geometry perimeter trigonometry induction ratio circumcircle

[quote="Amarjit"]Why would a hexagonal polygon have a greater area than a pentagonal one, when both have perimiters of, say, 500? My hunch is that a hexagon is composed of more triangles than a pentagon, giving more area; but I do not know whether this is the primary explanation.

Many thanks[/quote]
I'll assume that you mean the polygons are regular polygons. Let $P$ be the common perimeter between polygons $\mathcal{P}_n$ and $\mathcal{P}_m$ with $n$ and $m$ number of sides respectively, where of course $n,m\geq3$. Now we proceed to find the length of the apothems of both polygons and then find the area of the polygons. For now we can find the area of $\mathcal{P}_n$ since the area of $\mathcal{P}_m$ can be found similarly. The length of one side of $\mathcal{P}_n$ is $s_n=\frac{P}{n}$. This means, using the right triangle with angle $\theta$ $(2n\theta=360^\circ)$ and side opposite to this angle is $\frac12s_n$, we have $\tan\theta=\frac{s_n}{2a}$, where $a$ is the length of apothem. So $a=\frac{s_n}{2\tan\theta}$. Now the area of the polygon $\mathcal{P}_n$ is
$A_n=\frac12 aP=\frac{Ps_n}{4\tan\theta}=\frac{P^2}{4n\tan\theta}=\frac{P^2}{4}\frac{1}{n\tan\left(\frac{180^\circ}{n}\right)}$. I'll leave the rest for someone else to come in and add comments to it. :P  :bye:

Err I think you just have to prove that

$\frac {nx^2}{4 \tan \frac {180}{n}} < \frac {\frac {x^2n^2}{n+1}}{4 \tan \frac {180}{n+1}}$

^^Induction
Where $n$ is the number of sides and $x$ is the measure of each side.

While it doesn't completely prove it, try this. Take a square with a perimeter of 8 and area of 4, and divide it into two right isosceles triangles to see that the diagonal is $2 \sqrt2 \approx 2.82$. The circumscribed circle has circumference $2 \pi \sqrt2$ and area $2 \pi$. Now take an octagon with a perimeter of 8, meaning side lengths of 1. You can divide it up into triangles to see that the diameter of the octagon is $\sqrt{4+2\sqrt{2}} \approx 2.61$.  While the diameter is smaller, you can find the actual area of the octagon to be $3+2\sqrt{2} \approx 5.82$. The circumscribed circle has circumference of $\pi \sqrt{4+2\sqrt2}$ and area of $\pi + \frac{\pi\sqrt2}{2}$. The more sides a polygon has for the same perimeter, the smaller the diameter, but the more of the circumscribed circle it fills up, meaning a larger area overall. At $n=\infty$, the polygon [b]is[/b] a circle, which has the maximum area for the perimeter, and the perimeter-diameter ratio changes slowly enough that the area always increases.


Find all values of $n \rightarrow n>3$ that minimize the value of the expression:

$\frac{\log2\cdot\log3\ldots\log n}{3^{n}}$

I thought the answer was $10$ but when I use $n=11$ the expression is smaller...hmmmm

[quote="E^(pi*i)=-1"]When you increase n by 1, you add a factor of 3 to the denominator.   The fraction will get smaller as long as log(n) is less than 3.  log(n) will equal 3 when n is 1000, so the answer is 1000.[/quote]Let $a_n$ be the expression to be minimized as $n$ varies. Your solution looks right to me, but we also have $a_{999}=a_{1000}$ since increasing $n$ from $999$ to $1000$ adds factors of $3$ to both the top and bottom. Afterwards, of course, the factor $\frac{\log{(n+1)}}{3}$ will be greater than $1$, and before $n=999$ the factor will be less than $1$.

edit: I never explicitly said it but my answer is $\{999,1000\}$.

[quote="bob123"]Actually, the answer is 3>n>1000[/quote]The set of solutions to that inequality is $\emptyset$ so I guess you're saying there is no minimum.  :D

Sorry, I knew what you meant, but your answer implies that $a_n$ (as defined above) is constant for all $3<n<1000$.

I had this on the CAML today, as well

I did not get it  :(

Umm... wow, I'm glad I didn't discover this too early... but anyways, I took the test on April 5, so many other people could've seen this before they took the test. I don't think we should talk about problems so soon.


what do you guys think the cutoff will be this year?

You may want to read the discussion here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265224[/url]

I figured predictions by people who wrote AIME II would be different from predictions by people who wrote AIME I so i made another thread


Disregarding the '':'', how many times from 12:00 noon until 12:00 midnight will the time on a digital clock with no leading zeros be a palindrome?


linear algebra
matrix
trigonometry
analytic geometry

Assume that:
\[ A=\left[ \begin{array}{cccccccc} 1-2\alpha & \alpha & & & & & \\ \alpha & 1-2\alpha & \alpha & & & \BIG {0}\\ \ & \alpha &1-2\alpha& \alpha \\ & & & \ddots \\ \BIG{0} & & & \alpha&1-2\alpha&\alpha& \ \\ & & & & \alpha & 1-2\alpha \end{array} \right] \]
Prove that if $A\in \mathcal M_n(\mathbb F)$ then all eigenvalues of this matrix are \[ \lambda_j=1-4\alpha\sin^2(\frac{j\pi}{2n+2}) \ \ \ \ \ j=1,2\dots,n \]

The field should be either $\mathbb{C}$ or $\mathbb{R}$, so that $\sin$ makes sense.

A hint: Add a row and a column to make this a circulant. Look for eigenvectors with the last coordinate zero, then translate back.

The essential element of jmerry's hint is to start with the eigenvectors rather than the eigenvalues, and to assume a simple form for the eigenvectors.

Most of this is trigonometric identities:

$\sin(x+h)-2\sin(x)+\sin(x-h)=(\cos h-1)\sin x=-2\sin^2\left(\frac h2\right)\sin (x)$

Now assume that we have an eigenvector $x$ such that

$x_k=\sin\left(\frac{\pi jk}{n+1}\right).$

Apply the matrix, and apply the trig identity above:

$(Ax)_k=\sin\left(\frac{\pi jk}{n+1}\right)\left[1-4\alpha\sin^2\left(\frac{\pi j}{2(n+1)} \right)\right].$

Since this works for all $1\le j\le n,$ we have $n$ independent eigenvectors and hence all of the eigenvalues.

Even though it may look like I haven't used all of jmerry's hint, I really have used it. circulant matrices are diagonalized by the Fourier transform - that is, their eigenvectors are simple harmonic motions (characters of the cyclic group). This is not quite a circulant, but the way around that is to think of boundary values. Because it's symmetric, it has real eigenvalues and we can use the real form of simple harmonic motions: sines and/or cosines. I actually saw it as adding two rows and two colums - right and left, top and bottom - but then we more or less identify the extra rows with each other.

New challenge: find a physical interpretation of this problem. Maybe something involving identical masses connected by identical springs to each other and to walls? Perhaps that 1 on the main diagonal is gravity? Perhaps there's a "small osciallations" linearizing approximation?

Well, Here is another solution. Suppsoe $A_n$ is the $n\times n$ matrix and $f_n(\lambda)=\det (\lambda I_n-A_n)$. Now you could easily using Leibniz Formula you see that $f_n(\lambda)=(\lambda+2\alpha-1)f_{n-1}(\lambda)-\alpha^2 f_{n-2}(\lambda)$.So if $\zeta,\mu$ are roots of equation $p(x)=x^2-(\lambda+2\alpha-1)x+\alpha^2$. Easily you see that $f_n(\lambda)=\frac{\zeta^{n+1}-\mu^{n+1}}{\zeta-\mu}$. If $f_n(\lambda)=0$ then $\zeta^{n+1}=\mu^{n+1},\ \zeta\neq\mu$. So there exist $k\in\{1,2\dots,n\}$ which $\frac{\mu}{\zeta}+\frac{\zeta}{\mu}=\omega^k+\omega^{-k}=2\cos(\frac{2\pi k}{n+1})$ which $\omega$ is $n+1$'s root of unity. Now everything is easy to prove.

[quote="Omid Hatami"] Suppsoe $A_n$ is the $n\times n$ matrix and $f_n(\lambda)=\det (\lambda I_n-A_n)$. Now you could easily using Leibniz Formula you see that $f_n(\lambda)=(\lambda+2\alpha-1)f_{n-1}(\lambda)-\alpha^2 f_{n-2}(\lambda)$.[/quote]

Sorry what is Leibniz formula?


Hi there
I am after a symbol like \hbar but with a d instead of an h. Much googling has lead to naught, can anyone help me out?

When looking for symbols the best place to look is [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=comprehensive]The Comprehensive LaTeX Symbol List[/url]. Searching for \dbar gives the code
[code]\def\dbar{{\mathchar'26\mkern-12mu d}}[/code](The ' before 26 is a single quote)
Put that in your document and then you can use \dbar in  math mode.


All multiples of 8 are in B.  800 is a multiple of 8, so will go in B.

The "official" solution is that you have a pattern going B, C, D, E, D, C, B, A, and find the solution, by dividing 799 (800-1 because 1 is not a term) by 8 and taking the remainder, which is 7, and looking at the 7th term in the pattern: B.
I used a different way and got the right answer: I divided 800 by 4, because there are 4 letters in each term. The result was even, and so I saw that the pattern would be the same as the 2nd or 4th row. Because 1 is not a term, 800 would be the second from last term in that row. Thus, 800 comes in "B".
I am wondering: are there practical applications for my method? In this situation, it is probably not as simple, but it seems useful.
Then again, AIME15's solution is probably better.

Your solution is better when we have more complicated patterns and such, mine is better for simple problems like this.

I also divided 800 by 4, to discover that 800 would be in the 200th row. Then, if you look at the numbers vertically, in the B column the pattern is that you add six to the previous number, then two, then six again, and so on. This gives the repeating pattern of ones digits 2, 8, 0, 6, 8, 4, 6, 2, 4, 0. Since it's a 10 digit long repeating pattern, you can match it with the ones digit of the row it is in. Thus, the 200th row in B would have the same ending digit as the 10th row, or 0. As such, 800 must be in B.

Does that make sense? It got me the correct answer, but it was definitely more labor intensive and not nearly as elegant as the solutions here.

[quote=yofro]Am I the only one that did row instead of column to get $200$?  :furious:[/quote]

[quote=peelybonehead]“five columns (A, B, C, D, E)”[/quote]

This thread is from 2008


blogs Harvard college MIT geometry topology HCSSiM

Does anyone know of any outstanding undergraduate mathematics programs for U.S. colleges/universities? The regular qualities (graduates are well-prepared for graduate school (and get accepted to good graduate schools), classes are well-taught, plentiful opportunities in terms of course offerings and research opportunities), are expected, but what I want to know is whether there are math programs out there that have qualities that make them special.

For example, are there mathematics departments that believe in mathematics so much, they actively recruit students from other disciplines to study mathematics? What schools have the greatest number of students entering with the intention of studying mathematics and graduating with a mathematics degree? What about unique opportunities/math programs that I haven't heard about?

Also...what are your views on studying math at liberal arts colleges?

I'm going to the University of Chicago next year, so keep in mind that I'm horribly biased.

There are two things that pop into my head that make the U of C's math program so special.  One is that there is no engineering program at the school.  Therefore, all the math/science graduates are doing theoretical stuff, which I think is pretty cool.  The other thing is the Honors Analysis class.  It's typically considered the hardest freshman math class in the country (some would argue that that role deserves to Harvard's advanced freshman math class).

Oh, and one last thing is that the college on the whole is famous, in all departments, for allowing undergraduates easy access to graduate level classes.

[quote="Magnara"]It's typically considered the hardest freshman math class in the country (some would argue that that role deserves to Harvard's advanced freshman math class).[/quote]

The key difference here is that Honors Analysis at U. Chic. is not really a course for freshmen, it's for freshmen and sophomores.  (At least, that's the sense I've gotten from a friend at UCh.)  Not that it isn't an incredible course, of course.

Anyhow, I think you should pay a lot of attention to the "normal" things.  The size and friendliness of the math department, for example, will probably make a much bigger difference in your life as an undergraduate than any special programs that might exist.  And the size and nature of the college or university as a whole will probably have an even greater affect.

As for the question of liberal arts colleges and math, what exactly do you mean?  Most of the best math departments in the country are at liberal arts schools.  Even if we're talking really small liberal arts colleges, there are some good math programs to be found.

Thanks for the replies! Magnara, I've been expecting a pro-UC reply from you! (I passed by your blog recently.) Congrats on the acceptance, and I hope you get into your class of choice!

