For each positive integer $ n$, let $ S_{n}$ denote the set consisting of the first $ n$ natural numbers, that is

$ S_{n}\equal{}\{1, 2, 3, 4, ..., n\minus{}1, n \}$.

(i) For which values of $ n$ is it possible to express $ S_{n}$ as the union of two non-empty disjoint subsets so that the elements in the two subsets have equal sums?

(ii) For which values of $ n$ is it possible to express $ S_{n}$ as the union of three non-empty disjoint subsets so that the elements in the three sunsets have equal sums?

For part (I) :

It`s true for $ n \equal{} 3$ and $ n \equal{} 4$ , and it`s trivial that if $ n \equiv 1 (mod4)$ or $ n \equiv 2 (mod4)$ the sum is odd .
We induction on $ n$ :
Assume it`s true for $ n$ , since $ (n \plus{} 1) \plus{} (n \plus{} 4) \equal{} (n \plus{} 2) \plus{} (n \plus{} 3)$ it`s true for $ n \plus{} 4$.
So for all positive integer in $ 4k$ and $ 4k \plus{} 3$ shape , we can express the $ S_n$ as the union of two non-empty disjoint subsets so that the elements in the two subsets have equal sums.  :wink:

It has same solution with part (i) , but in this part the induction hypothesis for $ n$ implies $ n\plus{}6$.  :lol:


geometry MATHCOUNTS AMC USA(J)MO USAMO trigonometry ARML

If a top math student was moving to your area of the state, which secondary school would you recommend?   What makes that school attractive?  If possible, provide information regarding:

1) school background

size, public/private, tuition, admission restrictions (school district, qualifying exams,...)

2) learning environment

curriculum options, quality of instruction, partnerships with nearby universities for course offerings, extra-curriculars, math/science teams, competitions, etc.

3) other features of possible interest to top students - i.e. music programs.

It would be nice to have benchmark information available on good schools for students and school administrators.  My recent conversations with school administrators indicate that most are not really aware of what needs to be done to attract and retain top students.

If you were coming to Winston-Salem, I'd recommend Hanes Middle School.  It's a public institution that houses the "highly academically gifted (HAG)" program for 6th, 7th, and 8th grades that serves Forsyth county.  At least one approved test that places you in the top 1% academically is required for admission.  The school is located in a relatively seedy section of the inner-city and some have avoided the school for that reason.  In addition to approximately 275 HAG students, there are roughly an equal number of students who live in the school district.  Bus transportation is provided county-wide.

HAG students have separate core classes (English, Math, SS, Science) that are accelerated or enhanced.  Electives are integrated with the neighborhood kids.   Roughly half of the HAG students are placed into Algebra I as 7th graders and Geometry as 8th graders.  A few kids have been permitted to accelerate further - with mixed results.  The math teachers are good.  Environmental Science is completed in 8th grade.  They have a good band program and a good Spanish teacher.  The small school size limits the number of electives available.

Hanes has the most active and competitive math team in the area. They have placed highly at MATHCOUNTS, subject-area state math contests, and AMC contests in recent years.

They also offer HAG students a nice field trip each year.  Seventh graders spend a week in Florida on a marine biology trip enjoyed by all.

I'm going to Ligon next year  :mad: I'll make sure they don't lose to west cary by 93 points at Algebra 1 this year

-P.S. I was in Algebra 1 NCMTS west cary team o-o

We ALMOST beat them in chapter mathcounts(no thx to me)<--thats a lie<--thats a lie<--thats a lie<--thats a truth

[quote="Karth"]Is secondary school high school?[/quote]

secondary school= school#2 = middle school

I was using the term "secondary schools" to refer to anything between "elementary" school (grades 1-5/6) and college.  These would include high schools, junior high schools, and middle schools.

I'd agree that the schools attracting the top math/academic talent in the state are pretty well known.  I am more interested in what these schools are doing to attract the talent.

It's clear that Wake County public schools are doing a better job developing top talent than Charlotte/Mecklenburg and other school systems in NC.  With Ligon, Enloe and several other good public schools to compete with, the private schools in Wake county don't seem to get many top students.  On the other hand, a larger chunk of the Charlotte/Mecklenburg talent appears to prefer private (Providence Day, Charlotte Latin, Charlotte Country Day...) or home schooling.  Hoggard seems to attract most of the math talent in the Wilmington area.  I don't know much about Burlington, Durham, or Buncombe schools.

In Winston-Salem, the high school academic talent is roughly equally spread out among half of the public schools and 1 private school while the middle school talent is concentrated at Hanes.  Many AP classes in W-S are held in one geographic location (the "Career Center) for students at all W-S public schools.  Those attending the Career Center are responsible for some of the transportation and use a travel period in the middle of the school day to move between schools.  On the plus side, this gives everyone in the county a chance to take relatively esoteric courses like Japanese that wouldn't have enough demand at just one school.  On the minus side, since the talent is spread out among several schools, there is much less incentive to provide for the needs of top students (except at the Career Center) - classes at the high schools tend to be watered-down compared with what could be provided at a good magnet school.  This is most noticeable with freshman/sophomore courses: Pre-Calc, Alg II, Geometry, and English.  Career Center AP courses attract good students and appear to be appropriately challenging.

not_trig is right about talent attracting other talent.  I'd really like to know what makes schools like Ligon and Enloe so attractive to talented students and why more cities aren't providing similar alternatives.

Some thoughts:

1. To be attractive to serious students, a school must have a mission to serve top students and it must design curricular and extracurricular activites to meet these needs.

2.  Strong arts programs, especially music, are needed to attract top math/science talent.  It's no accident that Enloe is an arts-focused magnet rather than a science/math magnet and still attracts the top Wake county math and science students.  Math/science/technology magnets seem likely to fail unless they have strong music/arts programs (is this true at TJ? I'm not sure about this one.)

3.  Full-blown IB programs do not appear to be attracting top talent in NC.  These programs have a reputation for high work loads that leave little time for other activities.  Top talent seems to want the freedom to choose extracurriculars to fill up free time rather than IB coursework.

4.  Most people don't want to see top students given more advantages than they already have.  This might explain why so few public schools like Enloe exist.

The top high schools are those that let students take classes at the nearby university.  Other issues are secondary.  Middle schools are another story.  A good middle school ought to have challenging all-around classes in English as well as mathematics.  My 7th grade English teacher assigned 2-page essays once a week.  She returned them to us too, with detailed comments and corrections.  I think having her as a teacher was the second most important thing to happen to me in middle school, behind discovering mathematics.  The most important way to cultivate mathematics talent in a middle school is to have enthusiastic, knowledgeable math coaches.  I'm not sure it is possible to 'attract' top math talent to a middle school, as few students know much about mathematics upon entering.  The middle school years are the most important though, I think, when it comes to convincing students to love mathematics.

[quote="n^4+4"]The top high schools are those that let students take classes at the nearby university.  [/quote]

This is a great option for those living near an affordable, high quality university and most schools will let students take college classes for graduation credit.

Top students may find math courses and the talent level at most of the 16 NC public universities and virtually all of the community colleges less challenging than their high school AP Calc classes.  This won't be a problem at Chapel Hill or State.  Many of the others have a difficult time attracting good math students to their math major programs and concentrate on developing math teachers instead.  Those living in Raleigh or Chapel Hill and possibly a few other places have good university math courses available.

Winston-Salem students can go to Wake for $ \$$3,000 per course.  They gave tuition breaks (half price) to high school students a couple of years ago but now charge full price.  For this reason, I doubt that many high school students are taking Wake courses.

@n^4 - sounds like your 7th grade English teacher was a gem.  I'm really disappointed at what my kids are learning about writing within the W-S Forsyth County schools.  In the rare event that papers are written feedback is not provided.  I had a better experience with writing and English in Ohio public schools in the 70's.  A good school really needs to have at least one good writing instructor (I wish I could aim higher but right now I'd be happy to see just one.)

[quote="n^4+4"]I'm not sure it is possible to 'attract' top math talent to a middle school[/quote]

Maybe it depends on what we mean by "top."  If we're talking about the top 1% or so within a community, then I'd say it's possible.

Within a community, most parents of smart kids are connected and find out the strengths and weaknesses of the schools.  Typically, they flock to the same small set of schools for reasons discussed in this thread.

[quote="gt59"]Top students may find math courses and the talent level at most of the 16 NC public universities and virtually all of the community colleges less challenging than their high school AP Calc classes.[/quote]

I find higher math courses at UNCW (which is probably very average for NC public universities mathwise) a good deal more challenging than AP calculus. Granted my definition of "higher math" means proof-based, which excludes multivariable calc, linear algebra, and differential equations at most schools, and also excludes all but a few hundred (at most) high school students nationwide. On the other hand, my AP calc BC class had some much brighter students than some math majors at the university.

[quote="gt59"][quote="n^4+4"]I'm not sure it is possible to 'attract' top math talent to a middle school[/quote]

Maybe it depends on what we mean by "top."  If we're talking about the top 1% or so within a community, then I'd say it's possible.[/quote]

AoPS always tries to get us to not complain about "low" high scores, and I think this falls in the same category of loosely defining "top math talent". I agree that it should be defined loosely, and the top 1% of math students are very bright. However, a lot of people on these forums are very annoyed by the lack of others who really have a *deep* love and understanding of math. We just need a peer group, and I think that is perfectly fair.

And that leads in to the original question. Hoggard is really a very good school, even in the state. If I complain about it, it's only because almost all of the top math people are in the triangle. It has very good math and science programs, apparently does very well in sports (I think... I never keep up with these things), and I'm under the impression that the humanities are taught poorly no matter where you are, so that doesn't make a difference. Actually, though, Latin is taught effectively (and I imagine the AP humanities as well). The teacher does not comply with the crazy modern idea of "let's just learn a bunch of random phrases so we can talk to people but not know what we're saying or why we're saying it, and while we're at it, let's learn how to say 'scuba dive'". Maybe that's how Latin is always taught? It is missing some AP humanities though (languages and history). And it's WAY overcrowded (almost 2000 people). Also, certain councilors/staff members and the principal are very supportive of education, which seems rare these days.

As for public middle schools, I guess Myrtle Grove and Roland Grise are the best. Besides WAAS, no private middle schools focus on education - there are a ton of religious schools and then there's Cape Fear Academy, which is just private for the sake of being private ($ \$\$\$$). I loved WAAS because of the peer group and you can do literally *anything* education-wise, but it's extremely tiny (50 kids in a couple trailers) and two out of four of the main teachers (that's right, FOUR) have left since I was there so I dunno what's happening now.

