MIT college Stanford Harvard Princeton function calculus

Ok: General idea: For the prospective scientist, does it make much of a difference attending the local state U? Looking at the university attended of university faculty in the major state universities - most have attended prestigious instutitions for undergrads, though there are some exceptions - such as David L. Goodstein of Caltech.

At this point, it's pretty much guaranteed that I'm going to the local state university (and am seriously considering entering it next year although my lackadaisical grades with 4 A-'s  may hurt me). My grades aren't all that exceptional and I have almost no EC's. Nor do I want to be more well-rounded - I think sports and music are a waste of time and am not particularly enthusiastic in non math/science related EC's. To top it all off, I'm a Chinese male. So is there any major decrease in quality between major u's and state u's? Should I just look forward to attending the University of Washington?  Matt McGann told me that I should follow my passion - that I shouldn't do things just to try to get into MIT.

What do you think you'll get at MIT as an undergraduate that you won't get at a state university? I don't mean to sound harsh, but if the only thing you can name is prestige and variations on that theme, then go to a state university. Prestige isn't worth $\$100000.

The atmosphere and the people at MIT vs a state university are probably extremely different.

I was also thinking - maybe the classes are more rigorous and taught more in the ways of problem-solving and proofs (while state universities probably only teach out of the mass-market textbook). It's been said that Caltech doesn't recommend people to skip its first-year calc course.

I think you forget that colleges aren't anything like high schools. Professors can choose to teach from whichever textbooks they want (except perhaps for very basic classes). If you take two classes with the same name from two different professors, you may end up with two rather different classes. (I have done this.) As a student at a state university, I can tell you that the only reason that most of my friends here didn't get into the prestigious universities is that they didn't sell their souls for high grades at SAT scores in high school. One of my friends was at an REU last year with many students from prestigious universities (among these students was Daniel Kane), and he didn't feel that he was lagging behind them.

If your long-term goal is to get a doctorate degree in math, then go to the best math school you can.

For the rest of us, who like math and would like to somehow use it in a career, I don't think a top math school is so important at all.  (I work as an actuary.)

You mentioned not being interested in non-math and science courses.  Don't be so sure.  There's lots to learn in other subjects, and there's lots to learn from non-math/science people.  You will be dealing all your life with people who are not math and science people, so it might help to be around some of them.  In particular, your math ideas often need to be sold to non-math type people.

Also, who knows what will interest you decades from now?  In my own case, I majored in math in the late 1960's, and although I work in a math field, I now write some fiction for fun and wish I had taken some creative courses.

Good luck!

I absolutely disagree. If you want to get a doctorate in math, then go to the best graduate school you can go to. But it doesn't matter where you go as an undergraduate. Save the money for later.

[quote="ComplexZeta"]But it doesn't matter where you go as an undergraduate.[/quote]

I'm not sure I'd agree with that statement stated so strongly. I think it matters some where you go as an undergraduate mathematics major, although student initiative as an undergraduate can make a second-tier university be a good experience for that student.

i dont know about your situation.. but i think for normal fokes, prestigious colleges are worth the $100000
reasons are as follows:
1. better supplies.. i am currently attending a physics class at caltech (basically i ask the teacher if i could listen to the lectures, but tats all i do, i dont have a book, or a lab kit, or hw, and its all free).. and the professor does all these experiments every lecture and talk about them.. then the other students do experiments at home with their lab kits.. for physics, i think the labs could really make the concepts stick to you..

2. better teachers.. i have take classes at community college, Cal state fullerton, and of course caltech... cal state teacher is about the same as the one at community college.. in fact, he used to be a high school teacher.. im not saying its bad or anything, but the teacher at caltech is so much better..

3. more attension.. smaller classes.. freshmen classes at caltech are like 100 people (max).. freshmen classes at berkley are like 800 people

4. "better" peers.. many your friends are going to be rich if you go to a prestigious college.. which means if you ever need a job, favor, or anything, they could easily provide it to you.. im not saying u should minipulate them, im just saying the way it is.

5. and last but definitly not least, bragging rights.. u said u were chinese.. prestigious colleges are all chinese parents care about.. no doubt your parents would dish out the money for those bragging rights...

with all that said, if you dont want to work ur butt off to get good grades, sat socres, and wut not, dont.. dont let trying to get into college(or your parents) control your life... i didnt realize that until this year.. and now that im so much more carefree, i feel so much better...
as great as a good college is, its not worth doing things you dont want to do

Using Caltech as your model for a "general elite college" is like using the Great Salt Lake to discuss major bodies of salt water.  You could make the case that it belongs in the group, but it's not a good example from which to make generalizations.  In particular, you seem to be comparing a single class at Caltech to a single class at a community college to a single class at a larger public university.  This doesn't strike me as a very strong basis from which to say, for example, that $\{\text{Ivy League}\} \cup \{\text{Stanford, MIT, Caltech,}\ldots\} >> \{\text{Everywhere else}\}$.  In particular, I don't think your points 1, 2 or 3 are true in general.

One of the things I've noticed in comparing different schools is that it may appear that, say, Harvard or Princeton has much more difficult courses than X school, but its largely a function of what level the course is at at.

I go to one of the better mathematics schools in Canada (mainly for financial reasons, as I did get into one of those big name schools but ended up choosing not to attend).  As a freshman, I'm taking a mix of 2nd, 3rd, and 4th year math courses.  Now, I suspect that these are basically 2nd or (lower level) 3rd year math courses at a 'better' university, but most of them are taught at a rigorous level.

I compared the syllabi and the problemsets ofmy Math 320 (Analysis II) course and Math 25 at Harvard.  They are pretty much of equal difficulty and scope, except that there's no algebra component in my analysis course.  Of course, I'm taking separate algebra courses, so that solves that problem.  Now, there's no comparable course to Harvard's Math 55 anywhere, but there's really not much you can do about that.

Unfortunately, I've had to take certain required math courses that were a pain in the butt, e.g, plug and chug multivariable calculus, linear algebra, discrete math.  But I'm satisfied that the 'real' math courses I am taking cover the same material in the same breadth as they would be covered at just about any school.

On the downside, I don't have access to other people that are as smart or smarter than me, and I'm the youngest person in all of my classes (there are people 4 years old er than me in my complex analysis course).  I also am not impressed by the caliber of my fellow students who are roughly the same age as me, so it curtails my desire to take courses in the humanities and so on, which is a shame, because I really enjoy English, among other things.

So there are benefits and disadvantages.  But in the final analysis, I really didn't feel compelled to pay roughly 80,000 for my undergraduate education when I get paid to go to school(on the order of 40,000 over four years, which isn't huge, but it is at least significant , I have research opportunities in the summer (I'll be doing some work on functional analysis), and so on.

[quote="kentliu"]well, point 1 and 2 could be argued since they are subjective, but point 3 is not. Private college do indeed accept less people and has higher faculty:student ratios.. which is very important. A student should know the teacher well enough to not be shy to ask a question or make a comment.[/quote]

If one wants an emphasis on teaching, good faculty:student ratio, good teacher interaction, and small classes, one should go to any of the number of schools that fall less into either of the two categories we're discussing.  These are the strengths of smaller liberal arts colleges (and perhaps Brown, as well, which seems as if it is in the Ivy League by accident), although some of those colleges certainly fall into the "elite institutions" category (Amherst), even if no one on this site seems interested in them.

Anyhow, that little aside aside, having lots of professors around does not equate to more access.  It has much more to do with the individual teacher.  One of my two math teachers this semester (who is also an important Dean -- the kind of person you would expect to be most intimidating) learned all 70 of his students names in the first few weeks and has been exceptionally welcoming.  In my other math course, the professor might not be able to connect my face and name, doesn't respond to e-mails, and practically runs out of the class.  There are 15 people in that class, but I've never spoken with the professor face-to-face.  So, even if all of my classes were 15 people or smaller, if all my teachers were like the second, it owuldn't matter.  There's more to do with the atmosphere of the school than there is with the student:teacher ratio in determining this kind of thing.  I imagine, for example, that the Honors Math program (which is the program most people who talk seriously about their chances of getting into Harvard, MIT, etc. for math would go to if they went to UMich) has very good access to the department -- it's designed to be special in a variety of ways.  Even at a school of 25,000 or whatever number of students, you'd come out okay.  (And, if you happen to be a Michigan resident, you'd come out with $75,000 to spare.)

[quote="kentliu"]
4. "better" peers.. many your friends are going to be rich if you go to a prestigious college.. which means if you ever need a job, favor, or anything, they could easily provide it to you.. im not saying u should minipulate them, im just saying the way it is.
[/quote]

*laughs* funny that this falls under one of the reasons i would NOT want to attend a "prestigious college"... maybe this doesn't apply to most, or even many of you, but my family cannot afford to send me to a "prestigious college" (ie- one that costs 25,000+ tuition). the majority of my peers are in approximately the same financial boat as i am. i think that it would be extremely intimidating and, frankly, scary to a point, to be thrown into a group of people who, for the most part, can afford pretty much anything they want with "daddy's money." i have worked extremely hard since i was legally able to in order to gain the experience as well as the money i needed to make college happen. and again, i think it would be extremely... [b][i]odd[/i][/b] to be in the presence of so many people who could not relate at all to my personal experiences.

[quote="bleumoose"]i have worked extremely hard since i was legally able to in order to gain the experience as well as the money i needed to make college happen. and again, i think it would be extremely... [b][i]odd[/i][/b] to be in the presence of so many people who could not relate at all to my personal experiences.[/quote]

The admissions officers at the top schools want more applicants like you. Seriously, it is embarrassing at all the top schools that the applicants come so disproportionately from the richest families in America, and all those schools would like more hard-working people from less wealthy backgrounds to apply and attend. I felt the same way you do about the famous schools when I was in high school--why should I go to a school that charges almost as much as my family's whole income? But now I wish I would have applied (to be sure, I wouldn't have been a very well prepared applicant) and seen what happened in terms of admission, and in terms of financial aid offers. Try it, and see what happens.

You should remember that the prestigious universities are the ones with the money to make college affordable.  Harvard just recently made serious cuts in the cost of tuition for students whose families make less than $\$ 60,000 (and especially less than $\$ 40,000) and Princeton has eliminated student loans.  In New York, yearly state funding determines the costs of college and the availability of financial aid, and the result is that tuition can jump suddenly.  (This is especially true in the City University of New York, less so in the State University.  Every time budget-cutting happens, CUNY raises tuition $\$ 500 and many people are forced to drop out because there's not enough aid to cover the difference.)  That doesn't happen to, say, Cornell.  It's true that there is a middle range of less-elite, less-old, or less-well-endowed private colleges and universities that would be too expensive due to their inability to provide adequate financial aid, but many of the top-notch colleges can.  Also, you can sometimes bargain with them (although it's not easy to do): if you get into State U. and MIT, tell MIT that you can't afford it and see if they'll give you a better financial aid deal.  Worst-case scenario is that you end up going to State U. like you're planning to, anyhow.

As for the more, interesting, point you were making -- I don't know how odd you would find it.  I think that you will find people are not in fact all that different than you, except in their flexibility to spend.  Many people at Harvard have significantly greater income than my family and plenty of people have significantly less, but I don't think that plays a major role in interpersonal interactions.  I guess this is a personal sensitivity, though, so only you can really decide if yo ufind it uncomfortable.

Thanks for the responses!

I see. So while the atmosphere is certainly very encouraging for the highly-capable student (while they will always be self-motivated, friendly competition will always help.), the classes really do vary from prof to prof (so there's really some element of luck).. And better universities doesn't always equal better profs - a Nobel prize does ot equate into a good teacher.

Perhaps one substitute for an environment of highly capable peers is a forum where the highly capable can hang out. Obviously, one such forum is AoPS. Another is the Physics Forums. But when one needs a quick question, it's a disadvantage.

There is one way the highly capable student could benefit, I'm thinking, however. If the Ivy-calibre student decides to enroll in a state institution, this student could potentially stand out distinguished - perhaps this makes profs more willing to talk to the student. But that's highly an element of luck and cannot be guaranteed.