[quote="JBL"]Anyhow, I think you should pay a lot of attention to the "normal" things.  The size and friendliness of the math department, for example, will probably make a much bigger difference in your life as an undergraduate than any special programs that might exist.  And the size and nature of the college or university as a whole will probably have an even greater affect.[/quote]

Agreed.  :) I'm also interested in what the mathematical culture is like, though.

[quote="JBL"]As for the question of liberal arts colleges and math, what exactly do you mean?  Most of the best math departments in the country are at liberal arts schools.  Even if we're talking really small liberal arts colleges, there are some good math programs to be found.[/quote]

That's what I want to know about. What liberal arts colleges have strong math programs? In another post, someone had mentioned that even as an undergraduate, students should consider schools that have a graduate program in mathematics. Can a school without a graduate program in math offer a good math program? Which ones? And what are some benefits of pursuing a math degree at a liberal arts college? The obvious one is that students will have a variety of opportunities, but strictly from a mathematical perspective (with graduate school in mind), what do liberal arts colleges have to offer that universities may not?

[quote="Magnara"]It's typically considered the hardest freshman math class in the country (some would argue that that role deserves to Harvard's advanced freshman math class).  [/quote]

I doubt this is possible.  I suspect that Math 55 at Harvard is unique because that's the only place you can get such a concentration of brilliant math types who you can teach that course to, and I doubt that the University of Chicago or anywhere else could offer a comparable course just because they don't have the raw material (e.g, students) to do so.

But don't take my word for it, look at some of the 55 problemsets after you're finished your analysis sequence, which will doubtless put you on a strong theoretical footing, which is exactly what a good analysis course should do.

You can get some answers to your questions, and some other great information, by reading the book [url=http://www.amazon.com/exec/obidos/tg/detail/-/082183455X/learninfreed]A Mathematician's Survival Guide: Graduate School and Early Career Development[/url] by Steven G. Krantz. The book discusses which undergraduate schools have strong math departments, and much more.

Let me begin by revealing my biases:  I went to a small liberal arts college, and followed this with a Ph.D. at a top-ten math department.

Mel asks: "What liberal arts colleges have strong math programs? In another post, someone had mentioned that even as an undergraduate, students should consider schools that have a graduate program in mathematics. Can a school without a graduate program in math offer a good math program? Which ones? And what are some benefits of pursuing a math degree at a liberal arts college? The obvious one is that students will have a variety of opportunities, but strictly from a mathematical perspective (with graduate school in mind), what do liberal arts colleges have to offer that universities may not?"

The biggest difference I've noticed between students who've come from small liberal arts colleges vs. larger places with graduate programs is that the small-liberal-arts graduates tend to have greater breadth of training and the graduate-program graduates tend to have greater depth of training.   There are advantages for each of these backgrounds... my peers who went to MIT as undergrads had already had much of the first-year graduate material, and so those classes were  much easier for them.  On the other hand, my small-liberal-arts peers already understood what combinatorics, number theory, representation theory, etc. are about and didn't need to take graduate classes in these areas in order to decide whether or not they wanted to pursue them.

Here's one way to evaluate whether a school you're interested in might have a strong math program (as a faculty member, this is one criterion I use, anyway) -- does it offer a full year of Abstract Algebra, a full year of Real Analysis, and at least a semester each of Topology and of Complex Analysis?  It turns out that not so many schools meet that criterion.

As to actually naming some small liberal arts colleges with strong programs... another way to measure this is to ask how many of their students have gone on for Ph.Ds in mathematics, since that seems to be what you're interested in.  I actually managed to get ahold of some data  for this at one point.  It's not terribly recent, but in the top bracket are included Pomona, Harvey Mudd, Amherst, Haverford, Reed, Oberlin, Swarthmore, Carleton, St. Olaf, Grinnell, Bryn Mawr, Williams, and lots of others.  If you do weighting by student enrollment or faculty size, some of the nontraditional schools like Hampshire and Antioch pop up as well.

As a previous poster mentioned, there are lots of other aspects to the college experience besides the reputation of the school and the available courses.  You should try visiting a few places and see what the atmosphere is like; while you might spend a great deal of your time doing mathematics, that certainly won't be all of it.  Many of my friends in college were anthropology majors.

Once again, thanks for all the replies! Tokenadult, I'll try to get a hold of that book. :)

Smbelcas, very thoughtful!

You mention that many liberal arts colleges have high PhD productivity rates. Do you think that it's because the math departments at these specific schools do a better job retaining their majors and interesting them in graduate school, or because more of their incoming students already had graduate school in mind? I'm sure it's a bit of both, and it's obvious that these math departments did not [i]discourage[/i] their students from pursuing a graduate degree. But to what degree can a math department at a LAC positively impact a somewhat undecided student to attend graduate school?

Also, I noticed that you're female and you listed a few women's colleges as some top schools. I agree whole-heartedly that these are very good schools. What are your views on attending a women's college for prospective math/science majors? I read that students, particularly female arts/humanities students at Haverford and Swarthmore often go to Bryn Mawr to take their science classes because they are more accessible.

I'm not trying to perpetuate any stereotypes or begin any Summers-inspired debate, but at my school, the only people I can find to entertain me mathematically are male, and my concern is that I don't want to go from being the only math-crazed student in high school to being the only math-crazed student in college. I'm not saying that females are not good at math, I'm simply saying that I have yet to find one that is extremely excited about it and is willing to play math with me anytime, anywhere. Do you suppose that this will be a difficulty at a women's college (or even a liberal arts college)?

[quote="smbelcas"]As a previous poster mentioned, there are lots of other aspects to the college experience besides the reputation of the school and the available courses.  You should try visiting a few places and see what the atmosphere is like; while you might spend a great deal of your time doing mathematics, that certainly won't be all of it.  Many of my friends in college were anthropology majors.[/quote]

Agreed. As a musician and athlete, the quality of music program and the extracurricular opportunities are very important to me. Also, I come from a very unique background culturally, so, the atmosphere of the school will also be very important to me. The down side is that I won't be able to visit the colleges/universities during the school year, and since my interest is in the students, classes, and professors and not the buildings/trees/etc., I doubt summer visits would contribute much to my understanding.

[quote="mel"]Smbelcas, very thoughtful!

You mention that many liberal arts colleges have high PhD productivity rates. Do you think that it's because the math departments at these specific schools do a better job retaining their majors and interesting them in graduate school, or because more of their incoming students already had graduate school in mind?[/quote]

I will be interested to hear Smbelcas's answer to this question. Meanwhile, I can suggest at least one other reason why liberal arts college ("LAC") graduates go on to graduate school more than, say, graduates of Ivy League universities. An undergraduate degree from an Ivy can be a ticket to a very lucrative job. By contrast, some of the same jobs are closed to graduates of Carleton (my grandfather's alma mater, and definitely a strong math school) and St. Olaf (my dad's alma mater, also a strong math school). So when an LAC says that many of its graduates go on to graduate school, that in part means that many of its graduates couldn't find satisfying work with only their undergraduate degrees. (My grandfather and dad did stop with the degrees they got from LACs and go to find good jobs, but the job market was different in that era.)

There are several aspects to high Ph.D. productivity rates.  One is how many students from a given school attempt to get Ph.Ds... another is how many succeed.  I don't know what grad school attrition rates are like these days, but when I was in grad school about 50% of the entering class actually earned Ph.Ds at my institution.

After thinking about your questions a bit---I'm rephrasing them as "What effects can math departments at small LACs have on the grad-school intentions of their students, and why are some departments better at this than others?"---it seems to me that it's all related to advising, in some sense of that word.  For example, at a small LAC students are likely to get to know many of the faculty in the department well, and vice-versa.  Once a student has declared an intention to go to grad school, the faculty can help him/her choose appropriate schools to apply to, and then can help him/her decide among his/her options.  And before that point, any of the faculty might approach a student and recommend that s/he consider graduate school, or that s/he apply to summer research experience programs, or that s/he take certain classes, etc.  All of these things make large differences in preparing students for graduate school.

Why some schools are better at this than others may have to do with the culture at these schools---once a few students make it known that they'll be attending summer research experiences, or planning to go to grad school, everyone else knows about it and younger students start to  expect that they should consider these options as well.  It may also have to do with whether any of the faculty in the department happen to follow the national scene for grad schools, for research programs, etc.  in order to be able to give advice that will help their students succeed.

Now, as for women's colleges... they're a great idea for some students and a terrible idea for others.  It just depends on who you are.  Likewise, at small schools there's a bit of luck-of-the-draw in who your peers are.  There might be no one as interested in math as you, simply because there might only be 8 math majors.  Or, among those 8 math majors, perhaps 4 are math fanatics!  At a large enough school, there's a larger chance of finding a few peers who fit whatever specific profile you're looking for, just because there are more people in the pool.  I don't think, for that purpose, women's colleges are any different from coed colleges (except that perhaps the math fanatics are less likely to hide their fanaticism at women's colleges).

Even visiting colleges during the summer will give you some idea about the academic atmosphere.  Often there are some faculty lurking about, and usually there are at least some science students doing research on campus.  You can even get a feel for the atmosphere by conversations you overhear while you're on a campus tour.  As you know you're interested in mathematics, ask the admissions people whether there'll be math faculty and/or students you can talk to when you visit.  They may be able to arrange special meetings for you.

As a preface, I went to Grinnell, transferred to Harvey Mudd, and did my graduate work at Dartmouth.....learned a lot of my math at Budapest.

Okay, that out of the way.

First, I may get attacked for it, but after working in math camps for about 10 years, and meeting lots of students who went on to other colleges, and meeting lots of students in Budapest [the two different times I went there], I have to suggest from my interactions and conversations that the teaching ability at the big-name schools is a bit weak.  The reason you get to be a professor of mathematics at Harvard is not because you teach very well!  In fact, at many of those schools you can get into trouble as a faculty member for putting too much energy into your teaching.

There are many liberal arts colleges that have strong reputations for having strong math students.  Carleton, St. Olaf, Feed, and [not really a liberal arts college] Harvey Mudd produce very strong mathematics teams.  I actually was not particularly impressed with the teaching at Mudd when I was there, but that is just me.

It is rather biased by my own history, but the one thing I would stress to ANYONE interested in getting a good math background is to go to a small college [color=green][size=150] that participates in the Budapest Semesters in Mathematics Program[/size][/color].  There is an insane amount of mathematics that can be learned there.  They let me take TEN classes at once [I do not suggest this in general....:)].

Lastly, to address the "get into a good grad school" issue.  Many of the top graduate programs are so difficult to get into that admission is more like throwing darts at a dart board.  A few years ago, MIT accepted 8 people a year into its math program.  Princeton and Harvard are similar.  And many of those people are not from the US.  In any event, I would say the preparation for getting into a strong graduate program is different if you go to a small college than if you go to a University, but as long as you recognize that and take steps to prepare early [by doing REUs and such], you should be okay.

Liberal arts [and other] colleges I know of who participate in the Budapest program:

[* represents colleges that traditionally have sent large numbers at some point in the past]

*Grinnell
*Harvey Mudd
*St. Olaf
Oberlin
*Reed
Bryn Mawr
Pomona
Harvard
U. of Chicago
Rutgers
Swarthmore {I think]
*Dartmouth

There are several others that send people...I just can't think of any more off the top of my head.

Heh, BSM, good stuff.  I hope to be going there next semester.

Only one complaint with your other comments: there are good teachers at the "big name" schools, but they are probably a smaller percentage of the teachers than at small liberal arts schools.  (Not having been to any s.l.a.s., I can't be certain of this.)  But the downside is that the opportunities to learn are often much more limitted.  You can't take graduate courses at a school without graduate courses to take.  Not that this is necessarily a problem -- I haven't taken any graduate math courses in 2 years at Harvard.

Two people from my university will be going to BSM next year. I was originally planning on going in Spring 2006 (since going in the Fall would mess up all my sequences) until I looked at the proposed course offerings for next year and saw that I'd be missing classes like algebraic number theory and homological algebra that aren't taught very often around here.

[quote="mel"]One question though, does the BSM offer courses and academic (not social/cultural) opportunities that an undergrauate cannot gain access to by taking graduate courses in large university?[/quote]

Well, Harvard (for instance) doesn't have a very good program in combinatorics at all, while BSM has a number of combinatorics-related classes.  I could also probably go and take combinatorics classes at MIT, which is very good at it.  BSM does offer quite a wide array of classes, though.  See http://www.renyi.hu/~dezso/budsem/ for syllabi of current and past courses and see http://www.stolaf.edu/depts/math/budapest/ for the main program page.