[quote="not_trig"][quote="Teki-Teki"]dude ligon pwns there can be no arguing.

especially now since the math team coach is Mr. Rohan Lewis who was a former USAMO qualifier in 1997 and did quite well at the Comprehensive Contests around that time (up to 1999)

but ligon has gone downhill since vivek and i left and will further decline after vivek's little bro arjun leaves after this year.

but yeah ligon has a bunch of awesome teachers and electives and really smart kids (although no REALLY smart kids except for maybe calvin deng) which hanes does not really seem to have.

and ligon's spanish teacher (at least, one of them - the one I had) is REALLY REALLY good.  their music programs are top notch.[/quote]

I agree... Ligon is definitely the best middle school in the triangle, academically. Being a magnet school, it enjoys a pretty good advantage, but more importantly, there are lots of smart people that go there. Nick Tobey is apparently good (although obnoxious :wink: ) and Calvin Deng is too. Next year, my little brother is going there... I hope he can [i]make[/i] their MC team. (as of now, he is doing pretty poorly... but he'll improve) Ligon has a lot of depth, and yes, has gone downhill since the departure of teki-teki (which was pretty helpful in Chapters '06... lol... Ligon was sad...) but I think Ligon continues to be the dominant [i]middle school[/i] in NC math.[/quote]

Not_trig, haven't you heard that Calvin is married with Nick

I hope I can make their MC team, even though I got regional first this year... An individual got 4th in regionals

[quote="PI-Dimension"][quote="not_trig"][quote="Teki-Teki"]dude ligon pwns there can be no arguing.

especially now since the math team coach is Mr. Rohan Lewis who was a former USAMO qualifier in 1997 and did quite well at the Comprehensive Contests around that time (up to 1999)

but ligon has gone downhill since vivek and i left and will further decline after vivek's little bro arjun leaves after this year.

but yeah ligon has a bunch of awesome teachers and electives and really smart kids (although no REALLY smart kids except for maybe calvin deng) which hanes does not really seem to have.

and ligon's spanish teacher (at least, one of them - the one I had) is REALLY REALLY good.  their music programs are top notch.[/quote]

I agree... Ligon is definitely the best middle school in the triangle, academically. Being a magnet school, it enjoys a pretty good advantage, but more importantly, there are lots of smart people that go there. Nick Tobey is apparently good (although obnoxious :wink: ) and Calvin Deng is too. Next year, my little brother is going there... I hope he can [i]make[/i] their MC team. (as of now, he is doing pretty poorly... but he'll improve) Ligon has a lot of depth, and yes, has gone downhill since the departure of teki-teki (which was pretty helpful in Chapters '06... lol... Ligon was sad...) but I think Ligon continues to be the dominant [i]middle school[/i] in NC math.[/quote]

Damien, haven't you heard that Calvin is married with Nick

I hope I can make their MC team, even though I got regional first this year... An individual got 4th in regionals[/quote]

Definitely, Ligon used to be the best. But now its math is a lot worse -- I'm predicting Calvin, Nick, Allen Yang, and Allan Jiang will be its top math students next year. A lot of kids still try to get on the MathCounts team, but they're not serious. Cuts to last year's team had to be made in November instead of February because no one would stop talking.
Ligon's gonna keep getting worse, since a lot of smart kids can't get in; so many more students are applying! Plus a lot of people don't want to spend 4 hours on the bus every day -- that's why I'm going to my base school instead of Enloe.

Ligon's electives are really good though, I guess they help kids become well-rounded. The arts, especially, are fantastic. The smartest kids still come for the math, but the arts programs attract all-around good students.

Like at any school, some teachers are really good and others, not so much. It's all based on luck, but I'd say almost all of the math teachers that the smart kids have are really good at math and/or love math.

BUT...Mr. Lewis is brilliant. He can explain pretty much anything, and he'll help the math team kids a lot :)

[quote="notsobright48"]
Definitely, Ligon used to be the best. But now its math is a lot worse -- I'm predicting Calvin, Nick, Allen Yang, and Allan Jiang will be its top math students next year. [/quote]

haha the latter still needs much work, but hopefully he'll do well :wink:

@not_trig: If he's anything like you he ought to be fine.  Again, Ligon is going downhill.  Or maybe it's just the whole state.

@gt59: Enloe is not really an arts-focused magnet; it may only seem that way in comparison to NCSSM.  Enloe is a good school for almost anything, mainly because it is a magnet school and THE flagship magnet school for Wake County.  For example, about one-sixth of the teaching staff is National Board certified, and about two-fifths have master's or doctorate degrees in their respective subjects.  As such, it has a much better reputation than Broughton (no offense, Yasha) or Southeast Raleigh.  In terms of being a cool-looking school, though, Broughton wins hands down.

Also, the IB Programme is not as bad as it seems - I still will manage to go beyond the furthest limits of mathematics at Enloe before I graduate while pursuing my IB Diploma.  It also gives students a little broader background, which may be useful later in life.

@notsobright48:  It's at most an hour bus ride each way, and you can be productive on the bus.  It's less than half an hour if you drive.

I would also like to say to those who have raved about the fact that Calvin Deng is 10 years old.  Wow.  Amazing.  I was ten in 6th grade.  So was Vivek.  So were probably half the people on NC ARML A-Team.  Get over it.  The only reason you think he is amazing is because there is nobody else on his level at his age.  When Vivek and I were at Ligon, people knew we were both pretty darn good. (Vivek was a little better in middle school; we're about even now.)  People did not think we were amazing because either of us could be compared to the other.

Also, anybody know why Nick Tobey didn't go to ARML?  I have a guess...but I won't say what it is.  (for politeness' sake - hopefully he won't turn out like somebody who did go to ARML this year - A teamers should know who I'm talking about.)

Mr. Lewis is quite impressive; what less would you expect of a USAMO qualifier?

Also, I would like to ask if anybody here realizes how lucky North Carolina is, in that the state has a fairly weak teacher's union.  Although this may not seem like much, mainly because NCers hardly hear about the teacher's union, it is actually almost essential to providing a good gifted education system.  I know that in some states with strong teacher's unions almost all positions are assigned by seniority: in other words, if a teacher who has taught longer than you have wants your job, that teacher gets it.  No questions asked.  In our state, however, the weak teacher's union system allows the principals and county/district officials to have more influence in appointing teachers, which is how schools like Ligon, Enloe, Hanes, etc. get many of the very best teachers.  I went to a neighborhood school in elementary school for K-3 but transferred to Hunter, a magnet elementary school, for grades 4-5.  The difference was extremely noticeable.  At Hunter, and especially in 5th grade (i believe my math teacher that year went on to win some statewide teaching award) my teachers were excellent, and my 5th grade math teacher in particular helped jump-start my accelerated education.  In a state with a strong teacher's union, this math teacher would never have gotten a post in which she could have helped so many talented kids simply because she was quite young.  This in turn discourages magnet schools, because there is no point in sending talented kids to a school where they will not necessarily be exposed to teachers who can really nurture said talent.

Did anybody else in the Raleigh area see the fashion article in the Life section of the N&O?  Disgusting.  It made me laugh.  It almost made me not want to read the comics.  I wonder why all of the people featured were white females.  Is nobody else sexy enough?  Or are all males and minorities simply not fashionable enough?  Another side note: Why were there no Enloe upperclassmen?  Freshmen have no idea of how people dress at Enloe.  Vivek will attest to that.  Also, I would like to see any one of the Enloe freshmen try to pass Healthful Living's physical education section wearing that.  I would laugh my guts inside out.

Oh man North Carolina how you have changed.

Speaking of which:

(a) whoa ten years old I used to be 10

(b) the fashion article came out the day before I arrived, which is unfortunate, but all the letters to the editor complained about it the next day

(c) huh is Nick Tobey this 10-year old, because often parents don't let really young kids go due to being-really-young (I qualified in 7th grade but didn't go because my mom was all like dude you almost died at state math contest)

(d) what why have we clearly established that ligon is the best I BET THEY HAVE NOT WON STATE MATHCOUNTS FOR THE PAST FIVE YEARS or like whatever else you use to quantify middle-school mathematical success

(e) dude has east not been mentioned yet in this thread?

@MysticTerminator: In response to (e), I didn't mention East Chapel Hill because of the weird districting over there.  Chapel Hill has its own school system, and so does the rest of Orange (?) County.  I didn't know how your guys' schools handle stuff like this, and I didn't want to be misleading.

Also, in response to (d), Ligon won in 2004, and placed second in 2005.  I left after that, so I don't really know how they did.  Also, Ligon produced quite a few AIME qualifiers around those years.  I don't know how else you would quantify middle-school mathematical success.

I agree with you on (a), (b), and I see your point in (c).


function
calculus
derivative
integration
real analysis
real analysis unsolved

[quote="cimabue"]Assume $f\in C^{\infty}(\mathbb R)$ is a smooth function such that:
a) $f^{(k)}\in L^2(\mathbb R)$, $\forall k \geq 0$  (the $k^{th}$ derivative)
b) $f$ is rapidly decreasing, meaning $\sup_{t\in\mathbb R}|t|^N|f(t)|<+\infty$, $\forall N\geq 0$.

Prove that $f^{(k)}$ is rapidly decreasing, $\forall k\geq 0$.[/quote]
It is  enough to prove that $f'$ is rapidly decreasing at $+ \infty$ if $f$ is rapidly decreasing at $+\infty$ and $f'' \in L^2$.
So let $x > 0$, without loss of generality large enough such that $A: = \sup_{t > x}|f(t)| \le \left(\int_{\mathbb R} |f''(t)|^2 dt \right)^{1/2} =: B$. We must show that for any $N$ there is a $C_N$ independent of $x$ such that $x^N|f'(x)| \le C_N$.

By Taylor's Theorem, for any $0<h \le 1$, $f(x+h) = f(x) + hf'(x) + \int_0^h (h-s)f''(x+s) ds$.
Therefore, $f'(x) = \frac{f(x+h)-f(x)}{h} - \frac{1}{h} \int_0^h(h-s)f''(x+s) ds$, and by Cauchy-Schwarz
$|f'(x)| \le \frac{2}{h}\sup_{[x,x+1]}|f(s)| + \sqrt{h} B$. Set $h = \left(\frac{\sup_{[x,x+1]}|f(s)|}{B}\right)^{2/3} \le 1$, then one deduces
$|f'(x)|\le 3 \left(\sup_{[x,x+1]}|f(s)| \right)^{1/3} \left(\int_{\mathbb R} |f''(t)|^2 dt \right)^{1/3}$
Thus
$x^N|f'(x)| \le C_N$ with $C_N = 3 \left(\sup_{[x,x+1]}|f(s)|s^{3N} \right)^{1/3} \left(\int_{\mathbb R} |f''(t)|^2 dt \right)^{1/3}$, which is a finite number.  ;)


calculus
integration
LaTeX
trigonometry
function
real analysis
real analysis unsolved

hi, before i post my question, i'm new here so i was wondering if someone could tell me how to insert the math symbols in the text box. It would make it a lot easier to pose the problem, thanks.

You put the code in between dollar signs. There is a full guide on this site; click the LaTeX link on the sidebar. Most things are pretty intuitive: it's just a backslash followed by what you want to write, or sometimes a contraction of it: \int, \sum, \theta, and \sin are some good examples. You do superscripts with a caret ^ and subscripts with an underscore _. Don't forget to include braces { } when you want to do something that's more than one character in a subscript or superscript. Some functions take their arguments in braces, too, like \sqrt{ax + b} = $\sqrt{ax + b}$ or \frac{1}{5} = $\frac{1}{5}$.

You can mouse over other people's symbols to see what code generated them.