If my school doesn't enforce a 7th period on me, I might consider taking an EPGY Stanford course, which would then result in... (providing that I get a '5' on the BC Calc exam this year as a sophomore).

11th grade: Multivariable calculus
12th grade: Differential equations and maybe Complex Analysis

Which would then have me starting at some pretty complex math at the beginning of my undergrad year - and I would definitely finish undergrad math before my senior year in college. But are universities flexible in granting credit to Stanford EPGY courses?

And could the undergrad potentially take grad-level courses as an undergrad?

I'm taking graduate level courses as an undergrad, so yes, it is possible.

You should take Linear Algebra before you take a Diff Eq course though.  From what I've seen (and this suspicion is validated by their choice of textbooks) most of the EPGY courses save for maybe Topology and Real Analysis are not on the level of math course for elite students at an elite school.  This is similarly true of the Complex Analysis book they use (it's nominally the textbook used at my university, but our problem sets and most of the material are drawn from Ahlfors.)

For example, if you have taken a procedural multivariable calculus course, vector calculus course, or linear algebra course at a local university, it may teach you some useful skills but  comparable courses at an elite university will be far more theoretical and place only a modicum of weight on being able to calculate dozens of banal integrals, or eigenvalues, or whatever.

thanks to tokenadult and jbl for the self-esteem boost. will consider it. most definitely. as they say, you increase your chances by submitting an application, right? still worried about being in debt. number one rule of the family: DO NOT GET IN DEBT. EVER. but i digress.

[quote="Simfish"]There is one way the highly capable student could benefit, I'm thinking, however. If the Ivy-calibre student decides to enroll in a state institution, this student could potentially stand out distinguished - perhaps this makes profs more willing to talk to the student. But that's highly an element of luck and cannot be guaranteed.[/quote]
I think this is true -- many state universities have special programs to attract high-calibre students.  I know I got a bunch of letters from colleges due to my PSAT scores, and I know the U. Michigan has an Honors Math program, for example.

[quote]...11th grade: Multivariable calculus
12th grade: Differential equations and maybe Complex Analysis
Which would then have me starting at some pretty complex math at the beginning of my undergrad year - and I would definitely finish undergrad math before my senior year in college. But are universities flexible in granting credit to Stanford EPGY courses?
And could the undergrad potentially take grad-level courses as an undergrad?[/quote]
Schools vary greatly in flexibility with respect to graduate classes.  At Harvard, every undergrad can take math grad classes basically as soon as they want.  On the other hand, some schools do require their students to take introductory freshmen classes (Harvard does this, although I think they're relatively more rigid than most schools), so even having taken an advanced course in high school won't 100% push you up a level.  (Just for reference, I took multi, diff eq, and a graduate complex analysis course before coming to Harvard and I'll have basically duplicated all of that here by the end of next semester, probably.)  I think other schools are probably more rigid about the grad courses and less rigid about the freshmen courses, in general.  You can always go and talk to someone at the math department of schools that you visit (or e-mail them, if you don't visit) in order to find out what their views on these things are.

[quote="bleumoose"]thanks to tokenadult and jbl for the self-esteem boost. will consider it. most definitely. as they say, you increase your chances by submitting an application, right?[/quote]
Right-o -- if you won't be able to afford it, that's okay: it's happened to plenty of people before you, and smart individuals turn out perfectly well without going to an Ivy League school.  But there's no point in deciding you can't do it without at least giving it a shot.

[quote="ComplexZeta"]The complex analysis text used in the EPGY class (I'm assuming it's still Brown and Churchill) is a frequently used book in undergraduate complex analysis classes (although it isn't being used in the undergraduate complex analysis sequence here this year).

Under the right circumstances, it is possible for a undergraduate to take graduate classes. In math, I more or less take exclusively graduate classes.[/quote]

Brown and Churchill is not a suitable textbook for a true complex analysis course, inasmuch as its somewhat lacking in rigor.  Its fine for an introductory course to complex analysis and so forth, but it is particularly weak on applications to number theory and so on.

[quote="ComplexZeta"]Well, that's what an undergraduate complex analysis course is supposed to do. It is supposed to teach complex analysis, not number theory. Leave the rest for the graduate course. How is Brown and Churchill lacking in rigor? So what if they don't prove Picard's Theorem? It is not an important proof for undergraduates to know.[/quote]

Brown and Churchill does, however, include a large number of physics applications, and they, for the most part, leave out the epsilons and so forth.  They also don't have certain useful theorems, such as the homotopy version of Cauchy-Goursat.  It handles certain other things poorly, such as branches, its worth noting.

Maybe I overstated my case by a bit (I'm still a bit peeved that I shelled out 170 CDN for it), but I feel it to be a classic (my dad studied with the 2nd edition and now I'm using the 7th) more because of its accessibility than its actual depth.

These aren't necessarily just my opinions, either; my Complex Analysis prof strongly dislikes Churchill and Brown.  I think that its also partly terminology too; the first course involving complex numbers at my university is called Complex Variables and it covers the usual mappings, series, integrals, and residues; the second one is called Applications of Complex Analysis and it redoes the first one in rigor with either applications to number theory or physics, depending on prof.

That's basically the same as what we have. The undergraduate class is called complex variables (or introduction to complex variables or something like that), and the graduate class is called complex analysis (or introduction to complex analysis). They cover the same sorts of topics (at least for a while), but the graduate class is more in-depth. But I don't see much of a difference in the name between complex variables and complex analysis. This year, we are using [i]Theory of Functions of a Complex Variable[/i] by Markushevich (all three volumes) for the first two quarters of the graduate class. I'm not sure what the book is being used for the undergraduate class.

I see lots of $\epsilon$s in Brown and Churchill.

Branches are covered somewhat poorly in Brown and Churchill; that is true.

[quote="tokenadult"][quote="ComplexZeta"]How is Brown and Churchill lacking in rigor? So what if they don't prove Picard's Theorem? It is not an important proof for undergraduates to know.[/quote]

It would be interesting to know whether this is the response of undergraduate students at all schools, as a reality check on the answers being given in this thread to the original poster's question.[/quote]

The obvious answer to this, I guess, is that there aren't enough people here who have used Brown and Churchill to comment.  Obviously Simon and I have some experience with it, JBL might (through work in high school, not university), and I can't think of anyone else in the AoPS community who would have used a text like that.

As for EPGY, based on the description of their courses and of the math courses listed on their website, I suspect that for the most part the ones offered through EPGY are the equivalents of the non-math major math courses at Stanford; courses where there is still a lot of mathematical content but where ther target audience is a mix of engineers, physics, chemistry, and stats majors, and only the occasional math major.  I'm not saying that its not a good thing to have taken some procedural, plug and chug type courses in multivariable calculus and the like, but the courses would not mirror the typical math sequence for math majors at a school like Harvard or Princeton.

Of course, there's nothing wrong with that, but it's good to know.

Well, yesterday I ordered a copy of the textbook used at Harvard and Princeton for graduate complex analysis classes. (This is the text by Stein and Shakarchi.) I think it's probably a bit more advanced than Markushevich, but I'll see when it comes.

For the one-variable course I'm particularly fond of Complex Analysis in One Variable by Narasimhan.

I've only used Ahlfors.  This coming semester I may take the undergraduate complex analysis here which uses "Complex Analysis" by Theodore Gamelin, at least this semester.  I had thought that the graduate course here used Ahlfors, but it may depend on who's teaching.

Hm. I saw somewhere on the Harvard website that Stein and Shakarchi was being used, but that may have been in the past. Anyway, they definitely used it at Princeton last semester. Ahlfors is a really nice book, but it's way too expensive!


inequalities
induction
inequalities unsolved

Let $k$ and $l$ be two given positive integers and $a_{ij}(1 \leq i \leq k, 1 \leq j \leq l)$ be $kl$ positive integers. Show that if $q \geq p > 0$, then \[(\sum_{j=1}^{l}(\sum_{i=1}^{k}a_{ij}^{p})^{q/p})^{1/q}\leq (\sum_{i=1}^{k}(\sum_{j=1}^{l}a_{ij}^{q})^{p/q})^{1/p}.\]

Denote $r= \frac{q}{p}$ and $b_{ij}=a_{ij}^{p}$. Then $r \geq 1$ and the inequality is equivalent to:

$(\Sigma^{l}_{j=1}(\Sigma^{k}_{i=1}b_{ij})^{r})^{\frac{1}{r}}\leq \Sigma^{k}_{i=1}(\Sigma^{l}_{j=1}b_{ij}^{r})^{\frac{1}{r}}$.

We shall proceed by Mathematical Induction on $k$. For $k=1$, we have equality. For $k=2$ we have Minkowski's Inequality.

Suppose that for some $k \geq 3$ the inequality holds for $k-1$. Then by the induction hypothesis we have

$\Sigma^{k}_{i=1}(\Sigma^{l}_{j=1}b_{ij}^{r})^{\frac{1}{r}}\geq (\Sigma^{l}_{j=1}(\Sigma^{k-1}_{i=1}b_{ij})^{r})^{\frac{1}{r}}+(\Sigma^{l}_{j=1}b_{kj}^{r})^{\frac{1}{r}}$.

Using the induction assumption for $k=2$ with:

$B_{1}= \Sigma^{k-1}_{i=1}b_{ij}$, $B_{2}=b_{kj}$

we have

${(\Sigma^{l}_{j=1}(\Sigma^{k-1}_{i=1}b_{ij})^{r}})^{\frac{1}{r}}+(\Sigma^{l}_{j=1}b_{kj}^{r})^{\frac{1}{r}}\geq (\Sigma^{l}_{j=1}(\Sigma^{k}_{i=1}b_{ij})^{r})^{\frac{1}{r}}$.

Q.E.D.


Prove that:

If $ n\equal{}p_{1}^{a_{1}}p_{2}^{a_{2}}\dots p_{k}^{a_{k}}$ is a prime decomposition of $ n$, then $ n$ has $ (a_{1}\plus{}1)(a_{2}\plus{}1)\dots (a_{k}\plus{}1)$ divisors.

$ f|n\implies f \equal{} p_{1}^{m_{1}}p_{2}^{m_{2}}\dots p_{k}^{m_{k}}$ s.t. $ 0\leq m_{i}\leq a_{i}\forall i$

The set $ \{ 0, 1,\ldots , a_{i}\}$ has length $ a_{i}\plus{}1$, so we have $ a_{1}\plus{}1$ choices for $ m_{1}$, $ a_{2}\plus{}1$ choices for $ m_{2}$, ... $ a_{k}\plus{}1$ choices for $ m_{k}$, and so a total of $ \prod_{i \equal{} 1}^{k}(a_{i}\plus{}1)$ divisors.

Note that divisors here includes $ n$ itself and $ 1$.

Follow up: Find an expression for the sum of these divisors.

@Sludgethrower:
[hide]$ \prod_{i \equal{} 1}^{k}(\sum_{j \equal{} 0}^{a_{k}}p_{k}^{j})$

because basically you're adding up all the possibilities for taking different numbers of each prime

I hope I didn't make a typo.[/hide]

Yes...but you can simplify it further...

I stole this out of [i]Elementary Number Theory - Clark[/i]:

For
$ n\equal{}p_{1}^{e_{1}}p_{2}^{e_{2}}...p_{r}^{e_{r}},r\geq 1$,

$ \sigma (n) \equal{}\left(\frac{p_{1}^{e_{1}\plus{}1}\minus{}1}{p_{1}\minus{}1}\right)\left(\frac{p_{2}^{e_{2}\plus{}1}\minus{}1}{p_{2}\minus{}1}\right) ...\left(\frac{p_{r}^{e_{r}\plus{}1}\minus{}1}{p_{r}\minus{}1}\right)$

Where $ \sigma (n)$ is the sum of the divisors.


A group of Mathcampers went on a 3.5-hour hike on Mt. Rainier. In any continuous one-hour period during their hike, they covered exactly 2 miles. Does it follow that they hiked exactly 7 miles total? If not, what are the minimum and maximum distances they could have walked? [i]Prove your answer.[/i]

Are we allowed to post questions? People can still apply for the waitlist...

