Putnam Harvard college Princeton MIT Stanford Columbia

Here we go! The mail came today.

(I have to do some data processing with this, so information is going to trickle out over the afternoon. Be patient with me.)

First, the headlines.

[b]Winning Teams:[/b]

1. Harvard (Zachary Abel, Iurie Boreico, Arnav Tripathy)
2. Princeton (Peter Z. Diao, John Pardon, Adrian Zahariuc)
3. MIT (Qingchun Ren, Xuancheng Shao, Yufei Zhao)
4. Stanford (Young Hun Jung, Nathan Pflueger, Jeffrey Wang)
5. Caltech (Jason Bland, Zeb Brady, Brian Lawrence)

[b]Putnam Fellows:[/b] (In alphabetical order)

Brian Lawrence, Caltech
Seok Hyeong Lee, Stanford
Arnav Tripathy, Harvard
Bohua Zhan, MIT
Yufei Zhao, MIT

[b]Elizabeth Lowell Putnam Prize:[/b]

Victoriya Krakovna, Toronto

[b]Group N1:[/b]

Iurie Boreico, Harvard
Adam Hesterberg, Princeton
William Johnson, Univ. of Washington
Cedric Lin, Univ. of British Columbia
Anton Malyshev, Princeton
John Pardon, Princeton
Qingchun Ren, MIT
Oleg Rudenko, Florida Atlantic
Colin Sandon, MIT
Jacob Steinhardt, MIT
Alex Zhai, Harvard

[b]Group N2:[/b]

Zachary Abel, Harvard
Thomas Belulovich, MIT
Jason Bland, Caltech
Gabriel Bujokas, MIT
Konstantin Matveev., Toronto
Nathan Pflueger, Stanford
Aaron Potechin, Princeton
Dong Uk Rhee, Waterloo
Adrian Zahariuc, Princeton

[b]Honorable Mention Teams:[/b]

Duke
Michigan
Rochester
Toronto
Waterloo

Scores ran somewhat higher this year than last. Out of 3627 takers, the mean score was about 9.5 and the median score was 1. The five Putnam Fellows scored 117, 110, 108, 102, 101. (And, before anyone asks: the question of which of the Fellows matches which of those scores is not public information. I don't know it. The Fellows don't know it themselves. The faculty advisers at the Fellows' schools don't know it. It's never going to be announced. So don't ask.)

The minimum score to be a Putnam Fellow was 101 (as opposed to 82 last year).

The minimum score for group N1 was 87 (as opposed to 71 last year).

The minimum score for group N1 was 80 (as opposed to 64 last year).

The minimum score for Honorable Mention was 60 (as opposed to 45 last year).

The minimum score for Group I was 54 (as opposed to 42 last year).

The minimum score for Group II was 40.7 (as opposed to 30.3 last year).

The minimum score for the top 500 was 22 (as opposed to 21 last year).

This year the "top 500 list" had 496 names, with the lowest rank being 473. Last year, that list had had 455 names, with the lowest rank being 431.5

47.2% of all people who took the exam scored zero.

is there any mcgill students on the list? Thanks

I was HM, 66 pts.
I wonder what rank that corresponds to...

McGill:

Ioan Filip, Maksym Radziwill, and Thu Troung were in the top 500.

Canadian students in the top 200 were confined to just three universities: Toronto, Waterloo, and UBC.

======

A score of 66 was a rank of 54. There were three people who scored exactly that.

Who were the Stanford HMs?

Also, Stanford did way better than I expected. I never knew Seok Hyeong Lee was so good.

Stanford HM's:

Jeffrey Chen
Jackson Gorham
Marcello Herreshoff
Young Hun Jung
Yong Suk Moon
David Philipson (who was on the SoCal ARML team as a 12th grader)
Jeong Soo Sim
Kiat Chuan Tan
Jeffrey Wang (now there's someone with a lot of posts on AoPS)

Oh, and Albanian Eagle: you were HM from Caltech. So also were
Yakov Berchenko-Kogan
Sam Elder
Benjamin Zax

You're one of them - and one of three HM's that were ever on the Southern California ARML team. (Yeah, I know, you were an 8th grader at the time.)

The full list of MIT HM's:

Hansheng Diao
Michael Forbes
Andrei Frimu
Jeremy Hahn
Shaunak Kishore
Tony Liu
Ye Luo
Haitao Mao
Kevin Modzelewski
Thanasin Nampaisarn
Eric Price
Anand Rajagopalan
Krishanu Sankar
Xuancheng Shao
Jason Trigg
Alexandr Zamorzaev

You posted 50 out of the top 75.  Can you post the other 25 and their schools?  Thanks.

I compared those problem difficulty numbers to 2007, by calculating average scores. The difference is huge; of twelve pairs of corresponding problems, 2007 was easier only in B1 and B3. On the hard end, 2007's A5, A6, and B6 were all harder than any 2008 problem.
This only applies to the top students; at the level of ordinary takers, the tests were quite similar in difficulty. The easy problems A1-A3 and B1 did require some proof-writing skill.

OK, the rest of the HM's:

Farzin Barekat, British Columbia
George Boxer, Princeton
Kevin Carde, Harvard (comes from my area but never on our ARML team)
Robert Cordwell, New Mexico (was on our ARML team once - long story)
Peter Diao, Princeton
Samuel Dittmer, Brigham Young
Zhou Fan, Harvard
Simion Filip, Princeton
Daniel Hermes, Michigan
Matthew Jordan, Maryland
Viktoriya Krakovna, Toronto
Rosen Kralev, Harvard
Mikhail Lavrov, Duke
Xue Liu, Princeton
Keenan Pepper, Florida State
Aaron Pollack, Duke
William Shore, Brown
Yi Sun, Harvard
Kenneth Tay, Princeton
Andrei Ungureanu, Princeton
Dmitry Vaintrob, Harvard
Neal Wadhwa, Harvard
Kun-Chieh Wang, Purdue
Tengyao Wang, Princeton
Gjergji Zaimi, Caltech (I implied him but didn't actually name him above).

Wow, I was on an internship and had to write the Putnam in a different University.  Apparently that somehow cost me a space on the team (I have been told that I was on the team).  I suppose that the organizers have trouble locating my paper for some reason...
I don't know my score but I suspect it might be lower than the person who replaced me, so at least this might have worked to the team's benefit I think

I'm also curious about Caltech people in the Top 500, as well.

Can someone please post the results of students from [b][color=red]Can[/color][color=white]ad[/color][color=red]ian[/color][/b] schools?  :wink:

[quote="MysticTerminator"]
allowing an all-hindsight team (or just adding the top three scores) gives a sizable advantage to larger colleges over smaller ones...[/quote]

First of all, I doubt this is true. But secondly, even if it were, so what? Doesn't pretty much every Team Math contest work by adding the scores of the top X students? No one complains about it at any other math contest.

I think they should take the sum of the top 4 or maybe 5 students from each school. That rewards schools with a few individual students of excellence AND a depth of strong students. This whole thing about picking students to represent the school beforehand just makes the school rankings on the Putnam meaningless to me. It is not an accurate representation of the strength of the top students from each school who take the Putnam.

Actually, I hear that the original purpose of the this ranking system is to prevent elite schools (namely Harvard) from taking place first every single time. Indeed, during the initial years of the competition, Harvard was the only "top" school competing, and the organizers wanted to add some randomness and unpredictability into the school ranking.

And the tradition stuck.

Yeah, the team ranking system is not about being "fair" or "right". It's just history and tradition.

In any case, MIT has been pretty consistent in its team rank in the past three years. Some people say that we always "pick the wrong team." I disagree with that notion -- when your school has THAT many good people (just look at the number of MIT students in each category, and compare them to, say Harvard or Princeton), it becomes next to impossible to predict correctly each time. Our system, which just picks the top ranking students from the previous year, is basically the only fair and reasonable thing to do. (For example, holding a selection exam would NOT count as reasonable, given how much effort it would involve here.) It is fair also because it awards the students who did well in the previous year. When it finally comes down to it, what matters on exam day is simply luck. And I was pretty lucky this year.

U. Virginia:

Wuttisak Trongsiriwat was in Group II. That's it for the top 500.

Canadian universities: Toronto and Waterloo were both Honorable Mention teams, meaning top 10 but not top 5. Victoriya Krakovna of Toronto was awarded the Elizabeth Lowell Putnam Prize.

All of the top 200 students in Canada came from Toronto, Waterloo, and the University of British Columbia. The ranks at these places:

Toronto: N2, H, I, I, II
Waterloo: N2, I, II, II, II, II, II, II
UBC: N1, H, II

That makes the single highest scoring Canadian Cedric Lin of UBC.

In the top 500, UBC had 5 more, Toronto 5 more, and Waterloo 14 more. In addition, the following colleges placed people in the top 500:

U. Calgary (1)
Carleton U., Ottawa (1)
U. Guelph (1)
McGill (3)
Memorial U. of Newfoundland (1)
U. Quebec (2)
Queen's U. (1)
U. Sherbrooke (1)
Simon Fraser (1)
U. Victoria (3)
U. Western Ontario (1)
U. Winnipeg (1)

As for Caltech: there were 11 people in the top 200 (P, N2, H, H, H, H, I, I, I, I, II) and 8 more in the top 500. I don't particularly feel like listing them all, other than to note SoCal ARML alums Jeff Manning (I), Yingkun Li (I, with his name misspelled as "Yingkum"),  and Isaac Chao (top 500, see three posts up), as well as Yasha Berchenko-Kogan (H) and Tim Black (top 500) who have been helping with the team this year.

Kent, have you considered just scanning the results and uploading them?

Because it contains mailing addresses and I'm not going to upload that.

Hi... I was wondering if there is anyone from Hobart and William Smith Colleges on the list.  The faculty member in charge of our Putnam team spent the semester in Italy, so none of us have heard about scores yet.  It's driving me (and other people) crazy...

Hobart and William Smith didn't have anyone in the top 500.

The schools in the state of New York that did:

NYU
Columbia
U. of Rochester
Cornell
CUNY Hunter College
RPI
Bard College
Vassar College
SUNY Binghamton
Rochester Inst. Tech.

As to whether anyone at your school got 10 or 12 or 20 points - you'll just have to wait to find that out later.

Thank you for letting me know!  Hopefully we will hear about results soon.

Hi Professor Merryfield,

I know this is incredibly late, but would you be able to give specifics about who at Cornell placed in the Top 500 (for the 2008 competition of course; are the 2009 competition results even released yet?)? Thanks so much.

(I'm hoping to take the Putnam in 2010, albeit without nearly enough background to do so...)

Cornell:

Group II (rank between ~100 and ~200)
Wei Quan Julius Poh
Zheng Tham
Seth Weinberg

The rest of the top 500:

Kai Fong Ernest Chong
Daniel Collins
Theodore Ni
Pakawat Phaltinonkiat
John Sheridan
Stephen Wolf

As for the 2009 results: check the date on which this topic was started. You've still got two more months to wait.


What is the greatest number of distinct positive integer factors that a positive integer less than 20 can have?

Using brute force, we find that both 18 and 12 have 6 factors. Our answer is $ \boxed{6}$.

There should be a better method...

Well using the formula that a number that factors to:

$ {p_1}^{n_1}{p_2}^{n_2}{p_3}^{n_3}{p_4}^{n_4}...$

has $ (n_1\plus{}1)(n_2\plus{}1)(n_3\plus{}1)(n_4\plus{}1)...$ factors.

For you to have more than 6, you would need a number of the form $ x^6$ to have 7 (obviously the smallest is 64, greater than 20) 8,9,10,etc. AS the list goes on it becomes much less likely that you find one, so 6 is the answer. 8 is the most likely candidate, but you would either need $ xy^3$ the smallest of which is 24, or $ xyz$ the smallest of which is 30(where x y and z are prime numbers)


Let n be a positive integer. Show that the product of $ n$ consecutive positive integers is divisible by $ n!$

Consider the binomial coefficient $\binom{m}{n} = \frac{m(m-1)...(m-n+1)}{n!}$.
It is an integer for all m, so we are done for positive sequences of integers. Clearly this proof works also for strictly negative sequences, and in the case where one of the integers is 0, the product will be 0, and hence divisible by anything.

Let the consecutive +ve integers be (r+1), (r+2)......(r+n)
Then we have

$ (r\plus{}1)(r\plus{}2)(r\plus{}3).......(r\plus{}n)
\equal{}\frac{r!(r\plus{}1)(r\plus{}2)(r\plus{}3).......(r\plus{}n)}{r!}
\equal{}\frac{(r\plus{}n)!}{r!}
\equal{}\frac{(r\plus{}n)!n!}{r!n!}
\equal{}\frac{(n\plus{}r)!n!}{n![(n\plus{}r)\minus{}n]}
\equal{}C(n\plus{}r,n)n!$

Hence the result!

But it's so clear, since when we have $n$ consecutive integers each of them are different in $(mod k)$ where k=<n so at least one term must be $0$ in $(mod k)$

[quote="mavropnevma"]Although your way to express it is not altogether correct, it is true that for any $1\leq k \leq n$ at least one term is divisible by $k$. However, this does not guarantee the product of the terms is divisible by $n!$; think about it ... for example, looking at the numbers $3,4$, for any $1\leq k \leq 4$ there is one of them divisible by $k$, but $4! = 24 \nmid 12 = 3\cdot 4$.[/quote]
yes, understood.

[hide=Pretty much the same as post 3 but whatever]Suppose our consecutive integers are $k,k+1,..k+n-1$. It follows that $k(k+1)...(k+n-1)=\frac{(k-1)!*k(k+1)...(k+n-1)}{(k-1)!}$. Hmm we recognize this as $\frac{(k+n-1)!}{(k-1)!}$. Let's multiply the top and bottom by $n!$, this gives us $\frac{(k+n-1)!n!}{(k-1)!n!}$. This is equivalent to $\frac{(k+n-1)!n!}{((k-1+n)-n)!n!}=\binom{n+k-1}{n}*n!$. Thus it is divisible by $n!$.[/hide]

Let $n$ consecutive positive integers be $m+1,\dots ,m+n(m\in \mathbb N_{0},n\in \mathbb N)$.Then
$\prod_{k=1}^{n}(m+k)=n!\binom{m+n}{n}$
Therefore the proof is completed.$\blacksquare$ :coolspeak:

[hide = wrong]
There exist $n$ distinct residues for any $\pmod{k}$ where $k \geq n$. However for $k < n$, there must exist two numbers in the same residue, which implies that at least one number is $0 \pmod{k}$. Thus, the expression is divisible by anything less than $n$ so it remains to prove that it is divisible by $n$. However, this is clear since every residue is taken (by the $n$ consecutive numbers) so we conclude that $n!$ is a factor of the expression.