In a triangle with sides of lengths a, b and c, it holds (a+b+c) (a+b−c) = 3ab. The measure of the  angle opposite the side of length c is:

(A)15◦ (B) 30◦ (C) 45◦ (D) 60◦ (E) 120◦

[hide="Solution"]By Law of Cosine's, we get that $ c^2 \equal{} a^2\plus{}b^2\minus{}2ab\cos\theta$. However, we know that $ (a\plus{}b\plus{}c)(a\plus{}b\minus{}c) \equal{} a^2\plus{}2ab\plus{}b^2\minus{}c^2\equal{}3ab$, so $ c^2 \equal{} a^2 \plus{} b^2 \minus{} ab$. Therefore, $ \cos \theta \equal{} \frac{1}{2}$, so $ \theta \equal{} 60^\circ$. The answer is $ \boxed{D}$.

[/hide]


salam dostane aziz. baba ye peygham baraye bahse ghabli bezarid bebinam moshkel az mane ya az shoma(eradat darim). rassi ye masaleye jaleb: sabet konid $ \sum_{n\equal{}0}^\infty (\minus{}1)^n\equal{}\frac{1}{2}$  :!: jalebe na?  :P

khob bebin serie $ \sum_{n \equal{} 0}^\infty (x \minus{} 1)^n$ ye serie hendesie ba ghadre nesbate $ r \equal{} x \minus{} 1$ va jomle avvalie $ a\equal{}1$ pas haselesh mishe:

$ \frac {a}{1 \minus{} r} \equal{} \frac 1{2 \minus{} x}$

khob hala gharar bede $ x \equal{} 0$ mishe hamoon $ \frac 12$.

baba eival  :) gol kashti
ye esbate sade ham dare ke age seri ro besoorate $ 1\minus{}1\plus{}1\minus{}1\plus{}...$ bast bedim va oon ro be $ A$ nesbat bedim darim:
$ A\equal{}1\minus{}1\plus{}1\minus{}1\plus{}... \to A\equal{}1\minus{}(1\minus{}1\plus{}1\minus{}1\plus{}...) \to A\equal{}1\minus{}A \to 2A\equal{}1 \to A\equal{}\frac{1}{2}$

Ham khode saal va rahe halhaii ke neveshtin ham gheire daghighand va ham galat resp.  Serie $ \sum_{n\equal{}0}^{\infty}(\minus{}1)^n$ mitoone be $ \infty$ ham meil kone chon hamgaraye motlagh nist, va dar rahe halle Babak age $ x\equal{}1$ bezarim chi mishe? deghat konid ke baze hamgaraii $ |x\minus{}1|<1$ pas mojazim.

Hala age man serio in tori daste bandi konam chi? $ (1\minus{}1)\plus{}(1\minus{}1)\plus{}\ldots\equal{}0\plus{}0\plus{}\ldots\equal{}0$.

are,ba inike soal soale ghabele ghabooli nist moafegham,va inke har seri ke hamgerae motlagh nabashe mishe bahash azin kara kard,masalan serie $ 1\minus{}\frac 12\plus{}\frac 13\minus{}\frac 14\pm\ldots$,ba hamoon raveshi ke gofti mishe in serio be har adadi meil dad,vali in aslan dalil bar vageraish nemishe (tebghe azmoone leibnitz hamgeras)

javabe babak manteghie age $ n\equal{}0$ bashe $ a\equal{}1$ ke jomleye avvale va $ r\equal{}\minus{}1$ has. hala man migam $ x$ ro hazf konim va hamoon ke too masale bood dar nazar migirim, ye serie hendessei ke $ S\equal{}\frac{1}{1\minus{}(\minus{}1)}\to S\equal{}\frac{1}{2}$
agar bazeye hamgrayi ro bedas biarim darim $ \lim_{n\to \infty}|\frac{(x\minus{}1)^{n\plus{}1}}{(x\minus{}1)^n}|\equal{}x\minus{}1\to |x\minus{}1|<1 \to \minus{}1<x\minus{}1<1 \to 0 <x<2$
 :)
khob age seri $ \sum_{n\equal{}0}^\infty (x\minus{}1)^n$ bashe va $ x\equal{}0$ ghara bedim yek serie motenaveb mishe va hamgerast va $ x\equal{}2$ gharar midim ke seri va gara mishe dar natije $ [0,2)$ bazeye hamgerayist.

akhe moshkel injas ke vakhti mishe rajebe serie hendesi hokm kard ke ghadre motlaghe ghadre nesbatesh az 1 kamtar bashe,yeni $ |r|<1$...

hala namishe az lafze "serie hendesi" sarfe nazar konim

dar vaghe in yek serie motenaveb has ke bazeye hamgerayish [0,2) has
age az ghazihay va ahkame riazi sarfe nazar konim be nazar rae halle man doroste,  :oops: na?

[quote="BaBaK Ghalebi"]khob bazam farghi nemikone,chon khodetam dari migi bayad dashte bashim $ |x \minus{} 1| < 1$ pas $ 0 < x < 2$ yeni khode $ 0$ ghabele ghabool nis...[/quote]
man goftam $ [0,2)$ yani adade $ 0$ joze baze has

Akhe age in seri bekhad hamgera bashe bayad jomalatesh Cauchy bashan ke nistan maloome ! faseleye ye jomleye zoj ba ye jomleye fard 1 e ke nemishe be delkhah koochikesh kard .

age az seri begzarim darim $ 1\minus{}1\plus{}1\minus{}1\plus{}... \to A\equal{}1\minus{}1\plus{}1\minus{}1\plus{}... \to A\equal{}1\minus{}(1\minus{}1\plus{}1\minus{}1\plus{}...)$
$ \to A\equal{}1\minus{}A \to 2A\equal{}1 \to A\equal{}\frac{1}{2}$ doroste :maybe: ?

$ \frac{1}{2}$ ro ghabool darid? age ghabool darid $ \frac{1}{2}$ chi hast?

Bebin bayad avval sabet koni oon donbale seri hamgerast bad mitoony esmesho $ A$ bezari chon dar gheyre in soorat momkene ye adad nashe ... ye mesale dge mizanam . masalan $ 1\plus{}2\plus{}4\plus{}\cdots$ in cheghad mishe ? age az oon rahi ke to rafti mohasebe kardi mohasebe koni mibini ke $ 1$ mishe ke be voozooh ghalate .

[quote="shahimian"]javabe babak manteghie age $ n \equal{} 0$ bashe $ a \equal{} 1$ ke jomleye avvale va $ r \equal{} \minus{} 1$ has. hala man migam $ x$ ro hazf konim va hamoon ke too masale bood dar nazar migirim, ye serie hendessei ke $ S \equal{} \frac {1}{1 \minus{} ( \minus{} 1)}\to S \equal{} \frac {1}{2}$
agar bazeye hamgrayi ro bedas biarim darim $ \lim_{n\to \infty}|\frac {(x \minus{} 1)^{n \plus{} 1}}{(x \minus{} 1)^n}| \equal{} x \minus{} 1\to |x \minus{} 1| < 1 \to \minus{} 1 < x \minus{} 1 < 1 \to 0 < x < 2$
 :)
khob age seri $ \sum_{n \equal{} 0}^\infty (x \minus{} 1)^n$ bashe va $ x \equal{} 0$ ghara bedim yek serie motenaveb mishe va hamgerast va $ x \equal{} 2$ gharar midim ke seri va gara mishe dar natije $ [0,2)$ bazeye hamgerayist.[/quote]

dar inja esbat shod. hamgerayie seri $ [0,2)$ has. va in dalile bar hamgera boodane serist [b]amma yek serie motenaveb[/b]. khob dar $ S\equal{}\frac{1}{2\minus{}x} \to x\equal{}0 \to S\equal{}\frac{1}{2}$!!!  :ninja:

na,sabet shod ke too $ (0,2)$ hamgerae,rajebe khode noghte $ 0$ chizi sabet nashode.

babayi pesare gol mage peyghame ghabli ro nakhondi, mige:
age seri  $ \sum_{n \equal{} 0}^\infty (x \minus{} 1)^n$ bashe va  ghara bedim $ x \equal{} 0$ yek serie [b]motenaveb[/b] mishe va [b]hamgerast[/b]


geometry
3D geometry
geometric transformation
reflection
floor function

Let $n$ be a positive integer. A child builds a wall along a line with $n$ identical cubes. He lays the first cube on the line and at each subsequent step, he lays the next cube either on the ground or on the top of another cube, so that it has a common face with the previous one. How many such distinct walls exist?

It's a pretty difficult combinatorics problem.

I'm assuming the wall has no gaps in it...

Correct me if I'm wrong...it's the number of compositions of $n$ into positive integers, because we have $n$ blocks and $k$ stacks of blocks with a total of $n$ cubes among these $k$ stacks.  Then ...and the number of compositions of $n$ into $k$ positive integers is $\binom{n-k+k-1}{k-1}=\binom{n-1}{k-1}$...then your final answer is $\sum_{k=1}^{n}{\binom{n-1}{k-1}}=\binom{n-1}{0}+\binom{n-1}{1}+...+\binom{n-1}{n-1}=\boxed{2^{n-1}}$...

That's not so hard...

Unless I misinterpreted the problem?  When shobber said "it has a common face with the previous one", does that mean that any given cube is adjacent to the one placed before it?

[quote="K81o7"]Unless I misinterpreted the problem?  When shobber said "it has a common face with the previous one", does that mean that any given cube is adjacent to the one placed before it?[/quote]

Regardless of whether or not the requirement exists that each cube must be adjacent to the one placed before it, any "wall" produced by your formula is still possible to build (because if the requirement does exist, you can simply build left to right).

Consider this solution: label each block $1,2,...,n$. Place block 1 on the floor. Block 2 may go in one of two places: on top of block 1, or next to it. Similarly, block 3 may go on top of the rightmost stack (block 2), or it may go on the floor in a new stack. This continues until block $n$ is placed, either on top of block $n-1$ or on the floor to the right of it. This gives a total of $1 \cdot 2 \cdot 2 \cdot ... \cdot 2$ with $n-1$ 2's, and so the total is $2^{n-1}$.

Edit: Unless of course, reflections of walls do not count as distinct. In this case, it might be a bit tougher...

No, if each cube must be next to the previous cube, then it must either be directly above or below that cube, or at the same height...

For the sake of simplicity, I defined the building process as starting at the left and working right. As a result of that, blocks can only be placed to the right or on top of other blocks. This makes counting easier.

[quote="maokid7"][hide="um what about"]
${{n}\choose{n}}+{{n-1}\choose{1}}+{{n-2}\choose{2}}+\dots=\sum_{k=0}^{{\lfloor{\frac{n}{2}\rfloor}}}{{n}\choose{n-k}}$

since u can line them in a straight line, place 1 block on top of the other $n-1$, place 2 on top of the remaining $n-2$ and so on.

i am not sure how to find a closed form for this though.
[/hide][/quote]

But the wall can be more than 2 blocks high.  So when you stac, say, three blocks on top of the other $n-3$, there are more than $\binom{n-3}{3}$ ways to do that...

It's easier to look at it the other way I think...

Yeah, xsquardster's solution is essentially building the composition piece by piece: each step, adding either an object, or a divider and an object.

Sorry for bumping this thread, but is the answer $ 2^[n\minus{}1]$ correct? Can someone explain it to me better?