Yeah, I don't think we are... I'm pretty sure people can still apply.

Er,
yeah,
my friend has yet to send out her application lol.

btw,
do the problems change year to year?

The questions do change each year. In fact, you can find the questions from the last couple years online.

[hide]The minimum is 6 miles, and the maximum is 8 miles.

First, note that since the group covers two miles in every continuous one-hour period, the group covers two miles in each of the first three hours, and thus six miles in the first three hours. Now, in the last half hour, the group must cover at least zero miles (I was told to assume that the group is never moving backward), and at most two miles, since the last half hour is part of the last hour, a continuous one-hour period in which exactly two miles are covered. Thus, the group travels at least 6 and at most 8 miles.

It is left to show that for all $ 6\le d\le8$, it is possible for the group to travel exactly $ d$ miles. We construct such a scenario. Break up the 3.5 hour period into seven half-hour periods, and have the group going $ 2(d\minus{}6)$ miles per hour at all times during the each of the first, third, fifth, and seventh periods, and $ 2(8\minus{}d)$ miles per hour during each of the second, fourth, and sixth periods. The total distance traveled will be $ \dfrac{8(d\minus{}6)\plus{}6(8\minus{}d)}{2}\equal{}d$ miles. Note that any continuous one-hour period consists of half the time in an odd-numbered period (going $ 2(d\minus{}6)$ miles per hour), and the other half in an even-numbered period (going $ 2(8\minus{}d)$ miles per hour). Thus, the total distance it travels in this period is $ \dfrac{2(d\minus{}6)\plus{}2(8\minus{}d)}{2}\equal{}2$ miles. Hence, as $ 0\le d\minus{}6,8\minus{}d\le2$ for all $ 6\le d\le 8$, all these $ d$ are achievable, and we're done.[/hide]

[hide]
\item Let $ f(t)$ be the distance in miles covered after $ t$ hours.  Then what we are given in the problem translates to $ f(t+1) = f(t) + 2$ for $ 0 \le t \le 2.5$.

Thus,
\begin{eqnarray*}
f(3.5) &=& f(2.5) + 2 \\
&=& f(1.5) + 4 \\
&=& f(0.5) + 6
\end{eqnarray*}

Now, note that $ f(0)=0$ and $ f(1) = 2$.  Assuming that the hikers never walk backwards, we must have $ f(0) \le f(0.5) \le f(1)$, so $ 0 \le f(0.5) \le 2$.  Thus, we have \[ 6 \le f(3.5) \le 8.\]

Clearly, the lower bound is possible if $ f(0.5) = 0$, so that for the first 30 minutes of each hour, the hikers rest, and for the second 30 minutes of each hour, the hikers walk at 2 miles per hour.
Finally, the upper bound is possible if $ f(0.5) = 2$, so that for the first 30 minutes of each hour, the hikers walk at 2 miles per hour, and for the second 30 minutes of each hour, the hikers rest.
[/hide]

Heh.  I'm gonna be compulsive now.

[quote="CatalystOfNostalgia"]
It is left to show that for all $ 6\le d\le8$, it is possible for the group to travel exactly $ d$ miles.
[/quote]

[quote]If not, what are the minimum and maximum distances they could have walked?[/quote]

not required

[quote="archimedes1"]
Clearly, the lower bound is possible if $ f(0.5) \equal{} 0$, so that for the first 30 minutes of each hour, the hikers rest, and for the second 30 minutes of each hour, the hikers walk at 2 miles per hour.
[/quote]

4

[quote]In any continuous one-hour period during their hike, they covered exactly 2 miles.[/quote]


In right triangle ABC, the lengths of its legs CB and CA are [i]a[/i] and [i]b[/i], respectively. Find the distance from point C to the point of incircle which is the closest to point C.

Note that this point is the intersection of $ \widehat{ACB}$ bisector and the incircle(and use pythagora's)

 :)


Let $a,b,c>0$.
Prove that:
$\sum \frac{ab+ac-bc}{b^2+c^2}\ge \frac{3}{2}$.

I have seen this inq in the forum:

We can say that $ab+ac-bc\le ba+bc-ac\le ca+cb-ab$ and $b^2+c^2\ge c^2+a^2\ge a^2+b^2$, becouse those are equivalent to $c\ge b \ge a$. Now by Cauchy in Engel form
$\frac{b^2+c^2}{ab+ac-bc}+\frac{c^2+a^2}{bc+ba-ac}+\frac{a^2+b^2}{ca+cb-ab}\ge\frac{6(a^2+b^2+c^2)}{ab+bc+ca}\ge 6$

Beat, are you sure you solved the right problem? ;)

See http://www.mathlinks.ro/Forum/viewtopic.php?t=26102 anyway...

  darij

I said: i have seen this inq which is not as hard as this but it is similar! Nevermind..


Pascal\u0027s Triangle
combinatorics proposed
combinatorics

We obtain it by summing the pascal triangle (11, 121, 1331,...) with this...
2  4
2  6  4
2  8  10  4
2  10  18  14  4
2  12  28  32  18  4
Triangle above is obtained by summing the identical triangle like this...
1  2
1  3  2
1  4  5  2
1  5  9  7  2
1  6  14  16  9 4
While this is obtained by summing pascal triangle (11,121,1331,...) with this
0  1
0  1  1
0  1  2  1
0  1  3  3  1
0  1  4  6  4  1
And now it's easy to find that the desired formula is $ \frac{(n)(n\minus{}1)(n\minus{}2)(n\minus{}3}{8}\plus{}\frac{(n\minus{}1)(n\minus{}2)(n\minus{}3)}{3}$.


If all alligators are ferocious creatures and some creepy crawlers are alligators, which statements must be true?

All alligators are creepy crawlers.
Some ferocious creatures are creepy crawlers
Some alligators are not creepy crawlers

Is there any method to solve these problems? I've heard you can use a venn diagram or something of the sort...

All we can deduce are the following (bottom diagram; do you see why?):

From this, we see that statement one can be false: it is possible that all alligators are creepy crawlers, but not necessary, as seen in the second diagram.

Statement three need not be true either, as shown in the bottom right diagram.

However, $ \boxed{\text{statement two}}$ must be true, since there are definitely some alligators that are creepy crawlers and all alligators are ferocious creatures, so some alligators need to be both creepy crawlers and ferocious creatures.

Hmm, well by just thinking logic it makes sense, but by the rules of logic, the contrapositive of given statements must be true. Contrapositives are switched and negated.

Givens: If all alligators are ferocious creatures and some creepy crawlers are alligators.

Contrapositives:
All ferocious creatures are not alligators OR Not all ferocious creatures are alligators (that one makes more sense).

Some alligators are not creepy crawlers.

And that's the third one! Hmm, by just thinking logic, the second one does make sense. Whatever, I'm sticking with my contrapositives.

The second statement must be true though. If all alligators are ferocious, and you choose at least one alligator to be creepy crawly, that means that alligator, which is ferocious, is a creepy crawly.

Do contrapositives even work with this kind of stuff? A contrapositive says from A -> B one can infer ~B -> ~A. But what's A in this case? What's B?

Buzzer11 is right. A and B aren't well defined here--for example, "B" has two possibilities, as shown in my diagram, each yielding different answers.


I was wondering why people on FTW show up as in the forum index, on the list for "Who is Online" page. This is obviously not important at all, but I was just wondering.

The "Who is Online" is for the forum, so those on FTW are just on the forum index (usually).


Let n be a power of 2 and x_1,...,x_n real numbers with sum 0. Then
 sum sin (2x_i)<=(2ctg (pi/n))(sum (sinx_i)^2).

[i]Gazeta Matematica, 1993, proposed by Vasile Cartoaje[/i]

I have tedious proof.

Using identity sin2xi=1/2(1-cos(2xi)) we transform initial inequality to the following
\sum cos(pi/n-2xi) \leq n*cos(pi/n).

Now we will prove statement:
if \sum yi=(2k+1)pi then \sum cos(yi) \leq n*cos(pi/n).

We can WLOG suppose 0 \leq yi < 2pi (since cos(yi+2pi)=cos(yi)).

Suppose there are yi and yj such that 0< yi \leq pi and pi \leq yj < 2pi. Replace these values to
   *) 0 and yi+yj (if yi+yj \leq 2pi)
  **) yi+yj-2pi and 2pi (if yi+yj>2pi).
The sum cos(yi)+cos(yj) have increased.
After some number of such replacements we may sure all yi \in {0,2pi} or lie in one of [0,pi] or [pi,2pi].

Suppose WLOG yi \in [0,pi] for all i=1,2,3,...,n.

a) Suppose [pi/2,pi] contains two or more of yi ==> \sum cos(yi) \leq (n-2) \leq n*cos(pi/n) (because cos(x)>1-2x/pi).
b) Suppose [pi/2,pi] contains yk only. Let yl \in [0,pi/2]. Replace yl and yk to
   *) yl+yk-pi/2 and pi/2 (if yl+yk \leq pi)
  **) yl+yk-pi and pi (if yl+yk > pi)
The sum cos(yl)+cos(yk) have increased.
If new yk equals pi then \sum cos(yi) \leq (n-2).
If new yk equals pi/2 then \sum cos(yi) \leq  n*cos((2k+1)pi/n) /*Jensen's inequality*/ \leq n*cos(pi/n).
c) Suppose all yi \in [0,pi/2] ==> \sum cos(yi) \leq  n*cos((2k+1)pi/n) /*Jensen's inequality*/ \leq n*cos(pi/n).

Is it correct?

I will check the solution tomorrow, but I'm convinced it is true.

Well, I checked the solution and all I can say is: you're fantastic! It is splendid and the inequality is very difficult.


trigonometry
probability
summer program
Mathcamp
symmetry
ceiling function
AMC

This is to complement the complex numbers and trig marathon started by PenguinIntegral. More practice in counting/probability problems is needed, so let's get to it. I'll start off with an easy one.

In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible?

[quote="rnwang2"]This is to complement the complex numbers and trig marathon started by PenguinIntegral. More practice in counting problems is needed, so let's get to it. I'll start off with an easy one.

In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible?[/quote]

[hide="uhhh..."]$\frac{\binom{10}{5}}{2}=\boxed{126}$

not exactly intermediate[/hide]

somebody else can post the next one.

[quote="anirudh"]Bob and four friends have to make themselves into groups of three. How many distinct groups are possible?[/quote] How can 5 people make groups of 3? :maybe:
Again, not exactly intermediate.

Ok. I think this should be considered intermediate.

Three players are each dealt, in a random manner, five cards from a deck containing 52 cards. Four of the [i]dealt[/i] cards are aces. What is the probability that at least one person receives exactly two aces in their five cards?

EDITED

Here's a problem...from the MathCamp qualifying quiz, it's #1 and I could do it so it isn't that difficult.

How many ways can we color each small square in an n by n square grid either black or white so that it has lines of symmetry on its vertical and horizontal axes and its two long diagonals?

Let $k=\lceil \frac{n}{2}\rceil$.
For every grid with size k x n exists grid with size n x n, which has symmetry on its vertical axes.
But because the n x n grid must have also horisontal axes symmetry, the k x n grid also should have.
but there's a bijection between the k x n grids with this symmetry and all the k x k grids.
But the original grid must have diagonal symmetry too, and after some observations, we get that then the k x k grid sould also have diagonal symmetry.
Cut the k x k grid by one of it's diagonals, by leaving all the diagonal boxes to one of the triangles. That triangle will have 1+2+...+k boxes.
get this triangle. It chould look like this:
1000...00 ->k
0100...0
....
000...01
...
000
00
0
Now, this triangle to have symmetry means that the two triangles above and below the "1"-line should be symmerical. So it's enough to count the points in one of these triagles, including the ones.
If k is odd, it will look like this:
0
00
....   0
....   00
....   0
00
0
|
V
k
, which makes exactly $(\frac{k+1}{2})^{2}= \frac{1}{4}(k+1)^{2}$ boxes.
If k is even, then the triangle will be:
0
00
....   0
....   0
00
0
|
V
k
So we're looking for the sum:
$\sum_{i=1}^{\frac{k}{2}}2i = \frac{1}{4}k(k+2)$.