[quote="JBL"][quote="mel"]One question though, does the BSM offer courses and academic (not social/cultural) opportunities that an undergrauate cannot gain access to by taking graduate courses in large university?[/quote]

Well, Harvard (for instance) doesn't have a very good program in combinatorics at all, while BSM has a number of combinatorics-related classes.  I could also probably go and take combinatorics classes at MIT, which is very good at it.  [/quote]

*cough* Richard Stanley *cough*

Magnara,
Congrats on UChicago! I know a lot of awesome math people there, including Laci Babai and Paul Sally! I'm applying next year myself.

[quote="mel"]Once again, thanks for all the replies!

The Budapest Semesters in Mathematics programs sounds fantastic, and it appeases my worries of a LAC not being mathematically-intense enough. :)

One question though, does the BSM offer courses and academic (not social/cultural) opportunities that an undergrauate cannot gain access to by taking graduate courses in large university?[/quote]

The crown jewel of the BSM program is its offering in the combinatorics arena, along with the special class "Conjecture and Proof."

BSM offers [at last check] 4 -6 combinatorics classes, of those the Topics in Graph theory [or "Advanced Graph Theory" I forget] and the second course in combinatorics are probably hard to find equivalents for stateside.  Combinatorics II is essentially a hypergraph theory course and advanced graph theory is....advanced graph theory.

"Conjecture and Proof" is a special problem-sovling course that is mimicked by problem solving courses at US colleges, but probably not paragoned.

Also, the Advanced algebra course at Budapest has a different flavor than most advanced algebra courses, since it has a large focus on G-sets.

[quote="mel"]Also...what are your views on studying math at liberal arts colleges?[/quote]

I majored in math at a small liberal arts college and I have a few thoughts you might want to consider.

Pros: Small classes, friendly atmosphere, close contact with professors.

Cons: Limited selection of courses.  Limited choice of professors.

The limited selection of courses was significant for me.  Several of the advanced math courses at my school were offered only every other year.  Because I spent junior year abroad in a non-math-related program, I missed the opportunity to take several upper-level math courses.

I just checked the course offerings at several small LACs and verified that most small colleges have this problem.  Courses like Abstract Algebra, Real Analysis, Complex Analysis are usually not offered every year.  Moreover, at small colleges, the same professors teach the same subjects year after year.  If there's a particular professor you don't care for, you're still stuck with him if you want to study his specialty.

These days, many students take college level math courses (beyond Calculus) while still in high school.  If this is true for you, be sure to check the course offerings at schools you're interested in.  If there are not plenty of advanced courses, you might run out of classes before you graduate.

Overall I would rate my college experience highly.  Professors at small LACs put a high priority on teaching, advising, and involvement with students.  I became good friends with the chairman of my math department and kept in touch with him after graduation.  When I went to graduate school (in comp sci), on the other hand, I was pleased with the selection of courses but disappointed with the quality of the teaching.  Professors there placed research first, graduate students second, and undergraduates third.

My recommendation?  Choose a LAC with a large number of math majors and an extensive selection of math courses.  Or choose a university with a relatively small number of graduate students, where professors have time for undergraduates.  And definitely visit the schools you're interested in.  Stay overnight if possible.

Thanks for all the anecdotal tidbits! Now I'm definately looking forward to college/university!  :lol:

[quote="wild octagon"]I just checked the course offerings at several small LACs and verified that most small colleges have this problem.  Courses like Abstract Algebra, Real Analysis, Complex Analysis are usually not offered every year. [/quote]

Yeah, this was one of my concerns when I was considering schools. I feared being in my junior/senior year and needing/wanting an advanced class that wasn't offered because the professor's on leave or something. The "solution" I found around this was to consider colleges with strong cross-registration programs with strong schools. Of course that always brings up the question: are there dark secrets (like cross-registered students have difficulty getting into popular classes) that I should know about?

I go to a (non LAC) school with a smallish math department, and I'm planning on graduating in 3 years (I'm currently a first-year student).  Now, 300-level courses are in fairly regular rotation here (I got credit for all of the 100-level courses and 200-level courses already) but 400-level and graduate courses are only offered once every other year.  What this has forced me to do is plan out my schedule ahead of time and make sure I take 400-level courses every time they are offered.  For example, this semester, I took

Macm 201 - Discrete Mathematics II
Math 252 - Vector Calculus
Math 320 - Analysis II
Math 424 - Applications of Complex Analysis

and some Econ course

Now, I could have taken Discrete and Vector calculus at nearly any time (they are degree prerequisites :( ), but there was a 400-level Galois Theory course and a 400-level Calculus of Variations course being offered, so I should have substituted at least one of them for Vector Calc and Discrete, because now I will have to wait another whole two years for them (unless I take the graduate-level Groups and Rings course, which I probably will do instead)

So if you do go to a school with a smaller math department, just check the course rotation and plan out a tentative schedule as to what you'll be taking.  Of course its possible that you'll fall in love with, say, algebraic topology and want to take a number of courses in that, but if you don't plan ahead, you might find yourself with whole semesters where you have no math courses to take.

[quote]For example, are there mathematics departments that believe in mathematics so much, they actively recruit students from other disciplines to study mathematics? What schools have the greatest number of students entering with the intention of studying mathematics and graduating with a mathematics degree? What about unique opportunities/math programs that I haven't heard about? [/quote]

I think it has already been  mentioned here but Duke does a  very active recruting (They will fly potential students to NC -  including students from abroad)  Many good math students are there  as AB Duke scholars (with full tution merit  scholarships)

(Caltech, Duke, Washington U.  Rice are,  I think,  a few of top well-known schools who  award  merit scholarships - and may  invite exceptional  high school students  to the campus to check out the school and programs . Most of the Ivy's  and many other top schools, on the otherhand,  I believe,  do not have special 'honors' pograms  or merit based scholarships at the undergradute level.

[quote="blahblahblah"]I doubt this is possible.  I suspect that Math 55 at Harvard is unique because that's the only place you can get such a concentration of brilliant math types who you can teach that course to, and I doubt that the University of Chicago or anywhere else could offer a comparable course just because they don't have the raw material (e.g, students) to do so.

But don't take my word for it, look at some of the 55 problemsets after you're finished your analysis sequence, which will doubtless put you on a strong theoretical footing, which is exactly what a good analysis course should do.[/quote]

Caveat: I'm a current third-year mathematics concentrator at Chicago.  In any case, Honors Analysis at Chicago (Math 207-209) is almost directly comparable to Math 55 at Harvard in terms of difficulty and material covered.  We used the same text books (e.g., Baby Rudin, Spivak's [i]Manifolds[/i], etc.), and though Chicago might be a smaller school, don't be mistaken about the quality of its math students.  Something like 10% of its undergraduates concentrate in math, so there's no dearth of brilliance here.  I had the fortune of taking 207-209 in a year when Paul Sally taught all three quarters, and it was definitely what sold me on mathematics.

Also, speaking of BSM, I'm currently taking a combinatorics class with Laszlo Babai, one of the programs' founders and directors.  You don't need to go all the way to Budapest if you want to be taught discrete math by esteemable Hungarian mathematicians -- just come to Chicago.  ;)   Mostly kidding.  I know people here who have done BSM and they have nothing but praise for it.


A right circular cylinder of radius 8 is inscribed in a cone with height 10. (The lower base of the cylinder lies with in the base of the cone, and the upper base lies in the surface of the cone. The sum of the cylinder’s height and the cone’s radius is 12. Find the surface area of the cylinder.

[hide="Hint"]Do it two-dimensionally and use similar triangles.[/hide]

do you mind explaining it a little more?

[hide]Is the answer $160\pi$?[/hide]

[size=75][color=darkred]Make sure you hide your answers.
-nebula42[/color][/size]

[hide]We have two similar triangles as shown. Let the radius of the cone be y, and the height of the cylinder be x. Therefore, we have the height of the top cone 10-x and the radius of the cone 12-x. Since the radius of the cylinder is 8, we can use similar triangles. We have $\frac{12-x}{8}=\frac{10}{10-x}\rightarrow x=2.$ Therefore, the surface area of the cylinder is $16\pi(2)+64\pi(2)=160\pi$ and this answer is pissing me off because it's not one of the answer choices.[/hide]


geometry
3D geometry
analytic geometry
graphing lines
slope
integration
calculus

Hey.
I have the following problem: I need to find a moment of inertia of a cone that say, has the slope of r/h where r is the radius @ x = h.
so it is expresseded by this equation:

$ y = \frac{r}{h}x$

So since the IoM =\rho\int_{} r^{2}  dV I did the following but somewhere something went horribly wrong:
I decided to integrate the volume across slices to simplify the integral, thus giving me:
$ y = \pi * R^{2}$
$ V = \int_{0}^{h}\pi * \frac{r^{4}*x^{4}}{h^{4}}* \rho dx$
$ IoM = \frac{\pi * \rho * h^{4}*h}{5}$

Which is half as much as i should have for the IoM.
Any ideas where i made a mistake?
Thx[/code]

You're confoundig distance of an volume element with the radius of the "slice".
[hide="hint"]  You can integrate the moment of inertia of the element $ \;{{\rho \pi r^{2}x^{4}}\over{2 h^{2}}}dx \;$ [/hide]


Let $a(n),b(n),c(n)$ be the number of 'good' words with length $n$ and ending by $xx$, by $xy$ and by $yx$ respectively.
Note that $p(n)=a(n)+b(n)+c(n).$

It is easy to see that
$a(n+1)=c(n),b(n+1)=a(n)+c(n)$ and $c(n+1)=b(n).$
Thus :
$p(n+1)=p(n)+c(n)$ (1)
and $c(n+3)=c(n+1)+c(n)$ (2)

From (2), and $c(2)=c(3)=1,c(4)=2$, we deduce that $c(n)$ is periodic modulo $2$ with period $7$.
Direct checking shows that $p(n)$ is periodic modulo $2$ with period $7$, that $p(2)=3$ and that $p(7)=p(8) = 0 \mod [2]$.

Thus, the desired $n$ are those of the form $n=7k$ for a nonnegative integer $k$.

Pierre.

:mad:  :mad:  :mad:
You are so fast! All my work is vain.

[quote="pbornsztein"]
From (2), and $c(2)=c(3)=1,c(4)=2$, we deduce that $c(n)$ is periodic modulo $2$ with period $7$.
Direct checking shows that $p(n)$ is periodic modulo $2$ with period $7$, that $p(2)=3$ and that $p(7)=p(8) = 0 \mod [2]$.

Thus, the desired $n$ are those of the form $n=7k$ for a nonnegative integer $k$.

Pierre.[/quote]

OK, i don't understand this part...doesn't from (2) follow, that n=7k+4,7k+5 or 7k? You wrote only 7k???


what is AM-GM (exaclty)and how is it useful? can you gimme some problems applying AM-GM?

Brief Introduction of AM-GM

Let $a$ and $b$ two postive numbers. Then:

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

That's about it.

Now, derivation and other stuffs are what it makes it important.

One problem would be 1983 #9 about $9 sin x$ or something like that.

[quote="Silverfalcon"]One problem would be 1983 #9 about $9 sin x$ or something like that.[/quote]
Do you mean from AIME?

Edit: looking at the problem, how would you apply the inequality to this?

i think the problem is

find the minimum value of (9x^2sin^2 x   +4)/(xsinx) for x  defined in (0,TT)

 (9x^2sin^2 x + 4)/2 is am

gm is then sqrt(36x^2sin^2 x)

am>= gm

9x^2sin^2 x + 4 >= 2*(6xsinx) = 12xsinx

then 12xsinx / (xsinx) gives 12

wait one thing i dont get is.... how do you find am? i dont know where the 4 came from.

it is from the problem

oops,
my bad, i had typo on my last entry, i'll edit it

AM-GM can be applied to a whole list of nonnegative numbers.  AM stands for arithmetic mean and is probably what you would think of if I told you to find the average of a set of numbers.  GM stands for geometric mean.  The inequality looks like this if you are applying it to the set $\{a_1,a_2,\ldots,a_n\}$ where $a_i\ge 0$ looks like:

\[\frac{a_1+a_2+\cdots+a_n}{n}\ge\sqrt[n]{a_1a_2\cdots a_n}.\]


[quote="diophantient"][quote="zigga_plz"]Factor this polynomial.

x^4+1[/quote]
[hide="for real"]$a^{4}+b^{4}=(a^{2}+b^{2}+ab\sqrt{2})(a^{2}+b^{2}-ab\sqrt{2})$
[/hide]
[hide="unreal"]
$x^{4}=-1$
$x^{2}=\pm i$
$(x^{2}+i)(x^{2}-i)$[/hide][/quote]

the unreal can be factored further...you gave the factorization up to "gaussian integers"

Consider: \[\prod_{all \ \pm \ choices}\left(\pm 1 \pm i\right) = (2)(2)=4\]
We just scale down to find that $r_{k}\in\left\{\frac{1}{\sqrt{2}}(\pm 1\pm i)\right\}$ are the four roots.