Suppose we make a systematic procedure of creating each wall:

(1) Put down a block
(2) For each block, either put a block on top of this block or on the ground to the right of this block

Note that there's exactly one way that follows the above procedure format for building any valid wall. Thus, the number of possible walls is equal to the number of different ways we can go about the above procedure.

We have 1 way to place the first block (on the ground), and 2 ways to place any subsequent block (on top of the previous block i.e. in the same column, or to the right of the previous block i.e. in a new column).

Hence, $ 1 \cdot 2^{n\minus{}1} \equal{} 2^{n\minus{}1}$

Oh, I seem to have misunderstood the problem. When it says "has a common face with the previous one" I thought it meant the $ nth$ cube placed has to have a common face with the $ n\plus{}1th$ cube placed. Therefore, if you placed the second cube on top of the first cube, you could only place the third cube above the second.

Can anyone solve my interpretation?

There are $ n$ cubes that you can pick to be the "corner cube", where picking the first to be the corner cube makes a column, picking the last makes a horizontal row, and picking anything in between makes a sort-of L-shape. This gives you $ n$ walls.

No, I don't think $ n$ is correct.

For example, there are 3 ways of placing 3 blocks - a horizontal row, a vertical row and a backwards L shape.

However, with the backwards L shape, we can place the 4th block either on top of the 3rd block, or on top of the first. Therefore, there are 5 different walls of 4 blocks.

@immortaltechnique: it seems as if you are insisting on placing the second block to the right of or on top of the first, because otherwise, there are four possibilities for three walls: vertical column, horizontal row, forwards-L or backwards-L, where the forwards-L is achieved by going to the left and then up.


search
probability
AMC
USA(J)MO
USAMO
AMC 10
AMC 10 A

If it is allowed, can someone with the solutions manual post the answers (just the letters) for the AMC 10A?

I thought it would be a good idea to put them in a post with this title so it is easy for people in the future to find them quickly with the search function.

[quote="Yongyi781"]The numbers in the bag are now:

1 1 2 2 3 3 4 4 5 5 6 6

Forget about the dice. The method is equivalent to selecting two of these 12 numbers.

The first number doesn't matter. For the second number, there are exactly 2 in the bag that add, with the first number, to 7, out of 11.

Hence probability = $ \frac 2{11}$.[/quote]

actually, i disagree with that, you are ignoring the last step, which is rolling the dice that have the new numbers stuck on it. what you are saying is that once you have picked the two complements, the probability that each are rolled on their respective dice is 1 which is clearly not true... but i might be mistaken, i am not very strong in probability.

Your contest manager has a Contest Manager's Envelope with the solutions to all the problems on the 2009 AMC 10 A and 2009 AMC 12 A.  Of course, the correct letter choices are also included.

Please ask your contest manager to see the printed solutions.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="cognos599"]actually, i disagree with that, you are ignoring the last step, which is rolling the dice that have the new numbers stuck on it. what you are saying is that once you have picked the two complements, the probability that each are rolled on their respective dice is 1 which is clearly not true... but i might be mistaken, i am not very strong in probability.[/quote]
Roll the dice then put the numbers on randomly. This clarifies it.

[quote="cognos599"]oh  :wallbash_red: , i FAILED, i did 23 of them and i was 100% sure i got all of them so  i went for broke and guessed the other two. Missed both of them AND one that i thought i had gotten right. so 132 for me....[/quote]

BAD idea. By guessing them, you're expected to get a perfect only 1 out of 25 times. This was what I kind of did last year and also ended up with 132.

But if you just even guess one of them right, the sacrifice of the 3 points you would gain from leaving those two blank would be worth it!

132 and 135 are not that different....

Tell that to those people who missed USAMO by 0.5 points last year.

Also, you might not have gotten either one right. I would've guessed too though :P

i cant believe the  :censored: took up ppls test booklets
i so want to check my answers right now

Hey guys,

How do you do number 4 and 7 on the 2009 AMC 10A?

4. Eric plans to compete in a triathalon. He can average 2 miles per hour in the 1/4-mile swim and 6 miles per hour in the 3 -mile run. His goal is to finish the triathlon in 2 hours. To accomplish his goal what must his average speed in miles per hour, be for the 15-mile bicycle ride?

AND

7. A carton contains milk that is 2% fat, an amount that is 40% less fat than the amount contained in a carton of whole milk. What is the percentage of fat in whole milk?

I know this is sad, but whatever. Thanks for your help!

rts2007

They've been posted around this forum, check their [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=257448]respective[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=257666]threads[/url].

Hey Aime15 and Nikeballa96,

Thanks for answering my questions. OMG! I don't know how I could have missed those problems!!!!! Any :wallbash_red:  :wallbash_red: ways, thank you!

rts2007

4.   Eric can complete the wsim in (1/4)/2= 1/8 of an hour.  He can complete the run in 1/ of an hour.  This leaves 2-1/8-1/2=11/8 hours tofinish the bik.  His average speed would be 15/(11/8)=120/11 mph.

7.Whole mile =x% fat.  Then 60% of x =2

so  x= 2/.6 =  20/6 = 10/3


In the new mathlinks format, where is the list that tells the most recently updated competitions in the resources?

On the portal page (now the main redirect goes to forum because I am still working on the portal page).

It's a simpler view for now, you can access it by clicking on the big mathlinks logo on the top-left corner, or alternatively by clicking here: http://www.mathlinks.ro/Forum/portal.php


superior algebra
superior algebra unsolved

Let G be a group. Let ~ denote the relation on G given by a~b if and only if there exists an element X in G (possibly depending on a and b) such that b=xax^-1.
prove that ~ is an equivalence realtion on G.
 This is what I have but I need help with the transitive step:
(1) Reflexive.

Is a ~ a for every

Obviously a = eae^(-1) (where e is the identity element).
Hence a ~ a.

(2) Symmetric.

If a ~ b, then does b ~ a?

Starting with a ~ b,
we have b = xax^(-1) for some
Hence xbx^(-1) = a,
i.e., a = xbx^(-1),
which proves that b ~ a.

(3) Transitive.

If a ~ b and b ~ c, then does a ~ c? how do I show this?

b = xax^-1 and c = yby^-1 = yxax^-1y^-1 = yxa(yx)^-1


Έστω ένα τετράγωνο χωρίο $ E$ του επιπέδου με εμβαδό $ 1$ (τα νούμερα στην πραγματικότητα δε μας νοιάζουν, τα βάζω μόνο και μόνο for concreteness).

Έχουμε μια απειρία παικτών, που θα τους λέμε [i]ιδιοκτήτες[/i]. Κάθε ιδιοκτήτης διαλέγει με τη σειρά του ένα χωρίο $ E_n\subseteq E$ που θα το λέμε [i]οικόπεδο[/i]. Βέβαια, επειδή οι ιδιοκτήτες δε συνεννοούνται μεταξύ τους όταν διαλέγουν και δεν ξέρουν τι έχει διαλέξει ο προηγούμενος, αυτά τα οικόπεδα μπορεί κάλλιστα να επικαλύπτονται. Επίσης, για να μην είναι οι ιδιοκτήτες εντελώς άπληστοι, o πρώτος ιδιοκτήτης μπορεί να διαλέξει ένα οικόπεδο με εμβαδό το πολύ $ \frac {1}{2}$, ο δεύτερος το πολύ $ \frac {1}{4}$ κλπ (first come first serve).

Τα προβλήματα αρχίζουν όταν ένα σημείο ανήκει σε περισσότερους από έναν ιδιοκτήτη. Τα χειρότερα προβλήματα τα έχουμε όταν ένα σημείο ανήκει σε άπειρους ιδιοκτήτες :!:

Να δειχθεί ότι η πιθανότητα ένα σημείο να ανήκει σε άπειρους ιδιοκτήτες είναι $ 0$.

Cheerio,

Durandal 1707

@ mods: αν πιστεύετε ότι το συγκεκριμένο post θα καθόταν πιο καλά σε κάποιο άλλο section, please feel free to move it :)

Φαντάζομαι κάθε ιδιοκτήτης διαλέγει "τυχαία" το οικόπεδο του.  :P

Σχετικά με το αν πρέπει να μεταφερθεί σε άλλο section: Αν γνωρίζεις ένα πολύ σημαντικό λήμμα της θεωρίας μέτρου, τότε το πρόβλημα γίνεται μια απλή άσκηση. Αν δεν το γνωρίζεις, τότε ουσιαστικά αν το λύσεις αποδεικνύεις και αυτό το λήμμα. Το πρόβλημα ουσιαστικά απευθύνεται σε αυτούς που δεν γνωρίζουν το λήμμα. Αυτό το πρόβλημα θα ήθελα να μου είχε δοθεί μόλις άρχιζα να διδάσκομαι θεωρία μετρου. Θα ζοριζόμουν αρκετά και είτε το έλυνα (που δεν είναι αυτοσκοπός) είτε όχι, μόλις διδασκόμουν το λήμμα θα το εκτιμούσα περισσότερο. Δυστυχώς όμως έχουμε συνηθίσει να δίνουμε αυτά τα προβλήματα σαν ασκήσεις μετά την διδαχή της θεωρίας. (Έχει φυσικά τον σκοπό του εφ'όσων ελέγχεται αν ο φοιτητής έχει κατανοήσει την θεωρία.)

Γνώμη μου: Σίγουρα δεν είναι σχολικό πρόβλημα ή προβλημα ολυμπιάδας. Είναι φοιτητικό πρόβλημα. Αλλά όπως λέω και πιο πάνω αν το δώσεις σε ένα φοιτητή που γνωρίζει και καταλαβαίνει το λήμμα, τότε δεν εξυπηρετεί κανένα σκοπό. Άρα, παρόλο που είναι πάρα πολύ ωραίο πρόβλημα, δεν βλέπω πιο από τα ήδη υπάρχοντα sections είναι κατάλληλο. Ίσως να πρέπει να δημιουργήσουμε νέο section/subsection με τίτλο "Φοιτητικά προβλήματα που μπορούν να λυθούν με σχολικές γνώσεις"

[quote="Demetres"]Γνώμη μου: Σίγουρα δεν είναι σχολικό πρόβλημα ή προβλημα ολυμπιάδας. Είναι φοιτητικό πρόβλημα. Αλλά όπως λέω και πιο πάνω αν το δώσεις σε ένα φοιτητή που γνωρίζει και καταλαβαίνει το λήμμα, τότε δεν εξυπηρετεί κανένα σκοπό. Άρα, παρόλο που είναι πάρα πολύ ωραίο πρόβλημα, δεν βλέπω πιο από τα ήδη υπάρχοντα sections είναι κατάλληλο. Ίσως να πρέπει να δημιουργήσουμε νέο section/subsection με τίτλο "Φοιτητικά προβλήματα που μπορούν να λυθούν με σχολικές γνώσεις"[/quote]

Μωρέ ξέρεις τι? Αυτός ήταν ακριβώς ο σκοπός μου, να φτιάξω ένα παιχνιδάκι που οδηγεί πρακτικά στην απόδειξη του Borel-Cantelli, αποτέλεσμα που είναι μεν elementary αλλά έχει πολύ χρήσιμες ιδέες.