Let's see when \lceil \frac{n}{2} \rceil will be even:
$2q = \lceil \frac{n}{2}\rceil$
$2q \geq \frac{n}{2}> 2q-1$
Multypying by 2:
$4q \geq n > 4q-2$, so n = 4q or 4q+3.
And, of course, if n = 4q+1 or 4q+2, k will be odd.
now the answer is:
$\begin{array}{l}2^{\frac{1}{4}k(k+2)}\mbox{, if }n=4q \mbox{ or }n=4q+3\\ 2^{\frac{1}{4}(k+1)^{2}}\mbox{, if }n=4q+1 \mbox{ or }n=4q+2 \end{array}\mbox{, where }k=\lceil \frac{n}{2}\rceil \mbox{ and }q\mbox{ is integer.}$

[quote="rnwang2"]Is this problem too boring?

[hide="Solution"]15 cards are dealt from the pack, 4 of which are aces and 11 of which are other cards.

With 4 aces and 3 players we could have

$\Rightarrow$ Player 1: 1 ace, Player 2: 1 ace, Player 3: 2 aces
$\Rightarrow$ Player 1: 2, Player 2: 2, Player 3: 0
Player 1: 1, Player 2: 3, Player 3: 0
Player 1: 4, Player 2: 0, Player 3: 0

(or any permutations of these).

The distributions shown with arrows give at least one player exactly 2 aces.

Probabilities of the first arrowed distribution are

$\frac{3\cdot\dbinom{4}{1}\cdot\dbinom{11}{4}\cdot\dbinom{3}{1}\cdot\dbinom{7}{4}\cdot\dbinom{2}{2}\cdot\dbinom{3}{3}}{\dbinom{15}{5}\cdot\dbinom{10}{5}\cdot\dbinom{5}{5}}= \frac{50}{91}$.

Probabilities of the second arrowed distribution are

$\frac{3\cdot\dbinom{4}{2}\cdot\dbinom{11}{3}\cdot\dbinom{2}{2}\cdot\dbinom{8}{3}\cdot\dbinom{5}{5}}{\dbinom{15}{5}\cdot\dbinom{10}{5}\cdot\dbinom{5}{5}}= \frac{20}{91}$.

Total probability $= \frac{(50+20)}{91}= \boxed{\frac{70}{91}}$[/hide]

Someone please post another problem.[/quote]

You said in the problem that 4 of the deck cards are Aces, not four of the dealt cards.  That's why the probability you got is so high.

Since no one has posted a problem...

Two teams A and B play a sequence of games against each other in the world series of baseball, and the first team who wins a total of 4 games becomes the winner. If the probability team A wins any particular game against team B is $\frac{1}{3}$, what is the probability that A wins the world series?

ya thats what i thought so i got this...not sure if its correct
[hide]
$\binom{6}{3}\cdot\left(\frac{1}{3}\right)^{4}\cdot\left(\frac{2}{3}\right)^{3}$
** Series goes until 7 games
$\binom{5}{3}\cdot\left(\frac{1}{3}\right)^{4}\cdot\left(\frac{2}{3}\right)^{2}$
** Series goes until 6 games
$\binom{4}{3}\cdot\left(\frac{1}{3}\right)^{4}\cdot\left(\frac{2}{3}\right)^{1}$
** Series goes until 5 games
$\binom{3}{3}\cdot\left(\frac{1}{3}\right)^{4}\cdot\left(\frac{2}{3}\right)^{0}$
** Series ends in 4 games

When u sum them up u get:
$\boxed{\frac{119}{3^{6}}}$

P.S. i may have made an error..please let me know
[/hide]

yup thats what i did..so is my final answer right..like i may have amde an error..

i guess somebody else can post the next problem (if this one happens to be solved)

[quote="Fraxia"][quote="dasherm"][hide]
We have a string of length 7 of As and Bs, with at least 4 As.
(Instead of various strings of various probabilities, we have a bunch of essentially equal ones whose probabilities add up to the desired amount since the probability of AAAAXXX = the probability of AAAABBB+AAAABBA+...)

With 4 As:

$(\frac{1}{3})^{4}*(\frac{2}{3})^{3}*\binom{7}{4}=\frac{280}{3^{7}}$

With 5 As:

$(\frac{1}{3})^{5}*(\frac{2}{3})^{2}*\binom{7}{5}=\frac{84}{3^{7}}$

With 6 As:

$(\frac{1}{3})^{6}*\frac{2}{3}*\binom{7}{6}=\frac{14}{3^{7}}$

With 7 As:

$(\frac{1}{3})^{7}=\frac{1}{3^{7}}$

Adding these, we get $\frac{379}{2187}$[/hide][/quote]

 :maybe: I don't think that's how it's played. The series ends as soon as either team wins 4 games.[/quote]

[scathing sarcasm]And yet, somehow, it's the right answer (.173296).[/scathing sarcasm]  :wink:

We can just pretend that they finish playing out the series to all 7 games.  Instead of having a single string of like AAAA, for example, with some weird probability, we have AAAA and then all possible outcomes afterwards (AAA, AAB, ABA, ABB, ...).  These outcomes add up to the probability of AAAA.  I just thought it would be easier to have all strings be the same length than different lengths.

Oops, should have read your explanation in the beginning  :blush:

Off the top of my head; I hope there's a nice, not-way-too-complicated solution to this. If there isn't, drop it and post another problem  :P :

Alex and Alexis play a set of tennis. Each has a 1/2 probability of winning each point, regardless of the server. There are no deuces (if a game reaches a deuce, it is a scratch and is redone - points won as well as the game itself are stricken from the record), and it is considered a tie if they reach 6-6. What is the probability that Alexis wins despite winning fewer points?

[quote="rnwang2"][quote="JavaMan"][hide]
Think of it as a string. Fix the last A. Before the last A, there are always 3 other A's, being mixed with 0, 1, 2 or 3 B's.

$\binom{6}{3}\cdot (\frac{1}{3})^{4}(\frac{2}{3})^{3}+\binom{5}{3}\cdot (\frac{1}{3})^{4}(\frac{2}{3})^{2}+\binom{4}{3}\cdot (\frac{1}{3})^{4}(\frac{2}{3})^{1}+\binom{3}{3}(\frac{1}{3})^{4}(\frac{2}{3})^{0}$
[/hide][/quote]
Yes, this is correct.

[b]SM4RT[/b]: This result $\approx 0.1732$ does not match your answer $\approx 0.1632$. I'm sure it was just a little computation error, though.[/quote]

Indeed it was..i think the answer is
[hide]
$\frac{379}{3^{7}}$
[/hide]

[quote="PenguinIntegral"][quote="bpms"][quote="The Zuton Force"][hide="Comment"]
There are 15 cards dealt out in total out of 52, so the expected value of aces dealt out is 4(15/52)<2. It seems somewhat counterintuitive that the probability that ONE person with EXACTLY 2 aces is so high... I'm going to work this out myself sometime just because xD. In the meanwhile...
[/hide]

A small electronic gadget has an 8 pixel by 8 pixel display pixel display. The manufacturer attaches a "dead-pixel" policy to the device - if there are at least 4 dead connected pixels or at least 8 dead pixels in total, then the device can be either returned or exchanged.

Pixels are directly connected if they are adjacent - that is, they share a side (corners don't count). In addition, pixels are transitively connected (i.e. if A is connected to B and B is connected to C then A is connected to C).

If each pixel has a 1/16 chance of being dead, then what is the probability that the product can be replaced/returned?[/quote]
Perhaps trig and this should be stickied.
[hide] Perhaps a solution with inclusion exclusion is the simplest way:
Using casework there is a $\frac{297}{65536}$ chance of four pixels in a row.

The probability of 8 pixels anywhere is $\frac{\binom{64}{8}}{16^{8}}$ which is $\frac{3 \cdot 899 \cdot 3599 \cdot 57}{2^{29}}$. I'm starting to think I'm doing this the wrong way.

I'll tackle the probability of both happening in a minute.[/hide][/quote]
Wrong, they don't have to be in a row :P  I think you just need to tackle all the shapes and use PIE.

BTW, this is an example of a good problem. Very simple, yet very complex. I'm still hoping there is an elegant solution....[/quote]
I know they don't have to be in a row. I counted for all quadronomoes (sp?) . The both case is starting to look ugly, can someone tackle it. After that this is just going to boil down to PIE.

[quote="mdk"]v, w are roots of z^1997 = 1 chosen at random. Find the probability that |v + w| >= root(2 + √3).[/quote]

[url=http://tmp.mathideas.org/problems/2006/8/3.pdf]Solution[/url]

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent of to slay a troublesome dragon. What is the probability that at least two of the three knights sit next to each other?

This is probably wrong.

[hide]We can find P(no knights sit next to each other) and subtract from 1.

We choose our three knights, $k_{1}, k_{2}, k_{3}$, $\binom{25}{3}$.
We set knight one, $k_{1}$ on a fixed point at the head of the round table.
Then only the permutations of $k_{2}$ and $k_{3}$ matter.
So far we have $2\binom{25}{3}$.
Then we can choose 3 knights at random from the remaining 22 and place them in some order, one to the right of each of the knights - i.e. pairing these with the previous three. That's $\binom{22}{3}$.
So we're just putting boxes in bins, making sure no bins are empty.

There are 19 remaining, so there are 3 pairs of knights and 19 knights to put between them.

So $\binom{m+n-1}{m-1}= \binom{21}{2}$.

So the answer is $1-2\binom{25}{3}\binom{22}{3}\binom{21}{2}$ over $24!$. That looks way too complicated.[/hide]

I believe the answer should be between 0 and 1.

Yeah, It should.

[hide="Please let this be right..."]Let's find the probability that no knight that is sent is sitting next to another knight that is being sent.

There are $\binom{25}{3}$ choices without restrictions.

Let's say that one of the knights is King Arthur, who is always sitting at the head(how convinient). Let's say that the second knight is two seats away from the King. There's 2 ways to do that. Then there are 20 ways to choose the third knight.

Let's say that the second knight is 3 seats or more away from the king. There are 2 ways to do that. Then there are 19 ways to pick the third knight.

So, altogether, there are 20*2+19*3=97 ways to pick the knights so that no two are sitting together.

$\frac{97}{\binom{25}{3}}=\frac{97}{2300}$

$1-\frac{97}{2300}=\frac{2203}{2300}$

Siince 97/2300 is in lowest terms, 2203/2300 is in lowest terms.[/hide]

That poor king.... :rotfl:

You're making this too complicated.

[hide]
There are $\binom{25}{3}=2300$ ways to choose three knights.

There are 25 ways to choose two knights that are adjacent to each other.  There are then 23 knights to choose to be the third.

But we have overcounted the case where all three knights are adjacent.  Again there are 25 ways to do this.

So the answer is $\frac{25*23-25}{2300}=\frac{550}{2300}=\frac{11}{46}$.
[/hide]

Wasn't this from AIME '83 :)

[hide]
Pr(2 sit next to each other) + Pr(3 sit next to each other) = 1 - Pr(None sit next to each other)

We'll find Pr(None sit next to each other) by considering three cases.

The first can be placed anywhere:
* The next two are placed 2 spots away from the first
* One of the next two are two spots away from the first, and the last isn't
* None of the next two are placed two spots away from the first.

$1-(\frac{2}{24}\cdot\frac{1}{23}+\frac{2}{24}\cdot\frac{19}{23}+\frac{20}{24}\cdot\frac{19}{23}) = \frac{11}{46}$
[/hide]

I suppose I should post the next problem:

2003 University of Maryland Math Competition Part I #24

How many 7-digit sequences of 0's and 1's are there that have a block of 3 successive 0's but do not have a block of 4 or more successive 0's?

A bit easy, perhaps.

[hide]Casework on where the 000 is.

