Putnam
AMC
AMC 12
AMC 12 A
AMC 12 B
Harvard
college

According to statistics,

3733 students took the most recent Putnam exam
83,728 students took the most recent AMC12A and 44,303 students took the most recent AMC12B (these are the most poular high school math competitions)

Comparing these numbers, we can see that a MUCH MUCH higher number of high school students take math competitions in comparison to the number of college students who take math competitions. Of course, this seems logical because many high school students are "forced" to take the AMC's, and because there are many more other activities to do in college than there are to do in high school. Nevertheless, the different between the numbers above is simply HUGE. What do you folks think about this?

First and foremost, that people don't want to get up early in the morning (at least on the West Coast) and spend 6 hours on an exam which they are likely to get a score of 0 on around the time of the year when they should be preparing for exams?

Of course, if 100,000 or 200,000 students took the Putnam, they'd never get it graded. :D

Note that it's a one-step process. This is the Olympiad level, with no preliminary or qualifying round, and that makes it pretty terrifying for an inexperienced student to subject himself to.

The Putnam IS growing. It was over 3500 this year; it was closer to 2500 just 4 or 5 years ago. I think two things have been happening. One is that at the places where 50 or 100 students took the exam (Harvard, MIT, Caltech), even more students are taking it now than did a few years ago. On the other end, the number of schools offering the exam has increased. As an example, my college (CSU Long Beach) didn't offer the Putnam between about 1980 and 1999; I had it re-instituted in 1999 and have continued it since, adding a 1-unit course to our curriculum to help students prepare for it.

Here at CSULB, I don't advertise it outside of recruiting people to take the 1-unit course, and I mostly get them based on personal connections. (I'm the undergraduate advisor, and I do have a few ARML team alums here.) We've been having about 7 people a year take it. So some people get zeros - I don't care; they've still gotten something out of the experience of preparing for it.

In comparison to the MAA/AHSME/AMC exams, there is very little incentive to take the Putnam competition unless one has some chance of reaching the top 200.

At university, time is scarcer than in high school. The exam comes toward the end of semester, when final examinations occur, as well as graduate school application deadlines (and some GRE exams and interviews for summer jobs).   The high-school competitions occur several months later, in mid-semester.   The Putnam is 6 hours long (about 7 including the intermission), not 3 hours.  The exam is on a weekend, so is not an improvement over spending time at class that morning, and is certainly not an improvement over the "weekend leisure options" available at university.

Moreover, anything other than a top 200 Putnam performance is worthless as a credential. This is because individual scores are not verifiable, and there are no subsequent exams.  The MAA/AMC, on the other hand, publish the list of the top several thousand scorers. Whereas above-average AHSME scores are of value as a university admission credential even (and perhaps especially) for those not continuing in mathematics, anything other than a publically documented high performance (top 200) on the Putnam is unlikely to generate any future advantage for the examinee.

I guess that the Putnam examinees consist of 100-200 people with a serious chance of scoring (say) 40 points, a few hundred more taking the exam just to try it out or see how they'll do, and most of the rest taking it purely as a hobby.  The "silent majority" of hobbyists can always find the problems on the Internet or in print, to solve if and when they please.

[quote="fleeting_guest"]In comparison to the MAA/AHSME/AMC exams, there is very little incentive to take the Putnam competition unless one has some chance of reaching the top 200.[/quote]

I [i]highly[/i] disagree with this statement. There can be many incentives to taking the exam, even for people who don't think they will win any prize or recognition. As Kent said, a lot of people get something out of it just by preparing for the exam.

The reasons for the dispairity in the number of people taking the AMC and the Putnam should be obvious though. First of all, the number of people who take the type of math covered on the AMC in high school is much larger than the number of people who have studied the kind of math covered on the Putnam exam.

Secondly, they're two completely different kinds of exams. The AMC is 75 minutes, multiple choice, and the Putnam is 6 hours, proof style. It would be [i]more[/i] appropriate to compare the number of students taking the Putnam to the number of students who [i]would[/i] take the USAMO if it were openly available. That number would probably be more comparable to the number of Putnam participants.

[quote="fleeting_guest"]In comparison to the MAA/AHSME/AMC exams, there is very little incentive to take the Putnam competition unless one has some chance of reaching the top 200. [/quote]

And, precisely how do you measure this "chance" without participating in the contest?

[quote="fleeting_guest"]  The exam is on a weekend, so is not an improvement over spending time at class that morning, and is certainly not an improvement over the "weekend leisure options" available at university. [/quote]
And, just when do you think a 6-hour exam can be held if not during the weekend?

[quote="fleeting_guest"] Moreover, anything other than a top 200 Putnam performance is worthless as a credential. This is because individual scores are not verifiable, and there are no subsequent exams.  [/quote]
The top 500 scores are very much verifiable.

[quote="fleeting_guest"] I guess that the Putnam examinees consist of 100-200 people with a serious chance of scoring (say) 40 points, a few hundred more taking the exam just to try it out or see how they'll do, and most of the rest taking it purely as a hobby.  The "silent majority" of hobbyists can always find the problems on the Internet or in print, to solve if and when they please.[/quote]
In my humble opinion, for what it's worth, there may be more to life that just trying to score 40 or more points on the Putnam. For instance, there may exist something called "joy" or maybe "the thrill of challenge" beyond the humdrum existence of college life where getting an A in a math course may just be as normal as brushing one's teeth; in such a case, some people may seek something more in things like, say, The Putnam Math Contest. Of course, we may all differ in our opinions. That's, however, everyone's prerogative.

[quote="tnn"]For Putnam, does the number of contestants matter that much ?
You know, all the good students are competing.
If 1000000 more students take Putnam, do you think the average, the median, or Top 100 scores will go up ?[/quote]

My guess is that the the average and the median would go down and the top 100 scores might increase by at most a few points, which I still suspect may not be entirely true.

The point is not that those are reasons to not take the Putnam, they are reasons as to why many people do not take the Putnam who reasonably could.

[quote="Xevarion"]I think the reputation of the Putnam as being inaccessible to the average person is really detrimental to participation. I tried to get a lot of my friends to take the test, but many of them declined saying "I don't have enough time to prepare, so I won't do well. Its not worth it." Or "I don't know enough about writing proofs." But they were getting as good grades as I was in the same math class (or close, at least)! I couldn't convince many of them that the test might be worth taking.[/quote]

 :(  :o  :maybe:

[quote="ComplexZeta"]To add to that, it tends to be (at least at my university) held during the weekend before finals, so most people just want to cram for finals. [/quote]
At mine, there has most often been one week of regular classes afterwards. This year, we have two weeks - the Putnam is on Dec. 2 and our finals don't start until December 18.

At [url=http://www.tamu.edu]my university[/url] we hold Math Contest for freshmen and sophomores in the spring, with modest cash prizes for top scorers. Being easier than Putnam, the contest serves as a gateway. It helps a bit, but we could/should have more contestants.   :(

[quote="ComplexZeta"]To add to that, it tend to be (at least at my university) held during the weekend before finals, so most people just want to cram for finals. Another problem we have is that the professor in charge of it refuses to advertise it outside of CCS since he just figures everyone will get a zero (which is not true, although many will). Having said that, we have 7 take it my freshman year and 15(?) my sophomore year. I don't know if that's a trend or an isolated event though[/quote]

By CCS do you mean the College of Creative Studies at UCSB?

Yeah. Hi Ben. How is the Gamma function?

I wonder why people are uncomfortable with getting zero or scoring low. As one of my friends said this morning, 5 minutes before the exam: "They won't give you a 'dishonorable mention', you know."

First off, the only thing that matters are the problems, and what you can do with them: at worst, you'll spend six hours being very puzzled, which is always good because it stimulates the mind. And second, if you do well, you'll benefit from it, if you don't, no one will care. The only bad thing about the Putnam is not taking it.

How true.

I have friends taking the same math class as me who do not want to go, just because they think they wouldn't know enough math and would end up getting zero. Different perspective I guess.  :P

There's nothing wrong with this discussion - it's "in bounds" - but I'd still rather you all chilled out until tomorrow evening.

According to the proctor MIT had a record number of people taking the exam this year.  I believe 156 signed up on time to take the take.  But there were people who signed up late who were allowed to take the spots of no-shows.  So I'd guess that about 140-150 people showed up in all.

ComplexZeta said (some time ago),

[quote]Actually the Putnam committee publishes lists of the top 500 every year, so that makes it a bit easier to use it as a credential.[/quote]
They do?  How can one find such a list, particularly from quite a few years back (say, like 1988 [could be a catchy saying - "Study like it's 1988; party like it's 1999"])?  Does anyone know?

[quote="Xevarion"]I believe they're published in American Mathematical Monthly. I don't know which issue of the year it would be, though.[/quote]
C'mon, Xev, you know how to use JSTOR, don't you?  The 1989 announcement (for the 1988 exam) was
"The William Lowell Putnam Mathematical Competition."  [i]The American Mathematical Monthly[/i], Vol. 96, No. 8. (Oct., 1989), pp. 688-695.

It includes a list of the top 104 individuals (including my freshman-year physics professor; go figure!), but nothing about the top 500.  The 1965 (or '69?) announcement I stumbled across noted that one could write to the MAA for the list of the top 500; I don't know if this is still the case.

[quote="JBL"]C'mon, Xev, you know how to use JSTOR, don't you?  The 1989 announcement (for the 1988 exam) was
"The William Lowell Putnam Mathematical Competition."  [i]The American Mathematical Monthly[/i], Vol. 96, No. 8. (Oct., 1989), pp. 688-695.[/quote]
I have exams right now. I'm busy. :P I'm pretty sure they publish the top 500 now in AMM, because I saw the 2005 ones in one of the Math Club's issues (there's a small chance that it was only the top 200 and I misremember). I don't know when they started doing that and I'm not going to check. :P

The top 500 list (which could actually be the top 473, or top 514, or wherever the scores break) is part of the package mailed to Putnam supervisors in March. It contains mailing addresses and expected graduation years, so it could be used for recruiting purposes - I imagine that some agencies and schools obtain this list (maybe paying for it?) for their own pupposes. I do remember once seeing a piece of recruiting mail from a company that consults with the NSA that I'm guessing was triggered by this list.

I only have mine back to 1999 (and that's assuming I would know where in my office to find them, which is not a given.)

Dr. Merryfield,

Thanks for the information.  (And may you long enjoy the feeling of wild freedom that I imagine your office exudes.)

I just today found out that Santa Clara University has lists of the top 500 scorers archived, but that they wouldn't have time to search for a given year's list until March or so.  I should have offered for them to buy me a round-trip plane ticket from Chicago to Santa Clara so that I could spare them the burden of searching through the archives.

I'm curious - according to a short article for the school paper written by a professor at my college back in 1989, I placed in a "multi-way tie" for 459th on the 1988 exam.  Would that mean that if 43 or more people tied for 459th, all those people would then [i]not[/i] be listed?  If so, can a playoff for all those who tied be arranged (complete with cable television coverage, of course)?

I'm sure that what he meant was that your [i]rank[/i] was 459.

I have no idea of the details from your year, but here's a hypothetical way that that could have occurred: Suppose that 401 people scored 11 or better on the Putnam and that 115 people scored exactly 10. Then those 115 people share all of the ranks from 402 to 516; averaging all of those numbers together gives a rank of 459 for all of them. And if that were the case, the "top 500" list would include everyone with rank 459 but no one below that, making it in practice a "top 516" list. The "top 500" list never has exactly 500 names on it - sometimes it's more, and sometimes it's less.