@below
Yeah, I forgot to consider that. oops ._.

[quote=lucasxia01]Thus, the expression is divisible by anything less than $n$ so it remains to prove that it is divisible by $n$. However, this is clear since every residue is taken (by the $n$ consecutive numbers) so we conclude that $n!$ is a factor of the expression.[/quote]

Hello,lucasxia01. :lol:

You say that if $A$ is divisible by $1\le \forall k\le n$,then $n!|A$,don't you?
This is wrong.
Counterexample:$A=12, n=4$.$k|12(1\le \forall k\le 4)$,but $4!\nmid 12$.

Your sincerely.

Takeya.O :w00t:


calculus
integration
trigonometry
logarithms
LaTeX
function
calculus computations

$ \int_{0}^{\frac {\pi}{2}}\sqrt {(\frac {1 \minus{} \sin x}{1 \plus{} \sin x})}\ dx.$

Multiplying top and bottom by $ 1 \minus{} \sin x$, we get:
$ \int_{0}^{\frac {\pi}{2}}\frac {1 \minus{} \sin x}{\cos x}dx \equal{} \int_{0}^{\frac {\pi}{2}}(\sec x \minus{} \tan x)dx$

Then we get:
$ \ln|\sec x \plus{} \tan x| \minus{} \ln|\sec x|$ evaluated from $ \pi/2$ to $ 0$=
$ ln|1\plus{}\sin x|$ evaluated from $ \pi/2$ to $ 0 \equal{}$
$ \ln 2$

[quote="kunny"][quote="mismatchtea"]$ \int_{0}^{\frac {\pi}{2}}\sqrt {(\frac {1 \minus{} \sin x}{1 \plus{} \sin x})}\ dx.$

Multiplying top and bottom by $ 1 \plus{} \sin x$, we get:[/quote]

Please edit the part.[/quote]

Thanks kunny...

$ \text{\LaTeX}$ tip: \left( and \right) are better parenthesis for tall symbols like fractions. Other enclosing marks can be modified by \left and \right as well.

That's a rather unsatisfying form- writing a bounded continuous function as the difference of two unbounded functions. It's also an incorrect antiderivative at the end; the two logarithms should be subtracted.

Now, can you evaluate that and get a number? What are you going to do about that $ \infty-\infty$?

My recommendation: use a different identity in the initial simplification. In particular, look for a half-angle identity.

$ \int_0^{\frac {\pi}{2}} \sqrt {\frac {1 \minus{} \sin x}{1 \plus{} \sin x}}\ dx \equal{} \int_0^{\frac {\pi}{2}} \sqrt {\frac {(1 \minus{} \sin x)(1 \plus{} \sin x)}{(1 \plus{} \sin x)^2}}\ dx \equal{} \int_0^{\frac {\pi}{2}} \left|\frac {\cos x}{1 \plus{} \sin x}\right|\ dx$

$ \equal{} \int_0^{\frac {\pi}{2}} \frac {\cos x}{1 \plus{} \sin x}\ dx\ (\because 0\leq x\leq \frac {\pi}{2})$

$ \equal{} \left[\ln (1 \plus{} \sin x)\right]_0^{\frac {\pi}{2}} \equal{} \boxed{\ln 2}$


trigonometry
limit
inequalities unsolved
inequalities

$Let A,B,C be three angle of triangle .Find the minimum value of the follow expression : \sum{\frac{sin A . sin B}{sin^{2}\frac{C}{2}}}$

Let $A=B=t$, $C= \pi-2t$. Then

$\frac{\sin A \cdot \sin B}{\sin \frac{C}{2}}\to 0 \ \textnormal{for}\ t \to 0$

$\frac{\sin C \cdot \sin A}{\sin \frac{B}{2}}= \frac{\sin C \cdot \sin A}{\sin \frac{B}{2}}= \frac{\sin t \cdot \sin 2t}{\sin \frac{t}{2}}\to 0 \ \textnormal{for}\ t \to 0$  (use  [b] l' Hospital [/b] )

Hence $\lim_{t \to 0}\sum_{cyc}\frac{\sin A \cdot \sin B}{\sin \frac{C}{2}}= 0$
Because the summands of $\sum_{cyc}\frac{\sin A \cdot \sin B}{\sin \frac{C}{2}}$ cannot be negative, the infimum is $0$.

EDIT: If $A,B,C \not= 0$, all summands are positive. Hence $\sum_{cyc}\frac{\sin A \cdot \sin B}{\sin \frac{C}{2}}$ does not achieve its infimum.

[quote="Do Tien Dang"]To Pi-J:
  oh..! I think so !
To hunter_cvp_ic
  I think we can find the minimum of $\sum_{cyc}\frac{sinAsinB}{sin^{2}\frac{c}{2}}$
What do you think? :)[/quote]

oh yeah
sorry i'm wrong
 you right  dotiendang
  It's same to my problem
You can solve it ????


AMC
AIME
ARML
USAMTS
National Assessment

UAB -- February 19, 2005
[url=http://mualphatheta.org/]Mu Alpha Theta[/url] Log1 Ciphering -- March 3, 2005
Hoover High School -- March 5, 2005
[url=http://www.unl.edu/amc/e-exams/e7-aime/aime.html]AIME[/url] I -- March 8, 2005
[url=http://www.mathleague.com/]Ala-La-Miss Math League[/url]  Contest 5 -- March 8, 2005
[url=http://rocketcitymath.org/]Rocket City Math League[/url] Round Three --March 7-11, 2005
[url=http://www.asan.com/asa/asma1.htm]ASMA[/url] Contest #6 -- March 10, 2005
[url=http://www.mandelbrot.org]Mandelbrot[/url] Round Four -- March 14-18, 2005
Alabama State Written - March 19, 2005
[url=http://www.unl.edu/amc/e-exams/e7-aime/aime.html]AIME[/url] II -- March 22, 2005
[url=http://www.mathleague.com/]Ala-La-Miss Math League[/url]  Contest 6 -- April 5, 2005
[url=http://www.natassessment.com/]National Assessment & Testing[/url]: Collaborative -- April 13, 2005
Grissom High School -- April 16, 2005
Alabama State Ciphering -- April 23, 2005
[url=http://www.arml.com]ARML[/url] -- June 3-4, 2005
[url=http://mualphatheta.org/]Mu Alpha Theta[/url] National Convention in Hawaii -- August 1-6, 2005

If you know of any other contests that Alabama schools might be interested in, you can post them below and I will add them.

Students interested in tackling problems that are a step harder than typical Alabama math team tournament problems should consider working on [url=http://www.usamts.org]this year's USAMTS problems.[/url]

I would suggest participating in the USAMTS because it helps motivate spending the time working on the problems.  These problems are particularly good for students wanted to learn how to solve some of the harder AIME and also ARML problems.  It is also a great way to develop proof-writing skills as these problems are not as hard as those that appear on the USAMO.

[url=http://mualphatheta.org/]Mu Alpha Theta[/url] Log1 Topic Tests -- January 13, 2005
Gainsville -- January 22, 2005
[url=http://www.mandelbrot.org]Mandelbrot[/url] Round Three -- January 24-28, 2005
[url=http://rocketcitymath.org/]Rocket City Math League[/url] Round Two -- January 24-28, 2005
[url=http://www.unl.edu/amc/]AMC[/url]-10/12 Contest A --  February 1, 2005
[url=http://www.natassessment.com/]National Assessment & Testing[/url]: Four-by-four -- February 3, 2005
Auburn High School -- February 5, 2005
[url=http://www.unl.edu/amc/]AMC[/url]-10/12 Contest B --  February 16, 2005
[url=http://mualphatheta.org/]Mu Alpha Theta[/url] Log1 Ciphering -- March 3, 2005
Hoover High School -- March 5, 2005
[url=http://www.unl.edu/amc/e-exams/e7-aime/aime.html]AIME[/url] I -- March 8, 2005
[url=http://rocketcitymath.org/]Rocket City Math League[/url] Round Three --March 7-11, 2005
[url=http://www.mandelbrot.org]Mandelbrot[/url] Round Four -- March 14-18, 2005
Alabama State Written - March 19, 2005
[url=http://www.unl.edu/amc/e-exams/e7-aime/aime.html]AIME[/url] II -- March 22, 2005
[url=http://www.natassessment.com/]National Assessment & Testing[/url]: Collaborative -- April 13, 2005
Grissom High School -- April 16, 2005
Alabama State Ciphering -- April 23, 2005
[url=http://www.arml.com]ARML[/url] -- June 3-4, 2005
[url=http://mualphatheta.org/]Mu Alpha Theta[/url] National Convention in Hawaii -- August 1-6, 2005

You can use or edit it however you want.

Here's a series that we are taking from 'Math League'.

[url=http://www.mathleague.com/]Ala-La-Miss Math League[/url]  Contest 3 -- January 11, 2005
[url=http://www.mathleague.com/]Ala-La-Miss Math League[/url]  Contest 4 -- February 8, 2005
[url=http://www.mathleague.com/]Ala-La-Miss Math League[/url]  Contest 5 -- March 8, 2005
[url=http://www.mathleague.com/]Ala-La-Miss Math League[/url]  Contest 6 -- April 5, 2005

Hello, I'm back. (Missed USAMTS 3 :/)

Here's another competition we're doing:

[url=http://www.asan.com/asa/asma1.htm]ASMA[/url] Contest #4 -- January 13, 2005
[url=http://www.asan.com/asa/asma1.htm]ASMA[/url] Contest #5 -- February 10, 2005
[url=http://www.asan.com/asa/asma1.htm]ASMA[/url] Contest #6 -- March 10, 2005

Why do I get the feeling ASMA is just another terribly uninteresting math league?

Maybe from the lengthy and uninteresting introductory letter on the site. :) Yeah, there's definitely nothing to set it apart from a lot of the other math leagues we do. Whether you want to keep up with the dates is up to you... I like to know when something is coming up in math team, but it's not essential.

I'll keep it up of course.  I just had a moment of cynicism.

I noticed we didn't have the UAB math tournament (February 19th) up on the list.


function
limit
trigonometry
real analysis
real analysis solved

show that are the function below differentiable at X = 0

             X 2 sin 1/x  ,  X not = 0

             0 , X = 0

Verify it by definiton.

$f'(0) = \lim_{h\to 0} \frac{f(h)-f(0)}{h}=\lim_{h\to 0} \frac{h^2\sin\frac{1}{h}}{h} = \lim_{h\to 0} h\sin \frac{1}{h} = 0$.

 ====

$\lim_{h\to 0} h\sin \frac{1}{h} = 0$  because $sin\frac{1}{h}$ is bounded.

wooo thx ..

but does my way determine whenever it is differentiable at X = 0?

 -1  <= sin(1/x)  <= 1

 x 2 is always bigger then 0

therefore , -x 2  <=  x 2 sin(1/x) <= X 2

Hence when x get closer to 0, x2 and -x2 become very small in magnitude. Therefore any number in between will also be very small in magnitude. In other words, we have

Lim X--> 0 X 2 sin(1/x) = 0

No, you only prove that $f(x)$ is continuous at $x=0$.

what should i add in to prove that its differentiable at that point also?


How does  $ \sum_{k = 0}^n \binom{n}{k}^2$ = $ \2\binom{2n}{n}$?

I know that  $ \sum_{k = 0}^n \binom{n}{k}^2 = \sum_{k = 0}^n \binom{n}{k} \binom{n}{n - k}$, so I tried expanding the series out, but that seemed to get me nowhere.

Any help would be appreciated.

There are $ n$ rocks (numbered from $ 1$ to $ n$) of kind $ A$ and $ n$ of kind $ B$ (numbered from $ 1$ to $ n$ too). I wanna choose $ n$ from these $ 2n$ rocks, and to this I can choose $ k$ from $ A$ and $ n\minus{}k$ from $ B$, $ k$ going since $ 0$ to $ n$.
That gives $ \sum_{k\equal{}0}^n \dbinom{n}{k}\dbinom{n}{n\minus{}k}\equal{} \dbinom{2n}{n}$.

[quote="erudito"]There are $ n$ rocks (numbered from $ 1$ to $ n$) of kind $ A$ and $ n$ of kind $ B$ (numbered from $ 1$ to $ n$ too). I wanna choose $ n$ from these $ 2n$ rocks, and to this I can choose $ k$ from $ A$ and $ n \minus{} k$ from $ B$, $ k$ going since $ 0$ to $ n$.
That gives $ \sum_{k \equal{} 0}^n \dbinom{n}{k}\dbinom{n}{n \minus{} k} \equal{} \dbinom{2n}{n}$.[/quote]

Now I get it! I was stuck on getting my head wrapped around it, now I have to try to prove explicitly. It helps to have a good understanding of it first, thanks.

Another way to see this: $ \sum_{k\equal{}0}^n\binom{n}{k}\binom{n}{n\minus{}k}$ is the coefficient of $ x^n$ in the expansion of $ (1\plus{}x)^n\cdot(1\plus{}x)^n\equal{}(1\plus{}x)^{2n}.$

That's stated in algebraic (generating function-like) terms, but at heart it's really the same explanation as erudito gave.

[quote="Kent Merryfield"]Another way to see this: $ \sum_{k \equal{} 0}^n\binom{n}{k}\binom{n}{n \minus{} k}$ is the coefficient of $ x^n$ in the expansion of $ (1 \plus{} x)^n\cdot(1 \plus{} x)^n \equal{} (1 \plus{} x)^{2n}.$

That's stated in algebraic (generating function-like) terms, but at heart it's really the same explanation as erudito gave.[/quote]

That was an explanation that Gauss gave to me in another thread,with erudito's explanation I understand it a lot better and I now see where Gauss was coming from.


deci cum sa nu te lase parintzi la nationala :|?...mai bine interpretatzi-o ca intrebare retorica ^ ^

daka ma gandesc mai bine...as vrea nijte reply-uri :)

e, asa e maica-sa . il iubeste mult si nu vrea sa se desparta de el, sau ce stiu eu . nu s-o fi dus omu niciodata in vreo tabara, si acuma ar fi prima oara cand pleaca d-acasa ... si d-aia nu ii vine. ce, care e problema ta ca nu il lasa ?

:) nici o problema, dar imi place cum suna :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

This is probably a known result, but I couldn't find it here.

Prove that 2 k-cycles are conjugate in $ A_n$, where $ k\leq n \minus{} 2$

Hint: An odd permutation centralizes such a k-cycle.

Steve

If k<n-1 then there is a 2 cycle that commute with the k-cycle j(that one that transpose the two of the element out of the cycle). But the 2-cycle is an odd permutation. Than Cj the centralizer of j has an odd element, than is simple to see that half of Cj is even and half is odd( in general if you have a subgroup of Sn with an odd element is true the same thing). Then if we act by coniugacy with An we are using half element respect to Sn, but also the stabilizer is half, so the formula of the orbit stabilizer show that the orbit of j is the the same(simplifyng the 1/2). But two k cycle are always conjugated in Sn. Nice excercise.