Case 1: 000----, ----000
The number adjacent to the 0's must be a 1, then the other 3 can be whatever, so $2(2^{3})=16$. However, note that we have counted 0001000 twice, so a total of $15$ for this case.

Case 2: -000---, --000--, or ---000-

The two numbers adjacent to the string of zeroes must be 1's, the other 2 can be whatever. $3(2^{2})=12$ for this case.

27 in total[/hide]

[quote="cincodemayo5590"][b]Problem #... 10-ish?[/b]
If a stick is broken at three points, what is the probability that the four pieces form a cyclic quadrilateral?[/quote]

Maybe you don't want the cyclic part? Though perhaps any four side lengths that can form a quadrilateral can also from a cyclic one.

[quote="cincodemayo5590"][b]Problem #... 10-ish?[/b][/quote]
#11.

@ paladin - Is it possible to prove your idea?

[quote="JavaMan"]Wasn't this from AIME '83 :)[/quote]
yes, it was.
[hide="a hint for the current problem"]A generalization of the TIT says that in an n-gon if the side lengths are $a_{1}$, $a_{2}$, ..., $a_{n}$ then
$\sum^{n-1}_{k=1}a_{k}> a_{n}$[/hide]
[hide="another hint"]Geometric probability[/hide]

[quote="cincodemayo5590"][b]Problem #... 10-ish?[/b]
If a stick is broken at three points, what is the probability that the four pieces form a cyclic quadrilateral?[/quote]

[hide="solution"]
let the stick have length $n$. in order to form a quadrilateral, the only thing that needs to happen is none of the four pieces have a length exceeding $\frac{n}{2}$ because then it can easily be proven that the sum of the lengths of the three shortest pieces exceeds that of the longest piece.

I like coordinates, so I am going to solve using 3-D space coordinates.

let $x$, $y$, $z$ be the lengths of three of the sticks. obviously, none of $x$, $y$, $z$ can exceed $n$, so we are bounded in the first octant by the space $x+y+z<n$, with intercepts at $(n,0,0)$, $(0,n,0)$ and $(0,0,n)$. By pigeonhole, only one of the pieces could possibly exceed $\frac{n}{2}$ in length. therefore, we have know that all of $x>\frac{n}{2}$ does not work (same goes for $y$ and $z$). each one of those plane cuts gets rid of one eighth of the tetrahedron formed by the original plane and the first octant. but we also have to make sure that $x+y+z>\frac{n}{2}$.

so we cut out a bunch four little tetrahedrons out of a total of 8 and get a probability of $\frac{1}{2}$.

remembering the problem just like this with the triangle, this answer makes sense because as more cuts are made in the stick, the less likely it is that one of them had a length over half of the length of the stick.
and if paladin's cyclic quad idea is correct, we are all set.[/hide]

Well can anyone provide a contradiction?

I guess I'll post a new problem  :lol:

[size=184]Problem 12[/size]

Consider $10$ people sitting around a circular table. In how many different ways can they change seats so that each person has a different neighbor to the right?

Since no one seems to be able to do this one...
Here's a new one (from this year's hmmt):
How many 5-digit numbers $\overline{abcde}$ exist such that all of the digits are the sum of the digits to their immediate left and right?

What do you do for the first and last letters? Only look at the digit to the right, and left, respectively?

(btw, I don't know if this was intentional, but the HMMT problem is slightly different, it only stated that $b=a+c$ and $d=c+e$)


Let $ABC$ be a triangle, and let $M$, $N$, $P$ be points on the segments $BC$, $CA$, $AB$. Prove that the area of triangle $MNP$ is greater or equal to the smallest of the areas of the triangles $ANP$, $BPM$, $CMN$.

[On the picture below, this means $S_2 \geq \min \left\{S_1,S_3,S_4\right\}$.]

[quote="Mathx"]Let $ABC$ be a triangle, and let $M$, $N$, $P$ be points on the segments $BC$, $CA$, $AB$. Prove that the area of triangle $MNP$ is greater or equal to the smallest of the areas of the triangles $ANP$, $BPM$, $CMN$.[/quote]

Let $ABC$ be the large triangle, and let $D,E,F$ be points on $BC,CA,AB$ respectively, and let $A',B',C'$ be the midpoints of $BC,CA,AB$, respectively. We want to prove that at least one of $AFE,BDF,CED$ has an area not larger than that of $DEF$.

If we can find a pair of segments among $(B'C',EF),(C'A',FD),(A'B',DE)\ (*)$ which do not intersect, then we are done. Then we will have a situation like the following one: both $E,F$, say, are on the same side of $B'C'$ as $A$, in which case the area of $DEF$ will be as large as that of $AFE$.

If, on the other hand, all three pairs in $(*)$ intersect, then the area of $DEF$ will be at least as large as that of $A'B'C'$ (this should be an easy exercise; just move the points $D,E,F$ around on $BC,CA,AB$ and see what happens), so the sum of the areas of $AFE,BDF,CED$ does not exceed three times the area of $DEF$, and we are, again, done.


Prove that A is a perfect square if and only if for every natural number n, at least one of the numbers (A+1) - A, (A+2) - A, ..., (A+n) - A is divisible by n.

- If A = m 2  is a square.
Let n be a positive integer.
Let A = qn+r be the euclidean division of A by n. Then, for i = 1,...,n the n numbers A+i are n consecutive integers. It follows that the numbers (A+i) 2  give each of values 0 2 , 1 2 , ...,(n-1) 2 mod[n] in some order. In particular, there exists i such that (A+i) 2  = m 2  mod[n], that is (A+i) 2  - A is divisible by n.

- Conversely, suppose that for each positive integer n there exists i in {1,...,n} such that (A+i) 2  - A = 0 mod[n].
If A = 1, then a is a perfect square...Now let's suppose that A  \geq 2.
Let p be a prime divisor of A, and t  \geq 1 be the exponant of p in the prime decomposition of A. Using the property for n = p (t+1), we deduce that there exists i in {1,...,p (t+1) } such that (A+i) 2 - A = 0 mod[p (t+1) ].
Since (A+i) 2 - A = A(A+2i-1) + i 2 , it follows easily that i 2 = 0 mod[p t ].
Thus A + 2i - 1 =/= 0 mod[p], so that A(A+2i-1) =/= 0 mod[p (t+1) . Therefore p (t+1) does not divide i 2 from which it follows that t is the exponant of p in the prime decomposition of i 2 . Then t is even.
It follows that each of the exponant in the prime decomposition of A is even, that is A is a square.

Pierre.


number theory
prime factorization
number theory unsolved

find all positive integers $ n$ such that
$ 4^n\plus{}n^4$ is a prime

[quote="tmy1018"]find all positive integers $ n$ such that
$ 4^n \plus{} n^4$ is a prime[/quote]
Obviously $ 2 \nmid 4$, so $ n \equal{} 2k\plus{}1$
Then $ 4^n \plus{} n^4 \equal{} 4 \cdot (2^k)^4 \plus{} n^4$ (Huh.. :maybe:  Who said Sophie Germain? :o )

solved
$ n \equal{} 1$ is the only one when $ n$ is not divisible by $ 5$
use $ mod 5$

but what about $ n\equal{}5k$

[quote="tmy1018"]solved
$ n \equal{} 1$ is the only one
use $ mod 5$[/quote]
Nice. So odd implies divisible by $ 5$ and even divisibility be $ 2$. Anyway.. To classic problem:P

lets also try this,for even $ n$ obviously its not a prime but for odd $ n$: $ 4^{n}\plus{}n^{4}\equal{}4^{n}\plus{}n^{4}\plus{}2\times2^{n}\times n^{2}\minus{}2\times2^{n}\times n^{2}\equal{}\left(2^{n}\plus{}n^{2}\right)^{2}\minus{}2^{n\plus{}1}\times n^{2}$
$ \equal{}\left(2^{n}\plus{}n^{2}\plus{}2^{\frac{n\plus{}1}{2}}\times n\right)\left(2^{n}\plus{}n^{2}\minus{}2^{\frac{n\plus{}1}{2}}\times n\right)$ which is composite for $ n>1$ and the only prime is for $ n\equal{}1$

[quote="sinajackson"]lets also try this,for even $ n$ obviously its not a prime but for odd $ n$: $ 4^{n} \plus{} n^{4} \equal{} 4^{n} \plus{} n^{4} \plus{} 2\times2^{n}\times n^{2} \minus{} 2\times2^{n}\times n^{2} \equal{} \left(2^{n} \plus{} n^{2}\right)^{2} \minus{} 2^{n \plus{} 1}\times n^{2}$
$ \equal{} \left(2^{n} \plus{} n^{2} \plus{} 2^{\frac {n \plus{} 1}{2}}\times n\right)\left(2^{n} \plus{} n^{2} \minus{} 2^{\frac {n \plus{} 1}{2}}\times n\right)$ which is composite for $ n > 1$ and the only prime is for $ n \equal{} 1$[/quote]
Which is the exact same as mine, no?

Mathias, his just looks better :P
Here is a thought.
What about $ 6^n \plus{} n^6$ ? Clearly $ n \equal{} 1$ works, which yields $ 7$. Are there any others? Maybe we can change this to $ a^b \plus{} b^a$ where $ a \plus{} 1 \equal{} prime$. This will always have a solution for $ b\equal{}1$, but any others?

mod 7 -> $ n^6 \equiv 1 (mod7)$ $ 6\equiv \minus{}1 (mod7)$ ,, n must be odd ,  $ 6^{2k\plus{}1}\equiv \minus{}1 (mod7)$ :)

[quote="Pablo09"]$ n^6 \equiv 1 (mod7)$[/quote]
Unless $ 7 | n$.  This is exactly the "easy" approach that fails for the original problem.  The following can be readily established in order for $ 6^n \plus{} n^6$ to be prime:

- $ n \equiv 1, 5 \bmod 6$
- $ n \equiv 0 \bmod 7$
- $ n \equiv 0, 1, 4 \bmod 5$.

Unfortunately, this doesn't really help.  A Sophie Germaine-like factorization does not (to my knowledge) exist for this problem, so it is possible it might be much harder.

Edit:  The 'closest' cases so far: when $ n \equal{} 35$ the prime factorization is $ 9007 \cdot 18374762213 \cdot 10387041650411$ and when $ n \equal{} 49$ the prime factorization is $ 11117 \cdot 1211796729704717409420101354375941$.  It appears that it might be possible to control some fairly small factor, but I am not at present sure what form that factor might take.  The numbers satisfying the above conditions rapidly become too large to compute with, though.


If $m,n \in N \neq 0$ and $ \sqrt{23} - \frac{m}{n} > 0$ then prove that:
$ \sqrt{23}- \frac{m}{n} > \frac{ \sqrt{6}-1}{mn}$.

cheers! :D :D :cool:

Let k =  \sqrt 6 - 1.
- If 0 < m/n  \leq (k \sqrt 23)/(5-k) then :
 \sqrt 23 - m/n  \geq (5 - 2k) \sqrt 23/(5 - k) = 2,83.... > k  \geq k/mn.

- If k \sqrt 23/(5-k) < m/n <  \sqrt 23 then 5m > k(n \sqrt 23 + m).   (1)
Moreover, 23n 2 - m 2 > 0.
In another hand, direct checking shows that for each integer a, we have a 2 =/= -1,-2,-3,-4 mod[23]. It follows that 23n 2 - m 2  \geq 5.
Thus : n \sqrt 23 - m  \geq 5/(n \sqrt 23 + m) > k/m from (1).
That leads to \sqrt 23 - m/n > k/(mn).

Pierre.


algebra
polynomial
function
geometry
3D geometry
tetrahedron
inequalities

Prove that one cannot embed the infinite binary tree in $ \mathbb{Z}^d$ for any $ d\geq 1$.

Is this just a consequence of the fact that the ball of radius $ n$ in $ \mathbb{Z}^d$ is of size $ P(n)$ for some $ d$-degree polynomial $ P$, while the ball of radius $ n$ in the infinite binary tree is of exponential size?  Or am I misunderstanding something?

It's a consequence of a lot of things, mostly related to how fast the boundary of the binary tree blows up :).  I was (am) just curious what sort of answers people would (will) give.