There's a very good chance that "rank 459" would be the lowest possible score for being on the "top 500" list; such a score would almost always be a multi-way tie. I suppose that if the 500 cutoff happened to hit a very rare score, like 6 or 7, then it could only be one or two people. (You can only get a score like 6 or 7 by getting partial credit on 3 to 5 problems without full or mostly-full credit on anything - that's hard to do.)

To address the original poster's question: Fierytycoon, I'm not sure if you're in college but if you are then I'm sure you realize that the Putnam comes at a pretty annoying time for a lot of students. Personally, I was VERY close to simply choosing not to take the Putnam for the sake of an extra 4 or 5 hours of sleep, and this is knowing that odds are I'd get an HM at the worst. The only thing that drove me out of bed the day of the competition (I was up till 4 the night before working, yes, on a Friday night, because it's that time of year at MIT...) was a sense of respect to the Putnam, USAMO, and all those other contests that have done so much for me as a mathematician and as a person.

Along those lines, I had a lot of friends who were going to take it, but simply decided not to in the end because they were just too tired. Maybe this is just an MIT thing (or maybe it's more of a factor at MIT than at other schools) but the whole "everyone is tired and it's near finals" argument definitely applies to my school.

Personally, I'm glad I had the willpower to get up that day. However, I would be lying if I said I didn't pay the price in terms of sleep and energy for the rest of the term, since the Putnam is obviously going to be mentally draining, and hurts your productivity pretty badly for the rest of the day, so you do in fact lose a day of work.

The median score is either 0 or 1 out of 120.  Now why would you (even you on this board) want to take an exam if you think that you are likely to get a 0 or a 1 out of 120?  How is it a valuable experience to look at a list of problems, all of which are impossible to you?  I don't see the point.

On the other hand, there are many students who *might* be able to score positively but don't take the exam.  To offer some insight on this, I'll use myself as a case study:  I liked math competitions in high school, but I never took the Putnam in college.  Why?  I had various reasons that were interrelated.  There are two fun aspects of doing math competitions: the satisfaction of winning (in a broad sense), and the act of doing the problems.  I knew I wasn't going to "win" the Putnam or even come close, so if I were to do it, then it would have to be "for the love of the game."  Math is fun when it's challenging, but not when it's effectively impossible.  I always imagined the Putnam to be too difficult to be fun.  On the other hand, this might have been remedied with sustained Putnam practice, but I preferred to spend my time on "real math" rather than "Putnam math."  One big difference between high school and college (for me at least) was that in high school, competition math was my only outlet for interesting math.  In college the situation was reversed, with classroom math being far more interesting than competition math.  Hence, the fun factor of the Putnam was low for me.  Besides for fun, one might take the Putnam because it's useful or beneficial in some way.  I never saw the Putnam this way, especially compared to learning "real math."

There's a line in the film, Bring It On, that goes, "Cheerleaders are dancers who have gone retarded. What you do is a tiny, pathetic subset of dancing."  I know that it's an unpopular point of view on this board, but sometimes I want to say that "competition math is a tiny, pathetic subset of math."  The level of emphasis placed on competitions in high school is totally out of whack, and then in college the Putnam rightfully fades into the background.  This is evident in some of the posts in this thread.  How many of you would have skipped the AMC because it was "too early in the morning" or because you "had exams the next week?"

yenlee, I definitely agree with you about the relationship between high school and college contests vs. classwork. I loved contests in high school mainly because there I didn't have any other good ways of finding interesting mathematics. My teachers weren't going to be of much use to me in that regard. But in college I spent very little time preparing for the Putnam because, well, why should I spend a lot of time studying calculus and linear algebra when I could be taking advanced courses that I find far more interesting?

Nonetheless, I enjoyed taking the Putnam, especially this past year (which is my senior year). I stopped caring much about my result and was just able to spend a day thinking about the problems. I really like doing that, but it's hard to do that except on Putnam day because I'm usually just way too busy. (I took 8 classes, or 30 units, last quarter. In general, I'd recommend not doing that unless you have a much better attention span than I do. I don't usually take quite that many, but I generally take 6 or 7. So it's hard to find days that I can just devote to thinking about a few problems.) I got a credible, but not terrific, 46 points, which is fine with me.

If you make putnam global instead of restricting it to US and Canada, you can easily get 100k competitors.


geometry
ratio
similar triangles
geometric transformation
perpendicular bisector
geometry proposed

On the sides $ AB$ and $ AC$ of the acute triangle $ ABC$ construct externally the similar triangles $ ABM$ and $ ACN$ such that $ \angle M\equal{}\angle N\equal{}90^{\circ}$ and $ \angle MBA\equal{}\angle NCA\equal{}\angle BAC$. From the midpoint $ A_1$ of the side $ BC$ construct the perpendicular $ A_1A'$ on $ MN$. Similar we obtain the lines $ B_1B'$ and $ C_1C'$. Prove that $ A_1A'$, $ B_1B'$ and $ C_1C'$ are concurrent.

call S the symmetric of A wrt $ A_1$. We have $ \angle MAN \equal{} 180 \minus{} \alpha$ so MANS is a cyclic quad. Call $ D: MN \cap AC$ and call K the midpoint of $ B_1C_1$.
We have $ \angle A'A_1B_1 \equal{} 90 \minus{} \angle NDC \equal{} 90 \minus{} (90 \minus{} \alpha \plus{} \angle DNA) \equal{} \alpha \minus{} \angle MSA \equal{} \angle KA_1C_1$.
So $ A_1A'$ is the $ A_1$-symmedian of $ \triangle A_1B_1C_1$. Then $ A_1A'$, $ B_1B'$, $ C_1C'$ concur at the symmedian point of $ \triangle A_1B_1C_1$.

The tangent at B and C to (ABC) meet at T; it’s well known that AT is the symmedian of trinagle ABC. See that a spiral similarity around B maps M to A and A1 to T, the ratio being 1/cosA, the same ratio of a spiral similarity around C will map N to A and A1 to T, hence A’A1 is the perpendicular bisector of MN. Further see that AT and MN are perpendicular (from AMDN cyclic and AT symmedian), therefore A1A’ is symmedian in the medial triangle A’B’C’.

Best regards,
sunken rock


What is the units digit of $ \left \lfloor \frac{10^{2000}}{10^{100}\plus{}3} \right \rfloor$?

Let $ x \equal{} 10^{100}$.

Then our expression is $ \left\lfloor\frac {x^{20}}{x \plus{} 3}\right\rfloor$. We can perform synthetic division to find the units digit. However, that is tedious, so we reverse the process and start with the remainder. By the Remainder Theorem, we have $ r \equal{} ( \minus{} 3)^{20} \equal{} 3^{20}$.
In the process of synthetic division, we have to multiply the constant term (after division) by $ \minus{} 3$, and then add it to the constant term of $ x^20$, which is $ 0$, to that. Thus, our constant term (after division) is $ \frac {3^{20}}{ \minus{} 3} \equal{} \minus{} 3^{19}$. We want to find the units digit of this, which is
$ \minus{} 3^{19}\equiv9^9\cdot \minus{} 3\equiv( \minus{} 1)^{9}\cdot \minus{} 3\equiv\boxed3\pmod{10}$.

[quote="math154"]LHowever, that is tedious, so we reverse the process and start with the remainder. By the Remainder Theorem, we have $ r \equal{} ( \minus{} 3)^{20} \equal{} 3^{20}$.
In the process of synthetic division, we have to multiply the constant term (after division) by $ \minus{} 3$, and then add it to the constant term of $ x^20$, which is $ 0$, to that. Thus, our constant term (after division) is $ \frac {3^{20}}{ \minus{} 3} \equal{} \minus{} 3^{19}$.[/quote]

I don't understand this part of the solution.  Could you please be a little bit more specific?  Thanks!

$ \frac{x^{20}}{x \plus{} 3} \equal{} \frac{x^{20} \minus{} 3^{20}}{x \plus{} 3} \plus{} \frac{3^{20}}{x \plus{} 3} \equal{} x^{19} \minus{} 3x^{18} \pm ... \minus{} 3^{19} \plus{} \frac{3^{20}}{x \plus{} 3}$.  The terms with an $ x$ in them have many zeroes and the remainder term is much less than $ 1$, so the units digit of the desired expression is the units digit of $ \minus{}3^{19}$.


algebra
polynomial
calculus
integration
algebra unsolved

Let p(x) be a n-degree polynomial. For how many consecutive integer values of x must p(x) be an integer in order to guarantee that p(x) is an integer for all integers x?

Answer: $n+1$. Hint: use Lagrange interpolation polynomial and the fact that the product of  $k$ consecutive integers is divisible by $k!$.

You can also get that (n+1) works just by using 'discrete derivatives' i.e. considering f_n(x) = f_(n-1)(x) - f_(n-1)(x-1), and using the fact that if it is degree n this becomes constant in n steps -- making each other value constructable.  (Just a common way to skip lagrange)

And of course for the counterexample.

consider the polynomial  :rt2:(x-1)(x-2)...(x-n). This is integral (0) at 1,2,3,...,n, but is irrational for all other integers.

you may also use the fact the the binominal coefficients $C_{x}^k$ where $k=0,1,2,...,k$ are integer-valued and form the base in the polynomial space, which degree $\leq k$, so every such polynomial could be represented in the form
$P_{k}(x)=\sum_{i=0}^{i=k}c_iC_{x}^{k-i}$ where $c_i$-some real numbers, then you may use that the this polynomial takes integer values at the points $x=n,n+1,...,n+k$ to prove that all the $c_i$ integers


How do we solve diophantine equations like the one below?

[hide="problem"]
Find one pair of values $ (x,y)$ satisfying the equation $ 3x^2 \equal{} 5y^3$, where $ x$ and $ y$ are integers.
[/hide]

[b]
thanks in advance[/b]

[hide="An approach"]Let's seek an answer of the form $ (3^a 5^b, \; 3^c 5^d)$

There are 2a+1 factors of 3 on the LHS and 3c factors of 3 on the RHS. Thus, we need any solution to 2a+1=3. (a,c)=(1,1) works just fine.

There are 2b factors of 5 on the LHS, and 1+3d factors of 5 on the RHS. Thus, we need any solution to 2b=3d+1. (b,d)=(2,1) works just fine.

Thus, $ (x,y) \equal{} (3^1 5^2, \; 3^1 5^1) \equal{} (75, \; 15)$ is a solution.[/hide]

[b]Follow-up:[/b] [i]Characterize the solution set of this equation[/i]

The general solution is:
$ x\equal{}k^3\cdot 3^{3t\plus{}1}\cdot 5^{3r\plus{}2}$ and $ y\equal{}k^2\cdot 3^{2t\plus{}1}\cdot 5^{2r\plus{}1}$ for any $ k,t,r$ nonnegative integers.

A general technique can be found [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=141316]here[/url] along the lines of The Zuton Force's solution.  For $ p \equal{} 3, 5$ we have to solve the linear Diophantines $ 1 \plus{} 2a \equal{} 3b$ and $ 2c \equal{} 1 \plus{} 3d$ and for $ p > 5$ the linear Diophantine $ 2e \equal{} 3f$.

that's OVERKILL for such a simple problem.
i don't understand your blog post, not a single bit.

I mention the general technique to put this problem into a class of problems that Engel calls those with "automatic" solutions.  What I think he means is that problems like this have solutions that are immediately deducible from the particular constants chosen.