[hide="or"]$|\sqrt{i}|=|\sqrt{cis(90)}|=cis(45)=\frac{\sqrt{2}}{2}(1+i)$ and same for the others.[/hide]


Do we type the problem in our solution?

Also, do the problems get progressively more difficult as the rounds increase?

1) No, you just have to specify which problem it is, you dont have to write out the entire problem.

2) Yes.
Even though its debatable. The idea is that round 2 is harder than round 1, round 3 is harder than round 2, etc. Well, its more like round 2 becomes more proof oriented than round 1, so it depends on what youre better at. Round 3 has "harder" proofs, and round 4 should have the hardest proofs ideally.

[quote="Thunder365"]1) No, you just have to specify which problem it is, you dont have to write out the entire problem.

2) Yes.
Even though its debatable; The idea is that round 2 is harder than round 1, round 3 is harder than round 2, etc. Well, its more like round 2 becomes more proof oriented than round 1, so it depends on what youre better at. Round 3 has "harder" proofs, and round 4 should have the hardest proofs ideally.[/quote]

You need to edit your post as you talked about the difficulty of Round 2, which hasn't ended.

EDIT: fixed.


function
combinatorics unsolved
combinatorics

if G be a graph that the number of vertices is n and the number of   edges is m  and the number of  connected components is k then
$ m \le (n\minus{}k)(n\minus{}k\minus{}1)$
how we can prove it

It is false if $ n \equal{} k \plus{} 1$, when we can have as many as $ 1 > 0$ edges.  Also when $ n \equal{} k \plus{} 2$, when we can have up to $ 3 > 2$ edges.  And the bound is weak when $ n > k \plus{} 3$:

Suppose the components have $ a_1, \ldots, a_k$ vertices.  Then certainly the number of edges is no more than $ \binom{a_1}{2} \plus{} \ldots \plus{} \binom{a_k}{2}$.  Because of the convexity (concavity?) of the function $ f(x) \equal{} \binom{x}{2}$, this sum is increased if we replace $ (a_i, a_j)$ with $ (1, a_j \plus{} a_i \minus{} 1)$.  This gives immediately that the maximal number of edges is attained if we have a graph that consists of an $ n \minus{} k \plus{} 1$-clique and $ k \minus{} 1$ isolated vertices.  In this case we have $ \binom{n \minus{} k \plus{} 1}{2}$ edges.


geometry
trapezoid
trigonometry
area of a triangle

In convex quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $P$. If the area of $\triangle APB$ is $9$ and the area of $\triangle CPD$ is $25$, what is the minimum possible area of quadrilateral $ABCD$?

[hide="maybe...?"]
As you decrease $\angle BPC$, the other two triangles get smaller and smaller, approaching 0.  So then the minimum area is 34?  Well, it could never be 34 theoretically.
[/hide]

Sorry that I wasn't clear about this- $ABCD$ is a convex quadrilateral, which means $P$ must be inside the quadrilateral.

deej21- The answer is greater than $34$ :D

Since the two triangles are similar, we can denote their bases $3y$ and $5y$ and their heights $3x$ and $5x$, and the height of the trapezoid is $8x$. We know that $xy=2$ and to find the minimum, we use AM-GM, so that $\frac{x+y}{2}\geq \sqrt{2} \Longrightarrow x+y=2\sqrt{2}$ for the minimum possible value. Thus $x=y=\sqrt{2}$ and the area of the trapezoid is $\boxed{64}$

Hmmm...I'm not sure if that's right though :? , seems the area could be less

I had the same line of thinking, that the smallest possible area for the quadrilateral would arise from a trapezoid, specifically an isosceles trapezoid. However, I did not know how to finish the solution.

Let $AP=a$, $BP=b$, $CP=c$, $DP=d$.  Now note $\sin{\angle{APB}}=\sin{\angle{BPC}}=\sin{\angle{CPD}}=\sin{\angle{DPA}}$, and the area of a triangle is half the product of two sides times the angle in between.  Let any of the sines above by $x$.  Then $\frac{18}{x}=ab$ and $\frac{50}{x}=cd$.  We want to minimize $bc+ad$.
We know $abcd=\frac{900}{x^2} \implies \sqrt{abcd} = \frac{30}{x}$, so by AM-GM, $\frac{bc+ad}{2} \ge \frac{30}{x}$  Hence, the minimum value of $ab+bc+cd+ad=\frac{128}{x}$, so the minimum area of the trapezoid [b]is[/b] $\boxed{64}$.  That proves it...

But something cool: note that the central angles don't matter.  :)

Yeah, $64$ is correct.

You can also use $\triangle APB\cdot\triangle CPD=\triangle BPC\cdot \triangle DPA$ to get $\triangle BPC\cdot \triangle DPA=9\cdot 25=225$.
Using AM-GM, we see that $\triangle BPC+\triangle DPA\ge 30$, which implies that the minimum possible area of quadrilateral $ABCD$ is $64$ ;)


geometry
parallelogram
trapezoid
Gauss
geometry solved

Let $ABC$ be a triangle, and $D$, $E$, $F$ the points where its incircle touches the sides $BC$, $CA$, $AB$, respectively. The parallel to $AB$ through $E$ meets $DF$ at $Q$, and the parallel to $AB$ through $D$ meets $EF$ at $T$. Prove that the lines $CF$, $DE$, $QT$ are concurrent.

Actually, the problem holds more generally for an arbitrary cevian triangle DEF. In other words, we have the rather general result:

[i]Let ABC be an arbitrary triangle, and let D, E, F be any three points on the lines BC, CA, AB such that the lines AD, BE, CF concur. Let the parallel to the line AB through the point E meet the line DF at a point Q, and let the parallel to the line AB through the point D meet the line EF at a point T. Then, the lines CF, DE and QT are concurrent.[/i]

[i]Proof.[/i] Let the parallel to the line AB through the point C meet the lines EF and DF at the points U and V, respectively. Since the lines AD, BE, CF concur, the Ceva theorem yields

$\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1$.

On the other hand, since the line CUV is parallel to the line AB, Thales yields $\frac{BD}{DC}=\frac{FB}{CV}$ and $\frac{CE}{EA}=\frac{UC}{AF}$, so that we have

$\frac{FB}{CV}\cdot\frac{UC}{AF}\cdot\frac{AF}{FB}=1$.

This simplifies into $\frac{UC}{CV}=1$. In other words, UC = CV, so that the point C is the midpoint of the segment UV.

Since DT || AB and UV || AB, we have DT || UV. Thus, if the line CF meets the line DT at a point R, Thales yields $\frac{TR}{RD}=\frac{UC}{CV}$. Since $\frac{UC}{CV}=1$, we therefore have $\frac{TR}{RD}=1$, too.

We have DT || AB and EQ || AB. Hence, DT || EQ. Thus, by Thales, $\frac{DQ}{QF}=\frac{TE}{EF}$. Together with $\frac{TR}{RD}=1$, this yields

$\frac{TR}{RD}\cdot\frac{DQ}{QF}\cdot\frac{FE}{ET} = 1\cdot\frac{TE}{EF}\cdot\frac{FE}{ET} = \frac{TE}{EF}\cdot\frac{FE}{ET} = \frac{ET}{FE}\cdot\frac{FE}{ET} = 1$.

Hence, by the Ceva theorem, applied to the triangle FTD, it follows that the lines FR, TQ and DE are concurrent. In other words, the lines CF, DE and QT are concurrent. $\blacksquare$

  Darij

i'll show that $CF$ bisects segment $QE$ (and analogously bisects segment $DT$). Since $DT//QE$, the midpoints of $DT$,$QE$, the point $F$, and the intersection of $QT$ and $DE$ are all collinear (Ceva). And the problem follows..

To prove that $CF$ bisects $QE$, let $S=QE\cap CF, P=BE\cap AD, R=DF\cap BE$. Since $(E,R;P,B)=-1$, then $-1=(FE,FR;FP,FB)=(FE,FQ;FS,FB)$. But then, using that $EQ//AB$, we have that the points $S$ and a point in the line at infinity are harmonic conjugates with respect to $EQ$. Therefor $S$ is the midpoint of $EQ$, and we are done.

[quote="darij grinberg"]Actually, the problem holds more generally for an arbitrary cevian triangle DEF. In other words, we have the rather general result:

[i]Let ABC be an arbitrary triangle, and let D, E, F be any three points on the lines BC, CA, AB such that the lines AD, BE, CF concur. Let the parallel to the line AB through the point E meet the line DF at a point Q, and let the parallel to the line AB through the point D meet the line EF at a point T. Then, the lines CF, DE and QT are concurrent.[/i][/quote]

I think this can be generalized a bit more still: instead of definining Q and T such, I believe we can pick any point G on line AB and then define Q = DF ^ EG, T = EF ^ DG, and the result should still hold. In your case, of course, G is simply the point at infinity corresponding to AB. So here is my attempt at proving this statement:

Take line ABFG to infinity. If P = AD ^ BE, we know CDPE is a parallelogram, but we are given that AD, BE, CF concur, so P lies on line CF, so the midpoint of DE is the midpoint of CF and lies on line CF. Now, we see that DQET is in fact another parallelogram, and DE ^ QT will be the center of this parallelogram, i.e. the midpoint of DE. As this lies on CF, we have the result that DE, CF, and QT concur. Does this work?

Edit: By the way, the problem statement is equivalent to "C lies on the Steiner line of trapezoid DTEQ". 2nd Edit: Oh wow now I see actually both the above solutions use Steiner line (without actually mentioning it) :D

[quote="MysticTerminator"]I think this can be generalized a bit more still: instead of definining Q and T such, I believe we can pick any point G on line AB and then define Q = DF ^ EG, T = EF ^ DG, and the result should still hold. In your case, of course, G is simply the point at infinity corresponding to AB. So here is my attempt at proving this statement:

Take line ABFG to infinity. If P = AD ^ BE, we know CDPE is a parallelogram, but we are given that AD, BE, CF concur, so P lies on line CF, so the midpoint of DE is the midpoint of CF and lies on line CF. Now, we see that DQET is in fact another parallelogram, and DE ^ QT will be the center of this parallelogram, i.e. the midpoint of DE. As this lies on CF, we have the result that DE, CF, and QT concur. Does this work?[/quote]

Indeed, this works fine! So we have another solution now...

[quote="MysticTerminator"]Edit: By the way, the problem statement is equivalent to "C lies on the Steiner line of trapezoid DTEQ". 2nd Edit: Oh wow now I see actually both the above solutions use Steiner line (without actually mentioning it) :D[/quote]

Hmm, I didn't even know that the thing was called "Steiner line"  ;)  anyway, of course I did use it implicitely.

  Darij

The only Steiner line I know (which also shows up with Google :)) is the parallel through $H$ to the SDimson line of a point $P$ (called the Steiner line of $P$ wrt $ABC$).

Sort of off-topic though :). Sorry about that :).

Ah, yes. My apologies. It is not called "Steiner line" as such - the theorem saying that the midpoints of the bases of the trapezoid are collinear with the intersection of diagonals and non-parallel sides is called "Steiner's theorem"

Hi Darij You have used Gauss theorem, I couldn not find . PLease post ity to me , u used but I coud not understend at the usage part of theorem.
    Abdurashid


The perimeter of an isoceles triangle is 38 centimeters and the two sides of the triangle are whole numbers in the ratio 3:8. What is the number of centimeters in the length of the shortest side?

There are two possibilities - the "8" in the ratio is one of the congruent sides and the "3" in the ratio is one of the congruent sides.

Suppose it was the 8. We would then have $ 19x \equal{} 38 \implies x \equal{} 2$. That means the side lengths would be 6, 16, and 16, which works. If the "3" was one of the congruent sides, it wouldn't work (try it.) Thus, the answer is $ \boxed{6}$.


Let A be the prod{ 1 \leq k \leq n-}sin(k \pi /n).
We will prove below that A = n/2 n-1 , from which it will easily follows that the desired limit is 2.

Now, let B = prod{1 \leq k \leq n-1} (e -ik \pi /n / (2i) )
and C  = prod{1 \leq k \leq n-1} (-1) n-1 (1 - e 2ik \pi /n ).

Since sin(k \pi /n) = (e -ik \pi /n / (2i) )*(-1)(1 - e 2ik \pi /n ), we have A = BC.

But :
B = (1/(2i)) n-1 *(e -i(1+2+...+n-1) \pi /n ) = ((e -i \pi /2 /(2i)) n-1  = (-1/2) n-1 .