Δεν ήθελα να το βάλω στο φοιτητικό γιατί απλώς θα έβγαινες και θα μου έλεγες Borel-Cantelli, και... χαμένος κόπος :) Από την άλλη, να το βάλω το ίδιο το Borel-Cantelli σαν ξερό λήμμα προς απόδειξη πάλι δεν έχει νόημα, γιατί δεν είναι και κάτι το εξωτικό και σπάνιο όπως το sunflower.

Οπότε, ναι... confusion :huh:  In retrospect, ίσως θα ταίριαζε στα γενικά θέματα, αλλά... eh, ας έχουμε κι ένα misplaced θέμα :)

Σε κάθε περίπτωση, any takers για το πρόβλημα? Που να μην έχουν δει κάτι παρόμοιο εννοείται  :P

Cheerio,

Durandal 1707


Please attach brief responses (maximum 250 words, typed or neatly handwritten) to both of the following prompts:

1)What extracurricular activities are you involved in? Describe the nature and extent of your involvement. Which one or two are most important to you?
2)Select a topic in mathematics that particularly interests you. What about this area intrigues you? Be as specific as you can.

This is part of the AMSP application....

Do I have to write 2 essays for each of the questions or

Just 1 essay answering 2 questions.

I would guess two essays, one for each question. And this isn't the appropriate forum; it should probably go under the "Other US Contests & Programs" forum, so can a mod please move it? Thanks.

[quote="fireemblem13"]isn't that where it is[/quote]

... it got moved already.

I wrote 2 separate responses to the questions

I did that for the early app, and got in, if thats worth anything :D


Hi guys.
I am looking for a book what its name is "a course an combinatorics".
Does anybody have this book?

There must be no book like this, maybe 'in' instead of 'an'?  :P

google book tells me this:

A course in combinatorics
By Jacobus Hendricus van Lint, LINT - WILSOM, VAN LINT - WILSON, R. M. Wilson
Published by Cambridge University Press, 1992
ISBN 0521006015, 9780521006019
530 pages
http://books.google.com/books?id=5l5ps2JkyT0C&printsec=frontcover&dq=a+course+in+combinatorics

Thx 4 ur notice to this book. I find this book so extensive (it already falls in my library  :lol: )

check the link~

[quote="daplikos"]Thanks for your attention to my request.
Have you found or had a Pdf version of this book?[/quote]

The newest edition of the book is only a couple years old, and there are probably not any legal .pdf versions available.

It's a great book though.

doesn't  Anybody have PDF format of this book?
These are the informations of this book:
Name: a course in combinatorics
author: J.H. Van Lint & R.M. Wilson
Publisher:Cambridge University Press


$\text{Show that if}$ $1 = as+bt$$\text{, then}$ $\gcd(a,b) =1$

:juggle:

[hide]Let $(a,b)=c$. Then, $c\mid a$ and $c\mid b$. Then, $a=ck$ for some integer k and $b=cq$ for some integer q. We know that $1=as+bt$. By substitution, we have $1=cks+cqt$, $1=c(ks+qt)$. Thus, $c\mid 1$ and $c=1$. QED[/hide]


Provide general methods for making the conversion

$ R\minus{}OH \longrightarrow R\minus{}CH_2CO_2H$

1.conversion to alkenes
2.oxo process(seperate the normal and iso products)
3.grignard synthesis with formaldehyde
4.oxidise to acid

is that correct :maybe:

Conversion to alkene? How about if the alcohol is tertiary? And isn't the oxo process an industrial process?

i think this method works.........

1.treat with SOCl2
2.treat with KCN
3.hydrolyse
4.resuce using NaBH4
5.Use SOCl2 .
6.treat with KCN
7.hydrolyse

the yield in $ \ce{SOCl2}$ or the Darzen's method are very poor for tertiary alcohols

[quote="valeriummaximum"]if so pardesi............
tell me a very good method for getting a tertiary halide from a tertiary alcohol.........[/quote]
why?cant we do elimination reaction for the tertiary alcohol and then do a markovnikoff addition of HBr?? :maybe:

srinitrs .........
it can rearrange also....
you might not have direct elimination after formation of the carbocation...........methyl shift or hydride shift can happen..............
i think you might not have considered this possibility

valeriummaximum yes i think that should o because i don''t atleast reemeber of any other method...and ever possible method i know has this restriction :D

To get a tertiary halide from a tertiary alcohol just use concentrated hydrogen halide.

hey Carcul,can you explain the mechanism of the reaction that you are mentioning for the reaction (using conc.HX)????

It's a SN1 mechanism. First, -OH is protonated by H+; then, a molecule of water, acting as a leaving group, departs from the molecule leaving behind a carbocation; finally, a chloride ion makes a bond with the cationic carbon.

i have a doubt............
will the carbocation formed undergo elimination or substitution?????????????

Substitution, because there is no strong enough base in the reaction medium.

Ok Carcul...
it need not eliminate..........
but can rearrange for greater stability................
so i think you cant get the R-Cl.......

why should it rearrange it's laready tertiary

The carbocation is already tertiary (because in my post above I was referring to the conversion of tertiary alcohols), so further rearrangement is very unlikely.

I am just asking ....................
there might be any conjugation in the system .............so you might not get the R-Cl
it should be a generic method.............

But a general method is one which should for all alkyl groups

yes that's what we are doing.ofcourse if u r synthesisng u know hwteher it is a tertiary/secondary/primary
then accordingly u can proceed either with darzen or with HX

The HCl reaction I gave above is of course only for tertiary alcohols which do not have nearby groups that might interfere. I didn't say it was general. For example, for R = Ph we cannot obviously do that.

cant you convert into a good leaving group (OTs for eg) and and then use SOCl2. wont the yield increase or does only the reate increase?

[quote="VARUNARASU"]cant you convert into a good leaving group (OTs for eg) and and then use SOCl2[/quote]

SOCl2 already converts the -OH group into a good leaving group when it reacts with alcohols.

[quote="pardesi"]isn't taht getting toooo specific.like not all syntheseis thaat we have proposed on this forum asked by carcul is that specific.ofcourse by adding different substances knowingly u can always have side reactions[/quote]
to the tertiary alcohol add Ph3P and DEAD(diethyl azo dicarboxylate)and your halogen and there it is your tertiary halide???am i right??? :D

Srini obviously has nved the rn well from the source he mentioned .
Utta Appel (Apple) rn use pannalam solluve polairruke .  :D   :rotfl:   :P

[quote="pardesi"]DEAD :D  -nice reagent ...hope it killed the problem :P[/quote]
i think this reaction is named as mitsinobu reaction thats what i read in one book! :rotfl:  :D

[quote="madness"]U may think you are smart but i already know that you have mentioned the source.
I too have read it   :P
 :D[/quote]
obviously madness everyone here must know about Jagadamba singh whats so big in that tell me whether it is a solution to the problem overwhich ranga and carcul are arguing over???? :D


geometry
area of a triangle
Heron\u0027s formula

Check [url=http://www.artofproblemsolving.com/Resources/Papers/Heron.pdf]this out[/url].  This cool proof by AoPS Community member Miles Edwards was just published in [i]American Mathematical Monthly[/i].

That is pretty cool proof. It is very clever (especially from transformation from $ r$ to $ uvw$), and not that hard to follow along. I'm going to sticky until end of this year so more people can have chance to look at it.

That was somewhat rigorous, here's a really cool proof I like: [url]http://mathforum.org/library/drmath/view/54686.html[/url]

I watched the movie Edward Scissorhands yesterday and it reminded me of Miles alot, then now I see this haha that was weird.  Anyway that is indeed a really slick proof of Heron's formula, good job Miles.


trigonometry
calculus
calculus computations

Say we need to write the trig expression: sin(arccos2x) algrebraically, my answer ended up being the square root of some algebraic expression. But that is irrelevant, because I am asking that, if we were supposed to consider the expression in terms of a triangle, the expression can only be a positive square root, since the negative wouldn't make sense. But the problem itself never mentioned anything about a triangle, do I still need to consider it in terms of a triangle and eliminate the negative square root, or do both the positive and negative roots apply?

That depends on whether the $ \arccos$ is the principal $ \text{Arccos}$, which only outputs angles between $ 0$ and $ \pi$.  For any particular $ x$, the choice of sign depends on where the $ \arccos$ falls in this range.

$ \arccos\alpha\in [0,\pi ],\forall \alpha\in [-1,1]$ and $ \sin x\in [0,1],\forall x\in [0,\pi]$.
So $ \sin (\arccos\alpha )\geq 0, \forall \alpha\in [-1,1]$


Hello,

Im pretty stuck on this one.
A spherical mothball shrinks in such a way that the radius decreases by 1 centimeter per month.  How fast is the volume of the mothball changing when the radius of the moth ball is 2 centimeter? (Assume the mothball remains spherical as it shrinks)

So I know
$V = (4Pi r^3 )/3$

where do I go from here?
Also, how do you display pi in latex? Couldn't get the fraction to work and im guessing that pi is the problem...
Thanks again,
Chris

Try (dollar) V=\frac{4 \pi r^3}3 (dollar): $V=\frac{4 \pi r^3}3$

Be sure to put a space between \pi and anything that starts with a letter. And it's case-sensitive: \Pi would get you $\Pi,$ and you don't want that.

In general, if you want a Greek letter, type (backslash) (name of letter). So \alpha\beta\gamma\delta\Gamma\Delta becomes $\alpha\beta\gamma\delta\Gamma\Delta.$

Now: if $V=\frac{4\pi r^3}3,$ then

$\frac{dV}{dt}=4r^2\frac{dr}{dt}.$

Of $\frac{dV}{dt}$ and $\frac{dr}{dt},$ which one do you already know and which one do you want to compute?

Kent,

 Thanks for the latex help. It makes perfect sense.

Ok so,

We want $\frac{dV}{dt}$ and we know $\frac{dr}{dt}$

$\frac{dr}{dt} = -1$

$\frac{dV}{dt} = 4(2)^2(-1)$

$\frac{dV}{dt} = -16 \pi cm^3$???

Yes you're right ( $-16 \, cm^3 / month$), the negative sign means that the volume decrease. Actually we use the chain rule $\frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt}$


analytic geometry
graphing lines
slope
symmetry
real analysis
real analysis unsolved

1. What is the minimum distance between the graphs of $y=e^x$ and $y=x$?
2. What is the minimum distance between the graphs of $y=e^x$ and $y=\ln(x)$?
3. What is the minimum distance between the graphs of $y=x^2+2x+2$ and $y=-x^2-6x-10$?

(all on real axes)

I figured out the first one, but can't seem to solve the next two.