The only really important part is the first statement, which is just a different way of phrasing The Zuton Force's solution.  We will use the more accepted syntax $ v_p(n)$ to denote the greatest power of $ p$ that divides $ n$ (for $ p$ a prime).  This is just the power of $ p$ in the prime factorization for $ n$.  For example,

$ n \equal{} 12 \implies v_2(n) \equal{} 2, v_3(n) \equal{} 1, v_5(n) \equal{} 0$
$ n \equal{} 21 \implies v_2(n) \equal{} 0, v_3(n) \equal{} 1, v_5(n) \equal{} 0, v_7(n) \equal{} 1$.

[b]Claim:[/b]  $ v_p(mn) \equal{} v_p(n) \plus{} v_p(m)$.

Hence if $ 3x^2 \equal{} 5y^3$, then it must be true that

$ v_3(3x^2) \equal{} v_3(5y^3) \implies$
$ 1 \plus{} 2 v_3(x) \equal{} 3 v_3(y)$.

This is the first linear Diophantine to which I refer, and it has general solution $ v_3(x) \equal{} 3t \plus{} 1, v_3(y) \equal{} 2t \plus{} 1$.  Similarly,

$ v_5(3x^2) \equal{} v_5(5y^3) \implies$
$ 2 v_5(x) \equal{} 1 \plus{} 3 v_5(y)$.

This is the second linear Diophantine to which I refer, and it has general solution $ v_5(x) \equal{} 3r \plus{} 2, v_5(y) \equal{} 2r \plus{} 1$.

For any prime $ p \neq 3, 5$ we have

$ v_p(3x^2) \equal{} v_p(5y^3) \implies$
$ 2 v_p(x) \equal{} 3 v_p(y)$.

This is the third linear Diophantine to which I refer and is the simplest of all; it has general solution $ v_p(x) \equal{} 3s, v_p(y) \equal{} 2s$.

[b]Claim:[/b]  $ n \equal{} m$ if and only if $ v_p(n) \equal{} v_p(m)$ for all primes $ p$.  (In other words, the prime factorization of a natural number uniquely determines it, which is obvious.)

paqui_2590's general solution then follows.  Do you understand how to solve linear Diophantines?  That's the underlying concept here.  Finding small solutions (which is sufficient to solve the original problem) with coefficients this small is trivial.  I am addressing the follow-up.

[b]paqui_2590:[/b]  $ k, t, r$ would uniquely characterize a solution if you added the restriction that $ 3, 5 \not | k$.  Otherwise, you can just write $ x \equal{} 75k^3, y \equal{} 15k^2$.


vector
linear algebra
matrix
linear algebra unsolved

Suppose U is a finite-dimensional real vector space and T is an operator on U. Prove that U has a basis consisting of eigenvectors of T if and only if there is an inner product on U that makes T into a self-adjoint operator.

An inner product is given by a symmetric positive definite matrix $ W.$ The inner product of $ x$ with $ y$ is then $ x^TWy.$

"$ U$ has a basis consisting of eigenvectors of $ T$" can be shortened to "$ T$ is diagonalizable."

"There is an inner product that makes $ T$ self-adjoint" is equivalent to there existing such a $ W$ such that $ WT \equal{} T^TW.$ After all, the inner product of $ x$ with $ Ty$ is $ x^TWTy$ and the inner product of $ Tx$ with $ y$ is $ x^TT^TWy.$ If these are the same, then $ x^TWTy \equal{} x^TT^TWy$ for all $ x$ and $ y,$ which in turn implies that $ WT \equal{} T^TW.$

=======

Suppose $ T$ is diagonalizable. Then there exists invertible $ P$ such that $ T \equal{} P^{ \minus{} 1}DP$ for $ D$ diagonal. Define $ W \equal{} P^TP$ and note that this $ W$ is symmetric and positive definite. Then:

$ WT \equal{} P^TPT \equal{} P^TPP^{ \minus{} 1}DP \equal{} P^TDP$ and

$ T^TW \equal{} T^TP^TP \equal{} P^TD^T(P^T)^{ \minus{} 1}P^TP \equal{} P^TD^TP \equal{} P^TDP$ since diagonal matrices are symmetric.

Hence, $ WT \equal{} T^TW$ and there is an inner product that makes $ T$ self-adjoint.

=======

Now suppose that $ WT \equal{} T^TW$ for some symmetric positive definite $ W.$ Note that $ WT$ and $ T^TW$ are transposes, so $ WT$ is symmetric.

$ W$ has a symmetric positive definite square root $ W^{1/2},$ and the inverse of that, $ W^{ \minus{} 1/2}$ is also symmetric positive definite.

$ W^{ \minus{} 1/2}(WT)W^{ \minus{} 1/2} \equal{} W^{1/2}TW^{ \minus{} 1/2}$ is symmetric since $ WT$ is symmetric.

Since symmetric matrices are diagonalizable, we can find $ Q$ such that

$ QW^{1/2}TW^{ \minus{} 1/2}Q^{ \minus{} 1} \equal{} D$ is diagonal, but then

$ [QW^{1/2}]T[QW^{1/2}]^{ \minus{} 1} \equal{} D$ and $ T$ is diagonalizable.


geometry ARML AMC AIME USA(J)MO USAMO algorithm

I just want to have everyone post in one topic so I can figure out what the ARML team will be like this year.

I'm quitting Chinese school so I can go to ARML. Say hello to the newest member of the D team!

im going
ill probably be on B team again, due to my awesome aime score(unless I can pwn those individual rounds)

pick me!
oh, btw albert, you wont be the first on the D team (even if you were bad which you arent contrary to what you continuously insist) last year i didnt sign up in time so i was on the alternate team with another SFBA and some people from Indiana
so ha

and junggi, i think ted will understand how horrible this AIME was (proof:  i did as well as you :lol: ) and adjust the teams accordingly

I'm excited. Eric invited me to the first meeting, so I'll be sure to make it there. I've joined the yahoo group, and I'm checking messages.

Any ideas on where/what team I'd go with a 118.5 AMC 12 and a 7 AIME?

I'm going to go...

Say hi (or boo) to the newest member of the Z team.
(There is less than a $0.1\%$ chance that I will actually get into that)

I highly doubt that 26 * 15 = 360 people will sign up for ARML. And man that would suck, cause I would get bumped down from the D team to the Z team.

So so far (in order of smartness by line):
Taylor Han, Charles Chen, Vishal Arul, Eric what's-his-last-name, DiscreetFourierTransform
[filler space]
[filler space]
[more filler space]
[even more filler space]
Albert Gu

How do they select the teams?

[quote="Ubemaya"]So so far (in order of smartness by line):
Taylor Han, Charles Chen, Vishal Arul, Eric what's-his-last-name, DiscreetFourierTransform
[filler space]
[filler space]
[more filler space]
[even more filler space]
Albert Gu

How do they select the teams?[/quote]
ALBERT YOU'RE NOT GOING TO BE ON THE D TEAM GET USED TO IT
i think matt(discreetfouriertransform) would be ahead of me (i really dont care though :D ) cuz i beat him by 2 points on BAMO but he beat me by 6 on the AMC 12 and 4 on the AIME :oops:

do the first three people you listed have AoPS accounts?  (this is probably a stupid question :lol: )
EDIT:  OMGOMGOMG i am so stupid
i thought that vishularul was one word i for some reason didnt realize it was vishal arul :rotfl:

hmm im gonna make some guesses
is junggi taylor han?
is ccy charles chen?  (lots of C's so maybe)

[quote="junggi"]omg why am i first in line for A team[/quote]
the AIME isn't everything :D
especially this one
anyways, those are just Albert's (incorrect- look at where he placed himself :P) predictions

yay i was right in at least one guess

[quote="DiscreetFourierTransform"]Yeah, I completely bmbed the bamo... haha that line made me think of stewart's theorem again... man + dad = bmb + cnc ;-) rofl waffles[/quote]

BOMB IN THE SINK=DEAD MAN :)

they should totally have a waffles smiley :rotfl:

A terroristic man and his equally terroristic dad built a C4 bomb in their kitchen sink.

I'm joining, even though I suck. Say hello to another member of the D team, lol.

My scores:
BAMO - didn't know it existed.
AMC 12A - 112.5
AIME - 5.

And I'm skiving off chinese school too.

[quote="Ubemaya"]READ THE POST AGAIN IT SAYS RANKED BY LINE SO EVERYONE ON THAT FIRST LINE HAS AN EQUAL CHANCE OF MAKING THE A TEAM.

Do they make the teams based on previous tests or by selection tests?
My scores:
BAMO - 18 :(
AMC 10A - 150 :(
AMC 12B - 124.5 :(
AIME - 8 :(
Hmm what else is important?[/quote]
dude you really think i'm as good as taylor???? :o
i reeally dont think so

anyways, how can you possibly be sad about a perfect on the 10?????? :roll:
i failed the 10- only got a 121.5 :rotfl:

today, vishal asked me if i had a aops account
i think i said no
hmm

while we're talking about vishal, he poked me in the neck today :ninja:

ooh, and he was like, "you have a blog?'
and he's posted stuff in it before

Me and David were on a band trip.

I have this feeling that the A team did a lot better today. [i]Just[/i] a feeling.

there were some good problems today
i singlehandedly got the first two on the team in about 3 minutes
then didnt get anything else
the power had more than a page of explanation, but once you figured out what the heck they were saying, it was pretty easy
i only wrote up #4 with andrea, but basically figured out everything else working with andrea and olen

EDIT:  i have the same number of posts as taylor :P

I wish I was there yesterday, but no, I'm stuck in Deustchland right now. If I get off the plane and immediately head for ARML practice, I might make it next week. But I'll probably be brain-dead. Who knows.

Anyone else think it would be cool to have some sort of flag/banner to bring to ARML?  Most Stanford dorms make one, and people like it.  So Nathan and I were having a little discussion of fun things to do, and that came up.  I can hopefully make one by next week if it sounds like a good idea to others.  Ideas for designs (must be easy to sew) and colors are appreciated.

Also, team cheers.  A few people tried to come up with them on the spot last year, but it didn't work.  If you have any ideas, try and develop them.

I'll probably send an email to the list about this soon.

That. Would. Be. 1337.

End of story, I think UBEMAYA wins.

/sarcasm.

You guys should try making a cool banner out of fractals  :lol:
I'd be more than happy to provide an original design.

As for the chant, here it is. Anyone from MIT should know this by heart anyways:
(note that it's modified)
e to the u, du/dx
e to the x, dx
Cosine, secant, tangent, sine
3.14159
Integral, radical, mu lamda!
Slipstick, Sliderule,
San Francisco Bay Area!

How about "Integral radical mu dA/Slipstick sliderule SFBA"?

I like the fractal idea for a banner, but it won't work if I'm going to be sewing it together.  I believe sewing a fractal onto a banner would require infinite time.

I think we might be able to say that the act of sewing a fractal on a banner is NP-complete.
Goodluck with that, and I like the change to the cheer ;-)

Edit: just dropped my laptop, and I notice that I completely broke the screen.

That's a costly error.
Darn.

I'll have to be more careful.

[quote="Ubemaya"]Banner design: See my avatar.

Cheer: TEAPOT TEAPOT TEAPOT TEAPOT ...

Problem solved.[/quote]

Woah those ideas are so cool! We should do that!!

C'MON PEOPLE AT LEAST HAVE A TEAPOT *SOMEWHERE* ON THE BANNER!!! AND THE WORD "TEAPOT" *SOMEWHERE* IN THE CHEER!!!

hmmmmmmm....

how bout...