[quote="Karlenskij"]Then if we act by coniugacy with An we are using half element respect to Sn, but also the stabilizer is half, so the formula of the orbit stabilizer show that the orbit of j is the the same(simplifyng the 1/2). [/quote]
I didn't really understand this last part of your solution and how do you take the conclusion.

Can't we prove that, supposing $ c$ and $ k$ are the k-cycle and $ c$ is disjoint with $ (12)$ wlog.

Then, as we know, $ g c g^{ \minus{} 1} \equal{} k$, with $ g \in S_n$

If $ g$ is odd, we take $ g' \equal{} (12)g \in A_n$ and $ g'cg'^{ \minus{} 1} \equal{} k$ ?

But this is probably false, because $ g$ doesn't need to commute with $ (12)$

:blush:  :blush:  :blush:

How silly.

After all this is very simple.

[quote="rmfgcampos"][quote="Karlenskij"]Then if we act by coniugacy with An we are using half element respect to Sn, but also the stabilizer is half, so the formula of the orbit stabilizer show that the orbit of j is the the same(simplifyng the 1/2). [/quote]
I didn't really understand this last part of your solution and how do you take the conclusion.

[/quote]
Sorry maybe i was'nt enough clear. The argument is: 0)if you take a k cycle j in Sn you know that all the other k-cycle are conjugated to him,conjugating with Sn.1) Now if k<n-1 there is a transposition, so an odd permutation, that commute with j.2) So call $ O_{S_n}(j)$ the coniugacy class of j in Sn.Call $ O_{A_n}(j)$ the conjugacy class of j just doing conjugacy with even permutation;call $ C_{S_n}(j)$ the centralizer of j in Sn; and $ C_{A_n}(j)$ the intersection between $ C_{S_n}(j)$ and $ A_n$. By the orbit stabilizer relation you have that $ |O_{S_n}(j)| \equal{} |S_n|/|C_{S_n}(j)|$ and $ |O_{A_n}(j)| \equal{} |A_n|/|C_{A_n}(j)|$. Using 1) you have that $ |C_{A_n}(j)| \equal{} 1/2|C_{S_n}(j)|$;putting these in 2) and considering that $ |A_n| \equal{} |S_n|/2$ you have that $ |O_{S_n}(j)| \equal{} |O_{A_n}(j)|$, but clearly by definition $ O_{A_n}(j)$ is inside $ O_{S_n}(j)$, then $ O_{S_n}(j) \equal{} O_{A_n}(j)$;using 0) you conclude that a k-cycle is always conjugated to j also conjugating with An, that is the thesis.


calculus
AMC
AMC 12
AIME
AMC 10
AMC 10 A
AMC 12 B

I think there will be some correlation, but it depends a lot on how hard the school/district pushes math advancement.

Class: Advanced/Multivariable Calculus
AMC12: 144 (A), 132 (B)
AIME: 10 (I)

BC Calculus

AMC10A: 150 AMC12B:137

AIME I: 10

Class: AP stat
AMC: bad
AIME: horrid  :(
USAMO: Expected value hopefully statistically significant difference  :)

Calc BC

AMC12B:118
AIME:12

However, I agree with paladin 8 that it depends much more on the district than the student:  once you have finished pre-calc, you are prepared to take the AIME test, and beyond precalc, the math class you take is heavily dependent on the school district you are in.

Multivariable Integral Calculus

AMC 12: 137
AIME: 11

I was fighting with the school to put me in the highest possible math class I could, and I'm sure I could survive a more difficult one. Here in New York State we have an extremely strict Board of Regents and a lot of people willing to enforce it.

Hmmm, as I just moved to this state at the beginning of the year, they refused to put me into calc BC, so I was stuck in stat.
So a lot does have to do with the school, since about every school offers ap calc and most are taking it apparently...

AP calculus BC

AMC12A 138, AIMEI 7
tho i attribute any success (if thats what you would call that  :oops:) to my Honors Precalculus with Trig class.  Best class ever.  And I wanna take it again.  And even tho I still know most of what was taught I'd still benefit a lot from it, it was just that good.  (and some credit to honors advanced algebra, there was a really nice book for that class)

I think it depends on how hard the teacher pushes problem solving (back to my precalc class, the teacher didn't tell us everything we need to know, and scores like 4/10 on quizzes didn't phase people.  the problem solving was tough (with questions that were self written, borrowed from AHSME's, and borrowed from other math contests) with short time limits and quizzes at least every other day)

heh sorry, I didn't mean to get that much in to it.  but ohh how I miss that class.

Alegbra II  (No Precalc/Trig or Calc yet)

AMC 10A: 144
AMC 12B: 130
AIME I: 10

Although I usually can't solve any problems that are purely trig (like AIME I #12, which now that I've seen the solution to seems much easier than I expected), I still often use trig in my solutions to geometry problems (like using Law of Sines/Cosines and using tangent for slope-related problems).  So it doesn't significantly hold me back, since I've taught myself most of the trig I need to know for competitions (except Temple Lightning round  :D ).  Yeah... white horse king and scorpius and jiangwei know what I'm talking about... :lol:

Algebra II/Trig
Grade: 9
AMC 10A : 130
AMC 12B: 112.5 (One bubbling mistake)
AIME II: 4 (One adding mistake, two mistakes reading the question)

MV calc/ Diff. equations

grade: 12

AMC12 A: 150

AIME I: 10

[quote="h_s_potter2002"]I was just wondering if you guys could list what math you're taking in school right now (i.e. Algebra 1, PRe-Calculus, or whatever), and your AMC/AIME scores, since I thought it'd be interesting to compare the two.[/quote]

Class: AP Calculus BC
AMC-12 Score: 123.0
AIME Score: 10

Having taken precalc really does help in terms of AMC-12 and AIME scores. However, I think, ironically, success in USAMO is not affected much by the math class you are in, as all the topics are not part of the high school curriculum (except maybe Graph Theory if you take enough CS classes).

al/geom, which is algebra and geometry in one  year
Grade: 7
AMC 10A: 122.5
AMC 12B: 114.5
AIME: 8
:blush: I know it's bad, but oh well

That's great for a 7th grader. Your AIME score is great for any grade.

Me:
Alg2 last semester. No math this semester.
146.5 AMC10B
7 AIME I

Advanced / Multivariable Calculus
AMC 12A:  126.5 (I was constipated  :( )
AMC 12B:  139.5
AIME I:  11

Really, if you find any correlation, it probably won't be cause-and-effect.  This is a prime example of the [i]cum hoc ergo propter hoc[/i] logical fallacy.  The real correlation is probably due to the fact that people who tend to be better at logical thinking and math in general push both for harder classes and for better AMC/AIME scores, not because their harder classes actually helped them get those scores.

Although there was a problem on the AMC 10A that was very easy to solve with cross products...  :)

This is TRUE:

Grade: 8
Class: Pre-Algebra/Concepts in Algebra I
Estimated AMC 12 score: 119
Estimated AIME score: 6

I don't get this spam going around either...

Grade: 12

Class: AP calculus (No BC stuff is covered, so if you want to take BC then independent study is required)

AMC 12A: 122
AIME: 8

Math Level: Precalculus
Grade: 10
AMC 12 A: 126.0
AMC 12 B: 137.0
AIME: 9

Quit your spamming people  ;)

Not funny anymore

Yeah, what's the point?

Grade: 10
Math class: Ehhh, not really any right now; I've basically taken just multi-calc and matrices in the last two years since taking BC calc in 8th grade...
AMC 12A: 133.5; AMC 10B: 150 :)
AIME I: 10

Math Class - BC Calculus
AMC 12A - 122.5
AMC 12B - 136  :)
AIME ------ 8

Senior
Linear Algebra.

12A: 121.5
12B: 133.5

AIME: 10

Math Class: Pre-Calculus
Grade: 10
AMC 12B: 123
AIME I: 9

[quote="tekno10m"]Grade: 11
Math Class: Algebra II
AMC12: 144
AIME: 10[/quote]
Woah, seriously?

[quote="keeyounglee"]Grade: $\infty$

Class: Grade $\infty$ Math

AMC12: $\infty$

AIME: $\infty-5$ (i fell asleep in the middle)[/quote]

lol  :lol:

Grade: 11
Math: Lie algebras, Representation Theory
AMC12: 100.5
AIME: 4

[quote="JSteinhardt"][quote="keeyounglee"]Grade: $\infty$

Class: Grade $\infty$ Math

AMC12: $\infty$

AIME: $\infty-5$ (i fell asleep in the middle)[/quote]

Grade: $\infty^{\infty}$

Class: Math $\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty}}}}}}}}$

AMC12: ${\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty}}}}}}}}}}}}}}}}}}}}$

AIME: ${\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty^{\infty}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$[/quote]

Wow, your AIME score is higher than your AMC score! That's awesome!

[quote="t0rajir0u"]If $\infty = \aleph_{0}$ then they're all equal, and given his grade and math level that's not so impressive  :rotfl:[/quote]

Math Class: I solved every problem ever in math so math is over...(think Wittgenstein and philosophy)

AMC Score: $\aleph_{\infty}$

AIME Score: $\aleph_{\infty}$

Ha!~ :P


Find a formula for $a_{k}$ if

$\sum_{k=1}^{n}\binom{n}{k}a_{k}=\frac{n}{n+1}$

*Edited

o whoops i made a typo in the problem. but i have fixed it.

[hide]
\[\boxed{a_{k}=\frac{(-1)^{k+1}}{k+1}}\]
The base case $k=1$ is clearly true. For the inductive step, if we can show
\[\sum_{k=1}^{n}(-1)^{k+1}\cdot \frac{n+1}{k+1}\cdot \binom{n}{k}=n\]
then we will be done, because the right hand side is equal to
\[(n+1)\sum_{k=1}^{n}a_{k}\binom{n}{k}=(n+1)a_{n}+\sum_{k=1}^{n-1}(-1)^{k+1}\cdot \frac{n+1}{k+1}\cdot \binom{n}{k}\]
But we have $\frac{n+1}{k+1}\cdot \binom{n}{k}=\binom{n+1}{k+1}$, so that
\[\sum_{k=1}^{n}(-1)^{k+1}\cdot \frac{n+1}{k+1}\cdot \binom{n}{k}=\sum_{k=1}^{n}(-1)^{k+1}\binom{n+1}{k+1}\]
\[=-1+(n+1)+(1-1)^{n+1}=n\]
as desired.
[/hide]


AMC
AIME
AMC 12
USA(J)MO
USAMO
AMC 10
ARML

I'm going into my sophmore year, and I'm not sure if I should take the AMC 10 or 12.  If I do take the AMC 10, then I'll probably have a better chance of making it to the USAMO.  However, I've heard that taking the AMC 12 has some advantages also.  Any comments or advice would be appreciated.   :mrgreen:

-interesting_move

The only difference it will make is whether you can make the AIME. It's probably easier to make AIME with the AMC10. The problems on the AMC12 are more challenging and thus possibly more fun, but in terms of qualifying for the USAMO, you have to qualify for the AIME first, so if you're worried about that, take the AMC10. If not, you have a 150 anyway if you're a sophomore, so do what you want.

[quote]If not, you have a 150 anyway if you're a sophomore, so do what you want.[/quote]
What did you mean by that?   :?

-interesting_move

As long as you meet the mininum AIME score of the juniors and seniors, you're in.

Fierytycoon

I was a Freshman last year but I also had the option of taking the AMC 10 or 12.  I took the 10, but looking back I wouldve taken the 12.  I scored something like a 118.  The passing score on the AMC 10 was either 120 or 126, while the AMC 12 was 100, and the AMC 10 and 12 had at least 10 questions that were the exact same.  Of course ultimately the choice is yours

I see what you mean ChunckyMonkey.  I barely made it to AIME last year with an AMC 10 score of 119.5 (the cutoff score was 119  :o ).  I guess it kinda depends on what your goal is (i.e. if you want to make it to USAMO or just AIME) for your decision on either AMC 10 or 12.

-interesting_move

[quote="interesting_move"]I guess it kinda depends on what your goal is (i.e. if you want to make it to USAMO or just AIME) for your decision on either AMC 10 or 12.[/quote]

how so?

I don't think interesting_move was correct since only AIME score matters for grade 10 and below. It may be easier to qualify for AIME if you take the AMC10 though, so if you are not certain that you will qualify for AIME, I would advise taking the AMC10 (but many others would disagree with me).

It really depends on your goals.  If you want to give a shot at the AMC 12 go ahead, if you make it to AIME great, if not you will have had practice so you can do better your junior and senior years.  If you really want to make it to AIME and beyond, then it might be better to take the AMC 10. I took the AMC 10 in eigth grade and got into AIME and I might be faced with the same dilema next year.  I guy from my school who could have taken the AMC 10 scored 99 or something on the AMC 12.  THERE IS ALSO POSSIBLY AN ADVANTAGE IN TAKING THE AMC 12, that is you can qualify for ARML if you score high enough on the AMC 12. I don't think you can qualify for ARML with the AMC 10, at least that's why some kids from my school took the AMC 12 when they could have taken the AMC 10.  Basically, it comes down to if you want to take a risk or not.  However, the AMC 12 is harder than the AMC 10.  There may be questions that are the same, but questions from the back of the AMC 10 will be close to the beginning on the AMC 12.

[quote="MarkSmith888"]THERE IS ALSO POSSIBLY AN ADVANTAGE IN TAKING THE AMC 12, that is you can qualify for ARML if you score high enough on the AMC 12.[/quote]

That depends on where you live. Some regions send only a single team, but others send many. In San Francisco Bay Area, anyone who wants to go may. I'm not sure how selection procedures work in other regions.

[quote="MarkSmith888"]THERE IS ALSO POSSIBLY AN ADVANTAGE IN TAKING THE AMC 12, that is you can qualify for ARML if you score high enough on the AMC 12. I don't think you can qualify for ARML with the AMC 10, at least that's why some kids from my school took the AMC 12 when they could have taken the AMC 10.[/quote]

I disagree.  Selection of the ARML team is [usually] made by the coach of the team.  A coach is a human, not a machine; he or she will be able to tell whether you're good enough for the team.  Usually, coaches prefer to use other contests like the AMC12 [u]as an aid[/u] to identify who the strongest math competitors are.  However, if a student were to score really high on the AMC10, say 150 or 144, then the coach will probably notice this and think "boy, maybe I should have that guy/girl on my team" or something to that effect.  I doubt you'd miss out on going to ARML if you're good enough but you didn't take the right test.