[hide="Number 1"]The line containing the shortest distance will always be perpendicular to the curve of $y=e^x$, and it will be perpendicular to $y=x$, so the tangent lines at $y=e^x$ and $y=x$ must be parallel. That means that $y=e^x$'s tangent line must have a slope of 1. $y'=e^x=1$ so $x=0$, and therefore $y=1$. Drawing a diagram, there is a 45-45-90 triangle with hypotenuse the line from $(0,0)$ and $(0,1)$, so the shortest distance is one leg of that triangle, which equals $\sqrt{2}/2$.
[/hide]

2. There is some symmetry between the two graphs. You may want to utilize that ;)

$P(a,a^{2}+2a+2)$
$y' = 2x+2$, the slope of the tangent line in $P : m_{t}=2a+2$.
$Q(b,-b^{2}-6b-10)$
$y' = -2x-6$, the slope of the tangent line in $Q : m_{t'}=-2b-6$
Equal slope: $2a+2 = -2b-6$, solved: $b=-a-4$
Point $Q(-a-4, -a^{2}-2a-2)$.
Then $PQ^{2}=4\left[(a+2)^{2}+(a^{2}+2a+2)^{2}\right]=f(a)$
Distance PQ has a minimum if $f'(a)=0$, solved numerically: $a=-1.31290\cdots$


Four straight lines intersect a circular region. The lines and circle are coplanar, and
two of the lines are parallel. What is the maximum number of non-overlapping regions
inside the circle?

...is there another away to find out beside drawing a diagram?

answer : 10

The first two paralllel lines give 3 regions. A 3rd one can split each of these two regions into halves, and the 4th line can split all three lines, thus adding 4 more regions. So 3+3+4=10.

Basically instead of looking at the diagram itself, we can easily solve this problem by first noticing like dragon96 said that the two parallel lines create three regions. Instead at splitting it in halves, an easier way is to look at the[b][u][i] number of intersections[/i][/u][/b] with the previous lines. Since the circle itself is included we have 1 region, then one line intersects  so we add one etc until it looks like 1+1+1 for the circle itself and the other two parallel lines. The last two line you just count the intersections. 1+1+1 we have 3 "lines to intersect" so we just add 3 more regions. 1+1+1+3 has 4 lines to intersect. The last line intersects 4 previous so we just add 4. 1+1+1+3+4 is 10.

In hindsight this explanation wasn't really necessary but this is my first time replying on AoPS so I hope it is helpful :)

You can also use this in other math competitions.
If it asks on a plane how many maximum regions are formed by 2 circles, 2 triangles, and 2 lines, you can do..
1 (the plane) + 1 (one line) + 2 (second line intersecting first) + 4 (first circle intersecting four points) + 6 (second circle intersecting 6 points) + 16 (first triangle) + 22 (second triangle) = 1+1+2+4+6+16+22=52

(if you have anything to add or if i made a mistake please reply)


I was just wondering how many people here know several languages.  By knowing I mean that you can understand most of a text in that language.

In my case it is three : Dutch (of course), French( eight years at school), English( five years but due to the language being easier, handier, and more frequently present in my daily life much better than my French)

I've had some German and I can understand some of it but I don't count it along?

My impression is that everyone here knows English, his own language, and usually even a third?

Japanese (my mother language) and English (2nd language).

Does half a year of Japanese (so far) count? >_>

[quote="fredbel6"]Is that difficult to go from Japanese to English, certainly with the different alphabet?  Or is romanji taught in schools like pin yin in china?[/quote]
Well, I'm afraid that I don't know "pin yan" in China, but I would say it is generally difficult to go from Japanese to English or vice versa(not only because the letters/symbols are different, but also because grammer and pronounciation are considerably different).

Cantonese (mother language)
Mandarin (2nd to 6th grade)
English (7th to 11th)
Japanese (9th to 11th)

5.  Persian, English, French, Latin, and almost Korean.

Question: Does LaTeX count?

Technically, I know 5, but in three of them, I can do no more than count to 10.

latex does not count
which in your case?  english spanish french?

C, C++, LISP, a bit of ASM, java, BASIC dialects, Ruby, Python, some perl, some C#, and varioius other scripting languages.

Chinese (Main Language), Shanghainese, English

I know English, weak French, very weak Hungarian, and the word for "ice cream" in more than a dozen other languages. :-)

I know english. that's it. I'd like to learn chinese, but haven't gotten around to it. I'm learning french.

I know English
I took Spanish for a year
I can say toilet paper and mushrooms in Italian (carta igenica y fungi)
I can say I and strawberry in German (ich erdbeere)
So that's really only 2 but I like to count that as 4

If you count dialects, I can understand about 5 chinese dialects (from suzhou, shanghai, beijing, etc.).

Spanish: I'm supposed to be learning this.  Put emphasis on the "supposed to."

French: took it in 1st and 2nd grade.... (un, de, twa is it?)

German: Was forced to learn a little.

English: Yes, ENGLISH, not ENGRISH (im not that FOB!!).

I know Spanish (perfect, because I´m Argentinian), and English (I´m in my 12th year).

Personal Opinion: I don´t think that it´s important to know several languages, I think that if you know your mother language and another (English or Chineese, because they`re spoken  mostly everywhere) it`s OK and enough.

(Y) English: Main language

Hindi: Used to learn it: my grandmother used to teach me when I was very young, but I couldn't really learn to write it at that age if I didn't regular speak it.  Later, I was taking a class, but we really didn't learn anything in that class.

Bengali: Mother tried to teach it to me for a while, but didn't really work out that well...

Japanese: Learning it in school right now, have been for 2.5 years, but I probably don't knowenough Kanji/vocabulary in general to understand a lot...

So that makes it..1?  1.5?

1. French-have been taking it for 3 years now
2. Chinese- have spoken it since I was like 3 years old
3. English (self-explanatory) :D

My native language- Korean

Second language - English

Third language - Chinese
( because Koreans were taught to do so and I was pretty good at it - although I can only write :blush: )

Fourth language - French
almost 3 years

Chinese (Mandarin)

English

Weak Spanish

If reading is at issue, in order of decreasing proficiency, my languages would be

1) English

2) Chinese

3) German

4) Koine Greek

5) OT Hebrew

6) any random Romance language

7) Japanese

8) Russian

or something like that. I've been able to read some texts in Dutch before, and other random Indo-European languages with Roman scripts.

Hm well concering dutch i guess it is a language that is a whole lot easier to read than speak.  I once saw an english book explaining dutch and i did not understand.  Word order, the fact that our infinitives end with -en, and the sch sound, as in school, it's not as in skating nor as in shooting, it is s+gggg

1st language -  spanish
2nd language - french
3rd language - english
i went on a trip to italy and knowing spanish and french it was amazingly easy. portuguese is harder to understand but i can communicate with someone that only speaks that language if he/she speaks slow enough :)

when you say second language, is that how they do it in school, or your personal ability?
Because the first language we learn is french as the other half of belgians speak it, but english is so much more easier and handy, and you catch it on easier than french from tv

I can speak 2 languages well, spanish and english. I can read portuguese too (really easy, being so similar to spanish) and i'm learning german, however, i can't understand it well yet

Random: This seems interesting: [url]http://ia.wikipedia.org[/url].  Confusing, but interesting.

only chinese and english (english better of course)

maybe going to learn spanish or something later

Chinese and English. But I may choose Russian or German as my third language in College.

[quote="ashwath.rabindranath"]English , Tamil , Hindi , Malayalam and Sanskrit.
I'm a bit weak on Malayalam though.[/quote]

You understand sanskrit well?I just mugged sanskrit throuhg 6th and 7th grade.I can't remember a think now.Only hindi and english for me.

I can read and understand any non-technical (apart maths :) and Computer science) book in
1) Polish
2) French
3) Russian
4) English
I'm quite happy with that :)
(What an egocentric post will you say, huh ?)

i no English and a certain language in India

Arabic:My mother tongue ;) btw algebra sugar coffee zero cotton banana etc come from arabic  :D
Français:The main foreign language in Morocco i start learning it when i was 7
Espanol:as i'm originally from Tangier 14km from Spain
English:learning it just for 3 yrs so i'm trying to improve my english with more practice
Deutch:started it this year
Hebrew:alphabet grammatical rules and some vocab it's a semitic language so sort of similar to arabic

1. spanish
2. english

Also, i think that if you know spanish it won't be so hard to understand french and portuguese, because they're very similar

[quote="JBL"]... and the word for "ice cream" in more than a dozen other languages. :-)[/quote]
JBL - Me too, and for reference for you all:  Here it
French   glace
Spanish helado
Italian gelato
Portuguese      sorvete
Rumanian        inghejata
German  Speiseeis
Dutch   roomijj
Swedish glass
Danish  flodeis
Norwegian     iskrem
Polish  lody
Czech   zmrzlina
Serbo-Croat, sladoled
Hungarian  Fagylalt
Finnish  jäätelö
Turkish dondurma
Indonesian      askrim
Esperanto       glacialo
Russian marozhenoye
Greek   pagoton'
Arabic  jilati, bouza
Hebrew  glidah
Yiddish gefroirene
Japanese         aisukurimu
 Swaliili       aisi krimi
Sanskrit – Himkshir (or may be himpayasanan?)
Rajastani -   Burf ki Malai
Hindi -Kulfi  (Icekreem - will work too:)
Kulfi (Technically a type of ice cream - yours tryly's faviorite) i - will work   in Marathi, Gujrati, Bangali, urdu, Bhojpuri, Nepali  and I have even used it (successfully ) in Kanada and Telgu. :)

(Of course, I can not compete with my wife, who knows, [i]every [/i] (it seems )  food item in every  language i can think of :))
Added Later: Thanks Puuhikki -  I edited  the list

[quote]I just [i]mugged [/i]sanskrit throuhg 6th and 7th grade.[/quote]
I thought "mugged" is IIT lingo only :)  Interesting! Now it's use is common. BTW one of the  best way to like a language, is pick a book which is really interesting and read it.. (My father gave me a copy of Surya Sidhant -  book about astronmical calculation but written in Sanskrit  -   which made me like Sanskrit more than  the class room lessons)

1. English - second language, but since i speak it all the time, it is my best language
2. Mandarin - okay at speaking, but not that well in reading
3. Spanish - reading is poor (2 years), but I can still read it.

[quote="puuhikki"]Finnish, English, Swedish, German. I also learnt some words Hungarian when I visited there. And once I read one mathematical paper in Estonian and I understood that. However I can't produce Estonian on my own.

[quote="Gyan"]
Hungarian fagylall
Finnish jaatelo
[/quote]

Fagylalt and jäätelö I think.[/quote]

A friend of mine wants to go to Erasmus next year in Oulu.  She doens't know any finnish right now.  She does know dutch, english, french and turkish.  Do yo think it is doable to learn finnish that fast?
An erasmus student from hungary told me last year that hungarian , estonian and finnish are related, but that's it, and all other languages are totally different

To the person collecting 'ice cream' in all languages , in dutch it is 'roomijs' not 'roomijj'

Well, other people learns languages faster than others. It took several years for me to learn English so well that I can speak and write English at intermediate level. I still do mistakes in my English. However, a friend of mine can learn a new language in a couple of months and he can speak about ten different languages. So it depends very much how fast she learns new languages. Of course it is fast to learn some most important words like kiitos (thank you) and anteeksi (I'm sorry, excuse me), but it takes time to learn Finnish well. But in one year I think it is possible to learn at least basics of Finnish. For me Finnish is of course the easiest language I know.  :roll:

Hungarian, Estonian and Finnish are indeed related to themself. Many Estonian word sounds like a Finnish word but the meaning is not the same. And some Hungarian words are almost like Finnish words. For example

Hungarian - Finnish - English
autó - auto - car
hal - kala - fish
vaj - voi - butter
vér - veri - blood
viz - vesi - water

[quote="fredbel6"]A friend of mine wants to go to Erasmus next year in Oulu.  She doens't know any finnish right now.  She does know dutch, english, french and turkish.  Do yo think it is doable to learn finnish that fast?
An erasmus student from hungary told me last year that hungarian , estonian and finnish are related, but that's it, and all other languages are totally different[/quote]

Knowing those four languages won't provide help in the form of cognates or similar grammatical structures (or at least, if they do, only by chance).  On the other hand, it's impressive how much of a language one can learn quickly by submersion.  I suggest your friend take some sort of intensive language class, and be ready to be confused by everythign in Finnish for a while :-)

I speak ( as well as read)... (very well, though not perfectly except in english)

English - native tongue
Hebrew
Aramaic

But i understand some:
Latin
Spanish
French
Yiddish
Japanese

And technically i can read most scripting languages..

i know english (native language)
im decent in french
i can read hebrew but not for understanding and i dont know conversational hebrew
i used ot take chinese but i failed i know lots (more than i can count on my hands when not using binary) of words but cant speak meaningful sentences of read or write.  of the three i read the best.