Now let P(x) = 1 + x + ...+ x n-1 . It is well-known that the roots of P are the numbers (e 2ik \pi /n  for k = 1,2,...,n-1.
Thus C = (-1) n-1 P(1) = (-1) n-1 n, and we are done.

Pierre.

Your answer is right!
But I want to find another sol don't use complex(Because I really hate it  :D )


inequalities
integration
inequalities unsolved

I think this was posted before, not sure though.

It was already discussed, but I don't think the following solution was posted:
  It can be deduced from the following well-known inequality:
  If f is convex on [a,b] then f((a+b)/2) \leq  \int (a-b) f(t)dt/(b-a) \leq  (f(a)+f(b))/2. Just take f(x)=x^n.


convert benzene to napthaldehyde (with minimal use of chemicals as possilble)

Are we allowed to use other organic starting materials besides benzene?

[hide="Answer"]1) Benzene + Br2/Fe -----> Bromobenzene
2) Bromobenzene + NaNH2 + Furane, then H+ -----> alpha-naphthol

I think you can now finish the synthesis by yourself.[/hide]


I'm going to post sets of ARML problems.  Once all (or most) of the problems from one set is solved I'll go ahead and post another set.  All of the problems will come from past contests.

\[ \text{Set 1}  \]

1. (1991 I1) Consider the system of 25 equations in 26 variables: [img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=638[/img]

Compute $A+Z$.

2. (1989 T8)  Compute the length of the tangent segment from the origin to the circle that passes through the points (3,4), (6,8), and (5,13).

3. (1990 I5) Compute the integer $k$, $k>2$, for which \[ \log(k-2)! + \log(k-1)! + 2 = 2\log k!.  \]

4. (1991 I8) The number $85^9-21^9+6^9$ is divisible by an integer between 2,000 and 3,000.  Compute that integer.

5. (1994 T3) Consider the equation \[ 100\cdots00_b+100\cdots00_{b+1} = 100\cdots00_{b+2},  \] where each term contains $n$ zeros.  For how many $n, 2\le n\le 100$, will a solution exist for the equation?

[hide="1"]We subtracting two equations and then we add the next equation, etc.

So we have $1-2+3-4+5-5...23-24+25=\boxed{13}$[/hide]

[hide="4"]I just looked at it using mods for different numbers using trial and error, i found it to be divisible by $2^{6}, 5, 7$, so the answer is $\boxed{2240}$ and there aren't anymore divisors since the question specifies the interval  :lol: [/hide]

[hide="2"]You have $(x-5)^2+(y-13)^2=(x-6)^2+(y-8)^2=(x-3)^2+(y-4)^2$, From here we can get $5y-x=47$ and $8y+6x=75$, solving for this, we have $y=\frac{357}{38}$ and $x=\frac{-1}{38}$....gotta go now but i think this is how you do it, we need to eventually find the radius and then we can make a right triangle. [/hide]

i just looked up 2, so i wont answer, but if anyone needs a [hide="hint"]power of a point[/hide]

[quote="arkmastermind"][quote="joml88"]

3. (1990 I5) Compute the integer $k$, $k>2$, for which \[ \log(k-2)! + \log(k-1)! + 2 = 2%Error. "logk" is a bad command.
!.  \]

[/quote]

it should be, 3. (1990 I5) Compute the integer $k$, $k>2$, for which \[ \log(k-2)! + \log(k-1)! + 2 = 2\log{k}!  \]

small latex error, but it makes a big difference  :)

$k$ will b $5$, btw - just condense the left side into one log and go from there...[/quote]

ahh..good, I thought there would be a typo because I was wondering since the 2s cancel out, is it even possible?  :D

5.
[hide]The given expression is equivalent to $b^n+(b+1)^n=(b+2)^n$. Thus, n=2 by Fermat's Last Theorem; an example is b=3. Therefore, the answer is [u][b]1[/b][/u].
I have yet to see a nontrivial (but readily understandable) application of Fermat's Last Theorem.[/hide]

[quote="Iversonfan2005"]Thankfully the typo was fixed :P[/quote]

Yeah, sorry about that :oops:

[quote="Iversonfan2005"]BTW, does anyone know where the ARML site for SC will be this year? I heard it will be in Florida :surf:[/quote]

It's at Penn State.  There won't be a Florida site until next year at the earliest.  Don't worry though, I'm glad that we aren't going to Florida.  Not because I don't like Florida, but because it would have only been a few teams.  At Penn State, there are a ton of people...it's an amazing experience :)

[quote="Iversonfan2005"]Thankfully the typo was fixed :P
[hide="solution3"]This simply becomes $(k-2)!(k-1)!(100)=(k!)^2$

Since this is an ARML problem and time is limited, it's good to guess and check. Obviously $k>2$ and our best bet is to try $5$, since we need a $5$ in the factorization because of the $100$, and lucky enough, $\boxed{5}$ works.  :D [/hide]

BTW, does anyone know where the ARML site for SC will be this year? I heard it will be in Florida :surf:[/quote]

we dont need to guess and check. note:

\begin{eqnarray*}(k-2)!(k-1)!(100)&=&(k!)^2\\ 100(k-2)!^2(k-1)&=&k!^2\\ 100&=&\frac{k!^2}{(k-2)^2\cdot(k-1)}\\ 100&=&(k^2)(k-1)\\ \end{eqnarray*}

so, $k=\boxed{5}$...yes i know we have a cubic there, but i think the answer is easier to find after quickly canceling out terms first. ;)

[hide="another hint for number 2"] note that (0,0), (3,4), and (6,8) are on a line...THEN use power of a point  :) [/hide]


I have a question:

how to prove this:   $ \Bar{A} \cap B \subseteq \Bar{A \cap B}$,

with A and B open sets.

Thanks!

What is the closure taken over in the second expression?

That is the closure of $ A \cap B$. :blush:

Take $ x \in \bar{A} \cap B$, i.e. $ x \in b$, and for any open $ U$ containing x, $ U \cap A \neq \emptyset$. It suffices to do it for small open sets, in particular we may assume that $ U \subset B$, since $ B$ is open. Then $ U \cap A \equal{} U \cap A \cap B \neq \emptyset$ implies that $ x \in \overline{A \cap B}$

Note that $ \overline{A}\cap Int(B)\subseteq\overline{A\cap B}$ for every $ A$ and $ B$, where $ Int(B)$ is the set of all interior points of $ B$.


inequalities
calculus
inequalities solved

If  0<=x,y<=1  prove  that
1/[sqrt(1+x^2)] + 1/[sqrt(1+y^2)] <= 2/[sqrt(1+xy)]

There  shoud be  a  differtent  solution...not  with  calculus  as  in  book....

Well, I do not know what kind of calculus is used in Titu's solution. Here is mine :

First, if x = 0 the desired inequality is 1/ \sqrt (1+ y 2 )  \leq  1, which is obviously true, with equality iff y = 0.

Now, assume that xy > 0.
Let f(t) = 1/ \sqrt (1 + e t ). It is easy to verify that f is concave on ]-oo, 0] (since f''(t) = e t (1+ e t) -5/2 (e t - 2)/4 ).
Let x 2  = e a and y 2  = e b, thus xy = e (a+b)/2 , and since 0 < x,y  \leq  1, we have a,b  \leq  0.
The desired inequality is now just Jensen's inequality...
And equality occurs iff x = y.

Pierre.

Well  its  little  different  than  Titu's  solution but  i  like  it!!
Thanks!!


$ x,y,z$ be positive integer number such that :
$ (x \minus{} y)(y \minus{} z)(z \minus{} x) \equal{} x \plus{} y \plus{} z$

Prove that $ 27 | x \plus{} y \plus{} z$

hello, do you mean
$ (x\minus{}y)(y\minus{}z)(z\minus{}x)\equal{}x\plus{}y\plus{}z$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you mean
$ (x \minus{} y)(y \minus{} z)(z \minus{} x) \equal{} x \plus{} y \plus{} z$?
Sonnhard.[/quote]
yes, i edited  :arrow:  :arrow:  :arrow:

First case : , if $ x\equiv y \ (\text{mod} \ 3)$ , then $ 3|x \plus{} y \plus{} z$ , so from here we have $ 3|2x \plus{} z$ and one easily get $ x\equiv y \equiv z \ (\text{mod} \ 3)$ so $ 27|(x \minus{} y)(y \minus{} z)(z \minus{} x)$

Second case :  if $ x$ and $ y$ are different in modulo $ 3$ then  from the above case , we conclude that $ x$ , $ y$ , and $ z$ are all different in modulo $ 3$ , thus $ (x \minus{} y)(y \minus{} z)(z \minus{} x)\equiv 1$ or $ 2 \ (\text{mod} \ 3)$ , but $ x \plus{} y \plus{} z\equiv 0 \ (\text{mod} \ 3)$ , so we have no solution in this case.

 :wink:

i think the problem given is wrong, the left is negative but the right is positive :wink:

How about $ x\equal{}15,y\equal{}18,z\equal{}21$¿  :wink:


calculus
integration
trigonometry
calculus computations


Each person in a ticket line gets either one or two tickets. You are the $ 51^{\mathrm{st}}$ person in line. Forty percent of the people in front of you get one ticket each, and the rest of the people in front of you get two tickets each. How many tickets will have been given out when you reach the ticket booth?

Being the 51st person in line, there are 50 people ahead of you. $ 40\%\times50\equal{}20$ people got 1 ticket, and the last 30 people all got 2 tickets. $ 20\plus{}2\times30\equal{}20\plus{}60\equal{}\boxed{80}.$


Show that if $ a, b, c \in [0, 1]$, then $ abc \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \geq 1/4$.

If this isn't true, let me know. I just found it when doing [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=277698]a problem[/url], and this would have finished it off.

If $ a\equal{}b\equal{}c\equal{}0$, then it's not true, so it might not be true for other values..

omg sorry, I deserve a slap in the face for that.

So you could factor this as :

$ 1\minus{}(a\plus{}b\plus{}c)\plus{}(ab\plus{}bc\plus{}ac) \geq \frac{1}{4}$

Other than that, I got notin.

Plugging in $ (a,b,c) \equal{} (1,0,0)$ causes the expression to equal $ 0$.

The following isn't really necessary to show that this is the minimum value of the expression, as its easy to show that the expression is strictly non-negative,  but you might find it interesting nonetheless.

Consider the function $ f(a,b,c) \equal{} abc \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$.

Suppose that we fix two of the variables and vary the 3rd. WLOG, lets fix $ b$ and $ c$, and vary $ a$. Then the function becomes a linear function of $ a$. If it becomes a linear function, then it will take its minimum and maximum at the end points, which are $ 0$ and $ 1$.

Thus, we can conclude that $ f(a,b,c)$ takes its maximum and minimum values when all of the inputs are either 1 or 0. Testing, we find that if we have at least two differing inputs, the function equals $ 0$. If all of the inputs are the same (all 0 or all 1), then the function equals $ 1$.

Thank you, Lulze. That'll help me later on (that is, considering values at the endpoints more often.)


find all strictly increasing functions $f: \mathbb N \rightarrow \mathbb N$, so that
\[ \forall n \in \mathbb N: f(f(n)) = 3n \]


inequalities
geometry
3D geometry
inequalities unsolved

prove that if 1>a,b,c>0
Then
1 / (a+b+c)  >  1/3+(1-a)(1-b)(1-c)
1/(a+b+c)=(a+b+c)^-1=a+b+c divided by 1
Good luck!

hello, do you meant $ \frac {1}{a \plus{} b \plus{} c} > \frac{1}{3} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$ if $ 1 > a,b,c > 0$?
Sonnhard.

$ f(a, b, c) \equal{} \frac {1}{a \plus{} b \plus{} c} \minus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$ is (strictly) convex in all variables and $ lim_{a, b, c \rightarrow 0} f(a, b, c) \equal{} \plus{} \infty$. Hence, in the "cube" $ [0, 1]^3 \minus{} \{0, 0 , 0\}$, $ f$ attains its minimum in one of the vertices. Now it is just case analysis.

(how can one put the $ a, b, c \rightarrow 0$ under the limit?)


floor function
modular arithmetic
number theory

Let $\pi (n)$ be the number of primes $\leq$ n

Prove:
$\pi (n) = \sum_{j=3}^n \left( (j-2)! - j \left\lfloor \frac{(j-2)!}{j} \right\rfloor \right)$

This doesn't quite work... for example, $\pi(4) \neq 3$. In particular, if $j$ is composite, $(j-2)!$ does not always contain all the factors of $j$. But, for $j > 4$ it does. You also need to start the summation at $j = 2$ because $2$ is prime.