TEAPOT! YAY!
GO SFBA!

you must admit it pwns

this chant does not make any sense

you are just chanting random mathematical objects

[quote="archimedes1"]One question,  why is it San Francisco Bay Area instead of just plain Bay Area?  Is there another Bay Area ARML group?[/quote]
[b]San Francisco Bay[/b] Area...   :wallbash:


MATHCOUNTS
AMC
USA(J)MO
USAMO
AIME
geometry
perpendicular bisector

Let ABC be an isoscel triangle(AB=AC), <A=80, and M is a point in the interior of ABC so that <MBC=30 and <MCB=10.Find the angle <AMC

Um...i dont think that there is enough information to solve this problem, maybe i am missing something but everything i tried cannont be solved without making some assuptions that well cant be assumed.

yes, there is enough information to solve this

That's because it's always unimpossible for you mr. unimpossible! you're just sooooooo amazing. note the obvious sarcasm. xD.
This shouldn't be in mathcounts ~_-...

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185447[/url]

They look similar don't they? Not exactly the same though, but you're given just as little information.  Other than the fact that it's isosceles of course.

I made this picture from Paint. Just working on the algebra now.

the algebras kinda impossible without trig/construction

i dont know you would take triangle ACM and construct a congruent triangle off of side AB with AB=AC

[quote="hm29168"]Whoops. I didn't draw it to scale, just kind of put it together really fast.  :blush: [quote]i dont know you would take triangle ACM and construct a congruent triangle off of side AB with AB=AC[/quote][/quote]

huh..no i was just trying to guess where to construct a trinalge since this problem cant be done regularly..like I was saying to take a triangle and put a congruent one on that off the other side so you will split up a section into two trianlges, ones isoscles and the other one is some other tirangle.

Like M'MA is 50 degrees, but I cant find M'MC.

But if the answer were 70 degrees, that means something is wrong with my drawing since it looks to be >90 degrees.

[quote="mathcrazed"]The solution is actually much easier than the USAMO problem simply because it's isosceles.

[hide="Solution"]Extend $ BM$ to a point $ P$ such that $ \angle BCP \equal{} 30^\circ$
Notice that triangle $ BCP$ is isosceles with $ BP \equal{} PC$.  We thus have that $ \angle BPC \equal{} 120^\circ$
Also note that triangle $ ABP$ is congruent to triangle $ APC$.  Thus, we have that $ P$ is the fermat point, where all angles at $ P$ are $ 120^\circ$.

Now, we know that $ \angle CAP \equal{} 40^\circ$ by triangle sum, and because we have that $ \angle BCM \equal{} 10^\circ$ as given, we can see that $ \angle PCM \equal{} 20^\circ\longrightarrow\angle PMC \equal{} 40^\circ$.

Since $ PC$ is shared by both triangle $ PMC$ and $ PAC$, the two triangles are congruent which implies that $ AC \equal{} MC$.  Thus, since $ \angle ACM \equal{} 40^\circ$, $ \boxed{\angle AMC \equal{} 70^\circ}$[/hide]
Nice problem.[/quote]
Cough look at the solution button cough.

To answer the question that the future had posted but just deleted  :wink:

Why is triangle APB congruent to triangle APC?

Since triangle $ BPC$ is isosceles, $ BP\equal{}PC$.  We are also given that $ AB\equal{}AC$. Thus, since $ \angle ABP\equal{}\angle ACP$, we have by side-angle-side that they are congruent.  Or you could simply use that fact that $ P$ must lie on the perpendicular bisector of $ BC$ since it is equidistant from both sides.  This obviously hits $ A$, which implies that it is also the angle bisector of $ \angle BAC$.

[quote="the future"]sorry i was looking at two triangles but i realized it so i deleted it...

anyways nice solution!

yea but this is like AIME 8+, not mathcounts[/quote]

AIME???????? [b]8+[/b]????????, r u joking???

uhh no i wouldnt be surprised if this was like USAMO #4 or something...thats why it should be moved

have you seen the problem yet, once  you try to draw  the diagram, youll see that angle chasing goes nowhere.

mathcrazed's solution is very unique

possibly USAMO #1 or #4.

those problems usually just need creativity, not as much of the amazingly awesome and unbeleivably difficult math that people think of when they hear "USAMO"

 :P

Yea... well it still shouldn't be in MathCounts...
I've seen a lot of harder problems on MC forums recently...

yes i wouldnt be surprised to see this problem on USAMO #4, becuase by angle chasing you get nowhere so it takes an intense soltuion like the one by mathcrazed(nice one by the way!!!) to solve it.  At first glanse i thought one couldnt even solve it and untill i saw mathcrazed solution i was stumped.

This is relatively simple in comparison to USAMO problems.  Because you are given the triangle is isosceles, it makes the problem much easier.

a nice problem indeed...I found it in a old book from an old books shop:D....and I smashed my head for 3 days to solve it...Here's my method:
Just make a construction( $ AR \bot BC$) and let $ {Q}\equal{}BM \cap AR$ with Q on AR.Then results that QBC isoscel => $ <QCM\equal{}20\equal{}<ACQ$. $ <MQC\equal{}120\equal{}<AQC$ so $ \Delta AQC \equiv \Delta QCM$ => $ AC\equal{}MC$ => $ \Delta ACM$ isoscel => $ <AMC\equal{}70$ .

I have one more...this one I think is a little bit easier...It depends..
[b]In the triangle ABC we have <C=90, and consider D on AC and K on BD so that <ABC=<KAD=<AKD.Prove that BK=2DC.[/b]

These problems are much harder than mathcounts level  :wink: Could a mod move this?

You're right...It was my mystake to post them here...


analytic geometry
calculus
calculus computations

A small insect at point  $(4, 3, 1)$ is  inside a cage in the shape of the hollow rectangular region $0 \leq x \leq 5 , 0 \leq y \leq 5 , 1 \leq z \leq 4$ Suppose the temperature at any point  $(x, y, z )$  in the cage is $T( x, y, z) = 30 + x^2 + 2y^2 - z^2$. In order to cool off as soon as possible, the insect flies from its position at the bottom of the cage such that it experiences the maximum possible rate of cooling at each of the points of its trajectory. What are the coordinates of the point at which the insect reaches the top of the cage?

Help on how to do this?

The insect should travel in the direction of the gradient, $\nabla T = <2x,4y,-2z>$.


videos
geometry
geometric transformation
reflection

First let me qualify this post by saying that I am asking this question mostly out of theoretical curiosity and not as a request for actual advice.  I've already finished and sent in most of my applications anyway.

Alright, on to the real point of my post.  I happen to be very, very good at certain video games, specifically the Guitar Hero and Rock Band video games.  To give you an idea, on Rock Band 2, which was released a few months ago, I have the highest score on the leaderboards on all ten of the ten hardest songs in the game on expert guitar, and I am ranked first in the world in terms of total score on Rock Band 2 expert guitar.

Obviously I spend a pretty decent amount of my time playing these games, so it's a significant part of who I am.  But if I talked about this on a college application, would it help, hurt, or not affect my admissions chances?

For the record, I have left this out of my applications.  I'm just curious.  :wink:

They might think that you are extremely addicted and prone to skipping class to play video games (which is probably not true :P ), but seeing as you probably have much more on your application, I don't think it would hurt. Of course, it's just my opinion. :wink:

Obviously if you have the highest score in the world in something, it won't hurt your application.

The fact that you show excellence and achievement in academics, and show that you are pursuing those interests seriously as well, will dispel any worries about you wasting time in college. In fact, it shows that you can have a serious, fun commitment and still manage your time.

Here are what I think are the potential positive impacts this can have on an application:

1. It shows you are highly talented
2. It shows you are being honest about what you've been doing
3. It makes your application more colorful
4. It shows you can seriously pursue many interests at once/time management
5. It shows that you can have fun/that you have a life

Having said that, don't worry or double-think yourself. I'm sure you have an impressive application without that, too. And if you feel like you should submit it separately, just make sure to do so in a way that will highlight 1-5 above (for example, in the form of an updated resume, in a separate hobbies section or something).

Yeah, I am also curiosed...
what happens if someone put "WCG Wacraft III Campeon" on the college application?
It is a international competition and is rigurous. How could college view it?

If you've won any [b]large[/b] (international or national) competition, then you can consider placing it on your application. But you'll have to think hard at what category that fits under. (Obviously, putting to show you're hard-working in academics would be self-contradictory).

But how would you prove that you are on the leaderboard? In my opinion, there are less benefits to go to all that work. Accounts are easily hacked on Xbox LIVE - and finding an admissions officer who understands video games is quite rare.

If you're MLG at any particular game - then you should definately put that on your applicatoin. It'll show that you balanced your academics with a job (they get paid a pretty nice amount too) as well as your interests.

@Matt: Have you ever considered joining MLG for Rock Band or Guitar Hero? It's too late now - I doubt you'll hit their ranks in time for the college applications but that should be something you should do in the future. Who knows where it'll lead you?

The OP asked a genuine question -- he has the best score in the world at something, and he wants to know if it counts, and I think it does. Win an international DDR championship, and then you get to ask the same question. In the meanwhile, don't expect DDR to affect your college apps significantly, and don't use it as an excuse to play DDR.

That's a really interesting question.  In general, if you are extremely good at something, you should say so, even if it's not something academic or a "respectable" extracurricular.  However, to say that you spend a ton of time on video games, frankly, does not look good.  Playing video games is widely considered frivolous at best, and a harmful waste of time and talent at worst.  I say this as someone who has interviewed college applicants.

As much as I hate to admit it, I think you did the right thing by leaving it off your application.  In terms of risk vs reward, the risk here is greater than the reward.  But here's an example of a situation in which it might be worth including:  (1) You have strong evidence that you are a very hard worker and not at all a slacker.  For example, if you have evidence that you spend long hours on less frivolous pursuits that are objectively time-consuming.  (For example, if you log long hours on debate, football team, community service, violin, etc.)  You'd want to present yourself as a dynamo and a perfectionist, not some guy who spends too much time on video games.  (2) Your video game accomplishments should be easy to understand for the uninitiated and clearly impressive.  For example, if you receive cash prizes or sponsorship deals, or were featured on MTV or something.  (3) You come off as socially well-adjusted.  (4) Don't make a big deal out of it.  Just mention it as one of the many things you do.

An appropriate place to mention it would be when an interviewer asks you about your hobbies.  And here's a general piece of advice for interviews:  Try to think of something interesting and intelligent to say about your hobbies, and think of how to present your hobby as something interesting for the uninitiated.

Helpful hint:  If your application suggests in any way that you are a "lazy smart person" (I'm sure you all know what I mean), DO NOT say anything about video games.

[quote="yenlee"][b]Playing video games is widely considered frivolous at best, and a harmful waste of time and talent at worst.[/b]  I say this as someone who has interviewed college applicants.[/quote]

Interesting; do most/all college interviewers have this viewpoint? I only play Xbox two hours [i]a week[/i] and if college interviewers believe that video games are a complete waste of time then I've lost all respect for them.

Playing video games does not mean you waste all of your free time. What about playing sports? Instruments? Colleges recieve hundreds of thousands of applications - many of them include impressive titles in sports or dazzling awards in music. Anything worthy in video games will clearly stand out. I agree that a main entry of video games would not be the best - but if you do couple that with other extracuriculars (volunteering, or even the music and sports, etc.) then you have one hell of an application.

