangents. Next, denote $PI = QI = x$. Then the semiperimeter is $x+37$, so the area is $\sqrt{(x+37)(x)\left(\frac{1225}{37}\right)\left(\frac{144}{37}\right)}$ and we hence have $\frac{\sqrt{(x+37)(x)\left(\frac{1225}{37}\right)\left(\frac{144}{37}\right)}}{x+37} = \frac{210}{37}$. Cross multiplying and squaring gives $210^2(x+37)^2=37^2(x+37)(x)\left(\frac{1225}{37}\right)\left(\frac{144}{37}\right)$, or $210^2(x+37)=37^2(x) \cdot \frac{420^2}{37^2}$, equivalent to $x+37 = 4x \implies x = \frac{37}{3}$. The perimeter of $\triangle ABI$ is $2\left(\frac{37}{3}+37\right) = \frac{296}{3}$, so the answer, upon dividing, is $\frac{8}{3} \implies \boxed{11}$.

Let $M$ be the midpoint of $CD$, and let $\omega$ be tangent to $AI$ and $BI$ at $E,F$ respectively. Then $\tan \angle AMD=\frac{24}{35}$ while $\tan \angle BMD=\frac{35}{6}$. It follows that $\tan \angle AMB=-\frac{37^2}{630}$ so that $\tan \angle IME=\frac{37^2}{630}$. Thus $IE=\frac{37}{3}$ and the answer is $\frac{8}{3}\implies \boxed{011}$.

[quote=hurdler]Well, note that from Heron's: $ A \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)( s \minus{} c)}$, that $ x \equal{} s \minus{} a, y \equal{} s \minus{} b, z \equal{} s \minus{} c$ from equal tangents. Also, from equal tangents, $ x \plus{} y \plus{} z \equal{} s$. So
\[ A^2 \equal{} r^2s^2 \\
\Longleftrightarrow r^2s \equal{} \frac {A^2}{s} \equal{} \frac {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}{s} \equal{} xyz \\
\Longleftrightarrow r^2 \equal{} \frac {xyz}{s} \equal{} \frac {xyz}{x \plus{} y \plus{} z}
\][/quote]

Was thinking of using Heron's with variables before plugging everything in... :|

Thankfully, the Heron's application with all the side lengths involved turned out to be clean.

[quote=abacadaea][hide="again i feel rword responding to my own post"]
First we see $ AB\equal{}37$ by the pythagorean theorem. We find $ CD \equal{} \frac {35*12} {37}$ using the area of the triangle. It follows by the Pythagorean theorem that $ BD \equal{} \frac {35^3} {37}, AD \equal{} \frac {12^2} {37}$. In triangle ABI, let the tangents from I have length a. We have $ s \equal{} 37 \plus{} a$.
By herons formula, $ [ABI] \equal{} \frac {35*12} {37}\sqrt {(37 \plus{} a)(a)}$. Since $ s \equal{} A/r$,
We have $ 37 \plus{} a \equal{} \frac {\frac {35*12} {37}\sqrt {(37 \plus{} a)(a)}} {\frac {35*12} {2*37}} \equal{} 2\sqrt {(37 \plus{} a)(a)}$.
So, $ \sqrt {37 \plus{} a} \equal{} 2\sqrt {a}$, and $ a \equal{} \frac {37} {3}$. We want $ \frac {2s} {37}$ which is $ 2\frac {\frac {37} {3} \plus{} 37} {37} \equal{} \frac {8} {3} \Rightarrow \boxed{011}$

[/hide][/quote]

back in the days where censors didnt exist

Let $\omega$ be the mentioned circle, and let $T = BI\cap\omega$ and $Q = AI\cap\omega$. We have $BT = BD = 35^2/37$ and $AQ = AD = 12^2/37$. Let $QI = TI = x$. Then, the semiperimeter of $\triangle ABI$ is $x + 37$, and the area is $\frac{12\cdot 35}{37}\sqrt{x(x+37)}$ by Heron’s. We now notice that $\omega$ is the incircle of $\triangle ABI$, so the inradius is $CD/2 = 35\cdot 6/37$. Now, we use the formula $A = rs$ in $\triangle ABI$ to get
\[ \frac{35\cdot 6}{37}(x + 37) = \frac{12\cdot 35}{37}\sqrt{x(x + 37)}\implies \sqrt{x+ 37} = 2\sqrt{x}\implies x = 37/3.\]
Hence, the perimeter of $ABI$ is $2(x + 37) = 2(37/3 + 37) = 8\cdot 37/3$, so the answer is $$(8\cdot 37/3)/37 = 8/3\to\boxed{11}.$$

First, note that $AB=37$; let the tangents from $I$ to $\omega$ have length $x$. Then the perimeter of $\triangle ABI$ is equal to \[2(x+AD+DB)=2(x+37).\] It remains to compute $\dfrac{2(x+37)}{37}=2+\dfrac{2}{37}x$.