[quote="mathmanman"]Well, actually, this formula is known, but here it indeed has some typos.
It follows from the fact (which itself follows from Wilson's theorem) that :
\[ (j-2)! \equiv a \pmod j, \ \& (j \geq 3)  \]
where $a = 1 \ or \ 0$, depending on whether $j$ is prime or not.
The good formula is thus :
\[ \pi (n) = 1 + \sum_{j=3}^n \left( (j-2)! - j \left[\frac{(j-2)!}{j} \right] \right) \ \& (n \geq 3),  \]
with $\pi(1) = 0$ and $\pi(2) = 1$.[/quote]

The formula still doesn't work. $\pi(4) \neq 4$. I was pointing out that $(j-2)! \equiv a \pmod{j}$ doesn't hold for $j = 4$ because $2! \equiv 2 \pmod{4}$.

I am sorry about this mistake, the formula doesn't work like this indeed:
[quote="Mafioso"]$\pi (n) = \sum_{j=3}^n \left( (j-2)! - j \left\lfloor \frac{(j-2)!}{j} \right\rfloor \right)$[/quote]
Though it's just a little difference, this might be the right one:
$\pi (n) = -1 + \sum_{j=3}^n \left( (j-2)! - j \left\lfloor \frac{(j-2)!}{j} \right\rfloor \right)$  with n>3 and $\pi (1) = 1, \pi (2) = 1$ and $\pi (3) = 2$.


sorry if i sound kind of stupid, but how do you give yourself a signature? i've seen lots of people with signatures put below their post... so how can you do that??  :oops:

Go [url=http://www.artofproblemsolving.com/Forum/usercp.php?mode=editprofile&sub=sig]here[/url]

Also [url=http://www.artofproblemsolving.com/Forum/index.php?f=145]here[/url] might be a better place to post that question

Also, you have to tell it to attach your signature [url=http://www.artofproblemsolving.com/Forum/usercp.php?mode=prefs&sub=posting]here[/url]

whats up with the caps?
hey and this should be under help and support.
mods plz move this!

please use proper grammar and spelling, not in all caps.
Also, please don't revive threads that are over a month old.


I was given the question......

"The Solution of the linear system :

-3x + y = -2
 3x + y = 10

.................   is ( x , y ) =  __________________-    "

I was told to find the answer I ...

"add 2y = 8
          y = 4       ...... now substitute y"

Why the hell do I add 2y-8? How am I supposed to know y = 4?

lol please help me

Did you know that you can add two equations to find the solution? Well, this is what they want you to do. When you add them you get:
$-3x+y+3x+y=-2+10$
$2y=8$
$y=4$
Then you substitute $y=4$ into any of the two equations:
$3x+4=10$
$3x=6$
$x=2$
So your answer is $(x,y)=(2,4)$.

That's amazing.

I understand it now (I think :o) )

Thank you sooo much.

I don't know if this was appropriate for Intermediate Level.

No, it wasn't supposed to be Intermediate, but since gizm0 is a new member, I didn't say anything  ;)


algebra
polynomial
function
algorithm
system of equations

Define a polynomial [tex]f(X)=(X-p)(X-q)(X-r)(X-s)(X-t)[/tex] [tex]=X^{5}-s_1X^{4}+s_2X^{3}-s_3X^{2}+s_4X-s_5[/tex], where [tex]s_1,s_2,s_3,s_4,s_5[/tex] are elementary symmetric functions of [tex]p,q,r,s,t[/tex].

Define a sequence [tex]a_n=xp^n+yq^n+zr^n+us^n+vt^n[/tex].

Then [tex]f(X)[/tex] is the characteristic polynomial of the linear recurrence sequence [tex]a_n[/tex].

So the sequence [tex]a_n[/tex] satisfies the relation:

[tex]a_{n+5}-s_1a_{n+4}+s_2a_{n+3}-s_3a_{n+2}+s_4a_{n+1}-s_5a_n=0[/tex] for all [tex]n\ge0[/tex] ------ (1)

(Looks like I am assuming that [tex]p,q,r,s,t[/tex] are different, actually not, the linear recurrence relation (1) is still true, it can be verified directly).

Therefore we have a system of linear equations in variables [tex]s_1,s_2,s_3,s_4,s_5[/tex] by letting [tex]n=0,1,2,3,4[/tex].

Solving for [tex]s_1,s_2,s_3,s_4,s_5[/tex] from the system of linear equations.

Once we have found [tex]s_1,s_2,s_3,s_4,s_5[/tex], we know what [tex]f(X)[/tex] is explicitly. Then we can solve for [tex]p,q,r,s,t[/tex] from [tex]f(X)=0[/tex], ..., and eventually find all [tex]x,y,z,u,v[/tex] once we have the values of [tex]p,q,r,s,t[/tex] (another system of linear equations).

So , I just described an algorithm for solving the system. I don't think I am really going to solve it :-)

I will illustrate my algorithm to solve a smaller system:

[tex]x+y=3[/tex]
[tex]px+qy=1[/tex]
[tex]p^2x+q^2y=3[/tex]
[tex]p^3x+q^3y=1[/tex]

I will follow up later.

I think I basically did the same thing as you, in a different form. I use [color=white]generating function[/color] instead, which is basically the same thing.

A solution, using a computer to solve the linear systems. This is essentially fuzzylogic's method, found independently.

[hide]

Let 2= f_0, 3=f_1, ..., 11595=f_9

Since it is a sum of five exponentials, it must satisfy a fifth-order recurrence

a*f_(n-4) + b*f_(n-3) + c*f_n-2 + d*f_n-1 + e*f_n = f_n-1, with

w^5 - e*w^4 - d*w^3 - c*w^2 - b*w - a = (w-p)(w-q)(w-r)(w-s)(w-t)

To find a,b,c,d,e, we solve the system

2a + 3b + 16c + 31d + 103e = 235

3a + 16b + 31c + 103d + 235e =674

16a + 31b + 103c + 235d + 674e = 1669

31a + 103b + 235c + 674d + 1669e = 4526

103a + 235b + 674c + 1669d + 4526e = 11595

The solution is a=1, b=-3, c=-1, d=5, e=1, so

w^5 - e*w^4 - d*w^3 - c*w^2 - b*w - a = (w+1)(w^2-3w+1)(w^2+w-1)

and p,q,r,s,t are -1,(-1- :rt5: )/2,(-1+ :rt5: )/2,(3- :rt5: )/2,(3+ :rt5: )/2 in some order. WLOG, p=-1, q=(-1- :rt5: )/2, r= (-1+ :rt5: )/2, s=(3- :rt5: )/2, t=(3+ :rt5: )/2.

Now that we know p,q,r,s,t, we solve

x+y+z+u+v=2

px+qy+rz+su+tv=3

p :^2: x+q :^2: y+r :^2: z+s :^2: u+t :^2: v = 16

p :^3: x+q :^3: y+r :^3: z+s :^3: u+t :^3: v = 31

p^4*x+q^4*y+r^4*z+s^4*u+t^4*v = 103

to find

x= -3/5, y= -1/2 - 8 :rt5: /5, z= -1/2 + 8 :rt5: /5, u= 9/5 -  :rt5: /10, v= 9/5 +  :rt5: /10

The roots p,q,r,s,t can be permuted, with the same permutation applied to x,y,z,u,v.

That accounts for all solutions, since the matrices here are invertible.[/hide]

[quote="jmerry"]A solution, using a computer to solve the linear systems. This is essentially fuzzylogic's method, found independently.
[/quote]

Wow, this is pretty cool. I am not sure whether [b]koopa[/b] has a better way to carry out the solutions which significantly beats this method.


let n be a positive integer!
determine minimum n>100 such that
 1²+2²+3²+4²+...+n²=A²;
A is a integer number!

we can think it more over!!
all value of n which is satisfied that the sum {1²+2²+...+n²} is a square integer number can be a sequence ?? and we can determine generality number?
I think it is hard!! :)

This was (almost) BMO 2, 1994 Q1

I remember finding a nice solution using Pell equations, although the first time I did it using modular arithmetic.

could you please give me the solution, SIMON?

This is a cubic Diophantine $ n(n\plus{}1)(2n\plus{}1) \equal{} 6A^2$, which is an elliptic curve.  Its set of integer points do not really form a sequence, as such.  It would be much easier to find the desired integer point than to find the set of integer points.


\[ 0 \le x,y \le 1 \\
x\plus{}y\equal{}1 \\
\Longrightarrow \\
x^y \plus{} y^x \ge 1 \\
x^{2y} \plus{} y^{2x} \le 1\]

Can someone teach me how to solve this problem :?:

$ x^y \plus{} y^x > 1$ for all $ x > 0,$ $ y > 0$ because $ x^y>\frac {x}{x \plus{} y}$ for all $ x > 0$ and $ 0 < y < 1.$
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42452

thank you :)

$ x^{2y} \plus{} y^{2x} \le 1$ is so difficult. :alien:

Bump! Who can prove this? Note that equality holds for $(x,y) \in \{(0,1),(1/2,1/2),(1,0)\}$.


summer program
Mathcamp
Support
geometry
AMC

I had this thought couple hours ago and I want to hear if anyone can help me. The first round of AMC is coming in about 5 months and it seemed that everybody is too busy to create a mock AMC. So, I was thinking of creating something like Mock AMC Committee where group of people work together to build a Mock AMC. It will be a lot easier and to see a various type of problems created by various people. Plus, I think this will be similar to real ones which are created in same ways (I suppose).

Anyone wants to help out on this? You don't really have to work hard on it. I'll be the chairman on this not because I'm better or smarter but since I have a lot of time :lol: and I like to do these "activities."

But as I said, I can't do this alone. I need other people and just for sidenote, you don't HAVE to be a great problem maker. Anyone who has time to spare can help out.

Thanks.

oooh, I could try, talk this weekend.  I write questions for my school's math club.  and amc level is fun.

sounds like a good idea...

did those MathCamp people ever post their problems?[/i]

I can help. I am writing weekly contests for my math club.

I can also help make problems, but how do we do that, do we just pm you the problems or something?

Sure, I'll try to make some, if I've got the time.

At this rate, we'll have more people helping out than people taking the test! :P

[quote="1234567890"]Okay, I'd rather take the test anyway. :)

Remove me as a contributor. I'm kind of busy now, so that's why.[/quote]

I wouldn't mind having many people in committee. But if you are busy, that's perfectly fine.  :)

Don't worry for having more people writing the test than people actually taking it. I know plenty of people who will want to take the test and even you don't get much people taking here, you can always take the copy and use for school practice.  :)

So, people who support this idea so far:

djshowdown2
yunorman
joml88
beta
Iversonfan2005
yif man12
and Silverfalcon (me)

So, we have 7 people now.. What I would like to do is volunteer for problems' # you want to do. Since we have 7 people which doesn't divide 25, I'll make 4 problems so each of you can make 3 problems.

Here is number of  4 problems I'll do:

#1
#2
#14
#15

Please sign up here for the problem number you want to do if you're one of those people up in the list? I'll be personally more appreciated if beta, yif man 12, or joml88 take up the later problems since they made good progress last year. This doesn't mean you shouldn't volunteer for later problems but I think they'll make good problems.

Thanks! :)

P.S. I forgot to add this but once you posted your number, you can work on to make the problem. Try to make it really AMC-12ish because I don't want to have really hard mock AMC-12 (i.e. this means on geometry, solutions using Ceva is not accepted because the real AMC-12 does not use Ceva on its problems). Also, try to make problems in different area (like one in NT, other one in Algebra, and other one in Combinatorics, etc).

Once you made the problem, please pm me the problems, choices (A,B,C,D,E) and solutions. Thanks.

[quote="Iversonfan2005"]I'll do #10, #11, #20 if that's okay[/quote]

I think it would be okay as long as nobody has comments on the problem #. So, so far, you're okay :D

[quote="amirhtlusa"]em...
any idea when you will be running this??[/quote]

This is going on now.. I received two problems from Iverson today and as soon as all people above sign up for the problem #, I'll set up the date that I need to receive problems, answers, and solutions.

What I'll then do is go through the problems and solve them by myself (which will take about a week since I'm pretty busy) and build up PDF document. I'll then send them to other members of the committee and ask them if they want to add anything to it. If all adjustments are made, I set up the date for the test like other Mock AMC and post the PDF test.

Hopefully this answered your question.

P.S. After the first test, I'm going to try to see if more people will want to involve in making problems. I'm not saying I don't have enough now but in case you get busy, it's always good to have somebody who can fill your space up. Or if we have enough, we can even decrease the number of problems each person has to do.  :)


[quote="subbi"]What else can we do with karma apart from rating,answering questions, watching videos, and reporting bugs?[/quote]

You can learn from it, like r15 said in the other thread, it's like your money to use the system.