Besides, there are usually other criteria for selecting a team.  AMC12 questions are very different from some of the questions in ARML (I believe that ARML has some much harder questions) and so the AMC12 wouldn't do a very thorough job of identifying who's really good at ARML.  In short, if you're good enough, you'll go; if you're not, well that depends on where you're from (you might go on B or C team, or you might not go).

I agree with Simon - different places might have different policies.  And again, there are other criteria than just the AMC12.  If you can prove you're good enough, then off you go to ARML.

[quote="MarkSmith888"]Basically, it comes down to if you want to take a risk or not.  However, the AMC 12 is harder than the AMC 10.  There may be questions that are the same, but questions from the back of the AMC 10 will be close to the beginning on the AMC 12.[/quote]

As for taking a risk, it is true that the AMC12 is harder than the AMC10.  However, if you just [i]study[/i] and work hard enough, you'll do well, regardless of which test.

If you're at the point where you think you could qualify for the AIME through the '10 but not through the '12, then - realistically speaking - you're probably not gonna make it past AIME and qualify for USAMO.  And what is the point of even getting to AIME [i]except[/i] to qualify for the USAMO?  Except maybe to improve a college application - but if you can still take the '10, then you have at least two years more to work on AMC and AIME.

So if you're a tenth grader or below, sure, the AMC10 might give you a better score and qualify you for the AIME, but if you take the AMC12 you'll get more practice and you'll probably have a better shot in the future of getting to AIME or USAMO.  Don't get greedy and think about the possiblity of getting to AIME with your AMC10 score.  If you can't do well on AIME, then what was the point of even taking it?  Instead, try out the '12...though your score will be lower, you'll get better at it.  And practice old tests, or read math books, or whatever it is you do to prepare!


Find all possible $ a$ values in system
$ a(a\plus{}b\plus{}c\plus{}d\plus{}e)\equal{}a^2\minus{}1$
$ b(a\plus{}b\plus{}c\plus{}d\plus{}e)\equal{}b^2\minus{}1$
$ c(a\plus{}b\plus{}c\plus{}d\plus{}e)\equal{}c^2\minus{}1$
$ d(a\plus{}b\plus{}c\plus{}d\plus{}e)\equal{}d^2\minus{}1$
$ e(a\plus{}b\plus{}c\plus{}d\plus{}e)\equal{}e^2\minus{}1$

I tried this and obtained
$ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{} a\minus{}\frac 1a \equal{} b \minus{}\frac 1b \equal{} c\minus{}\frac 1c \equal{} d\minus{}\frac 1d \equal{} e\minus{}\frac 1e$
so when $ a \equal{} b \equal{} c \equal{} d \equal{} e$
and then we can see
$ a(5a)\equal{}a^2\minus{}1$
solving it gives us $ a\equal{} \pm\frac{i}{2}$

the other four solutions I assume are obtained when
$ a \equal{} b \equal{} c \equal{} d  \equal{} \minus{}e$
$ a \equal{} b \equal{} c \equal{} \minus{}d  \equal{} \minus{}e$
$ a \equal{} b \equal{} \minus{}c \equal{} \minus{}d  \equal{} \minus{}e$
$ a \equal{} \minus{}b \equal{} \minus{}c \equal{} \minus{}d  \equal{} \minus{}e$


inequalities
number theory unsolved
number theory

[quote="countdownmath"]Let $ \alpha=r/s$ be a rational number written in lowest terms.

Show that there is exactly one rational number $ \frac{a}{b}$ satisfying the inequality $ |\frac{a}{b}-\alpha|<\frac{1}{sb}$.[/quote]
Let $ a \in \mathbb{Z}$ and $ b \in \mathbb{N}$ such that $ \left|\frac{a}{b}-\frac{r}{s}\right| < \frac{1}{sb}$, where we're admitting $ r \in \mathbb{Z}$ and $ s \in \mathbb{N}$. Then trivially $ |as-br| < 1$, yielding $ as = br$, i.e. $ a/b = \alpha$, as $ as-br \in \mathbb{Z}$. Sure you're posting in the right section? :maybe:


calculus
integration
trigonometry
logarithms
calculus computations


Prove or disprove the following inequality for positive real numbers $ a\leq b\leq c$ and $ r>1$: $ \sum_{cyc} a^r(b(a\minus{}b)^2\minus{}c(a\minus{}c)^2)\geq0$

[quote="bakerbakura"]Prove or disprove the following inequality for positive real numbers $ a\leq b\leq c$ and $ r > 1$: $ \sum_{cyc} a^r(b(a \minus{} b)^2 \minus{} c(a \minus{} c)^2)\geq0$[/quote]

try $ a \equal{} \frac {1}{10},b \equal{} \frac {1}{5},c \equal{} 1,r \equal{} 3$


inequalities
geometry
trigonometry
inequalities unsolved

[quote="JANKRI"]Please forgive me about posting this one, but I need a nice proof, thanks!
\[ s_{\alpha}^{2}\plus{}s_{\beta}^{2}\plus{}s_{\gamma}^{2}\geq 3\sqrt{3}A\]
$ s_{\alpha}$, $ s_{\beta}$ and $ s_{\gamma}$ are bisectors, and $ A$ is area of triangle.[/quote]
My ugly proof:
Let $ a\equal{}y\plus{}z,$ $ b\equal{}x\plus{}z$ and $ c\equal{}x\plus{}y.$
Then $ \sum_{cyc}s_{\gamma}^{2}\geq3\sqrt{3}A\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{2ab\cos{\frac{\gamma}{2}}}{a\plus{}b}\right)^{2}\geq\frac{3}{4}\cdot\sqrt{3(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(a\plus{}c\minus{}b)(b\plus{}c\minus{}a)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{ab(a\plus{}b\minus{}c)}{(a\plus{}b)^{2}}\geq\frac{3}{4}\cdot\sqrt{\frac{3(a\plus{}b\minus{}c)(a\plus{}c\minus{}b)(b\plus{}c\minus{}a)}{a\plus{}b\plus{}c}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{x(x\plus{}y)(x\plus{}z)}{(2x\plus{}y\plus{}z)^{2}}\geq\frac{3}{4}\cdot\sqrt{\frac{3xyz}{x\plus{}y\plus{}z}}.$
By Cauchy-Shwartz we obtain
$ \sum_{cyc}\frac{x(x\plus{}y)(x\plus{}z)}{(2x\plus{}y\plus{}z)^{2}}\equal{}\sum_{cyc}\frac{x^{3}}{(2x\plus{}y\plus{}z)^{2}}\plus{}(xy\plus{}xz\plus{}yz)\sum_{cyc}\frac{x}{(2x\plus{}y\plus{}z)^{2}}\equal{}$
$ \equal{}\sum_{cyc}\frac{x^{4}}{x(2x\plus{}y\plus{}z)^{2}}\plus{}(xy\plus{}xz\plus{}yz)\sum_{cyc}\frac{x^{2}}{x(2x\plus{}y\plus{}z)^{2}}\geq$
$ \geq\frac{(x^{2}\plus{}y^{2}\plus{}z^{2})^{2}\plus{}(xy\plus{}xz\plus{}yz)(x\plus{}y\plus{}z)^{2}}{\sum_{cyc}x(2x\plus{}y\plus{}z)^{2}}\equal{}$
$ \equal{}\frac{\sum(x^{4}\plus{}x^{3}y\plus{}x^{3}z\plus{}4x^{2}y^{2}\plus{}5x^{2}yz)}{\sum(4x^{3}\plus{}5x^{2}y\plus{}5x^{2}z\plus{}2xyz)}.$
By the way $ \frac{xy\plus{}xz\plus{}yz}{x\plus{}y\plus{}z}\geq\sqrt{\frac{3xyz}{x\plus{}y\plus{}z}}.$ Thus, it remains to prove that
$ \frac{\sum(x^{4}\plus{}x^{3}y\plus{}x^{3}z\plus{}4x^{2}y^{2}\plus{}5x^{2}yz)}{\sum(4x^{3}\plus{}5x^{2}y\plus{}5x^{2}z\plus{}2xyz)}\geq\frac{3(xy\plus{}xz\plus{}yz)}{4(x\plus{}y\plus{}z)}.$
But $ \frac{\sum(x^{4}\plus{}x^{3}y\plus{}x^{3}z\plus{}4x^{2}y^{2}\plus{}5x^{2}yz)}{\sum(4x^{3}\plus{}5x^{2}y\plus{}5x^{2}z\plus{}2xyz)}\geq\frac{3(xy\plus{}xz\plus{}yz)}{4(x\plus{}y\plus{}z)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(4x^{5}\minus{}4x^{4}y\minus{}4x^{4}z\plus{}5x^{3}y^{2}\plus{}5x^{3}z^{2}\minus{}14x^{3}yz\plus{}8x^{2}y^{2}z)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x\minus{}y)^{2}(2x^{3}\plus{}2y^{3}\plus{}3z^{3}\minus{}4xyz),$ which obviously true.

[quote="JANKRI"]Please forgive me about posting this one, but I need a nice proof, thanks!
\[ s_{\alpha}^{2}\plus{}s_{\beta}^{2}\plus{}s_{\gamma}^{2}\geq 3\sqrt{3}A\]
$ s_{\alpha}$, $ s_{\beta}$ and $ s_{\gamma}$ are bisectors, and $ A$ is area of triangle.[/quote]

Since $ s_{\alpha}^{2}\plus{}s_{\beta}^{2}\plus{}s_{\gamma}^{2}\geq s_{\beta}s_{\gamma}\plus{}s_{\gamma}s_{\alpha}\plus{}s_{\alpha}s_{\beta}$, the inequality from
http://www.mathlinks.ro/Forum/viewtopic.php?t=36329 is stronger. Arqady, do you see any nice approach to the latter one?

  darij

[quote="darij grinberg"]

Since $ s_{\alpha}^{2}\plus{}s_{\beta}^{2}\plus{}s_{\gamma}^{2}\geq s_{\beta}s_{\gamma}\plus{}s_{\gamma}s_{\alpha}\plus{}s_{\alpha}s_{\beta}$, the inequality from
http://www.mathlinks.ro/Forum/viewtopic.php?t=36329 is stronger. Arqady, do you see any nice approach to the latter one?

  darij[/quote]
No I don't see. I'll think about it.
Easy to show that
$ s_{\beta}s_{\gamma}\plus{}s_{\gamma}s_{\alpha}\plus{}s_{\alpha}s_{\beta}\geq3\sqrt3A\Leftrightarrow\sum_{cyc}\frac{\sin\alpha}{\cos\frac{\alpha\minus{}\beta}{2}\cos\frac{\alpha\minus{}\gamma}{2}}\geq\frac{3\sqrt3}{2}.$


MATHCOUNTS probability geometry rectangle ratio inequalities

1.
Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4.  If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?

2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can.  If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls.  Express your answer in terms if $\pi$.

3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass.  The rectangle has dimensions of 4 feet by 3.5 feet.  How many feet of weather-stripping will be needed to go around the glass?  Express your answer to the nearest integer.

4.
The numbers from 1 through 8 are separated into two sets, A and B.  The numbers in A are multiplied together to get $a$.  The numbers in B are multiplied together to get $b$.  The larger of the two numbers $a$ and $b$ is written down.  What is the smallest possible number that can be written down using this procedure?

5.
Eric's Florist Shop sells roses for 20 dollars per dozen, daises for 5 dollars per dozen, and chrysanthemums for 10 dollars per dozen.  At these rates how much would an arrangement of 9 roses, 6 daises, and 18 chrysanthemums cost in dollars?

6.
Given the equation $\displaystyle \frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ such that $m>n$ and $m$ is not a multiple of $n$.

7.
Use the pattern below to express $100^2$ in the form $a^2+b^2-c^2$.  Then find $a+b+c$.
\begin{eqnarray*}
12^2 = 8^2 + 9^2 - 1^2\\
14^2 = 10^2 + 10^2 - 2^2\\
16^2 = 12^2 + 11^2 - 3^2\\
18^2 = 14^2 + 12^2 - 4^2
\end{eqnarray*}

8.
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

9.
Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?

10.
What is the difference between the largest and the smallest values of $x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*}
2x-6 \le 6\\
4x+3 \ge 13
\end{eqnarray*}

1
the total area of the board is (16*3.1415.....).And as the total area of the part that is inside the 4raddi circle but not inside the 2raddi circle is (5*3.1415...) the desired prob. is 5/16

2
what is the height and raddi of the cylinder? assuming the cylinder's volume is v the answer is
(v-(64*4pi)


polynomial
Ring Theory
Gauss
algebra
calculus
integration
function

I tried to show that $ 5^{1/3}$ is not an element of $ \mathbb{Q}(2^{1/3})$ by trying to write $ 5^{1/3}$ as a lineair combination of $ \{1,2^{1/3},2^{2/3}\}$. But I cannot get to a contradiction.

Is $ x^3 \minus{} 5$ reducible over $ \mathbb{Q}(\sqrt[3]{2})$?  (Use Eisenstein.)

What is a quick way to see that there is a prime ideal $ \mathcal{P}\subset\mathbb{Z}[\sqrt[3]{2}]$ such that $ 5\in\mathcal{P}$ but $ 5\not\in\mathcal{P}^2$?

Is there a 'easier' way ?

[quote="t0rajir0u"]Is $ x^3 \minus{} 5$ reducible over $ \mathbb{Q}(\sqrt [3]{2})$?  (Use Eisenstein.)[/quote]
I only know eisenstein over $ \mathbb{Q}$.

LydianRain: By taking the ideal $ <5,\sqrt[3]4\plus{}3\sqrt[3]2\minus{}1>$. It's norm is $ 25$, thus it's square cannot contain an element of norm $ 125$.

ZetaX: How do we know it's prime?  And how do we finish the proof from there?  I don't think $ \mathbb{Z}[\sqrt[3]{2}]$ is a UFD, so we cannot use Gauss' lemma...

Amazigh: Here's Eisenstein for any commutative ring $ R$: If $ f(x)\equal{}a_n x^n \plus{} \ldots \plus{} a_0$ is a polynomial in $ R[x]$, and there is a prime ideal $ \mathcal{P}\subset R$ such that $ a_n\not\in \mathcal{P}$, $ a_i\in\mathcal{P}$ for $ i < n$, and $ a_0\not\in\mathcal{P}^2$, then $ f(x)$ is irreducible in $ R[x]$.

Proof: Suppose $ f\equal{}gh$, and reduce mod $ \mathcal{P}$.  Since $ R/\mathcal{P}$ is an integral domain, $ \overline{a_n} x^n \equal{} \overline{g}\overline{h}$ means that both $ \overline{g}$ and $ \overline{h}$ must be monomials of degree greater than $ 0$.  This means that the constant terms of $ g$ and $ h$ lie in $ \mathcal{P}$, so $ a_0\in\mathcal{P}^2$, contradiction.