A finitary version:  what's the smallest $ d$ such that the complete binary tree of depth $ n$ embeds into $ \mathbb{Z}^d$ (hence into a Cartesian product of $ d$ path graphs)?

[b]Edit:[/b]  So JBL's argument, which I think is probably the most natural one, gives the following strategy for determining a lower bound on $ d$.  Define the obvious metric; the ball of radius $ n$ in the complete binary tree of depth $ n$ centered at the root has $ 2^{n \plus{} 1} \minus{} 1$ elements, but any ball of radius $ n$ in $ \mathbb{Z}^d$ has exactly

$ [x^n] \frac {(1 \plus{} x)^d}{(1 \minus{} x)^{d \plus{} 1}}$

elements.  This occurs because the ball of radius $ n$ about the origin is the set of solutions to $ |x_1| \plus{} ... \plus{} |x_d| \le n$ in the integers, equivalently the solutions to $ |x_1| \plus{} ... \plus{} |x_d| \plus{} y \equal{} n$, which have generating function

$ \frac {(1 \plus{} 2x \plus{} 2x^2 \plus{} ...)^d}{1 \minus{} x}.$

Computation gives that the coefficient of $ x^n$ is

$ \sum_{k \equal{} 0}^d {d \choose k} {n \minus{} k \plus{} d \choose d}$

and we know that if $ d$ works, this quantity must be at least $ 2^n \minus{} 1$.  The leading term, as a polynomial in $ n$, has degree $ d$ and coefficient

$ \sum_{k \equal{} 0}^d {d \choose k} \frac {1}{d!} \equal{} \frac {2^d}{d!}$

(the same coefficient we get by the geometric approximation as $ 2^d$ tetrahedrons); this is a large underestimate and actually gives a decreasing answer if $ d$ too large, but a complete binary tree of depth $ n$ embeds into $ \mathbb{Z}^n$ in an obvious way, so $ d \le n$; then we can try to approximately solve

$ \frac {(2n)^d}{d!} > 2^{n \plus{} 1}$

or, by Stirling's approximation,

$ \left( \frac {2ne}{d} \right)^d > 2^{n \plus{} 1}$.

[b]Edit:[/b]  See other post.

Okay, so instead of doing that I'm going to bound in the [b]correct[/b] direction; note that

$ \sum_{k \equal{} 0}^{d} {d \choose k} {n \minus{} k \plus{} d \choose d} < \sum_{k \equal{} 0}^{d} {d \choose k} {n \plus{} d \choose d} \equal{} 2^d {n \plus{} d \choose d} < \frac {(2n \plus{} 2d)^d}{d!}$

and we'd like to find a function $ d \equal{} d_{min}(n)$ such that this is less than $ 2^{n \plus{} 1} \minus{} 1$ for sufficiently large $ n$.  Stirling's approximation gives the inequality $ d! > \left( \frac {d}{e} \right)^d$ for sufficiently large $ d$, hence the new bound we'd like to satisfy (dropping another factor of $ 2$) is

$ \left( \frac {2e(n \plus{} d)}{d} \right)^d < 2^n$.

Taking $ d \equal{} \frac {n}{r}$ for some constant $ r > 1$ gives

$ 2e(r \plus{} 1) < 2^r$.

The positive real root is somewhere between $ r \equal{} 5$ and $ r \equal{} 6$, hence we can take $ d_{min}(n) \equal{} \frac {n}{6}$.  I'm curious how tight this estimate is.

[b]Edit:[/b]  Incidentally, there is a Putnam problem that is equivalent to showing that $ [x^n] \frac{(1 \plus{} x)^d}{(1 \minus{} x)^{d\plus{}1}} \equal{} [x^d] \frac{(1 \plus{} x)^n}{(1 \minus{} x)^{n\plus{}1}}$.  The generating functions proof is straightforward, but I don't know a bijective proof.


A rectangular piece of sheet metal with an area of 1200 in^2 is to be bent into a cylindrical length of stovepipe having a volume of 600 in^3.  What are the dimensions of the sheet metal?

I know you have to write and solve a system of equations for this problem.

One of the equations is h*2pi*r=1200  (y=h and x=2pi*r).
Will someone please help me find the other equation and solve the system of equation?

Let the sides of the retangle be $a$ and $b$. Then, we have $ab=1200$

$V=h \pi r^2$

$2\pi r=a$
$r=\frac{a}{2\pi}$
$h=b$

Then,
$V=b \pi \left(\frac{a}{2\pi}\right)^2$
$600=\frac{1200}{a} \ \pi \left(\frac{a}{2\pi}\right)^2$
$1=\frac{a}{2\pi}$
$\boxed{a=2\pi}$ inches and $\boxed{b=\frac{600}{\pi}}$ inches


I have to organise a Maths club for some 9-year-old kids, can anyone suggests some interesting (and more importantly simple) topics or activities which is suitable.

Thank you very much! :P

what are these kids capable of?
like what level math...

I would just make it a fun.. well at least, interesting in a way.. and  learning..class!  :lol:

As in, no one bored and sleepy.

less math talk

if they are working on fractions
jump rope(to say 180) and every 10 say how much you have done
that is fun
i still do it
but not to practice fractions
then you can do sets of roping and
average what you do in 1 minute
and find the mode, and median

and you can also do pushups in other bases(saying what you would if it was expressed in base 10)

but i am not sure the level of math or energy these kids posses

I am not sure how capable these kids are, to be honest, I do not expect them to know any maths at all except some basic calculations (like addition and subtraction).
I was actually thinking of some amusing activities like filling in a magic square, some chess puzzles... :roll:
Can you think of any amusing activities (eg. puzzles or tricks) which is quite simple for the little kids?

Thank you very much! :P

mind games and games with small rewards were really fun for me when I was 9 years old.

What type of mind games were that? Can you please give me some examples?

Thank you very much! :P

anything that is enjoyable and appeals to the typical 9-year old.

I can't think of any examples... it's been too long...  :(

It probably depends on the types of students and varies.

My 4th grade teacher surveyed us.

I think permutations and trigonometry will be too difficult for the kids, algebra may be a good one, but I do not really want to teach the kids anything at all. I am hoping to do something different to what they will do in class. :)
Can you think of any maths puzzles or tricks?

Thank you very much! :P

[quote="now a ranger"][quote="funcia"]small algebra problems are very fun when you are small

my pre algebra gave us all these cool things at the end of the year  that i just loved at the time
there was a little bit of conic sections(not with the terminology of course) and some trig(with most of the terminology, not in radians however)
and there was more but that has been some time.... :(
challenging problems are really fun to do when you are little...and now that i think about it probably always
if the kids haven't done permutations, and combinations i say try starting with that(don't say the n things and all say i have 3 letters how should i organise them into 3 spaces 3*2*1 etc. unless they get factorials)[/quote]

are you kidding? I would never have understood permutations or any of that trig stuff at age 9. They don't know pre algebra ....hello..... Many might not like it TOO challenging  :P[/quote]

the way my book said it it was wonderfully understandable
i wasn't 9 at the time but i knew 9 yr old people in pre algebra

about puzzles: do you mean those puzzles you get at the airport
i'll ask a 9 yr old kid for you


inequalities
calculus
integration
function
calculus computations

Suppose I have the following improper integral:

$ \int_0^1\frac{dx}{\sqrt[3]{x^2\minus{}x^3}}.$

I know this one converges, but having $ 0\le x\le1$ I can't find the proper inequality to show convergence. How you guys do that work on finding such inequality so that comparison test applies?

Look at $ 0$ and $ 1$ separately. Can you write a simple equivalent of the integrand near each end?

The integral is improper for two reasons: $ 0^\plus{}$ and $ 1^\minus{}.$ That is, the function being integrated is singular ("goes bad") at both ends of the interval. The conventions surrounding improper integrals demand that we treat each of those cases separately. That is, pick sum convenient point in the interior (might as well be $ \frac12,$ can't think of anything better), write the integral as $ \int_0^{\frac12}\ \plus{}\ \int_{\frac12}^1\ ,$ and then deal with each piece separately.

As $ x\to0,$ $ x^2\minus{}x^3$ looks like $ x^2$ since $ x^3$ is much smaller than $ x^3.$ So the function is comparable to $ \frac1{\sqrt[3]{x^2}}\equal{}x^{\minus{}2/3}.$ Since $ \int_0^{\frac12}x^{\minus{}2/3}\,dx$ is integrable, that means the left hand piece converges by limit comparison.

As $ x\to 1,$ $ x^2\minus{}x^3\equal{}x^2(1\minus{}x)$ looks like $ 1\minus{}x$ since $ x^2$ goes harmlessly to $ 1.$ Then, we're going to compare the function to $ \frac1{\sqrt[3]{1\minus{}x}}\equal{}(1\minus{}x)^{\minus{}1/3}.$ Since $ \int_{\frac12}^1(1\minus{}x)^{\minus{}1/3}\,dx\equal{}\int_0^{\frac12}u^{\minus{}1/3}\,du$ converges, the right hand piece converges by limit comparison, and the whole integral converges.

[I don't need to bother to ask fedja if this is what he meant - it's obvious to me that this is exactly what he intended. Always think about the size of functions; always look for size or rate of growth/decay. Straighforward size or rate of growth issues govern this problem.]

Actually $ \int_0^1x^{\minus{}2/3}(1\minus{}x)^{\minus{}1/3}\,dx$ is a beta integral and can be expressed in terms of gamma functions. But that doesn't seem to be what the problem asks for.


[quote="kstan013"](add a poll or move this to the round table) :D  :wink:[/quote]

Well you can't add a poll because ther are an infinite number of problems  :lol: .

Then this should certainly go to the round table!

True, you can't add a poll, there would be an infinite number problems. But I like finding where a proof is wrong. Like finding how 1 is not equal to 2.


If $a+b+c=3$ and $abc=1$ a,b,c positive reals prove that

$\sum\dfrac{ab}{(a^2+b)(a+b^2)}\leq\dfrac{3}{4}$

:rotfl:
a+b+c=3 and abc=1 and a,b,c are positive immediately implies a=b=c=1...

[quote="silouan"]Yes it is very easy but complete the solution.[/quote]

your question really has no problem?

Silouan, I hope you understood what Soarer wrote, but in case you didn't: since

abc = 1, a + b + c = 3 and a, b, c > 0

we have equality in AM-GM, so that a = b = c = 1.

Hence your "sum" is EQUAL to 3/4.

I find this to Crux Mathematicorum. Ok this is the solution

I don't think such a problem can appear in Crux.

I don't see any solution either, but I don't think we need another one ;)

Silouan, are you sure that this inequality appeared in that form? :?


What is the number of square inches in the area of a circle with
a diameter of 10 inches? Express your answer in terms of $ \pi$.

diameter=10 so radius = 5

the area of the circle is $ \pi r^2$ when r=radius.

answer : $ 25\pi$


Two integers are randomly selected from the set of integers greater than $ \minus{}5$ and less than 5. The first is used as the $ x$-coordinate of a point and the second is used as they $ y$-coordinate. The point with those coordinates is graphed. What is the probability that the point is in the second quadrant? Express your answer as a common fraction.

A little bit of simple thinking gives $ \fbox{1/4}$. The two variables have the same allowed range, and random picking is used, so the point is equally likely to be in any quadrant.

Except some of the points are not in any quadrant :wink:

For the point to be in the second quadrant, $ x<0$ and $ y>0$.  The probability of $ x<0$ is $ \frac{4}{9}$ and the probability of $ y>0$ is also $ \frac{4}{9}$, so \[ \frac{4}{9}\cdot \frac{4}{9}\equal{}\boxed{\frac{16}{81}}\]


trigonometry
geometry
angle bisector
trig identities
Law of Cosines
similar triangles

In triangle ADC angle bisector DB is drawn. If AB=3, AD=6, and CD=8, find BD.

[hide]
From the Angle Bisector Theorem, $BC=4$.