Hello, rrusczyk!

Ragingg is too fast for me . . . LOL!

Quote:Find the area of a square given that there is a point on the circle inscribed in the square
which is 1 unit from the nearest side of the square and 2 units from the second nearest side.Code:
   *-*---+         This the upper-right quadrant
   |  P* |          of the square and the circle.
 r | /   |             Point O is their center.
   |/    *       Point P has coordinates (r-2, r-1).
   + - - *
   O  r

[hide]

Since OP = r, we have: (r - 2)^2 + (r - 1)^2 = r^2

. . . which simplifies to the quadratic: r^2 - 6r + 5 = 0

. . . which has roots: r = 1, 5.

The only 'reasonable' root is: r = 5.

Hence, the side of the square is 10 units

and its area is 100 square units.

[/hide]


geometry
MATHCOUNTS
analytic geometry
AMC
AMC 10
trapezoid
algorithm

So, how do you think you did on the AMC 10?

I think I got 150 only if I didn't make any stupid mistakes... This test was harder than last year.

EDIT: Arrghhh!!!!!!! After looking at the answers, I know I got 25 and 21 wrong!

i answered 20 questions, so I can miss up to one question and still get 120. I felt pretty confident about my answers, and I never guessed.

edit- after looking at other peoples' answers, I am pretty sure that I got them all right... so I should make it unless I bubbled in something wrong.

Except I don't trust that I did not mess up any questions.  Often, I just misread and put down the wrong thing or forgot to divide the whole thing by 2 or something like that.

You probably don't have that type of problem though.

[quote="Ignite168"]I got either a 138 or a 144, I did them all, I know I missed #25, and I'm not sure about my answer for #19.[/quote]
19 and 25 were the hardest ones for me. I answered 17 or 18 of the questions.  I probably got around 110 or 100.

#19 was not that hard IMO. Assume the paint brush length is 1, the center is 2, and then let's say the side of the triangle is x. So you have 4x+2 = 2x^2. Quad gives you it's $1%Error. "pmsqrt" is a bad command.
{2}$. So that means the length of the nonshaded area on the square is $2+rt(2)$. Then, you have to add on $\sqrt{2}$ for the shaded regions and you get C.

I think I put C on the test as there is no way to mess up and get $2\sqrt{2}+1$ as you already had the +2. If I messed up, I think I would've put $\sqrt{2}+2$ and that was not one of the answer choices. My recorded answers says I put A, but I think I put C, so I'll have to wait for my score.

25 was brute force multiples of 3, I missed 1980.

I think I got a 150, unless I bubbled something wrong.

WTF OMFG ACCORDING TO MY TEACHER I GOT A 138!!!!!! WTF!!!

 :cursing:

I bubbled #23 wrong and missed #25... :wallbash_red:

ARGGGGGGGGGGGGGGGHHHHHHHHHHHHHHHH!!!!!!!!!!!!!!!!!!!!!!!!!!!

(and I tied a certain 7th grader that I know)..................

I got a 141.  I omitted 2 geometry questions, and I feel kind of stupid now.  But since Harry didn't take it, that might be enough for an SC high score.

EDIT:

9th grade
I think I omitted #'s 15 and 24.

omg how do you get so good grades, I havent seen the scores yet, since my teacher that adminstered the test took away our booklets so we would not share answers.  from what i heard, only 2 people in our whole school passed the amc 10, and we had like 100 people take it

I did terribly on this AMC 10.  :noo:

After the test, I realized the problems were actually pretty easy. I'm just bad when there is time pressure.

132.
One problem I wasn't thorough enough, I didn't have time to be thorough because I lost 15 min on one problem because I didn't realize how much time I'd blown.
One problem I guessed on.
One problem I used complementary counting and FORGOT TO SUBTRACT FROM ONE

which one do you use complementary counting on?

You guys got pretty good scores.
I haven't taken AMC 10, yet, but 12, and I got 144.

[quote="tjhance"]a number is congruent to the sum of its digits mod 3, so if they n, S(n), and S(S(n)) are all the same, S+S(n)+S(S(n))=0 mod 3

start from 2006 because it obviously can't work 2007 or above.[/quote]

So, isn't it possible for n to be not a multiple of 3? Or does it have to be a multiple of 3 because 2007 = 3 * 3 * 223?

I'm confused, sorry.

[quote="123456789"][quote="Klebian"]18, i recommend that you look at my scanned image. I put this on a coordinate plane, with H at (0,0). I found the equation of HC, the equation of GA, then was able to find the intersection (which is point M). Then, you can use the shoelace algorithm to find the area of the quadrilateral as you know the coordinate of all 4 sides.

19, I just assigned the side length an arbitrary number, since we're just looking for ratios. s = 4, A = 16. Since half of the square is painted, each triangle is 1/8*A, or 2. I can't recall why, but I assumed that the base of the triangle is 2* the height. From here, it's fairly basic (assuming the assumption is correct, which it turned out to be). Try solving the rest.

22: A smart thing to do would be just take 2 or 3 sequences and see which numbers do divide, and which don't, and you can easily eliminate 7, 13, and 43. Then taking a sequence of 3 terms and looking at it algebraically (abc, bca, acb), you'll see that S = 111(a+b+c), and since a+b+c can be anything, the answer is 37 (as 37*3 = 111)[/quote]

Does anybody know why he assumed that the base is 2 times the height?

And how are #24, 25 done?[/quote]

If you stick the triangles together, you can make a square. Since the height is from the middle of the square to the side, that's half a side length.

Or you can cut a triangle up into 2 45-45-90 triangles...

I think a 132.  :mad: I'm so stupid! I can't believe it! I don't understand why I can't take it in the afternoon. All the tests I take are early in the morning or late at night, exactly the times when I'm not thinking right. I really need to take the AMC10B to just make myself feel better...

[quote="Treething"][quote="RC-7th"][quote="tjhance"]a number is congruent to the sum of its digits mod 3, so if they n, S(n), and S(S(n)) are all the same, S+S(n)+S(S(n))=0 mod 3

start from 2006 because it obviously can't work 2007 or above.[/quote]

S + S(n) + S(S(n)) = 0 mod 3 doesn't imply that S = 0 mod 3. In fact, it could be 0, 1, or 2. Just check S = 1 and S = 2.[/quote]

What you guys are talking about would make sense if you replaced  S + S(n) + S(S(n)) = 0 mod 3 with S + S(n) + S(S(n)) = 0 mod 9.[/quote]

except that isn't correct
if $n\equiv 1 \mod 9$ then $n+S(n)+S(S(n))\equiv 1+1+1\equiv 3 \mod 9$

so n+S(n)+S(S(n)) doesn't necessarily = 0 mod 9

[quote="tjhance"][/quote][quote="Treething"][quote="RC-7th"][quote="tjhance"]a number is congruent to the sum of its digits mod 3, so if they n, S(n), and S(S(n)) are all the same, S+S(n)+S(S(n))=0 mod 3

start from 2006 because it obviously can't work 2007 or above.[/quote]

S + S(n) + S(S(n)) = 0 mod 3 doesn't imply that S = 0 mod 3. In fact, it could be 0, 1, or 2. Just check S = 1 and S = 2.[/quote]

What you guys are talking about would make sense if you replaced  S + S(n) + S(S(n)) = 0 mod 3 with S + S(n) + S(S(n)) = 0 mod 9.[/quote][quote="tjhance"]

except that isn't correct
if $n\equiv 1 \mod 9$ then $n+S(n)+S(S(n))\equiv 1+1+1\equiv 3 \mod 9$

so n+S(n)+S(S(n)) doesn't necessarily = 0 mod 9[/quote]
Dude, n=3 mod 9 implies
n=1 mod 3.
Treething is right.

I think I got 144.  I filled in the wrong bubble for one of them...  :censored:  :wallbash_red:

Probably because I did #25 before #24 and switched them around on the answer form.  I only changed one of them, but not the other.

136.5  Missed 18, 19 and 25, I think.  Ugh, 18 made me feel stupid.

[quote="JAM"][quote="tjhance"]
except that isn't correct
if $n\equiv 1 \mod 9$ then $n+S(n)+S(S(n))\equiv 1+1+1\equiv 3 \mod 9$

so n+S(n)+S(S(n)) doesn't necessarily = 0 mod 9[/quote]
Except that your forgetting that $n+S(n)+S(S(n)) = 2007$, so it has to be $0$ mod $9$. [b]Then[/b] we use $n \equiv S(n) \pmod{9}$ to get $3n \equiv 0\pmod{9}\implies n \equiv 0\pmod{3}$[/quote]

Ah, right. I was thinking too general.

In this specific case, we know n+S(n)+S(S(n)) = 0 mod 9, so n would be 0 mod 3.

[quote="indianamath"][quote="JAM"][quote="tjhance"]
except that isn't correct
if $n\equiv 1 \mod 9$ then $n+S(n)+S(S(n))\equiv 1+1+1\equiv 3 \mod 9$

so n+S(n)+S(S(n)) doesn't necessarily = 0 mod 9[/quote]
Except that your forgetting that $n+S(n)+S(S(n)) = 2007$, so it has to be $0$ mod $9$. [b]Then[/b] we use $n \equiv S(n) \pmod{9}$ to get $3n \equiv 0\pmod{9}\implies n \equiv 0\pmod{3}$[/quote]

according to the number theory book, you can't divide congruences[/quote]

You can, but you have to be careful. In general if $ka \equiv kb \pmod{c}$, we don't necessarily have have $a\equiv b\pmod{c}$, but we do have the weaker condition $a\equiv b \pmod{\frac{c}{\gcd(k,c)}}$.

So since $3n \equiv 0 \equiv 3\cdot 0 \pmod{9}$ we have $n \equiv 0\pmod{\frac{9}{\gcd(9,3)}= 3}$.