Thanks, but since my algebra course does not discuss the "general case" of Eisenstein so I cannot use it. It should solve it with showing that no lineair combination of the basis over Q exists.

Perhaps you should do your homework yourself?

It is a UFD. Also, Gauss lemma doesn't need UFDs, at least we use something more general: for nomic polynomials it suffices to know that the ring is integrally closed, then any factorisation already happens in that ring.
We know that it is prime by construction (I got it by the characterisation of prime ideals containing $ 5$). The way it works:
Directly note that $ \mathbb Z[\sqrt[3]2] / < 5 > \cong \mathbb F_5[x]/<x^3\minus{}2> \cong \mathbb F_5[x]/<x\minus{}3> \times \mathbb F_5[x]/<x^2\plus{}3x\minus{}1>$. The last two rings are integral domains (in fact fields) as easily can be seen; especially, $ <x^2\plus{}3x\minus{}1>$ is a prime ideal in $ \mathbb F_5[x]$ and taking it's preimage in $ \mathbb Z[\sqrt[3]2]$, we get what we want.

OK, very clear explanation.  I will need to think about how to show integral closure (or unique factorization) but maybe this just means I should pick up an algebraic number theory book. :)

LydianRain , ZetaX:I think you are making a simple ques into a harder one.
Here is a solution using a little knowledge on Galois Theory. But I'm not sure if it's right, please help me check it.

$ y_{0} \equal{} 5^{\frac {1}{3}} ,x_{0} \equal{} 2^{\frac {1}{3}}$      $ E_{2} and E_{5}$ means the spliting fields of $ x^{3} \minus{} 2$ and $ x^{3} \minus{} 5$ on Q. If $ y_{0}\in Q(x_{0})$ than apparently $ E_{2} \equal{} E_{5}$ since their dimensions are equal. Set: $ g\in Gal(E/Q)\cong I_{3}$ is the generator, $ w \equal{} e^{\frac {1}{3}2\pi i}$. If $ y_{0} \equal{} a \plus{} bx_{0} \plus{} cx^{2}_{0}$, than $ g(y_{0}) \equal{} wy_{0}$. But we also have $ g(x_{0}) \equal{} wx_{0}$. Combine, we get $ aw \plus{} bwx_{0} \plus{} cwx_{0} \equal{} a \plus{} bwx_{0} \plus{} cw^{2}x^{2}_{0} ;w^{2}(a \plus{} bx_{0} \plus{} cx^{2}_{0}) \equal{} a \plus{} bw^{2}x_{0} \plus{} cwx_{0}$. Add up we get $ \minus{} y_{0} \equal{} 2a \minus{} bx_{0} \minus{} cx^{2}_{0} \equal{} 3a \minus{} y_{0}$,($ w \plus{} w^{2} \equal{} \minus{} 1$), which means a=0. By a=0, b=c=0 follows clearly.

I was just trying to follow t0rajir0u's solution, which is an elegant approach here. (though a pretty subtle one)

I don't think we can assume $ g(y_0)\equal{}\omega y_0$.  We need to be prepared for it to be any root of $ x^3\minus{}5$.  Fortunately, if $ g(y_0)\equal{}\omega^2 y_0$, your proof still works, and if $ g(y_0)\equal{}y_0$, we arrive at $ a\equal{}y_0$, certainly a contradiction.

More should be said on why $ a\equal{}0$ implies $ b\equal{}c\equal{}0$, though.  We really need to use some properties of $ m\equal{}5$ and $ n\equal{}2$, specifically that neither of $ mn$ and $ mn^2$ is a cube.  This follows quickly from $ y_0\equal{}bx_0\plus{}cx_0^2$ by cubing, though.

By the way, I don't think that we should talk about the "generator" of $ S_3$.  How about "an element of order 3"? :)

I think the best way to solve the problem is to use Kummer theory. Take $ K \equal{} \mathbb Q[\zeta_3]$, then by Kummer theory that question gets purely arithmetical in an UFD, the integers of $ K$. In general, this can be generalized to how $ x^{n_i} \minus{} p_i$ behave altogether.

The minimal polynomial of $ 2^{1/3}\plus{}5^{1/3}$ is $ \minus{}343\minus{}123x^3\minus{}21x^6\plus{}x^9$.
Then $ \mathbb{Q}(2^{1/3},5^{1/3})$ is an algebraic extension of $ \mathbb{Q}$ of degree 9. Thus $ 5^{1/3}$ is not in $ \mathbb{Q}(2^{1/3})$.

Well, how did you show that it is the minimal polynomial¿  :D

Thanks for your question Zetax....
1) I obtain  $ p(u)=-343-123u^3-21u^6+u^9$ as the minimal polynomial using LLL method (with Maple). In fact this polynomial is only a candidate.
2) Let $ a=2^{1/3}+5^{1/3}$; we must prove  $ a^3=3.50^{1/3}+3/5.50^{2/3}+7$ is a root of $ q(x)=-343-123x-21x^2+x^3$. $ x=y+7$ provides $ -1890-270y+y^3=0$ and $ y=z+90/z$ provides $ z^6-1890z^3+729000=0$. Then $ z^3=540$ (for example) and $ z=540^{1/3}$. Finally we obtain $ x=a^3$.
3) It remains to prove that $ p$ is irreducible. Assume it is not.
$ p(x)=(x^3+2x^2+10x+6)(x^3+6x^2+12x+6)(x^3+5x^2+4x+6) \;\;mod 13$
$ p(x)=(x^2+10x+1)(x+1)(x^2+10x+4)(x^2+4)(x^2+x+4)\;\; mod 11$.
Then $ p(x)=r(x)s(x)$ where $ r,s$ are monic polynomials of degree $ 3,6$.
In particular $ 3$ among the $ 9$ roots of $ p$ have a product which is an integer. Now we calculate the $ 84$ products $ 3$ by $ 3$ of approximations (with 30 digits for example) of roots of $ p$  (a little program !). We find no result which is near an integer.    QED
Remark: if $ a$ is of degree $ 3,6$ then its minimal polynomial is monic and the degree 0 coefficient is a divisor of 343. I think it is not difficult to find a bound for the other coefficients. More generally if we know $ a$ with many digits and if we know a bound for the coefficients of its minimal polynomial (we assume it is of degree $ \leq{6}$), then the LLL method permits to obtain the best candidate (among the polynomials of degree $ \leq{6]}$)to be the minimal polynomial; in our case this best candidate is not monic (!) and have very large coefficients.

Awards for this thread:

Easiest solution: mathiq
Most general solution: ZetaX (Kummer theory)
Most deceptively difficult solution: t0rajir0u
Most brute force solution: loup blanc

ZetaX: I have a ques about your #9 post. You successfully showed $ <5,4^{\frac{1}{3}}+3*2^{\frac{1}{3}}-1>$ is prime in $ Z[2^{\frac{1}{3}]}$ but you haven't show that $ Z[2^{\frac{1}{3} ]}$ contain all the integer of $ Q[2^{\frac{1}{3} ]}$ ,for which you write "integrally closed". But how to get it. Generally, what's the int base of the split fields of $ x^{n}-p=0$

Well, that's the tedious part  :P
One way would be to compute the discriminant $ d$ and then to manuelly check if we can find another base with discriminant dividing $ d$; although being a finite process, it's difficult to do in general and uses lots of computations. Another way is to show that all localisations at prime ideals are PIDs; this needs only be checked at primes dividing $ d$ and then for the maximal ideal of each such ring; this still is lots of works, but it generalizes much easier. By the way, I hope you don't want the splitting field of $ x^n\minus{}p$ but $ \mathbb Q[\sqrt[n]p]$ :D


Since there isn't a marathon for USAMO, perhaps I'd create one.

Expect the problems to be extremely easy the first few times, because a few of us (me, at least) have a hard time thinking of good problems.

Prove that for any $ n \times n$ square board, a game of tic-tac-toe where players will connect $ n$ in a row will end in a draw with perfect play, for all $ n > 2$.

[hide="Outline"]
We proceed by induction. It's well-known that a draw is inevitable for $ n\equal{}3$. Now assume it's true for $ n\equal{}k\minus{}1$; we show that it's impossible for $ n\equal{}k$.

To gain a winning position, a player must have at least two ways to get an $ k$-string at the end of their opponent's turn; hence two $ k\minus{}1$ strings are necessary. However, by our hypothesis, the other player can always stop the first one from getting a $ k\minus{}1$-string in a given $ k\minus{}1$ by $ k\minus{}1$ section of the grid at the end of their own turn; furthermore, each time the opponent places a mark outside of this grid, the other player simply places their own mark at the intersection of the two $ k$-strings not in the $ k\minus{}1$ by $ k\minus{}1$ square, thus prohibiting their opponent from winning.[/hide]

Let $ \, k_1 < k_2 < k_3 < \cdots \,$ be positive integers, no two consecutive, and let $ \, s_m \equal{} k_1 \plus{} k_2 \plus{} \cdots \plus{} k_m \,$ for $ \, m \equal{} 1,2,3, \ldots \; \;$.  Prove that, for each positive integer $ \, n, \,$ the interval $ \, [s_n, s_{n \plus{} 1}) \,$ contains at least one perfect square.

From 1994.

[quote]Let $ \, k_1 < k_2 < k_3 < \cdots \,$ be positive integers, no two consecutive, and let $ \, s_m \equal{} k_1 \plus{} k_2 \plus{} \cdots \plus{} k_m \,$ for $ \, m \equal{} 1,2,3, \ldots \; \;$.  Prove that, for each positive integer $ \, n, \,$ the interval $ \, [s_n, s_{n \plus{} 1}) \,$ contains at least one perfect square.

From 1994.[/quote]

[hide="My perhaps faulty solution"]
Since no two of $ k_n$ are consecutive, this means that they must each differ by at least 2.

We may look at two "worst" cases:

[b]Case 1[/b]

In the case $ k_1 \equal{} 1$, $ k_2 \equal{} 3$, $ k_i \equal{} 2i \minus{} 1$ etc., $ s_n$ will always contain a perfect square $ n^2$.

To rid us of this case, let $ k_n \equal{} 2n$ instead of $ 2n \minus{} 1$. This way, $ s_n$ = $ n^2 \plus{} 1$.

Adding another $ k_{n \plus{} 1} \equal{} 2n \plus{} 2$ will result in $ (n \plus{} 1)^2$ appearing in the interval, because the upper bound is then $ n^2 \plus{} 2n \plus{} 2$.

[b]Case 2[/b]

In the case $ k_1 \equal{} 2$, $ k_2 \equal{} 4$, $ k_3 \equal{} 6$, etc., $ s_n$ will be of the form $ n^2 \plus{} n$, which is less than $ (n \plus{} 1)^2$.

Adding another $ k_{n \plus{} 1} \equal{} 2n \plus{} 2$ will result in $ (n \plus{} 1)^2$ appearing in the interval, because the upper bound is then $ n^2 \plus{} 3n \plus{} 2$.

[b]In between Cases 1 and 2[/b]

For cases of $ k$ where exactly one pair $ (k_i, k_i \plus{} 1)$ from $ 1 \le i < k$ satisfies $ k_i \plus{} 3 \equal{} k_{i \plus{} 1}$ and the rest satisfy $ k_i \plus{} 2 \equal{} k_{i \plus{} 1}$, adding another $ k_{n \plus{} 1} \equal{} k_n \plus{} 2$ will result in $ (n \plus{} 1)^2$ appearing in the interval, due to the upper bound being $ n^2 \plus{} 2n \plus{} i \plus{} 2$, and the lower bound being $ n^2 \plus{} i < (n \plus{} 1)^2$ because $ i < 2n \plus{} 1$.

Since this cannot cover all cases, consider another third case:

[b]The general case[/b]

In any other cases, $ 2n < k_n$, but $ s_n$ will still be less than $ nk_n$, which means that the interval between squares is still less than $ \sqrt {nk_n} \plus{} 1 < \sqrt {\frac {k_n^2}{2}} \plus{} 1 \equal{} \frac {k_n}{\sqrt {2}} \plus{} 1 < k_n \plus{} 1$.

For any $ k_n$, $ k_{n \plus{} 1} \ge k_n \plus{} 2 > k_n \plus{} 1$, so the interval must contain at least one perfect square, QED.

[/hide]

I bet my solution, if correct, will only garner a 5/7 for bad presentation.

Alright, new problem! This one is combinatorics (I think.)

A bridge club is scheduling games for the first week of June. There are 20 people in the bridge club, and they have 10 games of 4 people each every week. Each player must play exactly twice.

No matter how the games are arranged, one will always be able to find a combination of 5 games with at least $ n$ players. Find, with proof, the highest value of $ n$.


vector
algebra
polynomial
linear algebra
matrix
abstract algebra
linear algebra unsolved

Let $f,g$ be endomorphisms of a $K$-vector space $V$. Then the characteristic polynomials of $fg$ and $gf$ are equal.

hmm interesting
this one bugged me some time ago, I can only do it when one of the two is invertible, let it be f,
then $det(fg-x I)= det(f^{-1}fgf-x f^{-1}If)=det(gf-xI)$
in the complex or real case, our professor told us something about a continuity argument...

For infinite fields $K$ you can always find infinitely many $\lambda\in K$ such that $f_\lambda=\lambda I-f$ is invertible. Now fix $x$. $\det(xI-f_\lambda g)-\det(xI-gf_\lambda)$ is a polynomial in $\lambda$ which is zero for infinitely many values of $\lambda$, so it's zero for $\lambda=0$ as well, which gives precisely $\det(xI-fg)=\det(xI-gf)$. The general case, when $K$ may be finite, follows from this one, since every field can be embedded in an infinite field.

we can also us a bloc multiplication in a matrix witchis definit by blocs...but the easiest why is to work in K(X) us grobber's idea.
The continuity argument may us that the set of invertible matrix is dense in the set of matrix .

Another version: Do this over one field of characteristic zero first, such as $\mathbb{R}$ or $\mathbb{C}$. In general, the coefficients of the characteristic polynomial are polynomials with integer coefficients in the entries of the matrix. By the case we have, these polynomials are identical for $fg$ and $gf$. That proves the desired result for all other fields, since we use the same polynomials to determine the coefficients.

It is true for any commutative ring. The coefficients of the characteristic polynomial of operator $U$ acting on module $M$ are traces of $U$ acting on exterior powers of the module.  But  $Tr (UV) = Tr(VU)$ over any commutative ring.