Use Stewart's Theorem to find $BD$.[/hide]

[hide]
Angle bisector theorem..
$\frac{3}{BC}=\frac{6}{8}\implies BC = 4$

Then use cosine rule twice, $x = BD$
$6^{2}= 3^{2}+x^{2}-2\cdot 3\cdot x\cdot \cos \theta$
$8^{2}= 4^{2}+x^{2}+2\cdot 3\cdot x\cdot \cos \theta$

Adding together..
$10^{2}= 5^{2}+2x^{2}\implies x = \boxed{\frac{5\sqrt{6}}{2}}$
[/hide]

[hide][img]6928[/img]

By the Angle Bis. Theorem, $\overline{BC}=4$.

Stewarts theorem states that in triangle $ABC$
[img]http://upload.wikimedia.org/wikipedia/commons/thumb/7/77/Stewart%27s_Theorem.svg/407px-Stewart%27s_Theorem.svg.png[/img]

$a (p^{2}+x y ) = b^{2}x+c^{2}y$

$AB=3$
$AD=6$
$CD=8$
$BC=4$

[img]6929[/img]

And then, by plugging into that theorem, the answer becomes \boxed{\dfrac{5\sqrt{6}}{2}}.[/hide]

or if you are totally cool and know the formula off the top of your head,

[hide]BD^2=6*8(1-[7/(6+8)]^2)[/hide]

Applying law of cosines on the bisected angle yields:

[hide]$cos{\theta}= \frac{3^{2}-6^{2}-x^{2}}{-2*6*x}= \frac{4^{2}-8^{2}-x^{2}}{-2*8*x}$

So $x = 6$.

I'm not sure how you guys got $\frac{5\sqrt{6}}{2}$.[/hide]

My bad. The answer at the end is supposed to be $6$. Miscalculation on my end :blush:

Exactly, the answer is $\overline{BD}=6.$

[hide="solution"]From the angle bisector theorem, BC=4. From Heron's theorem, the area of the triangle is $\frac{21\sqrt{15}}{4}$. We draw altitude DE to side AC.

$\overline{DE}*7=\frac{21\sqrt{15}}{2}$

$\overline{DE}=\frac{3\sqrt{15}}{2}$

We can find AE, and from that find BE, and from that find BD.

$\overline{AE}=\sqrt{36-\frac{135}{4}}=\sqrt{\frac{9}{4}}=\frac{3}{2}$

$\overline{BE}=3-\frac{3}{2}=\frac{3}{2}$

Since ADE and BDE are similar triangles, BD is 6.[/hide]

[quote="anirudh"][img]6928[/img][/quote]
Using anirudh's drawin'.

By angle bisector theorem, we can see easily that $\overline{BC}=4$. Now, by Stewart's theorem:

\begin{eqnarray*}\overline{BD}&=& \sqrt{\frac{{36 \cdot 4+64 \cdot 3}}{7}-12}\hfill \\ &=& \sqrt{\frac{{48(3+4)}}{7}-12}\hfill \\ &=& \sqrt{48-12}\hfill \\ &=& \sqrt{36}\hfill \\ &=& 6 \hfill \\ \end{eqnarray*}

Another theorem says that:

\begin{eqnarray*}\overline{BD}&=& \sqrt{\overline{AD}\cdot \overline{CD}-\overline{AB}\cdot \overline{BC}}\hfill \\ &=& \sqrt{6 \cdot 8-3 \cdot 4}\hfill \\ &=& \sqrt{48-12}\hfill \\ &=& \sqrt{36}\hfill \\ &=& 6 \hfill \\ \end{eqnarray*}

More direct  :rotfl:

[hide]using angle bisector and steward's theorem, 6[/hide]


linear algebra
matrix
induction
trigonometry
combinatorial geometry
linear algebra unsolved

Let $A$ be an $n\times n$ stochastic matrix and $\lambda$ an eigenvalue s.t. $|\lambda|=1$.
1. Show that $\lambda^{n!}=1$.
2. What is the minimal integer $k_{n}>0$ s.t. $\lambda^{k_{n}}=1$?

P.S: It might have already be posted on that forum, but I couldn't find it.

From a geometric point of view,it is easy to show that any
eigenvalue of a stochastic matrix $M \in M_{n}(\mathbb{C})$ with modulus $1$
is in fact a $k$-th root of the unity with $1 \leq k \leq n$.

Suppose that $v=(v_{1}, v_{2}, \ldots, v_{n}) \in \mathbb{C}^{n}$ is an arbitrary vector. Because $M$
is a stochastic matrix, if $w=(w_{1},w_{2}, .., w_{n})=Mv$, it is clear that every $w_{i}$ is
in the convex hull of the $\{v_{i}\}_{1 \leq i \leq n}$.
Then suppose that $v=(v_{1}, v_{2}, \ldots, v_{n}) \in \mathbb{C}^{n}$ is an eigenvector of $M$ corresponding
to the eigenvalue $\lambda$. Choose $i_{0}$ such that
$|v_{i_{0}}| = \max_{1 \leq i \leq n}|v_{i}|$. Then $|w_{i_{0}}|=|v_{i_{0}}|$ is in the convex hull of
the $v_{i}$. Hence $w_{i_{0}}=\lambda v_{i_{0}}=v_{i_{1}}$ with $1 \leq i_{1}\leq n$.
An induction shows that for any $k \geq 0$, there exists $1 \leq i_{k}\leq n$ such that
$\lambda^{k}v_{i_{0}}= v_{i_{k}}$
And because they are at most $n$ distinct possibilities for $v_{i_{k}}$, there exists
$k \leq n$ such that $\lambda^{k}=1$.

The same arguments shows that if $|\lambda| \leq \cos(\frac{\pi}{n})$ then there exists a stowhastic
matrix $M \in M_{n}(\mathbb{R})$ with eigenvalues $\lambda$.

Moreover, I think one can prove the following improvement:
if $Z$ is the number of zeroes in $M$ and $p$ is the integer part of $\frac{Z}{n-1}$ then
there exists $1 \leq k \leq p$ such that $M^{k}=I$.

PS: and of course I think the minimal $k_{n}$ that works is $GCD(2,3,..,n)$ because it clearly works, and one cannot  do better (because with permutation matrices for example )


geometry calculus AMC AIME MIT college Harvard

I am currently a freshman in high school. I thought i was doing well until, i learned that a really good person in my school didn't get into MIT. I don't have a lot so far, but i really need advice on what i should do in the next 2.5 years.

these are the high school courses i have taken/skipped so far:

Algebra I Honors (full year)
Geometry Honors (full year)
French I (full year)
French II (1 semester, we have only had 1 semester this year so far)
English 9 Honors (1 sem.)
Biology Honors (1 sem.)
Physics Honors (1 sem.)
Calculus AB-BC AP (1 sem.)

i am a 2-time AIME qualifier, and other than that, i have gotten 1st on many local math contests (2nd and 3rd on a couple)

as far as extra curriculars, i am an officer of our school math club?

i want to get into some ivy league school

preferably

MIT
HARVARD
PRINCETON

Zeroth of all, MIT (edit: snip) isn't an Ivy-league school.  (Most people seem to be totally clueless when it comes to the actual list of Ivy-league schools.  I've had people argue with me, for example, that Dartmouth isn't an Ivy.  Most people are also unaware that the Ivy League is a sports league.)

First of all, the fact that someone qualified was rejected from a good school isn't an indication that they weren't good enough.  After a certain point, the admissions decision is more about what applicants fit a school best than about any pretension at an objective standard of applicant quality.  So don't start panicking just yet.  (Full disclosure:  I was deferred from Brown.)

Second of all, here's your advice, and it's the same advice you'll find if you look at every other college thread here:  [b]find things to do outside of school that interest you, and do them well.[/b]  Every part of this statement is important.  "Outside of school" doesn't necessarily mean clubs and activities - it also means hobbies, for example.  "Interest" is [i]essential.[/i]  Never do something just because you think it'll look good; college admissions officers are smarter than this.  Doing them "well" means not being lazy about it; if it really interests you, you should give it your full attention.

Third of all, here's your second piece of advice:  [b]Name isn't everything.[/b]  You'd be surprised how great and specific the programs at less well-known schools are.  A friend of mine, for example, explicitly chose not to apply to MIT to go to UChicago, and he's extremely happy about his choice (not to mention about to take a summer job with Scott Aaronson, who is an MIT professor!  Imagine how cool you have to be to do that without actually going to MIT).  And I'll tell you right now that he's better at what he does than I am at what I do.

@t0r: Actually, Princeton is in the Ivy League. I agree with everything else you said, though.

To sort of tie the second and third points together, it'd be much smarter to figure out which schools fit you best and support your interests. Applying to all 8 Ivy-league schools is a waste of money (application fees) and time (application preparation/interviews). Decide which have programs that are interesting to you, offer good financial aid, and are situated somewhere you can stand to live for ~4 years (not only location, but food and dorm quality, job/volunteer/partying opportunities). Not only will you end up at a school where you can be happy, but doing this research will show to the school that you will fit in well and be able to contribute, as well as make the best use of the school's resources.

Whoops, I read that as Stanford and went with my automatic response.  :oops:

Hiya cognos, long time no see

Yea to agree with what everyone said, as an example, take the AMC because you love math, not becuase you think a 150 will look good on applications (worry about that stuff later).  Try to do activities that you like, not just ones that you think will help you with college applications.  Now, of course, you still have to do activities (so dont say, well I like sleeping, so I am going to do that), but try to do them for the right reasons.

Also focus on what you like.  I am not an admissions officer, but if I was, I would much more perfer a person that is dedicated in a few areas (math competitions>math research>math leadership), than a person that is scattered all over the place (science olympiad, debate, math club, key club, orchestra, basketball,...)-even if the latter has way more total hours.

Of course, if you enjoy doing those activities, still do them...but there are just too many people that make a "list" of their plans in high school and try to execute them to perfection-it may work for college admissions, but if you continue to follow that philosophy, your life will never live up to its potential (like Jay Gatsby-he had one goal to marry Daisy and he did everything for that goal).  High school is about exploring your own opportunities.

And same for colleges.  Now, just like you, I would love to have an opportunity to go to a top college like MIT.  However, if you do not get in, its alright.  Its not the end of the world (and dont let someone that got in try to brag to you or something-just show them up in 10 years at a reunion).  Just keep on trying, go to graduate school, surpass all those people that beat you in high school.  Remember that high school is suppose to prepare you for the rest of your life, not just for your undergraduate degree in college.

However, if you get into one of the top schools and you believe the atmosphere/curriculam fits you, then you should definetly go.  Youll meet a lot of very smart people and friends for the rest of your life at those top colleges.  And knowing you, I think you are at an excellent pace (i strongly think you will pass AIME this year as 9th grader).

Besides, why are you so worried now?  You are only a 9th grader.  I should be the one worried lol.  You can pm me if you want.

In short, don't feel compelled to do anything for the sake of your college applications. If you make the right moves for self-improvement, you'll find yourself being an ace applicant when the time comes around. Of course, even ace applicants get rejected.

Activities - If you have more diverse [i]serious[/i] interests, that will help you. Unfortunately, because everyone pads their resumes these days, you have the burden of proof to show that these are serious pursuits. Good ways of doing this are high levels of involvement, influence*, initiative*, and achievement. And to get into such a position, you need to first figure out what you're interested in. And to figure out what you're interested in, you need to try things out. Hence, the "diverse activities". That doesn't mean you should put every tiny thing you do onto your resume! Keep it down to things you either found particularly interesting, spent a long time on, or took seriously. Other stuff dilutes your resume. Now you may say, whatever, I have space for these. No, you don't! Use your space to substantiate the important things instead of just listing them.

Job - Get one. It doesn't matter if it's scanning items at Customer Service at Target. Internships do not count as jobs. While doing interesting internships can add a lot of flavor to your resume, you should still try to get a paying job with pay stubs (ie, teaching piano to family friends for a few Hamilton's doesn't count here). I guess, then, I should use the word "employment". You can use that word on your resume. Create a whole section for it!