The difference is that sports are fundamentally positive -- they build at least one of the following:

- Your health
- Teamwork
- Leadership
- Discipline
- Time management
- Other things I haven't mentioned

Someone who participates in a sport, even if they are not very good, will not be wasting their time. Similarly, music and several other things are fundamentally positive. There is research to show this. Also, colleges are interested in people to take part in music groups and athletic teams.

On the other hand, there is plenty of research showing the drawbacks of television and video games. So it is much harder to show that a video game is the best use of that particular chunk of time by some meaningful measure than it is for sports or music.

Does that mean that they're necessarily a bad thing? Obviously not. If you have a particularly strong talent in them, and still manage to be very competitive in everything else, then noone can criticize you. There are other situations as well -- say Bob is really into programming, and is interested in a career in game design or software design, and is an accomplished programmer, and also plays lots of video games. That is in line with a productive interest!

@7h3.D3m0n.117:  I think you misunderstood what I was saying.  Saying that video games are "frivolous at best and a waste of time at worst" is not at all a condemnation of playing video games or an attack on anyone who plays video games.  For example, you said that you play video games two hours per week.  Any true gamer would agree with me that this is quite a small amount of time spent on video games, and I would not characterize this as a huge waste of time.  Do I still think that it's frivolous?  Yes.  But does that mean that it reflects poorly on you?  No.  Playing video games for two hours per week is neither a good thing nor a bad thing.  It's like watching your favorite TV show.  Try reading my post again.  You'll see that I said absolutely nothing implying wild statements like, "Playing video games means that you waste all of your free time."

On the other hand, there are people who play video games something like 4+ hours a day.  That's something many would consider to be a "harmful waste of time and talent."  And therein lies the issue underlying the original question posed by matt276eagles.  If you're the best in the world at anything, that means one of two things:  (1) you've spent a lot of time and dedication becoming the best; or (2) the thing that you're the best at is silly and very few people care about being the best at it.  So when someone makes a credible claim of being one of the world's top Guitar Hero players, my honest natural reaction (inside my own brain) is simultaneously, "Wow, that's pretty cool," and, "You must spend a lot of time playing video games."  Similarly, if someone said that he was an accomplished pianist, tennis player, archer, tap dancer, or whatever, I would have the same thoughts.  Many people would say that practicing archery for hours on end is not a waste of time, but that practicing Guitar Hero for hours on end is a waste of time.  Personally, I'll stay neutral on the issue of archery vs Guitar Hero, mainly because I have no decent argument in favor of archery over Guitar Hero.  My previous post merely reflected the fact that there are many others who are far less sympathetic to video game playing than I am.

But maybe I'm just totally wrong.  I posed the question to my wife the other day, and her response was, "If you're world-class at anything, why wouldn't you put it on your application?  Leaving it out would be stupid."  The surprising thing to me was that my wife is the one who's never been interested in video games and has a generally negative opinion of them, while I'm the one who grew up loving them.  So you should probably just ignore what I say.

I think listing it is definitely the way to go.  It's a pretty awesome achievement (congratulations!) and if you write about it the right way I don't think it will have the same sort of connotations as being #1 in the world at, say, Call of Duty.

I think I would side with yenlee here - I look at this issue as a risk-reward situation, and it's a pretty significant risk for a fairly small reward, even to admissions officers who do look at these accomplishments positively.  Plus, video games doesn't really fit with the image that I want to portray on my applications, if that makes sense.  My early action decisions have gone well, so I don't think I'm going to change anything.  :)

The reason I think this is so interesting is, what makes being good at video games any different from being good at any other non-academic activity?  Just because our culture generally frowns upon video games?

7h3.D3m0n.117: MLG is really weak for Guitar Hero and Rock Band.  Almost all the GH/RB competition is on http://www.scorehero.com (absolutely amazing site).

[quote="matt276eagles"]The reason I think this is so interesting is, what makes being good at video games any different from being good at any other non-academic activity?  Just because our culture generally frowns upon video games?[/quote]

Unlike many other application-worthy activities, video games is pretty much a natural born skill. If you're rich, you can pay your way with lessons and spend hours practicing any particular thing. Violin, tennis, swimming, chess, or math, it doesn't matter. If you were born to suck at video games, then you're really not going to go from "terrible" to "world-class" no matter how much money you pay or how many hours you log.

I think the general stereyotype against video games in our society is unfair. It takes just as much talent as sports or music, but it doesn't get the credit.

And as for ScoreHero, if it gets the national attention like MLG, then that certainly works. I know that the ProCircuit has added new games every year to it's competition (recently, I believe Call of Duty 4 and Rainbow Six Vegas 2 was added). Super Smash Brothers Brawl was also added before as a special event - perhaps Guitar Hero or Rock Band could be added in the future.

[quote="7h3.D3m0n.117"][quote="matt276eagles"]The reason I think this is so interesting is, what makes being good at video games any different from being good at any other non-academic activity?  Just because our culture generally frowns upon video games?[/quote]

Unlike many other application-worthy activities, video games is pretty much a natural born skill.[/quote]

Umm... what? I see absolutely no evidence of this -- in both directions. Lots of "application-worthy" activities are natural born skill. If you are exceptionally good at something, e.g. playing a sport/instrument, solving math problems, etc., you probably have some natural born skill in that area. As for video games, I don't think being a great gamer is natural born skill any more than being a great pianist is. It requires a lot of effort and time investment and, of course, some natural skill.

I don't see many real distinctions between video games and other non-academic activities. People probably just think that being good at video games doesn't build the same skills as being good at other things, but I could disagree with that. I'm pretty sure I put DDR on my apps and its on the video game border.

[quote="paladin8"][quote="7h3.D3m0n.117"][quote="matt276eagles"]The reason I think this is so interesting is, what makes being good at video games any different from being good at any other non-academic activity?  Just because our culture generally frowns upon video games?[/quote]

Unlike many other application-worthy activities, video games is pretty much a natural born skill.[/quote]

Umm... what? I see absolutely no evidence of this -- in both directions. Lots of "application-worthy" activities are natural born skill. If you are exceptionally good at something, e.g. playing a sport/instrument, solving math problems, etc., you probably have some natural born skill in that area. [/quote]

Sorry for not being clear; it's true that many activities are natural born skills - but you can hardly pay for lessons in video games. If you're not naturally good at video games, no amount of practice or lessons will help (depending on the game of course; but most application-worthy games are like this). For example, I'm better than the average Halo 3 player. But no amount of time I log on will allow me to compete at the MLG or professional level.

On the other hand, being profficient at say, soccer, means that while you have some natural skill, you've most likely spent a lot of time playing for teams, going to practice, and getting coached.

7h3.D3m0n.117, actually you can pay for lessons in video games. Many Halo 3 MLG pros offer lessons for money.

I also have to disagree with you when you say "If you're not naturally good at video games, no amount of practice or lessons will help...". Practice helps everyone; it helps a lot. I've seen many people (including myself) who weren't gifted at Halo get better. It takes time and effort. I'm still pretty bad, but I understand what it takes to be a pro. I think it's definitely possible for most people to become pros if they are extremely determined and have the time.

[hide="my rant that turned out too long"]
If you don't believe me, let's compare Halo 3 MLG with another activity, say basketball. By looking at the qualities that pros must have, you should see the resemblance.

1. The Player
-In Halo 3, you must be proficient with several weapons, and in particular you would be expected to consistently get "4-shots" with the battle rifle. You must not have bad natural tendencies, which can be unlearned. For example, some people tend to charge in to the heat of battle, or they may travel in straight, predictable lines, both of which are easily fixed.
-In basketball (all sports), you must be proficient in some position(s) in which you have designated tasks. You may be expected to play good defense and to knock down most of your shots, among other things.

2. The Team
-In Halo 3, teamwork is the key. Good teams must have good strategies, or at least strategies that they can execute well. Good teams must also play well together. Players must understand each team member's play style.
-In basketball (all sports), teams run plays, and they can live or die by how well they run their plays. Good teams will execute plays better because of how their players tend to "gel" after awhile. This comes from knowing your team mates, and them knowing you.

2.5 Communication
This might as well be its own point since it it so important, but it really falls under the team.
-In Halo 3, "callouts" are crucial to success. Callouts let your team know what you can see on your screen. This information may give them some advantage in their locations, which benefits the team.
-In basketball, players must be on the same page. The player bringing the ball up the court needs to let the team know what play they should run. Players must notify each other of screens coming. In general, it helps to vocalize most things in game even if they aren't necessarily happening outside a player's field of vision, because it's not the players' responsibility to know who's looking where, but to aid each other when possible in order to benefit the team.

3. The Game
-In Halo 3, you must know everything about the game: the maps, the gametypes, the callouts, where weapons spawn, where people spawn, how long it takes for weapons to respawn, etc. You must be comfortable playing on each map, in any gametype, in any position.
-In basketball, you should obviously know how to play, and you must be comfortable playing.
[/hide]

in either case, the road to success would look something like this
-develop good fundamentals
-hone skills, become proficient in several things
-find a team, develop as a teamplayer
-mesh with your team, continue to develop as player
-???
-profit

This process isn't easy for anyone, even the naturally talented. You may be closer to the truth about certain videogames, including gh/rb, which tend to be more individual based and require sheer memorization (not that they don't require skill or hard work). But it's ridiculous for you to say in your examples that Halo 3 requires natural born talent and soccer requires hard work. The parallels between the two are infinite. On top of that, I think you should know that many Halo 3 teams do have coaches, and that MLG signed a contract with ESPN earlier this year! Not so different, are they?


limit
trigonometry
real analysis
real analysis unsolved

Let $f ( X, Y ) \in \mathbb{Q}[ X, Y ] .$ Prove that \begin{eqnarray*}\lim_{x \longrightarrow \infty}\frac{1}{x}\sum_{k \leq x}f ( \sin k, \cos k ) & \in & \mathbb{Q}\end{eqnarray*}

$f(\sin\theta,\cos\theta)$ is a trigonometric polynomial. That limit will turn out to be equal to the constant term of that particular polynomial. I know I've posted that proof somewhere here (probably in a problem to show that $\{e^{ik}\}$ is dense in the unit circle), but I don't have time to look right now.


Before they went into the math shop, Albert had 100 dollars and Beth had 50 dollars. At the shop, they spent a total of 88 dollars. After this, Albert had 3 times as much money left as Beth did. Who spent more money, Albert or Beth, and how many dollars more?

How would you do this problem?

[hide]We can make a system of equations:

$a=3b$ and $a+b=88,$ where $a$ is the amount of money Albert has after going shopping, and $b$ is how much Beth has after shopping.

We find that $a=\$66$ and $b=\$22.$

Albert spent $\$34$ and Beth spent $\$28,$ so Albert spent $\boxed{\$6}$ more than Beth.[/hide]

[hide="solution"]plug in variables for the equation...
$s=3r$ and $s+r=88$, where $s$ represents money albert has left and $r$ represents the same for beth. so, plug in 1st substitute first equation into teh second...
$3r+r=88$
$4r=88$
$r=22$
$s=3(22)$
$s=66$
We subtract this from money theu originally had, and albert spent $34$ dollars and beth spent $28$ dollars. so, albert spent $6$ more dollars than Beth.[/hide]


You collect trading cards, which have different scenes from a movie. For one movie, there are 90 different cards in the set, and you have all of them except the final scene. To try to get this card, you buy 10 packs of 8 cards each. All cards in the pack are different, and each of the cards is equally likely to be in a given pack. Find the probability that you will get at least one copy of the final scene. HINT: think backwards.