Let be $ a,b,c \in (0,\infty )$ such that $ abc\equal{}1$ . Show that :

$ 4\left(\frac{1}{a(1\plus{}bc)^2}\plus{}\frac{1}{b(1\plus{}ca)^2}\plus{}\frac{1}{c(1\plus{}ab)^2}\right)\le 1\plus{}\frac{16}{(1\plus{}bc)(1\plus{}ca)(1\plus{}ab)}$

$ \leftrightarrow 4\sum_{cyc} \frac {a}{(a \plus{} 1)^2}\leq 1 \plus{} \frac {16}{(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)}$
$ \leftrightarrow 20p \plus{} 20q \plus{} 4pq \plus{} 36\leq 36 \plus{} p^2 \plus{} q^2 \plus{} 20p \plus{} 20q \plus{} 2pq$
$ \leftrightarrow 0\leq (p \minus{} q)^2$


If F_n=2^(2^n) is the n_th Fermat number, find all values of n so that F_n

and F_(n-2) are twin primes.

First of all, I think F_n=2 2^n + 1, and second of all, does 'twin primes' actually mean that F_n-F_(n-2)=2? If it does, you have to be joking... :D  :D We don't even need to know they are primes.

Sorry very much. The correct test is :

If F_n=2 ^(2^n) + 1 is the nth Fermat number, find all values of n so that F_n and F_n -2 are twin primes.

The solution is very simple: Clearly one integer in any three consecutive integers is divisible by 3. Since F_n -1 is not divisible by three then either F_n  or F_n - 2 must be divisible by 3. Hence only F_1 -2=3 and F_1=5 are twin primes.

Otherwise : the number F_n -2 is never prime when n>1 because F_m
                 divides F_n-2 whenever n>m \geq 0 so the only twin primes
                 of the family are 3 and 5 for n=1.

Sorry again for the wrong test


geometry
incenter
analytic geometry
geometry proposed

Prove, that in non-isosceles triangle points $ I, K, E_C$ lies on one line, where:
$ I$ – incenter ;
$ K$ – Lemoine point;
$ E_C$ - midpoint of $ E_L E_R$
$ E_L$ – left Jerabek point: $ A_1  \in BC$, $ B_1  \in AC$, and $ C_1  \in AB$, $ E_LA_1\parallel CA$, $ E_LB_1\parallel AB$, $ E_LC_1\parallel BC$  and  $ E_LA_1 \equal{} E_LB_1 \equal{} E_LC_1$;
$ E_R$ – right Jerabek point: $ A_2  \in BC$, $ B_2  \in AC$, and $ C_2  \in AB$, $ E_RA_2\parallel AB$, $ E_RB_2\parallel BC$, $ E_RC_2\parallel CA$  and  $ E_RA_2 \equal{} E_RB_2 \equal{} E_RC_2$.

Using barycentric coordinates, we can show that $ E_L \equal{} (1/b, 1/c, 1/a)$ and $ E_R \equal{} (1/c, 1/a, 1/b)$, and $ E_C \equal{} X_{37} \equal{} (a(b \plus{} c), b(c \plus{} a), c(a \plus{} b))$, as listed in the ETC.  From here, it is a straight-forward calculation.

Can anyone tell me how to construct left, right Jerabek point?
thanks a lot in advance!

For example:
It’s easy to calculate, that $ E_LA_1 \equal{} E_LB_1 \equal{} E_LC_1 \equal{} E_RA_2 \equal{} E_RB_2\equal{} E_RC_2\equal{} (a^{ \minus{} 1}  \plus{} b^{ \minus{} 1}  \plus{} c^{ \minus{} 1} )^{ \minus{} 1}$


Let $f: \Re \rightarrow \Re$ be function satisfy: $\displaystyle |f(x+y)-f(x)-f(y)|\leq 2004; \ \forall x,y\in \Re$
Prove that there exists $g: \Re \rightarrow \Re$ such that:
$\displaystyle |f(x)-g(x)|\leq 2003$ and $\displaystyle |g(x+y)-g(x)-g(y)|\leq 1;  \ \forall x,y \in \Re$

Well  Supppose:$h(x)= lim \frac {f(2^nx)}{2^nx}$ when n goes to infinity
You can easily prove h exists and $\mid h(x)-f(x) \mid \leq 2004$ and
$h(x)+h(y)=h(x+y)$.So suppose $\displaystyle g=\frac {2003h(x)+f(x)}{2004}$ You can see easily g has the proble properties


Determine $ x,y,z>0$ for which $ x^3y\plus{}3<\equal{}4z, y^3z\plus{}3<\equal{}4x,z^3x\plus{}3<\equal{}4y.$

Let $ z \equal{} \min(x,y,z)$. Then $ z^4\le x^3y\implies z^4 \plus{} 3\le 4z$ which implies $ (z \minus{} 1)(z^3 \plus{} z^2 \plus{} z \minus{} 3)\le 0$. But if $ z\ge 1$ then $ z^3 \plus{} z^2 \plus{} z\ge 3$ and if $ z\le 1$ then $ z^3 \plus{} z^2 \plus{} z\le 3$ implying $ (z \minus{} 1)(z^3 \plus{} z^2 \plus{} z \minus{} 3)\ge 0$ in both cases. Thus $ z \equal{} 1$ and since equality must hold in $ z^4\le x^3y$ we get $ z \equal{} x \equal{} y \equal{} 1$.

EDIT: We can get the above directly using $ z^4\minus{}4z\plus{}3\equal{}(z\minus{}1)^2(z^2\plus{}2z\plus{}3)\ge 0$.

From $ z^4\plus{}3\minus{}4z\leq0$ we see $ (z^2\minus{}1)^2\plus{}2z^2\minus{}4z\plus{}2\equal{}(z^2\minus{}1)^2\plus{}2(z\minus{}1)^2\geq0$ (its equivalent to your factorization except easier to see at first)


abstract algebra
advanced fields
advanced fields unsolved

Show that $ \mathbb{Z}/(p^{e})$, p a prime, e>0, regarded as a $ \mathbb{Z} \minus{} module$, is not the direct sum of any two nonzero submodules.

it's enough to post in one forum ...
in the algebra forum, there is a full solution to your problem


When are we going to get our feedback for this? It was due on Nov. 10...

The remaining exams in Fall AoPS subject classes should be graded soon.  Some of the classes are nearly complete and graders will be working hard this coming weekend.

How about now? Sorry to be bothersome, but I'm impatient...

They've all been graded.  When I look at that class, I don't see your midterm at all.

Generally, as soon as we get a student's exam, we upload it to our server, so that you see your full solutions as a file in your My Classes page.  If you don't see it, then your solutions either were not uploaded or were misnamed when they were, which is rare.

I suggest emailing the course instructor and asking him to see if your exam was misfiled.  Sorry if there were any problems.


First of all: is there a typo here? Is the base for the c term's logarithmic exponent supposed to have base 11? And I'm supposing that the 125 should have been 25, so it's $ \log_{11}{25}$, right?

If so, then here is my solution:

We can see that whenever we have an exponent $ x^{n^2}$, we can make it into $ (x^n)^n$.

So that means $ a^{(\log_3{7})^2} \plus{} b^{(\log_7{11})^2} \plus{} c^{(\log_{11}{25})^2} \equal{} 27^{\log_3{7}} \plus{} 49^{\log_7{11}} \plus{} (\sqrt {11})^{\log_{11}{25}}$.

Note that $ 27 \equal{} 3^3$, $ 49 \equal{} 7^2$, and $ \sqrt {11} \equal{} 11^{\frac {1}{2}}$.

Then we can get $ 3^{3\log_3{7}} \plus{} 7^{2\log_7{11}} \plus{} 11^{\frac {1}{2}\log_{11}{25}} \equal{} 3^{\log_3{7^3}} \plus{} 7^{\log_7{11^2}} \plus{} 11^{\log_{11}{25^\frac {1}{2}}}$.

This is simplified to (with laws of exponents) $ 7^3 \plus{} 11^2 \plus{} 25^{\frac {1}{2}} \equal{} 343 \plus{} 121 \plus{} 5 \equal{} \boxed{469}$.

This problem took me waaaaaay too long  :blush: I actually did #1, tried #2 for about five minutes and got nowhere, moved onto #3, then spent another like... fifteen minutes on this one before I actually got it. I felt like facepalming when I realized how to do it >_>

That's the (algebraic) definition of a logarithm. It's the inverse of exponentiation.

[quote="worthawholebean"]Suppose that $ a$, $ b$, and $ c$ are positive real numbers such that $ a^{\log_3 7} \equal{} 27$, $ b^{\log_7 11} \equal{} 49$, and $ c^{\log_{11} 25} \equal{} \sqrt {11}$. Find
\[ a^{(\log_3 7)^2} \plus{} b^{(\log_7 11)^2} \plus{} c^{(\log_{11} 25)^2}.
\]
[/quote]

My incredibly lucky logic. $ a^{\log_3 7} \equal{} 3^3$, $ b^{\log_7 11} \equal{} 7^2$, and $ c^{\log_{11} 25} \equal{} 11^.5$  The log of 3 and 3 match up, log of 7 and 7, and log of 11 and 11, so we probably have to do something with 7 and 3, 11 and 2, 25 and .5. We can't multiply, because then we'll get a fraction, so let's go with $ 7^3\plus{}11^2\plus{}25^.5\equal{}469$

I spent a ton of time on this one, but logs are a weak point for me.

hahaha. i totally didnt know how to do this problem. cuz i didnt remember any of the log stuff i learned in pre calc. ><
so i just guessed & hoped that the "random" log rule i made up to cancel out the log & exponents worked .. and i guess it did :P

If you're really stuck in terms of remembering log rules, you can derive them all from the exponent rules - as exponentiation turns addition inside into multiplication outside, logs turn multiplication inside into addition outside.

hahaha i spent forever...

i actually derived it by letting log [size=50]3[/size] 7 = x
blabla bla
and so on
i got it right though so that's ok

haha 45 minutes


IMO 2004 have passed, so the problems of IMO shortlist 2003 can be released.
Where can I find the whole set of IMO shortlist 2003 problems?

[url=http://www.mathlinks.ro/Forum/viewforum.php?f=177]ISL03[/url]

you will (very) soon be able to find them on AoPS-MathLinks. Already almost all the questions are posted in different topics, so just search of ISL 2003 in problem source field, and you will find them.

I plan to put a pdf soon.


The squares of an infinite chessboard are numbered $1,2,\ldots $ along a spiral, as shown in the picture. A [i]rightline[/i]  is the sequence of the numbers in the squares obtained by starting at one square at going to the right.
a) Prove that exists a rightline without multiples of $3$.
b) Prove that there are infinitely many pairwise disjoint rightlines not containing multiples of $3$.

[quote="Andreas"]i) Note that the rightline of 6 contains the numbers $3^2 + 2, 5^2 + 3, 7^2 + 4, 9^2 + 5, ...$ that form the recursive series $2, 1$ in $\mathbb{Z}_{3}$.[/quote]

Yeah, and for the same reason taking all rightlines of numbers of the form $(6k+1)^2$ are free from multiples of 3.

Daniel


calculus
derivative
function
inequalities unsolved
inequalities

phuu, finally I finish the solution.My sol based on derivative of one variable function:
Since $ x \leq \frac{4-2y-z}{3}$ we get $S \leq f(y)=\frac{10}{3}y^2-(\frac{16-4z}{3})y+\frac43(z^2-2z+4)$ (1)
and $ S \leq g(z)=\frac43z^2+z(\frac{4y-8}{3})+\frac{10y^2-16y+16}{3}$ (2)
      We do with (1):
         $ f'(y)=0 \Longleftrightarrow y=\frac{4-z}{5}$ . If $ y>\frac{4-z}{5}$ then $ f'(y)>0$ then $ f(y) \leq f(1)=u(z)=\frac{10+4z^2-4z}{3}$
         $ u'(z)=0 \Longleftrightarrow z=\frac12$ . If z>1/2 then $ u(z) \leq u(1) = \frac{10}{3}$, if $ z \leq \frac12 $ ,hence x<z, and y <1 we have:
 		$ S \leq 3.(\frac12)^2+2+(\frac12)^2 < \frac{10}{3}$
      if $ y \leq \frac{4-z}{5}$ (3) then we do with (2):
	$ g'(z)=0 \Longleftrightarrow z=\frac{2-y}{2}$ .  If z> (2-y)/2 then $ g(z) \leq g(1)=v(y)=\frac{10}{3}y^2-4y+4$
	$ v'(y)=0  \Longleftrightarrow y=\frac35$ . If y>3/5 then $ v(y) \leq v(1) =\frac{10}{3}$,
if $ y \leq \frac35$ then $ S \leq 3.(\frac35)^2+2.(\frac35)^2+1 < \frac{10}{3}$.
	Finally, we have only to prove in the case $ z \leq \frac{2-y}{2}$ (4). From (3) and (4) we have $ 5y+z \leq 4$ and $ 2z+y \leq 2$
  $ \Rightarrow z+2y \leq 2$ . Adding with $ 2z+y \leq 2$ we have $ z+y \leq \frac43$ , hence  x<2/3 then $ S < 3*(2/3)^2 + 2y+z < 3*(2/3)^2+ 2 =10/3$ . We got the maximum is $ \frac{10}{3}$

Dear nttu,
Can you change your problem to: find max of 3x^n+2y^n+z^n and solve it?


This is my problem. I solved it.

Let $x$, $y$, $z$ be positive real numbers satisfying $x+y+z=3$. Prove that

a) $\sqrt x + \sqrt y + \sqrt z \geqslant x\sqrt y + y\sqrt z + z\sqrt x$.

b) $\sqrt x + \sqrt y + \sqrt z \geqslant \sqrt{xy} + \sqrt{yz} + \sqrt{zx}$.

c) $\sqrt x + \sqrt y + \sqrt z \geqslant xy + yz + zx$.

d) $\sqrt x + \sqrt y + \sqrt z \geqslant xy\sqrt z + yz\sqrt x + zx\sqrt y$.

Please enjoy!

These inequalities are nice, I agree. It is easy to "invent" them but not always that easy to solve them.

I solve the first one by homogonizing. Since $x+y+z=3$, I reduce the desired inequality to
\[ (x+y+z)(\sqrt x+\sqrt y+\sqrt z)\geq 3(x\sqrt y+y\sqrt z+z\sqrt x).  \]
But this is equivalent to
\[ x\sqrt x+y\sqrt y+z\sqrt z+x\sqrt z+y\sqrt x+z\sqrt y\geq 2(x\sqrt y+y\sqrt z+z\sqrt x).  \]
By AM-GM inequality, we have

$x\sqrt x+x\sqrt z+z\sqrt x\geq 3x\sqrt z,$
$y\sqrt y+y\sqrt x+x\sqrt y\geq 3y\sqrt x,$
$z\sqrt z+z\sqrt y+y\sqrt z\geq 3z\sqrt y$.
Adding the three inequalities, we are done.