* - Note that I did not say leadership. Someone can claim a 1m x 1m piece of land, but that doesn't make them powerful. Leadership alone gets you little mileage -- you have to show that you are a good/useful leader, for which you have to show influence and [b]initiative[/b].


This may seem silly, but it stuck in my brain like a bribe waiting to marry tomorrow. How long does it usually take for orders of books being processed and shipped to Canada? I am so excited about those books that i couldn't sleep well or eat well since i mailed the order.....

We usually ship within a day or two of receiving the order.  Mail to Canada, however, is erratic.  Sometimes it takes 8 business days, sometimes 18.  The customs process can really gum things up, particularly when shipping to Canada (no idea why this is the case, but it is).


Hi,

I'm currently a high school senior who is advanced in math and with some good math competition scores (e.g., USAMO qualification). I was wondering if there was any kind of part time job for someone like me which would use my mathematical background (i.e. better than say, a job at K-mart)

Note that I am planning to become a pure mathematician, and this is simply for some money I would like to make on the side.

Thank You

Well, I'm a freshman in high school in Georgia.

My school, Northview, offers some paid internships to students in junior and senior year with backgrounds like yours. You're given a catalog and you get to work under a professional. Don't know if this was what you were going for, but maybe you should look if your school has a program like that.

I wish you luck in pursuing your goal :)

Tutoring.  It's one of the only jobs you can get that has any relevance to your chosen future career, and it's one of the only jobs you can get in which you can use your math background.  It can also pay decently (compared to Kmart, perhaps).  The hard part is getting started finding clients.  For this you may have to use whatever connections you have (teachers, administrators, community groups, friends of parents, etc.)  If all else fails, you can try to find work at an SAT prep company.


Hint:
[hide]Use the Triple Angle Formula for $ \sin 3x$ and solve for $ \sin^{3}x$.[/hide]


How many triangles with all sides equal, one vertex at the origin and an altitude in the y-axis exists?

You sure there aren't any other restrictions? Because the answer right now would be infinite...


I know you can get them on this website, but is there somewhere I can get them cheaper? I can't seem to find any used book places that sell these.

[url=http://www.amazon.com/Art-Problem-Solving-Beyond/dp/1885875037]Amazon[/url] may have used copies when you look at it.

Well, books are by FAR cheaper than hiring a tutor or something...if you really want it, just buy it.


If there's one subject in school that I really struggle with, it is foreign languages. I have struggled through Spanish for five years and still I can only barely muster up the most basic of basic conversations in it. I strongly desire to be fluent in languages besides English, and would like to learn German and Chinese as well as Spanish. But the vocabulary won't stick in my head, I am constantly making mistakes in grammar, and my pronounciation is absolutely horrendous. Does anyone have some helpful tips for how to learn a foreign language?

First German sentence: My name is Pascal.

Mein Name ist Pascal. (German)
Menja sawutt Pascal (Myth or Darij write it in Russian charcters please.)  :D It seems I used German pronouciation.  :P

[quote="Myth"]Меня зовут Паскаль. (Russian)

And can anyone read russian letters? :) I don't think cyrillic is widely supported in west countries.[/quote]

At least in Eastern Germany Russian was the first foreign language. But somehow it seems my Russian fades away as I do not have to apply it too often. In general I feel Russian is a lot easier than English. In order to get some practise I request to change the forum from [b]English[/b] to [b]Russian[/b] language.  :lol: But as long as you, Myth, are writing such basic stuff as to Remike in another thread I don't have any problems in understanding...  :D

I think English is the easiest language, but not for me :(
Despite I like english very much, I am unable to learn it quite deeply. :( :( Even simple conversation force me to look into dictionary.
Another thing I can't understand is article. There are no articles in russian language.

Vocabulary doesnt seem to be a problem for me.  I think its the sentence structure because it changes in every language.  Feminine, masculine, tenses, and everything else.  I only know how to say a handful of phrases in different languages; Korean, German, Spanish, Chinese, Latin.  Nothing i can say in Chinese is appropriate though so that doesnt really count.  I'm taking Latin in school but all we've learned so far is stupid. The Roman slave works.  [i]Romanus servus laborat.[/i]  Always where under where  [i]Semper ubi sub ubi[/i]

I've taken French for the past 4 years, and we still are learning stupid things.  The main problem I have is remembering all the stupid vocabulary words for longer than it takes to get to the next test.  Also, the verb tenses are getting confusing (5 new tenses in the past 2 months...)

I'd like to learn Latin, although my school doesn't offer it.  Oh well.

I wish we did more vocab here, i did 9 years of French and my vocab is incredibly limited, but my sentence structure and conjugation skills are top notch, not that it matters since i dont know enough words to use:)

There is a Web page with some [url=http://learninfreedom.org/languagebooks.html]advice for language learners[/url], and I would advise finding books about favorite subjects (what are the chances that that would be math?) in the language you are studying, and also finding Internet broadcasting or shortwave broadcasting in the language you are learning. Passive, but REPEATED exposure counts for a lot in language learning.

I have some books about math in German, and LOTS of books about math in Chinese. French is the next language in which I will try to read math books.

Reading math in languages you doesn't know is easier
than you might imagine.  If you're familiar with the mathematical
material you end up absorbing some words and a general
"feel" for things.  That feeling may turn out to be deceptive
if you try and systematically learn the language later, but
I don't imagine the early exposure hurts.

I think, that for a full understanding and learning of a language, one must be familiar with the culture associated with it.
For example English can only be fully understood after a 'deep dive' into the culture of an English-speaking country, such as UK or USA, for example. By 'culture' I mean the day-to-day living as well as art, tv, literature, discussions and etc.
Therefore, to learn a language it's helpful to see TV series or movies in that language, read books, maybe even visit the country...


calculus
integration
function
real analysis
real analysis solved

I want to show that, given f:A-->R integrable, f(x)>0 --> integral of f over A >0, using Riemann integrals (rather than monotone convorgence theorem). How can i be sure that the infimum of the upper sums is not zero--or that the supremum of the lower sums is not zero, for that matter?

[quote="jmerry"]If the function is continuous at any point, the integral is clearly positive- you can stick a box under the graph. If we can show that a Riemann-integrable function is continuous somewhere, we will be done. There is a well-known theorem stating that the set of points of discontinuity has measure zero, but I'd like to stick with Riemann tools here.

Suppose $f$ is Riemann-integrable on the interval $[0,1]$. Choose a partition $P_1$ so that the lower sum and upper sum on $P$ differ by no more than 1. Then we can choose an interval of the partition $I_1$ with length $3\delta_1$ so that $\sup_{x\in I_1}f(x)-\inf_{y\in I_1}f(y)\le1$. Let $S_1$ be the closed middle third of $I_1$; then $|f(x)-f(x+t)|\le1$ for all $x\in S_1,t<\delta_1$.
Now choose a partition of $S_1$ with upper and lower sums differing by no more than $\frac12\delta_1$ and repeat the process; we find a nested sequence of closed intervals $S_n$ of length $\delta_n$ such that $|f(x)-f(x+t)|\le\frac1n$ for all $x\in S_n,t<\delta_n.$ This sequence has a nonempty intersection, and that point of intersection is a point of continuity.

This actually proves that the function is continuous at a dense set of points; we could have started with any subinterval.[/quote]

Thank you for the response. Regarding the theorem that the set of discontinuities of f has measure zero iff f is integrable, how could that help me here? I thought I could work with more elementary ideas, that is, since f is integrable, then its integral=sup{lower sums}=inf{upper sums}, then to somehow show that either of these quantities must be positive. Am I shooting in the dark here?

[quote="jmerry"]I don't see how to get at this one without using the fact that the function is continuous somewhere. That theorem felt a little too advanced here, so I found  an elementary proof that the function is continuous somewhere (which naturally extends to show that the function is continuous at a dense set.

You need to show that for some interval $I$, $\inf_{x\in I}f(x)>0$, and continuity at a point is the way to get there.[/quote]

Thank you. I might be missing something here, but a function like f(x)=x^3 is also continuous, but its integral is not always positive. In other words, I don't see why the continuity of the function implies that the integral is positive.

I apologize if I am missing something obvious.

Note that $f(x)>0$ was a condition of the problem.

What jmerry was finding was a way to get at this theorem:

Suppose $f$ is integrable on $[a,b]$ and also satisfies:

1) $f(x)\ge0$ on $[a,b]$
2) There exists a point $x_0$ such that
- a) $f$ is continuous at $x_0$, and
- b) $f(x_0)>0.$

Then $\int_a^bf>0.$

The short description of the proof: "stick a box under the graph."

Condition 1), the nonnegativity of the function, is vital - without that, the whole thing is nonsense.

[quote="Kent Merryfield"]Note that $f(x)>0$ was a condition of the problem.

What jmerry was finding was a way to get at this theorem:

Suppose $f$ is integrable on $[a,b]$ and also satisfies:

1) $f(x)\ge0$ on $[a,b]$[/quote]

Why is this not a strictly greater than?

[quote]Why is this not a strictly greater than?[/quote]
Becuase it didn't have to be for the particular theorem I was quoting to be true. That doesn't cause any trouble with your problems, since "positive" implies "nonnegative."

[quote="Kent Merryfield"]
Suppose $f$ is integrable on $[a,b]$ and also satisfies:

1) $f(x)\ge0$ on $[a,b]$
2) There exists a point $x_0$ such that
- a) $f$ is continuous at $x_0$, and
- b) $f(x_0)>0.$

Then $\int_a^bf>0.$

The short description of the proof: "stick a box under the graph."
[/quote]

Ok, I see why this is true. Thank you very much--that was insightful.

[quote="Evariste"]By the way, is it sufficient for me to say that the function is continuous 'almost everywhere' (i.e. its discontinuities form a null set) or must I show where its continuous as jmerry did?[/quote]
There's no clear or obvious answer to that question. That particular theorem is quite well-known among those who have progressed far enough in analysis, and there's one point of view that says that any "well-known" theorem is fair game.

Go back a re-read the first paragraph of jmerry's first post (#2 in this topic). To put what he said in perspective: I am finishing teaching a course right now that develops the Riemann integral, and deals with such topics as the uniform convergence of sequences of functions. I never did mention the notion of measure zero, and never did prove the theorem that has been quoted - interesting as it is, that theorem feels foreign to the courses in question. It's clear that jmerry's choice of methods was an aesthetic choice. He chose to confine himself to methods that are not beyond the course I am talking about.

[quote="Kent Merryfield"]There's no clear or obvious answer to that question. That particular theorem is quite well-known among those who have progressed far enough in analysis, and there's one point of view that says that any "well-known" theorem is fair game.

Go back a re-read the first paragraph of jmerry's first post (#2 in this topic). To put what he said in perspective: I am finishing teaching a course right now that develops the Riemann integral, and deals with such topics as the uniform convergence of sequences of functions. I never did mention the notion of measure zero, and never did prove the theorem that has been quoted - interesting as it is, that theorem feels foreign to the courses in question. It's clear that jmerry's choice of methods was an aesthetic choice. He chose to confine himself to methods that are not beyond the course I am talking about.[/quote]

Well, its not that difficult to show Riemann integrable --> measure zero. If you let B be the set discontinuities of f and write it as a countable union of B_i's where B_e={x in B | oscillation at x > e}, then given any e>0, the set B_(1/n) has measure zero, and since B is a countable union of sets of measure zero, it also has measure zero.


superior algebra
superior algebra solved

Is sqrt(2) the limit of a sequence of numbers of the form: n^(1/3)-m^(1/3) with n,m naturals?

I think that the problem should look like this:
Let A={n 1/3-m 1/3| m and n are naturals}

Is  \sqrt 2 in A' (the accumulation points of A - reals l such that there is a subsequence of A that converges to l)?

I think that we can give a recursive contstruction to this one.
n=1, m=0. Denote by d(n,m)=n 1/3-m 1/3.
If d(n,m) \leq  \sqrt 2 then increase n, else increase m.
This sequence converges to  \sqrt 2 because  (n+k) 1/3-n 1/3 tends to 0 as n tends to infinity for every fixed k. And d(1,0)< \sqrt 2.