What textbook/reference would you guys recommend to purchase to improve problem solving skills and proof skills?

Check out the "books" button on the top of the forum.

i suggest the aops books because if you work through them diligently, then you can get a lot out of them,

AOPS book 1 helps A LOT with geometry when you're starting out :)

Challenging Problems in Geometry is a really good resource for geometry/constructing proofs. The problems are similiar to that of the easier-mid level AIME geometry problems.

Also once you've done the AoPS books try the Art and Craft of Problem of Solving by Paul Zeitz. It's really helpful.


symmetry
geometry
circumcircle
trigonometry
cyclic quadrilateral
geometry proposed

Let $ A_1A_2$ be the external tangent line to the nonintersecting cirlces $ \omega_1(O_1)$ and $ \omega_2(O_2)$,$ A_1\in\omega_1$,$ A_2\in\omega_2$.Points $ K$ is the midpoint of $ A_1A_2$.And $ KB_1$ and $ KB_2$ are tangent lines to $ \omega_1$ and $ \omega_2$,respectvely($ B_1\neq A_1$,$ B_2\neq A_2$).Lines $ A_1B_1$ and $ A_2B_2$ meet in point $ L$,and lines $ KL$ and $ O_1O_2$ meet in point $ P$.
  Prove that points $ B_1,B_2,P$ and $ L$ are concyclic.

Since $ A_1K \equal{} A_2K \equal{} B_1K \equal{} B_2K$, $ A_1A_2B_2B_1$ is cyclic. Call this circle $ \omega$. Since $ \angle O_1A_1K \equal{} \angle O_1B_1K \equal{} 90^\circ$, $ O_1$ is the pole of $ A_1B_1$ wrt. $ \omega$. By symmetry, $ O_2$ is the polar of $ A_2B_2$ wrt. $ \omega$. Hence $ A_1B_1\cap A_2B_2$ is the pole of $ O_1O_2$.

An inversion wrt. $ \omega$ maps $ K$ to $ P$ so $ B_1K^2 \equal{} KP\cdot KL$. But then it follows from a SAS triangle similarity that $ \angle KB_1P \equal{} \angle KLB_1$. (1)

Let $ KL\cap B_1B_2 \equal{} Z$. It is clear that $ A_1A_2$ is antiparallel to $ B_1B_2$ wrt. $ \angle A_1LA_2$. So since $ LK$ is a median of $ \triangle LA_1A_2$, $ LZ$ is a symmedian of $ \triangle LB_1B_2$. Also $ KB_1 \equal{} KB_2$. Using those two facts, it follows by a well-known lemma that $ KB_1$ and $ KB_2$ are the tangents to the circumcircle of $ \triangle KB_1B_2$. But then $ \angle B_2B_1K \equal{} \angle B_1LB_2$. (2)

Subtracting (1) from (2) gives $ \angle PB_1L \equal{} \angle PLB_2$. But this implies that $ LB_2PB_1$ is cyclic, as desired.

---

I wouldn't call this an 'easy' solution. It uses too many high powered theorems. What is your solution and/or the official one?

Alternatively it can be proven that $ KL\perp O_{1}O_{2}$ which is easily proven by the equivalence with $ LO_{2}^2 \plus{} KO_{1}^2 \equal{} LO_{1}^2 \plus{} KO_{2}^2$ wherei denote the center of the circles respectively by $ O_{1},O_{2}$.To prove this relation just notice that $ LA_{1}\perp KO_{2}$ and $ KO_{2} \perp LA_{2}$ so $ LO_{1}^2 \equal{} LK^{2} \plus{} A_{1}O_{1}^2 \minus{} A_{1}K^2$ and the similar one so the statement to prove reduces to $ KO_{2}^2 \plus{} A_{1}O_{1}^2 \equal{} KO_{1}^2 \plus{} A_{2}O_{2}^2$ which is trivial.Now form this we have $ KO_{2}BP$ cyclic so $ \angle BPL \equal{} \angle KO_{2}B \equal{} \angle LA_{1}K$ and to finish the proof notice that $ B_{1}$ and $ B_{2}$ are the feet of the altittudes in triangle $ LA_{1}A_{2}$. :lol:

My solution isn't nice.
1-step. In this step we will show that $ O_1O_2\perp LK$.
It is well-known that $ A_2B_2\perp KO_2$.Assume that $ A_2B_2\cap O_1O_2 = T$ and $ A_2B_2\cap KO_2 = M$.
Then $ \frac {PK}{KO_2} = \sin \angle PO_2K$ and $ \frac {O_2M}{MT} = \tg \angle MTO_2$.Hence $ \frac {PK}{KO_2}\cdot\frac {O_2M}{MT} = \frac {LM}{LK}$ (1).
Since points $ P,T,O_2$ are collinear we have $ \frac {LP}{PK}\cdot \frac {KO_2}{O_2M}\cdot \frac {MT}{TL} = 1$ or $ \frac {PK}{KO_2}\cdot\frac {O_2M}{MT} = \frac {LP}{TL}$ (2).
Comparing (1) and (2) we have that $ \frac {LM}{LK} = \frac {LP}{TL}$ or $ LM\cdot LT = LK\cdot LP$ (3).
(3) means that $ K,P,T,M$ are concyclic, i.e $ \angle KPT = 90^\circ$.
Since $ \angle KPO_2 = 90^\circ$ points $ K,P,O_2,A_2$ are also concyclic.But $ K,B_2,O_2,A_2$ are concyclic.Hence $ K,P,B_2,A_2$ are concyclic.Then $ \angle LPB_2 = \angle B_2A_2K$ (4).
Since $ \angle LB_1B_2 = \angle B_2A_2K$ and from (4) we have $ \angle LPB_2 = \angle LB_1B_2$, i.e points $ B_1,L,B_2,P$ are concyclic.

[quote="Altheman"]
I wouldn't call this an 'easy' solution. It uses too many high powered theorems. What is your solution and/or the official one?[/quote]
This is a solution,that i gave at olympiad:
Since $ KA_1\equal{}KA_2\equal{}KB_1\equal{}KB_2$,then $ A_1B_1B_2A_2$ is the cyclic quadrilateral.Hence $ LA_1\cdot LB_1\equal{}LA_2\cdot LB_2$.Hence $ L$ lies on the radical axe of the circles $ \omega_1$ and $ \omega_2$,hence $ KL$ is a radical axe of these two circels.So $ KL\perp O_1O_2$.Now the problem became obvious,since
$ O_1B_1PKA_1$ is cyclic,hence $ A_1B_1PK$ is a cyclic quadrilateral,hence $ \angle LPB_1\equal{}\angle B_1A_1A_2\equal{}\angle B_1B_2L$ and we are done.
The official solution is longer and not so beautiful :blush:

[b][i][size=150][color=#960000]My Solution[/color][/size][/i][/b]
--------------------------
[b]Lemma [/b] [i]With problem conditions $KL$ is the [u]Radical Axis[/u] of the cicles $\omega_1$ and $\omega_2$.[/i]
[b]Proof [/b] Since $A_1K=A_2K=B_1K=B_2K$ that means $\angle A_1B_1A_2=\angle A_1B_2A_2=90^{\circ}$ that means the points $A_1,A_2,B_1$ and $B_2$ lie on the circle with diametr $A_1A_2$ and radius $A_1K=A_2K=B_1K=B_2K$. Then if we draw two tangents to the circles $LC_1$ and $LC_2$ then we have $|LC_1|^2=|LB_1|\cdot |LA_1|$ and $|LC_2|^2=|LB_2|\cdot |LA_2|$ We previously proved that $A_1,A_2,B_1$ and $B_2$ are cyclic and $|LB_1|\cdot |LA_1|=|LB_2|\cdot |LA_2|$. That means $|LC_1|=|LC_2|$ Which proves the [b]Lemma[/b].
----------------------

Since $KL$ is radical axis of $\omega_1$ and $\omega_2$ we have $KL\perp O_1O_2$. We also have $O_1B_1\perp KB_1$ and $O_2B_2\perp KB_2$ $\longrightarrow$ $O_1B_1PK$ and $O_2B_2PK$ are cyclic quadrilaterals. It follows that $\angle KB_1P=\angle KO_1P=\angle KA_1P$ ($A_1O_1PK$ is cyclic) , $\angle KB_2P=\angle KO_2P=\angle KA_2P$ ($A_2O_2PK$ is cyclic). Then $A_1B_1PK$ and $A_2B_2PK$ are cyclic quadrilaterals It follows that $\angle B_1A_1K=angle B_1B_2L=B_1PL$ Then $\boxed{ B_1,L,B_2,P \ \text{are cyclic}}$
-------------------

Too easy for P5.
Let $(KA_1O_1B_1)\cap (KA_2B_2O_2)=P'$. Since $A_1B_1A_2B_2$ is cyclic $LB_1\cdot LA_1=LB_2\cdot LA_2$, so $L-P'-K$ are collinear. Since $\angle O_1P'K=\angle O_1A_1K=90=\angle KA_2O_2=\angle O_2P'K$ we get $O_1-P'-O_2$ are collinear. So $P'=KL\cap O_1O_2$ so $P'\equiv P$. Miquel theorem on $\triangle LA_1A_2$ implies that $LB_1PB_2$ is cyclic. So done! :oops:


geometry
circumcircle
geometric transformation
rotation
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14530[/img]

 :(

The three parallel line from A,B,C wrt B1C1, C1A1, A1B1 meet at three intersection A2,B2,C2.
Construct with Peletier's theorem.

[quote="mr.danh"]Construct with Peletier's theorem.[/quote]
What is Peletier's theorem ? (I've never heard of it.)

Three parallel lines $ a_2 \parallel B_1C_1, b_2 \parallel C_1A_1, c_2 \parallel A_1B_1$ through $ A,B,C,$ respectively, pairwise intersect at $ A_2 \equiv b_2 \cap c_2, B_2 \equiv c_2 \cap a_2, C_2 \equiv a_2 \cap b_2.$   Circumcircles $ (X), (Y), (Z)$ of the $ \triangle BCA_2, \triangle CAB_2, \triangle ABC_2$ meet a a point $ P$ (by Miquel's theorem), you have to construct just 2, say $ (X), (Y).$ Perrpendiculars $ a_0, b_0, c_0$ to $ PA, PB, PC$ through $ A, B, C$ pairwise intersect at $ A_0 \equiv b_0 \cap c_0, B_0 \equiv c_0 \cap a_0, C_0 \equiv a_0 \cap b_0,$ such that $ A_0 \in (X), B_0 \in (Y), C_0 \in (Z)$ and the $ \triangle A_0B_0C_0 \sim \triangle A_2B_2C_2$ are spirally similar. $ P$ is the rotation center of their spiral similarity.

[img]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=14570[/img]

 :thumbup:


Asymptote USAMTS LaTeX blogs geometry vector

So, I was wondering... Is asymptote really worth learning for USAMTS? I currently use TeX, but instead of using Asymptote for diagrams, I just draw the diagram in Paint and then use \includegraphics{}. I find that it is short and easy, and the quality isn't bad either.

Of course, I do not know asymptote, so I don't know how much better coding diagrams in that language would be.

How much of a difference does asymptote make?

Well since I'm new to asymptote and starting USAMTS this is my take: make your pictures in Asymptote ONLY if you feel like you are comfortable enough to make decent diagrams. If you aren't then I would simply just use paint or another photo editor or software. (The time you spend learning Asymptote could rather be used to help you solve more problems)

Have you tried GeoGebra? It's like Geometer Sketchpad, only open source, which is great. I use that for diagrams. Just export it to an image file, crop it, and include it in your LaTex. Make sure the image file is in the same folder.

http://www.geogebra.org/cms/

Asymptote produces really good diagrams that are visually consistent with LaTeX, as well as being very easy to modify. If you already know C++ or similar, learn it, since the syntax is essentially identical; otherwise, it probably isn't worth the trouble. Myself, I enjoy the challenge of learning a new language, and I'm also a perfectionist in this aspect, so I've spent a lot of time tinkering with diagrams. Asymptote takes much longer than a similar GUI-based tool but often produces more precise, more modifyable, better-looking diagrams.

@ Piano
Is Geogebra the same quality wise as Geometers Sketchpad?

I have no idea, I've only used sketchpad a few times during school a few years ago.

Thanks for all the replies.  :lol:

I'll check it all out

Geogebra is pretty good quality. Personally I found it easier to use than Geometers Sketchpad. It also comes highly recommended (from rrusczyk's blog and from ACoPS V2). I never figured out how to get it to actually measure angles and lengths, but if you need to create a diagram for a geometry problem, it's wonderful.