For the second one, I put $s=\sqrt x+\sqrt y+\sqrt z$, then $s\leq 3$ (by Cauchy). From this relation, we have
by squaring \[ s^2=3+2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})  \]
So that the desired ineq becomes $s\geq \tfrac12(s^2-3)$ which simplifies to $(s+1)(s-3)\leq0$. This is true. Proof complete.

The third inequality is NOT new. I saw it somewhere.

What is your solution for the last one? I am too lazy for that. I believe that we should think of $(x+y+z)^2$.

Aha, pvthuan, ur solutions are good but not interesting, I'm sure. Let have a look on my solution of the first part, I mean part a).

By Cauchy, we follow that
\[ \begin{gathered} 3 \geqslant 2\sqrt{ab} + c, \hfill \\ 3\sqrt a \geqslant 2a\sqrt b + c\sqrt a. \end{gathered}  \]

Similarly, we can conclude that
\[ \begin{gathered} 3\sqrt b \geqslant 2b\sqrt c + a\sqrt b, \hfill \\ 3\sqrt c \geqslant 2c\sqrt a + b\sqrt c. \end{gathered}  \]
Thus, the conclusion of part a) follows from adding these above inequalities.

Is it nice?  :lol:

The second part, part b) is followed from the fact that
\[ \begin{gathered} a\sqrt b + \sqrt b \geqslant 2\sqrt{ab}, \hfill \\ b\sqrt c + \sqrt c \geqslant 2\sqrt{bc}, \hfill \\ c\sqrt a + \sqrt a \geqslant 2\sqrt{ca}. \hfill \\ \end{gathered}  \]
That's all..!

In order to prove the last part, u need to prove this inequality

\[ \sqrt a + \sqrt b + \sqrt c \geqslant 3\sqrt{abc} \geqslant ab\sqrt c + bc\sqrt a + ca\sqrt b.  \]

Please enjoy!  :roll:


ratio
trigonometry
geometry
similar triangles

Plz help me to find the exact value of cos 18 degree (w/o using calculator)

[hide="Geometric method"]Draw a 36-36-108 isosceles triangle $ ABC$ with $ \angle B \equal{} 108$. From $ B$ draw a segment $ BD$ with $ D$ on $ AC$ so that $ \angle ABD \equal{} 72, \angle DBC \equal{} 36$. Equal angles get that $ \triangle CDB \sim \triangle ABC$. Then draw the bisector from $ A$ in the isosceles $ \triangle DAB$. This makes a right triangle with an $ 18$ degree angle, and all the similar triangles and isosceles allow you to get the side ratio you need.[/hide]

i have tried using cos 36   and cos 72 but it's too complicated and confusing.

hello, using the fomula
$ {\cos(5x) = \cos(x)^5 - 10\cos(x)^2\sin(x)^2 + 5\cos(x)\sin(x)^4 = 16\cos(8x)^5 - 20\cos(x)^3 + 5\cos(x)}$.
Setting $ x = 18^{\circ}$ we get the equation
$ 0 = \cos(18^{\circ})(16\cos(18^{\circ})^4 - 20\cos(18^{\circ})^2 + 5)$ solving this we get the solution
$ \cos(18^{\circ}) = \sqrt {\frac {5}{8} + \frac {\sqrt {5}}{8}}$.
Sonnhard.

http://jwilson.coe.uga.edu/emt725/ssm4409/SSM4409.html

$ cos18^{\circ}$ can be found trivially using the method shown here for $ cos\frac{\pi}{5}\equal{}cos36^{\circ}$.

Presuming you know that $ \cos 36^\circ\equal{}\frac{\phi}{2}$ , then:

$ \cos(2\cdot 18^\circ)\equal{}2\cos^218^\circ\minus{}1\equal{}\frac{\phi}{2}$

$ \therefore\cos^218^\circ\equal{}\frac{\phi\plus{}2}{4}$

$ \therefore\cos 18^\circ\equal{}\sqrt{\frac{\phi\plus{}2}{4}}$

Where $ \phi\equal{}\frac{\sqrt{5}\plus{}1}{2}$


Is there a square number, such that the sum of its digits is $2006$?

[hide="hint"]
A number is congruent to the sum of its digits $\mod 3$
[/hide]


Given triangle $ ABC$ and two isogonal conjugate points $ E$ and $ F$. Denote $ A_1B_1C_1, A_2B_2C_2$ the pedal triangles of triangle $ ABC$ wrt $ E$ and $ F. B_1C_1\cap B_2C_2\equal{}\{A_3\}$, similar for $ B_3,C_3$. Prove that $ AA_3//BB_3//CC_3$

It's well known that $A_1,B_1,C_1,A_2,B_2,C_2$ lie on a same circle $\omega$ whose center is the midpoint of $\overline{EF}.$ Let $T$ be the 2nd intersections of $\odot(AB_1C_1)$ and $\odot(AB_2C_2),$ i.e. the projection of $A$ on $EF.$ Then $AT,B_1C_1,B_2C_2$ are pairwise radical axes of $\omega,$ $\odot(AB_1C_1),$ $\odot(AB_2C_2)$ concurring at ther radical center $A_3,$ i.e $AA_3 \perp EF.$ Similarly, $BB_3$ and $CC_3$ are perpendicular to $EF.$ The conclusion follows.

Thanks you dear [b]Luis[/b] for your nice solution.

[b]Lemma.[/b]
Let $ X,Y,Z$ be three points in a plane and $ l_1,l_2$ be two lines in plane $ (XYZ)$. Denote $ X_i,Y_i,Z_i$ the orthogonal projections of $ X,Y,Z$ on $ l_i$ where $ i \equal{} 1,2$. Then $ X,Y,Z$ are collinear iff $ \frac {\overline{X_1Y_1}}{\overline{X_1Z_1}} \equal{} \frac {\overline{X_2Y_2}}{\overline{X_2Z_2}}$.

The lemma is quite trivially so I will not present its proof.

Now to the solution.

It is known that $ A_1,A_2,B_1,B_2,C_1,C_2$ lies on a circle whose center is the midpoint of $ EF$. Define that circle $ (O)$.

Let $ C_0\equiv A_1B_2 \cap A_2B_1$ and $ C_a,C_b$ be the orthogonal projections of $ C_0$ on $ CA,CB$. We have $ A_1,A_1,B_1,B_2$ are concyclic so $ \triangle C_0A_1A_2$ and $ \triangle C_0B_1B_2$ are similar. This yields $ \frac {\overline{C_aA_1}}{\overline{C_aA_2}} \equal{} \frac {\overline{C_bB_1}}{\overline{C_bB_2}}$ so by the lemma, $ O,C_0,E,F$ are collinear. Now in the complete quadrilateral $ (A_1A_2B_1B_2,CC_3)$, the pencil $ (C_3A_1,C_3B_1,C_3C_0,C_3C)$ is harmonic, so $ C_0$ is the pole of $ CC_3$ with respect to $ (O)$. From this we deduce that $ OC_0 \bot CC_3$, but $ O,C,C_0,E,F$ are collinear so $ EF\bot CC_3$. Analogously $ EF\bot AA_3$ and $ EF \bot BB_3$. Therefore $ AA_3 \parallel BB_3 \parallel CC_3$. QED.


Well, I'll start off.  I thought this USAMO was surprisingly easy, but I failed to capitalize on it.  I thought #1 was essentially trivial once one noticed to investigate the radical axes, #2 became easy with casework, and #4 was simple induction.  #3a was okay as well.  If only I had had a few more hours...  :(

Oh, and if pythag is still browsing the MO forum, nice job  :lol: .  Blue MOP for sure if you really solved all of those.

[quote="Mewto55555"]Solved 2,4,6. I think I made an arithmetic mistake in my coordinate bash of #1. If that is the case, then tack that on there too!

Double-digits mayhaps? =O[/quote]

Completely solved?  Really?

If you know him at all, you'll know that he overestimates himself in general... :P I personally have no idea how well he did.

Guys guys, stop making me feel like I'll end up overestimating my score.

I [i]felt[/i] like I solved them. Of course, as number 6 felt easy, I should assume i didn't solve it :P

Still, I've got a pretty good shot at red MOP cutoff score (sadly, I am pretty sure I will not even make red MOP!)

Guess what?  I got  8  8  8  8  8  8.

If you've got nothing to say about your own scores or nothing productive to say about others', then don't say anything.

I [b]am[/b] from Missouri.  Don't let the flags fool you.  How can I be an American Samoa from Puerto Rico?

Same reason why I am not Japanese. Yet Japanese.

Do you even know where American Samoa is?

guys, I kinda like hope that you know that max is like,

kidding?

[quote="pythag011"]Dude max beat me by like a whole problem :O

I CONCUR THAT MAX IS BEASTLY.

/joinmaxfanclub[/quote]

lol, I call maxfanclub's treasurer

*sigh*

nostalgia for ubemaya's teapot appreciation club.

those were some *very* uh *interesting* days. (if anyone wants to look through his blog...)

[quote="mousy"]So What was the highest score?????  and what did pythag011 get?!?!?![/quote]

I got a negative zero.

No, not really.

Wenyu got the highest score with a 38.

Dude I think it would have been better if you didn't say his name in such a lowly forum as this one :(

pythag got the black mop cutoff. That's all I know.

[quote="alex960531"][quote="1=2"]pythag barely made black mop[/quote]

What was the black cutoff? and what did pythag get?[/quote]

27 and 27


Να αποδειχθεί ότι για κάθε $ n\in\mathbb{N^*}$ ισχύει ότι

$ 3^{n\plus{}2}\cdot{n!}>(n\plus{}1)^{n\plus{}2}$

[quote="kritikos"]Να αποδειχθεί ότι για κάθε $ n\in\mathbb{N^*}$ ισχύει ότι

$ 3^{n \plus{} 2}\cdot{n!} > (n \plus{} 1)^{n \plus{} 2}$[/quote]

Δεν το ξανακοιτάς λίγο; Πρέπει να έχεις κάποιο τυπογραφικό λάθος..

Λουκάς

Πολύ καλλή Δημήτρη!

Νομίζω ότι βγαίνει χρησιμοποιώντας τη σχέση:

$ ln(n!) \equal{} \sum_{2}^{n}lnn > \int_{1}^{n} lnx dx$

Θα το κοιτάξω για μια πλήρη λύση αργότερα, γιατί τώρα δεν έχω πολύ χρόνο.

Γεια σας,
νομίζω ότι δεν χρειαζεται να βάλουμε το n να ανήκει στο Ν*
Και για n=0 ισχύει η σχέση'
Αν κάνω λάθος μπορείται να μου πειτε?

[quote="theo_1"]νομίζω ότι δεν χρειαζεται να βάλουμε το n να ανήκει στο Ν*[/quote]
Φυσικά και δε χρειάζεται!!! Φτάνει μόνο να πούμε ότι ισχύει για $ n\in\{0,1,2,3,4\}$

Ελέγξτε το ρε παιδιά, για $ n\equal{}5$ δεν ισχύει.

σωστος ο ποντιος.κανενα τυπογραφικο ισως...για ν=5 αλλαζει η φορα..


LaTeX
abstract algebra
vector
calculus
group theory

hi,

i will have to make a mathematical speech in a few days. however, this is going to be the first time i have to do this in english, and i have some questions regarding the pronunciation of mathematical formulas. how do you say in english

[list]
[*]$a=b$ (esp. the "="...do you say "is", "equal", "is equal to",...?)
[*]$a^{2}$ (i assume "a square"?)
[*]$a^{b}$ ("a squared by b" or "a to the power of b" or ....?)
[*]$A\cap B$
[*]$A\cup B$
[*]$A\backslash B$
[*]$f(x)$
[*]$\sqrt[n]{a}$
[*]$a/b$ ("a divided by b" or "a over b"?)
[*]$a\leq b$[/list]

i know this is emberassing but one doesn't learn the pronunciation by reading papers  :(
and i don't learn this at school either.

thanks
robs

What I say:

[list]
[*]$a=b$ (esp. the "="...do you say "is", "equal", "is equal to",...?)
a equals b
[*]$a^{2}$ (i assume "a square"?)
a squared
[*]$a^{b}$ ("a squared by b" or "a to the power of b" or ....?)
I say a to the b-th, but I'm not sure that's standard.
[*]$A\cap B$
Quite non-standard, but I say A dumpout B
[*]$A\cup B$
A union B
[*]$A\backslash B$
Dunno
[*]$f(x)$
f-x, said in such a way that f is high and x is lower, f is on the inhale, x on the exhale of breath, with a good space between the f and x. Some people say f-of-x run together into one word.
[*]$\sqrt[n]{a}$
nth root of a
[*]$a/b$ ("a divided by b" or "a over b"?)
Either is common.
[*]$a\leq b$
a less than/equal to b, or occasionally I say a leck b [leck is my pronunciation of leq, i.e. the latex command :)
[/list][/list]

Some of mine agree with what solafidefarms posted- I'll just mention the differences.
$=$: "equals" or "is equal to", depending on context and the flow of the sentence.
$A\cap B$: intersection of $A$ and $B$, or $A$ intersect $B$.
$A\setminus B$: No standard. It doesn't come up often enough.
$f(x)$: $f$ of $x$, possibly run together.
$a\le b$: $a$ is less than or equal to $b$; if I need to shorten it, I'll drop the "or equal to" and accept the verbal inaccuracy, or switch to "$a$ is at most $b$".

There are a few odd things out there- I had a professor use "nabla" for $\nabla$ this quarter- I've never seen that except in $\LaTeX$ code.

$A^{b}$ is "a to the b power" or "a to the b"

[list]
[*]$a=b$ [b]a is equal to b[/b]
[*]$a^{2}$ [b]a squared[/b]
[*]$a^{b}$ [b]a to the power of b[/b]
[*]$A\cap B$
[*]$A\cup B$
[*]$A\backslash B$
[*]$f(x)$ [b]f of x[/b]
[*]$\sqrt[n]{a}$ [b]the nth root of a[/b]
[*]$a/b$ [b]a over b[/b] or [b]a divided by b[/b]
[*]$a\leq b$ [b]a is less than or equal to b[/b][/list]

[quote="robs"][list][*]$A\backslash B$[/list][/quote]I would say "$A$ exclude $B$"

For $A\backslash B$, I would probably say "The complement of $B$ in $A$" ...

[quote="Kent Merryfield"]$A\setminus B:$ I'd probably say "A but not B.", as illustrated below.

$x\in A\setminus B:$ "x is in A but not B."[/quote]

I would just say "A minus B", or the difference of A and B.  The latter is dangerously close to $A\Delta B$ (the symmetric difference), and it isnt clear whether we are referring to $B\setminus A$ or $A\setminus B$, but I think it gets the point across fairly well.

Context is everything of course - most ambiguities can be settled by something present in the context. But I have to point out that in an abelian group (which includes vector spaces), "A minus B" could be taken to be $A-B,$ which has a whole other meaning.