The probability of NOT finding the scene in a given pack is $ \frac{\binom{89}{8}}{\binom{90}{8}}\equal{}\frac{82}{90}\equal{}\frac{41}{45}$, and the probability of not finding it in any of the 10 packs is $ \left(\frac{41}{45}\right)^{10}$. So the probability of getting the scene is $ 1\minus{}\left(\frac{41}{45}\right)^{10}$.


Let  $ABC$ be a triangle and let $X,Y$ and $Z$ be points on the sides $[BC],[CA]$ and $[AB]$, respectively, such that $AX=BY=CZ$ and $BX=CY=AZ.$ Prove that triangle $ABC$ is equilateral.

Angle chasing? I fear this is not enough. My solution uses a more complicated approach:

[color=blue][b]Problem.[/b] Let ABC be a triangle, and let X, Y, Z be points on the segments BC, CA, AB such that AX = BY = CZ and BX = CY = AZ. Prove that triangle ABC is equilateral.[/color]

[i]Solution.[/i] We shall first prove:

[color=blue][b]Lemma 1.[/b] Let UVW and U'V'W' be two triangles satisfying VW = V'W' and WU = W'U'. Assume that $\measuredangle WUV+\measuredangle W^{\prime}U^{\prime}V^{\prime}\leq 180^{\circ}$. Then, we have $UV\leq U^{\prime}V^{\prime}$ if and only if $\measuredangle UVW\geq\measuredangle U^{\prime}V^{\prime}W^{\prime}$, and $UV\geq U^{\prime}V^{\prime}$ if and only if $\measuredangle UVW\leq\measuredangle U^{\prime}V^{\prime}W^{\prime}$.[/color]

[i]Proof of Lemma 1.[/i] Let V* be the point such that triangle UV*W is congruent to triangle U'V'W' and such that the points V and V* lie in different halfplanes with respect to the line WU (the congruence is possible, since WU = W'U'). The congruence of triangles UV*W and U'V'W' yields V*W = V'W', UV* = U'V', < WUV* = < W'U'V' and < UV*W = < U'V'W'. Thus, the inequality $\measuredangle WUV+\measuredangle W^{\prime}U^{\prime}V^{\prime}\leq 180^{\circ}$ becomes $\measuredangle WUV+\measuredangle WUV^{\ast}\leq 180^{\circ}$; but this inequality actually tells us that the point W lies inside the angle VUV*. Hence, either < UVW = < UVV* + < V*VW and < UV*W = < UV*V + < VV*W, or < UVW = < UVV* - < V*VW and < UV*W = < UV*V - < VV*W. But < V*VW = < VV*W (since triangle VWV* is isosceles, because V*W = V'W' = VW); thus, in both cases, we can say that $\measuredangle UVW\geq\measuredangle UV^{\ast}W$ holds if and only if $\measuredangle UVV^{\ast}\geq\measuredangle UV^{\ast}V$. But $\measuredangle UVV^{\ast}\geq\measuredangle UV^{\ast}V$ holds if and only if $UV\leq UV^{\ast}$ (since in a triangle, the greater side lies opposite to the greater angle; this fact was applied to triangle VUV* here), and this is equivalent to $UV\leq U^{\prime}V^{\prime}$ (because UV* = U'V'). Combining, we obtain that $\measuredangle UVW\geq\measuredangle UV^{\ast}W$ holds if and only if $UV\leq U^{\prime}V^{\prime}$. But since < UV*W = < U'V'W', we can rewrite this as follows: $\measuredangle UVW\geq\measuredangle U^{\prime}V^{\prime}W^{\prime}$ holds if and only if $UV\leq U^{\prime}V^{\prime}$. Similarly, we can show that $\measuredangle UVW\leq\measuredangle U^{\prime}V^{\prime}W^{\prime}$ holds if and only if $UV\geq U^{\prime}V^{\prime}$. This proves Lemma 1.

Now to the solution of the problem:

Consider the triangles ABX and BCY satisfying BX = CY, XA = YB and $\measuredangle XAB+\measuredangle YBC\leq 180^{\circ}$. (Actually, $\measuredangle XAB+\measuredangle YBC\leq 180^{\circ}$ follows from $\measuredangle XAB<\measuredangle CAB$, $\measuredangle YBC<\measuredangle ABC$ and $\measuredangle CAB+\measuredangle ABC<180^{\circ}$ since the sum of two angles of a triangle is always < 180°.) Hence, applying Lemma 1 to these triangles, we see that $AB\geq BC$ holds if and only if $\measuredangle ABX\leq\measuredangle BCY$. But $\measuredangle ABX\leq\measuredangle BCY$ rewrites as $\measuredangle ABC\leq\measuredangle BCA$, and this is equivalent to $CA\leq AB$ (since in a triangle, the smaller angle lies opposite to the smaller side). Hence, we have shown that $AB\geq BC$ holds if and only if $CA\leq AB$. Similarly, $AB\leq BC$ holds if and only if $CA\geq AB$, and similar assertions can be found by cyclic interchange.

Now, let's combine: First, $AB\geq BC$ holds if and only if $CA\leq AB$; then, $CA\leq AB$ holds if and only if $BC\geq CA$; finally, $BC\geq CA$ holds if and only if $AB\leq BC$. Hence, $AB\geq BC$ holds if and only if $AB\leq BC$. In other words, either both $AB\geq BC$ and $AB\leq BC$ hold, or none of $AB\geq BC$ and $AB\leq BC$ holds. But the latter case (none of $AB\geq BC$ and $AB\leq BC$ holds) is impossible; hence, both $AB\geq BC$ and $AB\leq BC$ hold. This yields AB = BC. Similarly, BC = CA. Thus, AB = BC = CA, so that triangle ABC must be equilateral. Problem solved.

Hope this was correct...

  Darij


Show that, for any nonnegative integer $n$, \[\sum_{k=0}^{n}k{n \choose k}^{2}=n{2n-1 \choose n-1}.\]

[quote="Farenhajt"]$k{n\choose k}=k\cdot\frac{n!}{k!(n-k)!}=n\cdot\frac{(n-1)!}{(k-1)!(n-k)!}=n{n-1\choose n-k}$

Therefore $S=n\sum_{k=0}^{n}{n\choose k}{n-1\choose n-k}$

The term of the latter sum is the number of simultaneous choices of $k$ elements from the first set of $n$ elements and $n-k$ elements from the second set of $n-1$ elements. Summing all the possibilites from $k=0$ to $k=n$, we actually get nothing else than choosing a total of $n$ elements out of the conglomerate set of $2n-1$ elements, i.e. ${2n-1\choose n}$, which is of course equal to ${2n-1\choose n-1}$, hence the result.[/quote]
Thanks. That's a nice solution.  :)

I myself got to \[S=n\sum_{k=0}^{n}{n\choose k}{n-1\choose n-k},\] but I couldn't figure out what to do after that.  :ewpu: :fool:
Now I just feel dumb.  :oops:

When products of binomial coeff's are involved in a sum, it's often a very useful trick to imagine a combinatorial situation in which such a sum would arise and then to simplify that situation.

Or you could use the fact that this is Vandermonde's theorem.

[quote="1234567890"][quote="Farenhajt"]Summing all the possibilites from $k=0$ to $k=n$, we actually get nothing else than choosing a total of $n$ elements out of the conglomerate set of $2n-1$ elements.[/quote]

I don't follow this......how is this true? :maybe:[/quote]

Imagine that the elements of the first set are painted white and the elements of the second set are painted black. Now put them all together and choose any $n$ of them. Among those $n$ you'll have $k$ white and $n-k$ black, where $k$ can be anywhere between $0$ and $n$, and any such choice of $n$ elements can be uniquely split into two parts, where the elements of the first part belong to the first set (i.e. they are white) and the elements of the second part belong to the second set (i.e. they are black). Does that clear the picture?

[quote="1234567890"]Yup, thanks!

Do you have any other good combinatoric sum bijection problems you could post?[/quote]
All right, here's another one (that looks very similar  :P ):
Show that \[\sum_{k=0}^{n}k^{2}{n \choose k}=n(n+1)2^{n-2}.\] I don't think this problem uses bijections though...


algebra
polynomial
function
linear algebra
puzzles

Hi, I've been trying to work out this puzzle for a while but can't solve it. If you could help me I'd be much appreciated.

Here's a number sequence:
111, 112, 21123, 1312, 2114, 121, 25262, 73123...
Find the rule and post two next numbers in the sequence.

I obviously haven't solved it but I have made some observations about it:

- the first number only uses digits for binary numbers; the second only for base 3 numbers; the n+1th only for base n numbers
- none of the digits is zero
- longer and shorter numbers seem to be mixed

Thanks.

Is there anyway to simplify that or is that as simple as it gets?

He's only remarking that arbitrary sequence problems are very annoying to most people on this board because there isn't a straightforward way to approach them, since no information is given about how the sequence might be generated.

The person who proposed the problem knows, and the way he did it is probably very simple.  The point is that there isn't a straightforward way to figure it out, since it could be [b]anything[/b] - recursion of a strange function, recursion with the digits, not a recursion at all...

Another way of putting it is that you could make the next two numbers anything that you wanted by using linear algebra to create an arbitrary polynomial that contains these points.  However, I'm sure that the person who created this sequence probably didn't have that in mind anyway.

What are the next two numbers in the sequence tho since my calculator won't let me put all of the function in ?

[quote="modeler"][quote="Hawk Tiger"][quote="modeler"]All sequence can be expressed by a polynomial

The function:
f(x)=(-113.521825396825)*x^7+(3659.56944444444)*x^6+(-48388.4027777778)*x^5+(337820.361111111)*x^4+(-1331011.44444444)*x^3+(2914152.56944445)*x^2+(-3208777.13095238)*x^1+(1332769)
satisfy your condition
f(1) = 111
f(2) = 112 ...

f(9),f(10) are extremely small negative number[/quote]
Did you use computer to solve this problem?[/quote]

I used a program named [hide="Graphmania"] which is created by me :D [/hide][/quote]
You can use Lagrange Interpolation Polynomial.

This is not a math qn! This is just a play with the alphabets.

We first notice that the following words:
one, two, three, four, five, six, seven, eight
have $3, 3, 5, 4, 4, 3, 5, 5$ letters respectively, which correspond to the number of digits of the terms.

Next, we notice that for each of the digit of each term, the digit actually counts the number of times the corresponding letter has been used.

For example, starting with the first term, none of the letters have been used, so letters o, n & e are each used for the first time, so the first term is $111$. for the 2nd term, since o has already been used in the first term, it is used for the 2nd time in 'two', while the letters 't' and 'w' are used for the first time.

The next two terms in the sequence would be $3448$ and $495$.