//I never figured out how to get it to actually measure angles and lengths//

Hmm? Doesn't geogebra have the measurements for all the objects in the side, where they list all the objects - like measure of angles, lengths of lines, areas of polygons? (Or maybe you're talking about something else?)

[quote="worthawholebean"]Asymptote takes much longer than a similar GUI-based tool but often produces more precise, more modifyable, better-looking diagrams.[/quote]
hehe it actually takes less time for me to use Asymptote because if I was to use a GUI-based tool, this perfectionist would spend hours trying to make every bit of the diagram perfect. :D

And then I'd find that I'd made a mistake and have to draw it all over again, which is easier done with Asymptote, in my opinion.

I think asymptote is much better if you actually spend time to learn it. When you use .png or other picture files and use includegraphics:

1) the picture has to be aloof from other things you wrote
2) the picture has to be cropped and resized by trial error
3) when you decide to use another image instead you not only have to draw the whole thing again, but also repeat (1) and (2)
4) being a vector-based drawing program, altering pictures can be less of a hassle on asymptote


I personally found numbers that made the sum a multiple of 7 and plugged in. It worked for me.

[quote="myc"]When the mean, median, and mode of the list:
        10, 2, 5, 2, 4, 2, x

are arranged in increasing order they form a non-constant arithmetic progression.  What is the sum of all possible real values of x?

I don't understand their solution...can someone help me? [hide]

"If $x\geq 4$, then $4$ is the median, $2$ is the mode, and $\frac{25+x}{7}$ is the mean, which must be 0, 3, or 6. Thus $x=-25,-4, or 17$, of which $17$ is the only one of these values greater than or equal to 4. "

Why does it have to be 6?  It has to be 3 ,because
let $y$= the mean.  then  $y>4$ or $4>y>2$ or $y<2$.
Why can't the mean be 5 or 1?  actually since the mean can be greater than 4, can't it be anything?
[/hide]

Thanks so much![/quote]
[hide="Explanation"] The mean, median, and mode make an arithmetic progression.  We know that the mode must be 2, since there are 3 twos.  Now, if $x\le2$, then the median is also 2.  Thus, the mean must be 2.  However, the problem stated that the arithmatic proggesion is non-constant, meaning that the terms of it cannot be the same.  Thus, x>2.

Now, if $2<x\le4$, then the median is x.  Now, the mean is \[\frac{10+2+5+2+4+2+x}{7}=\frac{25+x}{7}\]  Thus, 2, x, and $\frac{25+x}{7}$ must form an arithmetic progression.  Hence, \[\frac{25+x}{7}-x=x-2\Rightarrow 2x-2=\frac{25+x}{7}=2x-2\Rightarrow 14x-14=25+x\Rightarrow 13x=39\Rightarrow x=3\] so x=3 works.

Now, if x>4, we know that the median is 4.  Thus, since mean, median, and mode form an arithmetic sequence, we find that the mean must be either 0, 3, or 6.  Now, the mean cannot be 0 since all of the numbers are greater than 2.  Now, if x=3, then the mean is 4, so obviously if x>4, then the mean is higher than three.  Thus, the mean must be 6.  Therefore, we have that \[\frac{25+x}{7}=6\Rightarrow 25+x=42\Rightarrow x=17\]  We can conclude that our sum is 3+17=$\boxed{20}$ [/hide]


For any real value of $ n$ other than $ 0$, $ n^0\equal{}1$. So the answer is $ \boxed{1}$.

Also, this is too easy to be in this forum, although this is the lowest level forum. I think you would be able to do that without help.

OKay look you got it right and i am sure i dont need help for it

Yeah, I agree with PowerofPi, and also dont copy ppl's avatars..

I also agree with PowerofPi
You are very smart :lol:

[quote="AGAK"]99999999999999999999999999999999^0

 :D[/quote]

The answer is 0. any number to the power of 0 equals 1! :D

...except $ 0$.
$ 0^0\equal{}\text{undefined (depending on context)}$
A discussion can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=278190]here[/url].

The answer is 1 unless the base is 0 , As powerofpi said.

By the way who can explain to me why $ 0.\infty$ is an indetermination as well? :maybe:

Does that mean 0 followed by a decimal point then infinity? or does it mean $ 0\times\infty?$

Assuming it's 0 x infinity it's because infinity times anything is infinity and zero times anything is zero. So this is undefined because you cannot satisfy one rule without breaking the other. :noo:

[hide="Algebraic explanation"]

$ 0\times\infty$

$ 0\times\frac10$

$ \frac{0\times1}{0}$

$ \frac00$

$ \frac{0^1}{0^1}$

$ 0^{(1\minus{}1)}$

$ 0^0$

Which is undefined.  :wink:
[/hide]

I would like to point out some things. $ \infty$ is a [b]concept[/b] not a number.
$ \frac{1}{0}\neq\infty, \frac{1}{0}\equal{}\text{undefined}.$
From your explanation, it looks like you can "cancel" the $ 0$ in the denominator with the $ 0$ in the numerator, but you really can't.

Hi, are there any geometric interpretation to $ 0x\infty$ be undefined? , I know geometric interpretations to $ \infty \minus{} \infty$ and $ \frac {0}{0}$

Does that mean $ 0x\infty$ or $ 0\times\infty$?

$ 0\times\infty$, FlyAgaric.

I don't know of an exact geometric interpretation, but if you have, say $ f(x)\equal{}\frac{1}{x}$, the undefined part is when $ x\equal{}0$ because then you'd be dividing by 0. I'm guessing it'd be similar to that.

To 0 times infinity? Wouldn't it be 0 then?

0·∞=undefined. 0 times anything is 0. ∞ times anything is ∞. Therefore, 0·∞ is undefined.


floor function
calculus
integration
logarithms

If $ n$ is natural number and $ n^a$ is natural number too , prove that $ a$ is natural .

Your claim is wrong. Let $ n$ be $ 4$ and $ a$ be $ \frac{1}{2}$.

I meant for every $ n$ .Let me rewrite the problem :

Let $ a$ be positive number ,and $ n^a$ is natural number for every $ n\in \mathbb{N}$ .Prove that $ a \in \mathbb{N}$

[quote="darkdieuguerre"][hide="Proof"]Let $ k \equal{} a \minus{} \lfloor a \rfloor$. $ 2^k$ is integral if and only if $ k$ is zero.

Therefore, the fractional part of $ a$ is zero, so $ a$ is natural.[/hide][/quote]

$ 2^k$ is integral if and only if $ k$ is integral, not $ 0$.   :huh:

[quote="AdithyaGanesh"]$ 2^k$ is integral if and only if $ k$ is integral, not $ 0$.   :huh:[/quote]

Firstly, that's not right. $ 2^{\log_2 5} \equal{} 5$, yet $ \log_2 5$ is not integral.

I make my claim in the context that $ 0 \le k < 1$, $ 2^k$ is integral if and only if $ k$ is zero, because $ 2^0 \equal{} 1$ and $ 2^1 \equal{} 2$, and there are no integers between $ 1$ and $ 2$, exclusive, and we know that $ k$ is within those bounds because we defined $ k$ as the fractional part of $ a$.

[quote="darkdieuguerre"][hide="Proof"]$ 2^k$ is integral if and only if $ k$ is zero.

Therefore, the fractional part of $ a$ is zero.[/hide][/quote]

These do not follow.

Do you have an example where $ 2^k$ is integral where $ 0 < k < 1$? Look at my previous post responding to AdithyaGanesh; I am more explicit there.

[quote="darkdieuguerre"][hide="Proof"]$ 2^k$ is integral if and only if $ k$ is zero.

Therefore, the fractional part of $ a$ is zero.[/hide][/quote]

The first part is true. The second part does not follow from the first part. $ 2^k$ can be $ \frac{3}{2}$, not an integer, while $ 2^a$ is an integer.


inequalities
real analysis
probability
inequalities theorems

Let X be a random variable and let 0 < r < s be constants. Show that if E(|X|power s) < ∞ then E(|X|power r)  < ∞

I don't know what you call Jensen's inequality here. The result is obvious considering that:

$|X|^{r} \leq |X|^{s}$ on $X^{-1}([1,\infty[)$, whence $E(|X|^{r})<\infty$ (direct consequence of Lebesgue convergence theorem).

Now on $X^{-1}([0,1[)$, $|X|<1$ whence $E(|X|^{r})<\infty$ (because we work in a probability space, i.e with finite measure).

Conclusion: $E(|X|^{r})<\infty$.


inequalities
induction
inequalities proposed

Let $ n$ be an integer greater than $ 2$.

$ a_1,a_2,\cdots ,a_n$, $ b_1,b_2,\cdots ,b_n$ are positive real numbers such that

$ a_1\plus{}a_2\plus{}\cdots \plus{}a_n\equal{}1$

$ b_1^2\plus{}b_2^2\plus{}\cdots \plus{}b_n^2\equal{}1$

Prove　　$ a_1(b_1\plus{}a_2)\plus{}a_2(b_2\plus{}a_3)\plus{}\cdots \plus{}a_n(b_n\plus{}a_1)<1$

Multiple both sides with $ 2$. Then the inequality becomes:

$ 2a_1(b_1 \plus{} a_2) \plus{} ... \plus{} 2a_n(b_n \plus{} a_1) < 2$

But we have:

$ LHS \equal{} 2(\sum_{cyc}{a_1a_2} \plus{} \sum_{i \equal{} 1}^{n}{a_ib_i}) \leq 2\sum_{cyc}{a_1a_2} \plus{} \sum_{i \equal{} 1}^{n}{(a_i^2 \plus{} b_i^2)} \equal{} 2\sum_{cyc}{a_1a_2} \plus{} \sum_{i \equal{} 1}^{n}{a_i^2} \plus{} 1$

Now I will prove by induction that:

$ 2\sum_{cyc}{a_1a_2} \plus{} \sum_{i \equal{} 1}^{n}{a_i^2} \leq (a_1 \plus{} a_2 \plus{} ... a_n)^2 \equal{} 1$, for $ n > 2$

For $ n \equal{} 2,3$ it is true as equality. Suppose that it is true for $ n \equal{} k > 2$
Then for $ n \equal{} k \plus{} 1$ we have:

$ (a_1 \plus{} ... a_k \plus{} a_{k \plus{} 1})^2 \equal{} (a_1 \plus{} .. \plus{} a_k)^2 \plus{} a_{k \plus{} 1}^2 \plus{} 2a_{k \plus{} 1}(a_1 \plus{} a_2 \plus{} ... a_k) \geq a_1^2 \plus{} ... \plus{} a_{k \plus{} 1}^2 \plus{} 2(a_1a_2 \plus{} ... \plus{} a_ka_{k \plus{} 1} \plus{} a_{k \plus{} 1}a_1) \plus{} 2a_ka_1 \plus{} ... \geq a_1^2 \plus{} ... \plus{} a_{k \plus{} 1}^2 \plus{} 2(a_1a_2 \plus{} ... \plus{} a_ka_{k \plus{} 1} \plus{} a_{k \plus{} 1}a_1)$

And our proof is finished.

Equality we have only for $ a_i \equal{} b_i$ for $ i \equal{} 1,2,3...n$ and obviously from the induction for $ n < 4 \Rightarrow n \equal{} 3$ . So equality holds only for $ n \equal{} 3, a_1 \plus{} a_2 \plus{} a_3 \equal{} 1, a_1^2 \plus{} a_2^2 \plus{} a_3^2 \equal{} 1 \Leftrightarrow 2(a_1a_2 \plus{} a_1a_3 \plus{} a_2a_3) \equal{} 0$ which is impossible. So we have never equality.

[quote="NickNafplio"]ill prove by induction that:

$ 2\sum_{cyc}{a_1a_2} \plus{} \sum_{i \equal{} 1}^{n}{a_i^2} \leq (a_1 \plus{} a_2 \plus{} ... a_n)^2 \equal{} 1$, for $ n > 2$
[/quote]

I think  $ (\sum a_i)^2 \equal{} \sum a_{i}^2  \plus{}2 \sum_{i<j} a_{i} a_{j}  >  \sum a_{i}^2  \plus{}2 \sum_{cyc} a_{1} a_{2}$
so we don't need induction :wink:


Sasha has $ \$3.20$ in U.S. coins. She has the same number of
quarters and nickels. What is the greatest number of quarters
she could have?

The way i get this is by doing 320/30
The reason why i do this is because then it is like doing 3.20/ 25cents+5 cents
320/30 is 10 remainder 20 so the answer is 10