I usually do read $A\setminus B$ the same way as blahblahblah but when I talk about sets in a Euclidean space (the only example of an Abelian group that comes up in calculus courses), and I do not want to spend my time saying "the complement of $B$ in $A$" or some other long phrase, I usually do not hesitate to say "$A$ setminus $B$". It is not perfect English, of course, but it gets the message across without any ambiguities and I do not see why it is much worse than, say, the phrase "Latex this file", which almost everybody would use instead of the formally correct "Run the Latex program on this file".  :P

Hi
I've got another important question (for an english talk). How do you say
\[ R^*
\]
(e.g. for unit group or Kleene-star, etc)? Just "R star"? "R asterisk"? (the latter sounds strange)

Thanks!

"R star" is the standard way to say it, as far as I know (compare with "$ C^*$-algebra")


[quote="shobber"]In $\triangle{ABC}$, $CD \perp AB$ at $D$. $CE$, $DF$ and $DG$ bisects $\angle{ACB}$, $\angle{CDB}$ and $\angle{ADC}$ respectively with $E \in AB$, $F \in CB$ and $G \in AC$. If $\angle{GEF}=90^o$, then:
Prove that $\angle{ACB}=90^o$.[/quote]
We have $\angle{GEF}=\angle{GDF}=90^o$, because $CD \perp AB$ and $DF$ and $DG$ are bisectors of $90^o$ angles so $\angle{GEF}=1/2(\angle{BDC}+\angle{ADC})=90^o$.
So, $\angle{GEF}=\angle{GDF}=90^o$ hence $DEFG$ is cyclic. Now $\angle{EDF}=45^o$ hence $\angle{EGF}=45^o$. But $\angle{ADG}=45^o$ so $AB$ is parallel to $FG$ and so  $CD \perp FG$. $\angle{FDC}=45^o$, $\angle{DMF}=90^o$ where $M$ is pt of intersection of $FG$ and $DC$. So $\angle{DFG}=45^o$ so $\angle{DEG}=45^o$ so pt $D$ is same as pt $E$ because $\angle{ADG}=\angle{AEG}=45^o$. But if $ABC$ is isosceles with $BC=AC$ have always $\angle{FEG}=90^o$.
So $ABC$ is isosceles but $\angle{ACB}$ doesn't have to be $90^o$. ;)


Well, since the side length of the outer square is 8, half of it would be 4. We see that half the sidelength of the inner square is 4-3 = 1. Thus the sidelength of the inner square is 2, and its area is 4.


function
calculus
integration
calculus computations

suppose $ \{f_n\}_{1}^{\infty}$ is an ortho-normal set of functions on $ [a,b]$.  Given $ c>0$ and a real number $ d$, let $ g_n(x)\equal{}\sqrt{c}f_n(cx\plus{}d)$  Show that $ \{g_n\}_{1}^{\infty}$ is an ortho-normal set on $ [\frac{a\minus{}d}{c},\frac{b\minus{}d}{c}]$.

If you try calculating the inner product of $ g_n$ with $ g_m,$ and make the "obvious" linear substitution in the integral, what happens?


Determine a point $M$ interior to a triangle $ABC$  such that the product of the distances of $M$ to the side of triangle $ABC$ is maximum

Let $x,y,z$ be the distances form sides with lenghts $a,b$ and $c$. Then $ax+by+cz=2P$ ($P$ is area), and from AM-Gm inequality we have $2P\geq 3\sqrt[3]{abcxyz}$, or $xyz\leq \frac{8P^{3}}{27abc}=const.$ Hence the product is maximum when $ax=by=cz$, or $M$ is centroid of $ABC$.


geometry
incenter
circumcircle
geometry unsolved

[quote="pardesi"]Let $ I$ be the Incentre of $ \Delta ABC$ with circumcircle $ \omega$ let $ P \equal{} AI \cap \omega$ and $ Q \equal{} AI \cap BC$.Prove that $ PI^{2} \equal{} PQ . PA$[/quote]
I don't know anything about Harmonic Quadrilaterals, but here is another solution:
[hide="Solution"]
Notice that
\[ \angle BIP \equal{} \angle IAB \plus{} \angle IBA \equal{} \angle PAC \plus{} \angle IBC \equal{} \angle PBC \plus{} \angle IBC \equal{} \angle IBP\implies IP \equal{} PB
\]
Now, notice that $ \angle PBQ \equal{} \angle BPC \equal{}\angle PAC \equal{} \angle BAP$ and $ \angle BPQ \equal{} \angle APB$.  Hence, $ \triangle PQB\sim \triangle PBA\implies \frac {PQ}{PB} \equal{} \frac {PB}{PA}\implies PI^2 \equal{} PB^2 \equal{} PA\cdot PQ$. [/hide]

P is the midpoint of the arc BC then BP=CP
<BIP=<BAP+<ABI=<PAC+<IBC=<PBC+<IBC=<IBP  then  BP=IP
So if we make an inversion with center P and radius PI then
    B'=B and C'=C  then ω'=BC  then A'=Q then PQ·PA=PI²

[quote="pardesi"]Let $ I$ be the Incentre of $ \Delta ABC$ with circumcircle $ \omega$ let $ P \equal{} AI \cap \omega$ and $ Q \equal{} AI \cap BC$.Prove that $ PI^{2} \equal{} PQ . PA$[/quote]
this is my solution:(using harmonics) :lol:
Let $ (J)$ is the exircle in the angle $ A$,we have $ JB\perp BI$ and $ \angle ABI\equal{}\angle IBC$,so it's implies that:
$ (JIQA)\equal{}\minus{}1$.Notice $ PI\equal{}PJ$  thus $ PI^2\equal{}JP^2\equal{}PQ.PA$


If we can show that $a+b+c\geq ab+bc+ca$, then the problem is solved by holder:\[ \sum\frac{a}{\sqrt{b+c}}\cdot\sum\frac{a}{\sqrt{b+c}}\cdot\sum a(b+c)\geq\left(\sum a\right)^{3}\]

By Cezar Lupu.

Problem 33 --> http://web.mit.edu/~tmildorf/www/Inequalities.pdf

[quote="rem"]What do you mean by $a,b,c$?[/quote]
Sorry, I confused $x,y,z$ with $a,b,c$.

Problem:

Let $a,b,c\in\mathbb{R}^{+}$ such that $a+b+c+abc=4$

Prove that:
$\sum\frac{a}{\sqrt{b+c}}\geq \frac{\sqrt{2}}{2}(a+b+c)$

Solution:

$a+b+c+abc=4$ <==> $\sum\frac{a}{a+2}=1$

Let $\frac{a}{a+2}=\frac{x}{x+y+z}$ and so on, we have $a=\frac{2x}{y+z}$ and so on.

Now we have to prove:

LHS$=\sum\frac{x\sqrt{(x+z)(x+y)}}{(y+z)\sqrt{yz}}\geq 2\sum\frac{x}{y+z}=$RHS

By CS ineq., we get $\sqrt{x+z)(x+y)}\geq x+\sqrt{yz}$ and so on.

By AM-GM, we get $2\sqrt{yz}\leq y+z$ and so on.

LHS$\geq\sum\frac{x(x+\sqrt{yz})}{(y+z)\sqrt{yz}}=\sum\frac{x^{2}}{(y+z)\sqrt{yz}}+\sum\frac{x}{y+z}$

$\geq 2\sum\left(\frac{x}{y+z}\right)^{2}+\sum{x}{y+z}\geq \frac{2}{3}\left(\sum\frac{x}{y+z}\right)^{2}+\sum\frac{x}{y+z}\geq$ RHS

The Final Step use the Nesbit ineq.

[quote="andyciup"]If we can show that $a+b+c\geq ab+bc+ca$, then the problem is solved by holder:
\[\sum\frac{a}{\sqrt{b+c}}\cdot\sum\frac{a}{\sqrt{b+c}}\cdot\sum a(b+c)\geq\left(\sum a\right)^{3}\]
[/quote]Yes we can show that. [hide]It is equivalent to Schur. :)
Do you see how?  :D[/hide]

[quote="andyciup"]Thank you for the hint.Actually, I can prove that $a+b+c\geq ab+bc+ca$, by contradiction and by Schur. I use the fact that $a+b+c\geq 3$ and the fact that $a+b+c\leq 4$.
Do you see how? :)[/quote]
Do you mean you can prove $a+b+c\geq ab+bc+ca$ whenever  $3\leq a+b+c\leq 4$ and $a,b,c>0$? But that's wrong (take $a=b=c>1$)  :maybe:

As to my proof, nothing very creative: Using libra_gold's substitution, we have to prove $a+b+c\geq ab+bc+ca \Longleftrightarrow\sum\frac{2x}{y+z}\ge\sum\frac{4xy}{(x+z)(y+z)}$ $\Longleftrightarrow\sum x(x+z)(x+y)\ge\sum2xy(x+y)\Longleftrightarrow Schur$  :lol:

1. No, I mean :prove $a+b+c\geq ab+bc+ca$ whenever $a+b+c+abc=4$, which implies $3\leq a+b+c\leq 4$.
2.I don`t quite think that $x=\frac{2a}{b+c}$ etc.. is the right substitution, because \[ \sum\frac{2a}{b+c}+\frac{8abc}{(a+b)(b+c)(c+a)}=4\] is equivalent to \[ \sum\frac{a}{b+c}+\frac{4abc}{(a+b)(b+c)(c+a)}=2\] \[ \Longleftrightarrow \sum\s(a-b)(a-c)=0,\]which is obviously not allways true :P. To make that substitution, you need $xy+xz+yz+xyz=4$.

oops, you are right. I didn't verify libra_gold's substitution  :blush: but that means his solution is wrong as well. :( But both are OK for the changed condition then.  :)

andyicup: what is your solution?
In fact I have a weird solution.

Ok, here is my solution:
From Schur`s inequality we have \[ 2\sum ab-\sum a^{2}\leq\frac{9abc}{a+b+c},\]or equivalently \[ (a+b+c)^{2}+\frac{9abc}{a+b+c}\geq 4(ab+bc+ca).\]Now assume that $ \sum ab\geq\sum a$, and denote $a+b+c=s$. From the previous relation we get \[s^{2}+\frac{9abc}{s}\geq 4s.\] But because $a+b+c+abc=4$, we get $abc=4-s$, our relation becomes \[ s^{2}+\frac{36-9s}{s}\geq 4s\] \[ %Error. "Longrightleftarrow" is a bad command.
s^{3}-4s^{2}-9s+36\geq 0\] \[\iff(s-3)(s+3)(s-4)\geq 0,\]which is false!

Sorry, I've made a terrible mistake. This step of mine is wrong:

[quote]a+b+c+abc=4 <==> \sum\frac{a}{a+2}=1[/quote]

[quote=arqady]I think the following inequality is interesting enough.
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c+abc=4$. Prove that:
$$\frac{a}{\sqrt{a+3b}}+\frac{b}{\sqrt{b+3c}}+\frac{c}{\sqrt{c+3a}}\geq\frac{a+b+c}{2}$$[/quote]

wrong!!!

[quote=arqady]I think the following inequality is interesting enough.
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c+abc=4$. Prove that:
$$\frac{a}{\sqrt{a+3b}}+\frac{b}{\sqrt{b+3c}}+\frac{c}{\sqrt{c+3a}}\geq\frac{a+b+c}{2}$$[/quote]

Your inequlity is the same with the first inequality.
In fact, we can expand it to:
$\frac{a}{\sqrt{ma+nb+pc}}+\frac{b}{\sqrt{mb+nc+pa}}+\frac{c}{\sqrt{mc+na+pb}}\geq \frac{a+b+c}{\sqrt{m+n+p}}$
with $a+b+c+abc=4$ and $a,b,c>0$.

[quote=phuocchubeo]
Your inequlity is the same with the first inequality.[/quote]
Why? :maybe: Explain please.  ;)
[quote=phuocchubeo]
In fact, we can expand it to:
$\frac{a}{\sqrt{ma+nb+pc}}+\frac{b}{\sqrt{mb+nc+pa}}+\frac{c}{\sqrt{mc+na+pb}}\geq \frac{a+b+c}{\sqrt{m+n+p}}$
with $a+b+c+abc=4$[/quote]
Your inequality is wrong. Try $m=n=1$, $p=0$, $c=0$ and $a=b=2$.

[quote=arqady][quote=phuocchubeo]
In fact, we can expand it to:
$\frac{a}{\sqrt{ma+nb+pc}}+\frac{b}{\sqrt{mb+nc+pa}}+\frac{c}{\sqrt{mc+na+pb}}\geq \frac{a+b+c}{\sqrt{m+n+p}}$
with $a+b+c+abc=4$[/quote]
Your inequality is wrong. Try $m=n=1$, $p=0$, $c=0$ and $a=b=2$.[/quote]

Oh, I forgot the term $a,b,c>0$

[quote=arqady][quote=phuocchubeo]
Your inequlity is the same with the first inequality.[/quote]
Why? :maybe: Explain please.  ;) [/quote]

By $C-S$ inequality:
$LHS\geq \frac{(a+b+c)^2}{a\sqrt{a+3b}+b\sqrt{b+3c}+c\sqrt{c+3a}}\geq \frac{(a+b+c)^2}{\sqrt{4(a+b+c)(ab+bc+ca)}}$
We can prove $a+b+c\geq ab+bc+ca$ so $LHS\geq \frac{a+b+c}{2}$
Done!

[quote=arqady]Why? What is your counterexample?[/quote]

Arqady, Your result is right!

[quote=lwwwww][quote=Till]Let $x,y,z\in\mathbb{R}^{+}$ such that $x+y+z+xyz=4$

Prove that:
$\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{x+z}}+\frac{z}{\sqrt{x+y}}\geq \frac{\sqrt{2}}{2}(x+y+z)$[/quote]

Let $x,y,z\in\mathbb{R}^{+}$ such that $xy+yz+zx+xyz=4$

Prove that:
$x+y+z\geq 3$.[/quote]

We have:
$x+y+z+xyz=4$
$<=>\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=1$
Using C-S, we have: $LHS\geq\frac{9}{6+x+y+z}$
$=>6+x+y+z\geq 9<=>x+y+z\geq 3$


Show that the number of all possible ordered $ 4\minus{}$tuples of non-negative integers $ (n_1,n_2,n_3,n_4)$ such that $ \sum_{i \equal{} 1}^{4}n_i\le100$ is $ \dbinom{104}{4}$ .

[hide]
The number is equal to the number of $ (a,b,c,d,e)$ of nonnegative integers satisfying
$ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}100$,
therefore the number we want is $ _5H_{100}\equal{}\binom{104}{100}\equal{}\binom{104}{4}$ [/hide]