Observe $CD=\dfrac{12\cdot 35}{37}=\dfrac{420}{37}$, so the radius of $\omega$ is $\dfrac{210}{37}$. We may also compute $AD=\dfrac{12^{2}}{37}$ and $DB=\dfrac{35^{2}}{37}$ by similar triangles. Let $O$ be the center of $\omega$; notice that \[\tan(\angle DAO)=\dfrac{DO}{AD}=\dfrac{210/37}{144/37}=\dfrac{35}{24}\] so it follows \[\sin(\angle DAO)=\dfrac{35}{\sqrt{35^{2}+24^{2}}}=\dfrac{35}{\sqrt{1801}}\] while $\cos(\angle DAO)=\dfrac{24}{\sqrt{1801}}$. By the double-angle formula $\sin(2\theta)=2\sin\theta\cos\theta$, it turns out that \[\sin(\angle BAI)=\sin(2\angle DAO)=\dfrac{2\cdot 35\cdot 24}{1801}=\dfrac{1680}{1801}\]

Using the area formula $\dfrac{1}{2}ab\sin(C)$ in $\triangle ABI$, \[[ABI]=\left(\dfrac{1}{2}\right)\left(\dfrac{144}{37}+x\right)(37)\left(\dfrac{1680}{1801}\right)=\left(\dfrac{840}{1801}\right)(144+37x).\] But also, using $rs$, \[[ABI]=\left(\dfrac{210}{37}\right)(37+x).\] Now we can get \[\dfrac{[ABI]}{210}=\dfrac{4(144+37x)}{1801}=\dfrac{37+x}{37}\] so multiplying everything by $37\cdot 1801=66637$ lets us solve for $x$: \[21312+5476x=66637+1801x.\] We have $x=\dfrac{66637-21312}{5476-1801}=\dfrac{45325}{3675}=\dfrac{37}{3}$, and now \[2+\dfrac{2}{37}x=2+\dfrac{2}{3}=\dfrac{8}{3}\] giving the answer, $\boxed{011}$.

[img]https://cdn.artofproblemsolving.com/attachments/0/2/d80e14d512779a5aeb4ed1df3c1ef76c6f8d13.png[/img]


[quote="Virgil Nicula"]Please, know somebody how I can achieve one table ? If yes, you can offer me an example ? Thank you very much ![/quote]
Here's a website explaining how to use tables in $\LaTeX$:
[url]http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/Tables.html[/url]

[quote="anirudh"]You can use a tabular. I created one for my Mock MOEMS contest.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=629236#629236

Press the scores button to see the table.[/quote]
I forgot to turn in the answers. :blush:

Is it too late?


geometry
circumcircle
incenter
trigonometry
angle bisector
geometry solved

if$D,H\ and\ D',H'$ be intesection points of angle bisector and Altitude from points$A,B$ in triangle ABC respectively.
prove that $O,D,D'$ are collinear iff $I,H,H'$ be collinear.($O\ ,\ I$ are circumcenter and incenter of triangle ABC)

http://www.mathlinks.ro/Forum/viewtopic.php?t=50582

  darij

$O\in DD'\Longleftrightarrow \cos C=\cos A +\cos B\Longleftrightarrow I\in HH'$. I used the following [u]Lemma:[/u]

Let $d$ be a line and the points $X\not\in d,\ \{ A,B,C\}\subset d,\ M\in XA,\ N\in XB,\ P\in XC$. Then

\[ \overline {\underline {\left| \ M\in NP\Longleftrightarrow \dfrac {\overline {MA}}{\overline {MX}}\cdot \overline {BC} +\dfrac {\overline {NB}}{\overline {NX}}\cdot \overline {CA} +\dfrac {\overline {PC}}{\overline {PX}}\cdot \overline {AB}=0\ \right| }}  \]

[u]My solution.[/u]

$\blacksquare \ I\in HH'\Longleftrightarrow \dfrac {HB}{HC}\cdot {LA}+\dfrac {H'A}{H'C}\cdot {BL}=\dfrac {IL}{IC}\cdot AB\Longleftrightarrow$

$\dfrac {c\cos B}{b\cos C}\cdot \dfrac {b}{a+b}+\dfrac {c\cos A}{a\cos C}$ $\cdot \dfrac {a}{a+b}=\dfrac {c}{a+b}\Longleftrightarrow \cos A +\cos B =\cos C.$

$\blacksquare\ O\in DD'\Longleftrightarrow \dfrac {DB}{DC}\cdot \sin 2B +\dfrac {D'A}{D'C}\cdot \sin 2A =\sin 2C\Longleftrightarrow$

$\dfrac cb\cdot \sin 2B +\dfrac ca\cdot \sin 2A=\sin 2C\Longleftrightarrow \cos B+\cos A =\cos C.$


:)  :(  :?  :D  :P  :blush:  ;)  :lol:  :o  :mad:  :oops:  :roll:  :idea:  :?:  :!:  :|  :rotfl:  :huh:  :ninja:  :alien:  :yinyang:  :stretcher:  :starwars:  :ddr:  :bruce:  :football:  :gathering:  :winner_first:  :welcomeani:  :10:  :cleaning:  wow this is fun

:huh:  mathbomberII, are you [i]trying[/i] to get banned?? :?:  :?:

[quote="mathbomberII"] wow this is fun[/quote]

It's also spam. Locked.