This inequality is Romania SL seniors.Is it easy? :huh:

[quote=Inequalities Master]Let $ a,b,c,d\geq0$ s.t.$ a\geq b\plus{}c\plus{}d$.Prove that:
$ (a\plus{}b)(b\plus{}c)(c\plus{}d)(d\plus{}a)\geq20abcd$[/quote]

Let $f(a)= (a+b)(b+c)(c+d)(d+a)-20abcd$.
Hence, $f'(a)=(2a+b+d)(b+c)(c+d)-20bcd\geq(3b+3d+2c)(b+c)(c+d)-20bcd\geq6\sqrt{bd}\cdot4c\sqrt{bd}-20bcd\geq0$.
Thus, it remains to prove that $f(b+c+d)\geq0$ or
$(2b+c+d)(b+c)(c+d)(b+c+2d)\geq20(b+c+d)bcd$ or
$c^4+4(b+d)c^3+(5b^2-8bd+5d^2)c^2+(2b^3+2d^3-10b^2d-10bd^2)c+2b^3d+2bd^3+5b^2d^2\geq0$.
Since $4(b+d)\geq8\sqrt{bd}$ and $5b^2-8bd+5d^2\geq2bd$ and $2b^3+2d^3-10b^2d-10bd^2\geq-16\sqrt{b^3d^3}$ and $2b^3d+2bd^3+5b^2d^2\geq9b^2d^2$,
it remains to prove that $x^4+8x^3+2x^2-16x+9\geq0$ for $x\geq0$, which is obvious.


geometry
3D geometry
percent
percent increase

A cube has edge length 2. Suppose that we glue a cube of edge length
1 on top of the big cube so that one of its faces rests entirely on
the top face of the larger cube. The percent increase in the surface
area (sides, top, and bottom) from the original cube to the new
solid formed (i.e., both cubes together) is closest to: \\ \\
$ \text{(A)}\ 10 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 17 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 25$ \\ \\
Select the most appropriate letter.

Original surface area: $ 6\times{2^{2}} \equal{} 24$

New surface area $ 24 \minus{} 1$(area covered by new cube)$ \plus{} 5$(new area)$ \equal{} 28$

$ \frac {28}{24} \equal{} \frac {7}{6} \equal{} 116.666666666666666666$%, which rounds to $ 117$

C


AMC
USA(J)MO
USAMO
inequalities
geometry
number theory

I keep hearing about different MOP colors.  What do they mean, and how do you get into each level?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107664
Red=Freshman
Blue=Sophmore/Junior
Black=Seniors

Yeah,

Black = USAMO Winners (plus a few honorable mentions usually as well)
Blue = Next however many
Red = freshmen

Freshman can make any group but most of them only score high enough for red.  Seniors are only eligible for black.

Last year a few freshmen scored well enough to make Blue group but were placed in Red anyways.  None of them would have made Black though

[quote="SplashD"][quote="bpms"]
Red=Freshman
Blue=Sophmore/Junior
Black=Seniors[/quote]

I thought Red was only for freshman and black and blue were just dependent of your skill level[/quote]
Sorry, Blue should be <12th grade.
Black should be anyone.

[quote="13375P34K43V312"]If you're like a sophomore doing your first time at MOP and you feel like you got lucky making the blue cutoff, can you demote yourself to red even though you're not a freshman?[/quote]

No, but there have been sophomores in the red group.  Only the first time they had it though, because not enough freshman qualified for the USAMO.

I'm pretty sure that won't be happening anymore though, with the expanded USAMO and stuff.

[quote="Phelpedo"]Why would you? You'd probably learn more in Blue unless you were a 5th grader who accidentally guessed all the right answers.[/quote]

Say you studied inequalities only, but knew squat about geometry and little number theory.

What if it turns out that #1 is really trivial number theory, and #5 is an inequality that you can do easily since you studied them really hard? You get a 7 on both and end up with a 14, making a very low Blue cutoff, since #2, #3, and #6 were impossible questions that only like 3 people each solved. Then you make blue, but know you aren't good enough to succeed there.

Trust me :  it just doesn't happen.  If you can get into blue MOP, you can cope with blue MOP material.  And I doubt that the blue cutoff has ever been that low, by the way.

[quote="Boy Soprano II"][quote="Phelpedo"]Why would you? You'd probably learn more in Blue unless you were a 5th grader who accidentally guessed all the right answers.[/quote]
... which would be impossible.  Anyhow, that is contradictory to MOP mentality, and once you got there, you would know better.  Besides, you can prosper in MOP even if you barely made the cutoff.  For instance, uh, someone made MOP by 0 points and still got 8th place in red group.

By the way, I thought it was only Delong (hence singular freshman) who might have theoretically qualified for blue group?  At least, I think that might've been what Eric said one time, because I don't think any red MOPpers other than Delong and Eric would have almost been blue MOPpers.[/quote]

I think Delong got a 21 and Eric got an 18.


1. Solve 45x + 9x2 = 0

2. Solve for h; given A = 1/2bh

Please don't post the same problem twice (#1).

Anyway...

On 1, do you mean $45x+9x^{2}=0$?

[hide="hint for two, post what you can get on it"]$A=\frac12bh$.

You want to isolate $h$ on one side of the equation...[/hide]

[hide="1"]$9x^{2}+45x=0$

Divide by $9$ to get $x^{2}+5x=0,$ or $x(x+5)=0.$ This means $x=\{-5,\,0\}.$[/hide]

[hide="2"]To solve for a variable, all other numbers and/or variables must be placed on the other side of the equation/inequality.

$A=\frac{bh}{2}$

$2A=bh$

$h=\frac{2A}{b}$[/hide]

[quote="i_like_pie"][hide="1"]$9x^{2}+45=0$

Divide by $9$ to get $x^{2}+5=0.$ Then, $x^{2}=-5,$ or $x=\pm\sqrt{-5}.$ This means $x$ in an imaginary number. $x=\{-\imath\sqrt{5},\,\imath\sqrt{5}\}.$[/hide]

[hide="2"]To solve for a variable, all other numbers and/or variables must be placed on the other side of the equation/inequality.

$A=\frac{bh}{2}$

$2A=bh$

$h=\frac{2A}{b}$[/hide][/quote]

For 1, you solved the wrong equation  :wink:

[hide="1"]This is a quadratic equation.  Note that both terms in it have a common $9x$ factor, so we use the distributive property to factor it out.

$9x(x+5)=0$

Can you solve this know?

[hide="another hint"]The only way to make 0 the product of two numbers is for one (or both) of the numbers to be equal to 0.

The two numbers are $9x$ and $(x+5)$.[/hide][/hide]

[quote="Mark9182"]Sorry since I am a new member I am only able to post every 10 minutes.

Would that make the answer -5.

Also

That is exactly what I meant on that problem.  Thanks for the help.  The detailed explanations help tremendously.  On the hyperbola problem

The answer is
 x2/81  -  y2/40 = 1

You had a + instead of a -.  Could you check on that for me?[/quote]
[hide="a mistake you made with the first"]The answer could be 0 as well...[/hide]

[quote="b-flat"][quote="Mark9182"]Sorry since I am a new member I am only able to post every 10 minutes.

Would that make the answer -5.

Also

That is exactly what I meant on that problem.  Thanks for the help.  The detailed explanations help tremendously.  On the hyperbola problem

The answer is
 x2/81  -  y2/40 = 1

You had a + instead of a -.  Could you check on that for me?[/quote]

I am not sure what you are talking about...

$a^{2}+b^{2}=c^{2}$? Is that it?
If it is $a^{2}-b^{2}=c^{2}$, then the equation is an ellipse, not a hyperbola.[/quote]
No, he is talking about your post in the other topic:

[quote="b-flat"][hide="3"]It is best to start these with a picture of where the points are.

We know that the equation of a hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.

When $y=0$, we know that $x=\pm9$, so we can substitute this information into our equation and solve for $a^{2}$.

$\frac{(\pm9)^{2}}{a^{2}}-\frac{0^{2}}{b^{2}}=1$
$\frac{81}{a^{2}}=1\Rightarrow a^{2}=81$

We also know that the foci are located at $(\pm11,0)$.

The $x$ coordinate of the foci represents $c$ in the equation $a^{2}+b^{2}=c^{2}$
Substituting what we know from earlier, we have
$81+b^{2}=121$
$b^{2}=40$

Thus, the equation of the hyperbola is $\frac{x^{2}}{81}+\frac{y^{2}}{40}=1$.[/hide][/quote]
You put a $+,$ but Mark says it's supposed to be a $-.$

Ahh, yes, that was a typo.

Thanks, I will fix it (it should be +, which leads to the answer I gave).

On my answer key it says.

x2/81  -  y2/40 = 1

You definitely helped me figure out how to get there but I want to make sure that it is

x2/81  -  y2/40 = 1
and not
x2/81  +  y2/40 = 1

I edited it (see my previous post in this topic).

Welcome to AoPS!!

If you need any other help, etc, just post your problems, or you can PM me & I will get back with you.


If I had data from the entire class, and wanted to find the standard deviation for a lab, would I use the sample or population deviation, and why?

Thanks in advance!

sample deviation :rotfl:  :rotfl:  :rotfl:

Why would sample population be better?

Thanks!

Do you have data to account for every member of the population?

If so... daynnnnnnnng..........

....'nuff sed

[quote="feynman_wannabe"]Do you have data to account for every member of the population?[/quote]

That's what I'm not sure of.  Our class designed an experiment to shoot "rockets" (pipet bulbs).  We all measured the distances that our rockets went.  Do I have the population data or sample data if I have data from the *entire* class?

Thanks!

Population would refer to EVERY pipet bulb that could possibly be made. I doubt u have that data....Usually we always go with sample data..

Loving how this is so relevant to the forum. :rotfl:


algebra
polynomial
function
college contests

Hey,

Ex. 1.2.1
Prove that the product of four consecutive natural numbers cannot be the square of an integer.

What I need help is just critique of my proof. I want to make sure it is very rigorous (and ofcourse correct).

Attempt at a Solution:

Let,

$ {n \in \mathbb{N}}$
$ {\mathbb{N} = \{1, 2, 3, . . . ,\infty\}}$

$ {m \in \mathbb{Z}}$
$ {\mathbb{Z} = \{ - \infty, . . . , - 1, 0, 1, . . . , \infty\}}$

Prove that:
$ {n(n + 1)(n + 2)(n + 3)} \neq {{m}^{2}}$

(Proof by Contradiction)

Assume the Contrary:
$ {n(n + 1)(n + 2)(n + 3)} = {{m}^{2}}$

$ {{n}^{4} + 6{n}^{3} + 11{n}^{2} + 6{n}} = {{m}^{2}}$

Let,

$ f(n) = {{n}^{4} + 6{n}^{3} + 11{n}^{2} + 6{n}}$

Then, since $ f(n)$ is a fourth degree polynomial. Then for,
$ {{n}^{4} + 6{n}^{3} + 11{n}^{2} + 6{n}} = {{m}^{2}}$

there must exist some $ g(n)$ such that,
$ {f(n)} = {[g(n)]}^{2}$

If $ g(n)$ exists, then it must be a second degree polynomial--since $ f(n)$ is a fourth degree polynomial--fitting one of the following forms:

$ {(1) g(n) = a{n}^{2}}$
$ {(2) g(n) = a{n}^{2}} + c$
$ {(3) g(n) = a{n}^{2}} + b{n}$
$ {(4) g(n) = a{n}^{2}} + b{n} + c$

Consider case (1), this is not possible for $ f(n)$ since $ g(n)$ would lack a term of degree three.
Consider case (2), this is not possible for $ f(n)$ since $ g(n)$ would also lack a term of degree three.
Consider case (3), this is not possible for $ f(n)$ since $ g(n)$ would lack a term of degree one.
Consider case (4), this is the only possibility for $ f(n)$ since $ g(n)$ would contain terms of degrees: four, three, two, and one.

However, in order for,
${ g(n) = a{n}^{2}} + b{n} + c$

It must satisfy,
${ f(n) = {(a{n}^{2}} + b{n} + c)}^{2}$

Which when expanded is,
$ f(n) = {{a^2}{n}^{4} + {2ab}{n}^{3} + {(2ac + b^2)}{n}^{2} + {2bc}{n} + {c^2}}$

Where,
$ f(n) = {{n}^{4} + 6{n}^{3} + 11{n}^{2} + 6{n} + 0}$

Must match as follows,

$ a^2 = 1$
$ 2ab = 6$
$ 2ac + b^2 = 11$
$ 2bc = 6$
$ c^2 = 0$

However if,
$ c^2 = 0$

Then,
$ 2bc \neq 6$

Contradiction.

However, here is my question...

Is there anything I missed in this proof? Anything I should have done more rigorously?

Anything I left unaddressed?

Thanks,

-APSStudent

The proof is just wrong. Indeed, this way you can easily prove that $ n^3$ cannot be a square, which is certainly false (just take $ n\equal{}4$). Your approach raises an interesting question though: assume that all values of a polynomial with integer coefficients are perfect squares. What is the easiest way to prove that the polinomial is the square of another polynomial with integer coefficients?

Hey,

Thanks for the reply, fedja.

[quote="fedja"]
The proof is just wrong. Indeed, this way you can easily prove that $ n^3$ cannot be a square, which is certainly false (just take $ n \equal{} 4$).
[/quote]

Consider, $ n \equal{} 4$

$ f(4) \equal{} 840$

Then with proof by contradiction, does there exist some integer $ m$ such that,

$ 840 \equal{} m^2$

No. The closest is,

$ 29^2 \equal{} 841$

One less than a perfect square, which in fact is always the case for $ f(n)$, that is $ f(n)$ will always be one less than a perfect square for any natural number, $ n$.

What is your point?

[quote="fedja"]
Your approach raises an interesting question though: assume that all values of a polynomial with integer coefficients are perfect squares. What is the easiest way to prove that the polinomial is the square of another polynomial with integer coefficients?
[/quote]

Consider the following,

$ h(n) \equal{} [j(n)]^2$

I guess we would let,

$ h(n) \equal{} {{a_{1}}{n}^{j}} \plus{} {{a_{2}}{n}^{j\minus{}1}} \plus{} {\cdot}{\cdot}{\cdot} \plus{} {{a_{k\minus{}1}}{n}^{1}} \plus{} {{a_{k}}{n}^{0}}$

$ {{a_{1}}{n}^{j}} \plus{} {{a_{2}}{n}^{j\minus{}1}} \plus{} {\cdot}{\cdot}{\cdot} \plus{} {{a_{k\minus{}1}}{n}^{1}} \plus{} {{a_{k}}{n}^{0}} \equal{} [h(n)]^2$

However, I do not exactly see how this helps with my proof.

As to considering what is the easiest way to prove the polynomial is the square of another polynomial with integer coefficients. That is tricky, I'm not really sure.

Thanks,

-APSStudent

Read my post attentively! :roll: I said that your "proof" alllows to conclude (without any change in the argument) that $ f(n)\equal{}n^3$ can never be a square (indeed, there exists no polynomial $ g$ such that $ n^3\equal{}g(n)^2$: degree 1 is too small and degree 2 is too large). But $ f(4)\equal{}4^3\equal{}64\equal{}8^2$. So something in your argument should be wrong ;)

If you want a correct argument, follow up on that "one less than a perfect square" idea. That's much more specific than "not the square of a polynomial".

A polynomial sequence $ p(n)$ can have some values perfect squares without all of them being perfect squares. Even this one does, at $ n\equal{}0,\minus{}1,\minus{}2,\minus{}3$.

Hey,
[quote="fedja"]
Read my post attentively! :roll: I said that your "proof" alllows to conclude (without any change in the argument) that $ f(n) = n^3$ can never be a square (indeed, there exists no polynomial $ g$ such that $ n^3 = g(n)^2$: degree 1 is too small and degree 2 is too large). But $ f(4) = 4^3 = 64 = 8^2$. So something in your argument should be wrong ;)
[/quote]
Well your not very clear.

I let,
$ f(n) = {n(n + 1)(n + 2)(n + 3)} = {{m}^{2}}$

Now, if you want to say,
$ f(n) = m^3$

That is very different from,
$ f(n) = n^3$

That is another thing.
Now, following through with,

$ f(n) = {n(n + 1)(n + 2)(n + 3)} = m^3$

We would take the same approach as I described except, we let
$ {f(n)} = {[k(n)]^{3}}$

And follow through with the neccesary steps for the proof.
Additionally, if you wanted to show that,
$ f(n) = {n(n + 1)(n + 2)(n + 3)} = n^3$

Assuming if it was true that,
$ f(n) = [g(n)]^{2}$

(Which is not as proved by (Proof by Contradiction))

Then yes as you said, there exists no polynomial such that,
$ [g(n)]^{2} = {n}^{3}$

However, as your point is made we do find numbers that satisfy this,
$ {{(8)}^{2}} = {{(4)}^{3}}$

Although your point is made, I do not see how that in anyway questions the rigority of my proof.
As my proof states,

$ {{n}^{4} + 6{n}^{3} + 11{n}^{2} + 6{n}} \neq {{m}^{2}}$

Since,

$ {{n}^{4} + 6{n}^{3} + 11{n}^{2} + 6{n}} \neq {[g(n)]}^{2}$

Where is the problem?
=================================================================================
Thanks for the reply jmerry.
[quote="jmerry"]
If you want a correct argument, follow up on that "one less than a perfect square" idea.
[/quote]
Thanks for the tip. While I do already realize that,
$ f(n) \neq {m}^{2}$

And instead always equals,
$ f(n) = {m}^{2} - 1$

I did not like starting the proof from this statement. Since in order to have reached the above conjecture one would have had to create a table showing that for several terms of $ n$ the yield is always some perfect square minice one. Which I thought was not the most rigorous approach to take, since you had to have assumed it was true for all natural numbers, n.
[quote="jmerry"]
That's much more specific than "not the square of a polynomial".
[/quote]
Hmm, I don't exactly understand why there would be a preference between,

(Prove)
$ f(n) = {m}^{2} - 1$

versus

(Prove by Contradiction)
$ f(n) \neq {[g(n)]}^{2}$

The reason being is would not,

$ f(n) \neq {[g(n)]}^{2}$

imply,

$ f(n) \neq {m}^{2}$
[quote="jmerry"]
A polynomial sequence $ p(n)$ can have some values perfect squares without all of them being perfect squares. Even this one does, at $ n = 0, - 1, - 2, - 3$.
[/quote]
[quote="APSStudent"]
$ {n \in \mathbb{N}}$

$ {\mathbb{N} = \{1, 2, 3, . . . ,\infty\}}$
[/quote]

In other words, $ n$ is a natural number and therefore,

$ n \neq 0, - 1, - 2, - 3$

Thanks,

-APSStudent

Just as a side note not related to the proof. You should be a little careful with notation on $ \mathbb{Z}$ and $ \mathbb{N}$. It seems that you have included $ \minus{}\infty$ and $ \infty$. You don't want $ m$ or $ n$ to take those values. Most people just use the "$ \ldots$" without the "$ \infty$" to indicate that.

The whole point of fedja's post is that $ f(n) \ne [g(n)]^2$ does NOT imply that $ f(n) \ne m^2$.

His example was $ f(n) \equal{} n^3$; clearly $ f(n) \equal{} [g(n)]^2$ is not true ($ g(n)$ a polynomial with integer coefficients), but there exists $ n$ (specifically 4) so that $ f(n) \equal{} m^2$ (with $ m \equal{} 8$).

Your posts are extremely annoying to read because of all the white space. You don't need to give every equation its own line, and you really don't need to add an additional blank line before and after each equation.

You have proved the statement "The function $ f(n)\equal{}n(n\plus{}1)(n\plus{}2)(n\plus{}3)$" is not the square of a polynomial function." This does not imply the desired result, "None of the values of the function $ f(n)\equal{}n(n\plus{}1)(n\plus{}2)(n\plus{}3)$ at positive integers $ n$ are perfect squares." Both fedja's example and my note that $ f(0)\equal{}0$ is a perfect square indicate the problems with this step. If you try to justify it- there's just nothing there.

Now, if we use the more specific fact that $ f(n)\equal{}(n^2\plus{}3n\plus{}1)^2\minus{}1$, we're almost there. If $ a$ and $ b$ are integers with $ a^2\minus{}b^2\equal{}1$, $ a\equal{}\pm 1$ and $ b\equal{}0$. This means that $ f(n)$ can only be a perfect square for integer $ n$ if $ n^2\plus{}3n\plus{}1\equal{}\pm 1$ and $ f(n)\equal{}0$, which only occurs at $ n\equal{}0,\minus{}1,\minus{}2,\minus{}3$. None of these is a positive integer, and we are done.

Hey,
[quote="jmerry"]
Your posts are extremely annoying to read because of all the white space. You don't need to give every equation its own line, and you really don't need to add an additional blank line before and after each equation.
[/quote]
I've gone back and removed some of the blank lines, should be better now.
[quote="jmerry"]
You have proved the statement "The function $ f(n) \equal{} n(n \plus{} 1)(n \plus{} 2)(n \plus{} 3)$" is not the square of a polynomial function." This does not imply the desired result, "None of the values of the function $ f(n) \equal{} n(n \plus{} 1)(n \plus{} 2)(n \plus{} 3)$ at positive integers $ n$ are perfect squares." Both fedja's example and my note that $ f(0) \equal{} 0$ is a perfect square indicate the problems with this step. If you try to justify it- there's just nothing there.
[/quote]
Your note that $ f(0) \equal{} 0$, does not apply to this proof. Since in the beginning $ n$ was stated as a natural number. More clearly,

$ n \neq 0$

Additionally, this brings up an interesting detail about the perfect squares, noting that the only two consecutive perfect squares are: $ 0$ and $ 1$.
Addressing fedja's example, he is correct that had $ f(n)$ been unknown then indeed if,

$ f(n) \neq {[g(n)]}^{2}$
this does [u]not[/u] imply,
$ f(n) \neq {m}^{2}$

However, since $ f(n)$ is known. If,

$ {{n}^{4} \plus{} 6{n}^{3} \plus{} 11{n}^{2} \plus{} 6{n}} \neq {[g(n)]}^{2}$
does this imply that,
$ {{n}^{4} \plus{} 6{n}^{3} \plus{} 11{n}^{2} \plus{} 6{n}} \neq {m}^{2}$

I think that seems to be the part I am unclear on.
[quote="jmerry"]
Now, if we use the more specific fact that $ f(n) \equal{} (n^2 \plus{} 3n \plus{} 1)^2 \minus{} 1$, we're almost there. If $ a$ and $ b$ are integers with $ a^2 \minus{} b^2 \equal{} 1$, $ a \equal{} \pm 1$ and $ b \equal{} 0$. This means that $ f(n)$ can only be a perfect square for integer $ n$ if $ n^2 \plus{} 3n \plus{} 1 \equal{} \pm 1$ and $ f(n) \equal{} 0$, which only occurs at $ n \equal{} 0, \minus{} 1, \minus{} 2, \minus{} 3$. None of these is a positive integer, and we are done.
[/quote]
Right. Realizing early on that $ f(n)$ could be factored via the difference of perfect squares in to,

$ f(n) \equal{} (n^2 \plus{} 3n \plus{} 1)^2 \minus{} 1$

Would have seemed like the most direct approach. From there the details as you have shown are trivial.
However, I wanted to try my approach to the proof and see if it was just as rigorous.

It seems though, that mine's is not as specific. Does that mean it is not as rigorous?

Thanks,

-APSStudent

Your original argument is not a proof. Both fedja's example and mine are counterexamples to that argument; nothing you said rules them out. The hole is not fixable; just use a different argument.