If you prefer to think of it without mods, we know that $\frac{3n}{9}$ is an integer, and $\frac{3n}{9}= \frac{n}{3}$, so $\frac{n}{3}$ is an integer.


function
advanced fields
advanced fields unsolved

Fucntion f continious on the closed disk D={z:$|z|\leq1$} and analytic in its interior. All values of the function f on the boundary of the disk are real. Prove that f-constant

There is a short answer. $Imf$ is harmonic and vanish on the boundary, so vanish in the whole interior. That implies $f$ is constant


on the personal info application, is 300 the word limit for both prompts together or individually?

i believe 300 words for both of the prompts

sounds like youre having a hard time fitting everything in 300 words

its fine, you can go over 300 word limit (mine was 400 words)
theyd have to be really mean to not accept you because you didnt fit the thing in 300 words

For last year, I wrote 314, 517 words, respectively, for questions 1 and 2 in the personal information form. The point is to convey something in what you are writing. There is no reason for the word limits to be strict.


inequalities
geometry
inequalities proposed

Prove that if $ F_1$ is Torricelli point of triangle $ ABC$ with angles less than $ 120^{\circ}$ then
\[ AF_1\cdot BC\plus{}BF_1\cdot CA\plus{}CF_1 \cdot AB\geq 4\Delta\]
($ \Delta$ - area of triangle $ ABC$).

[quote="limes123"]Prove that if $ F_1$ is Torricelli point of triangle $ ABC$ with angles less than $ 120^{\circ}$ then
\[ AF_1\cdot BC \plus{} BF_1\cdot CA \plus{} CF_1 \cdot AB\geq 4\Delta\]
($ \Delta$ - area of triangle $ ABC$).[/quote]
Let $ AF_1\equal{}x,$ $ BF_1\equal{}y$ and $ CF_1\equal{}z.$ Hence, we need to prove that:
$ \sum_{cyc}x\sqrt{y^2\plus{}yz\plus{}z^2}\geq\sqrt3(xy\plus{}xz\plus{}yz),$ which is equivalent ( after squaring ) to
$ \sum_{cyc}2xy\sqrt{(x^2\plus{}xz\plus{}z^2)(y^2\plus{}yz\plus{}z^2)}\geq\sum_{cyc}(x^2y^2\plus{}5x^2yz),$ which is obviously true
because $ \sqrt{(x^2\plus{}xz\plus{}z^2)(y^2\plus{}yz\plus{}z^2)}\geq xy\plus{}\frac{(x\plus{}y)z}{2}\plus{}z^2$ is true.

[quote="limes123"]Prove that if $ F_1$ is Torricelli point of triangle $ ABC$ with angles less than $ 120^{\circ}$ then
\[ AF_1\cdot BC \plus{} BF_1\cdot CA \plus{} CF_1 \cdot AB\geq 4\Delta\]
($ \Delta$ - area of triangle $ ABC$).[/quote]

It is true for any point $ M$ inside $ \triangle ABC$ :

$ AM*BC \ge 2(S_{ABM}\plus{}S_{ACM})$
$ BM*CA \ge 2(S_{BCM}\plus{}S_{BAM})$
$ CM*AB \ge 2(S_{CAM}\plus{}S_{CBM})$

$ \Rightarrow AM*BC\plus{}BM*CA\plus{}CM*AB \ge 4\Delta$


calculus
integration
trigonometry
logarithms
calculus computations

Leisure for your tired brain...

$(1) \, \int_{0}^{\pi}\ln(1-\cos \theta) \, d\theta$

$(2) \, \int_{0}^{\frac{\pi}{2}}(\ln \sin \theta)^{2}\, d\theta$

$\int^{\pi/2}_{0}\ln(\sin t)\,dt =-\frac{\pi}{2}\ln 2$.

To demonstrate this, start by showing that

$\int^{\pi/2}_{0}f(\sin x)\,dx = \int^{\pi/2}_{0}f(\cos x)\,dx$.

[quote="Timestopper_STG"][quote="Carcul"]
$2 \cdot \frac{\pi}{2}\ln 2+4\int^{\pi/2}_{0}\ln(\sin t)\,dt = 2 \cdot \frac{\pi}{2}\ln 2-4 \cdot \frac{\pi}{2}\ln 2$[/quote]
I can't understand this line... :([/quote]

 :rotfl:

let $\textbf{I}=\int_{0}^{\frac{\pi}{2}}\ln \sin x\; dx=\int_{0}^{\frac{\pi}{2}}\ln \cos x\; dx$

(since $\int_{a}^{a+b}\textbf{f}(x)\;dx=\int_{0}^{b}\textbf{f}(a+b-x)\;dx\;\;\;\quad put\;\;\;a=0,\;\;\;b=\frac{\pi}{2}$)

now $\textbf{2I}=\int_{0}^{\frac{\pi}{2}}\left( \ln \sin x+\ln \cos x\right)\; dx=\int_{0}^{\frac{\pi}{2}}\ln \sin 2x\;dx-\frac{\pi}{2}\ln 2$

(since $\sin 2x=2\sin x \cos x$)

put $x=2t$ in the integral on RHS, and you get

$\int_{0}^{\frac{\pi}{2}}\ln \sin 2x\;dx=\frac{1}{2}\int_{0}^{\pi}\ln \sin t\;dt=\int_{0}^{\frac{\pi}{2}}\ln \sin t\;dt=\textbf{I}$

(since $\int_{a}^{a+2b}\textbf{f}(x)\;dx=2\int_{0}^{b}\textbf{f}(x)\;dx\;\;\;\quad \textbf{iff}\;\;\;\textbf{f}(a+2b-x)=\textbf{f}(x)$

$put\;\;\;a=0,\;\;\;b=\frac{\pi}{2}$)

so, $2\textbf{I}=\textbf{I}-\frac{\pi}{2}\ln 2\;\;\;\Rightarrow \textbf{I}=-\frac{\pi}{2}\ln 2$


http://www.cms.math.ca/MediaReleases/2009/imo-team

Robin Cheng
Jonathan Schneider
Danny Shi
Hunter Spink <-- C'est moi!
Chen Sun
Jarno Sun

Funny how the date of the media release is May 29...

Oh no Jarno, it was released half a day earlier :o

Lol, and on BIRS, it was released even sooner http://www.birs.ca/birspages.php?task=listpars&event_id=09ss007

[quote="soledad42"]BTW now we can celebrate the Calgary dominance this year. Hurray![/quote]

Blah! Adrian! Blah!  :omighty:

Yes, Adrian will always be remembered for the domination. Too bad he'll be graduating next year :(

:D   Congrats guys!  Nice to see some familiar names on the list  :P

Congrats guys! :D now its official :P

Strong team in my opinion! :!:

[quote="rofler"]Yes, Adrian will always be remembered for the domination. :([/quote]

  :clap: All hail to Adrian.  :omighty:

[quote="carpo"]Yo all come to Waterloo Seminar[/quote]

Are you going to be there?

[quote="cynthia1991"]wow~~~Congrats to all six of you. How many of you are from Calgary? 3?[/quote]

Yep. :)


When Jorge took his place in the marching band, he noticed that he was 10th from the front, 7th from the back, 3rd from the left, and 8th from the right in the rectangular formation. How many members were in the marching band?

Since he was 10th from the front and 7th from the back, that means there are 18 people in a column. Similarly, there are 12 in a row. Therefore, our answer is $ 18 \times 12 \equal{} 216$.

The answer was 160. This is because he was 10th from the front and 7th from the back, that means there are 16 people in a column. Similarly, there are 10 people in a row. Thus, our answer is 10 X 16 = 160

The answer is 160 because he was 10th from the front and 7 from the back, which is 10+7-1=16 people. Same for the row, 3+8-1=10. 16*10=160.


How many ways are there for 4 kids to pay for a 25 dollar item only using dollar bills if one  of the kids can only give 2-9 dollars and another can only give 3-8 dollars?

I know this is simple with generating functions but when I tried it with PIE, I couldn't get the same answer. Could someone show me how to use PIE in this?

[hide="Answer"]
720
[/hide]

2 to 9 dollars and 3 to 8 dollars you mean?

[hide="Comment"]
I can't see how to do it without generating functions except to get nasty with casework, i.e. the number of combinations if the first restricted kid has 2, and the second has 3, etc. There's not as many cases as it looks like there are.

Sorry, but I can't figure out how to use PIE with this problem...  :( [/hide]

The PIE I tried was:

#(without restrictions) - #(Kid 1 always breaks restriction) - #(Other kid always breaks restriction) + #(Both always break restriction)

Ball in urns was then relatively easy to use to calculate these separate cases but the answer doesn't match up. What am I doing wrong?

It's impossible to tell unless you actually show us what you've done, although I think in all likelyhood you've miscalculated some of the values in your PIE expression :P.  (They don't look like they should be particularly easy to express since "breaking restrictions" is not a nice condition for the given restrictions.)

Anyhow, there's no need for generating functions here:  two people paying whole dollar amounts can pay a total of $ k$ dollars in exactly $ k \plus{} 1$ ways, so the answer is just $ \sum_{i \equal{} 2}^{9}\sum_{j \equal{} 3}^8 (25 \minus{} i \minus{} j) \plus{} 1$.

hm JBL, the answer I got from your double sum is 673.  :huh:

Which one's right? Here's a brief of my generating functions solution:

[hide]
\[ (1 \plus{} x \plus{} x^{2} \plus{} \cdots)^{2}(x^{2} \plus{} x^{3} \plus{} \cdots \plus{} x^{9})(x^{3} \plus{} x^{4} \plus{} \cdots \plus{} x^{8})
\]

\[ \equal{} (1 \minus{} x)^{ \minus{} 4}(x^{5} \minus{} x^{11} \minus{} x^{13} \plus{} x^{19})
\]
So the coefficient of $ x^{25}$ is
\[ \binom{ \minus{} 4}{20} \minus{} \binom{ \minus{} 4}{14} \minus{} \binom{ \minus{} 4}{12} \plus{} \binom{ \minus{} 4}{6} \equal{} \boxed{720}
\]
[/hide]

hm can someone verify my answer in my original post? I'm not so sure my answer is right anymore.

[quote="Tenoreoz"]hm JBL, the answer I got from your double sum is 673.  :huh:

Which one's right? Here's a brief of my generating functions solution:

[hide]
\[ (1 \plus{} x \plus{} x^{2} \plus{} \cdots)^{2}(x^{2} \plus{} x^{3} \plus{} \cdots \plus{} x^{9})(x^{3} \plus{} x^{3} \plus{} \cdots \plus{} x^{8})
\]

\[ \equal{} (1 \minus{} x)^{ \minus{} 4}(x^{5} \minus{} x^{11} \minus{} x^{13} \plus{} x^{19})
\]
So the coefficient of $ x^{25}$ is
\[ \binom{ \minus{} 4}{20} \minus{} \binom{ \minus{} 4}{14} \minus{} \binom{ \minus{} 4}{12} \plus{} \binom{ \minus{} 4}{6} \equal{} \boxed{720}
\]
[/hide][/quote]
\[ (1 \plus{} x \plus{} x^{2} \plus{} \cdots)^{2}(x^{2} \plus{} x^{3} \plus{} \cdots \plus{} x^{9})(x^{3} \plus{} x^{3} \plus{} \cdots \plus{} x^{8})
\]

\[
= (1 - x)^{ - 4}(x^{5} - x^{11} - x^{13} + x^{19})
I don't think you get it right in this equation!Check again since it needs to be some simple bunches of calculating the coefficient!

[quote="Tenoreoz"]hm JBL, the answer I got from your double sum is 673.  :huh:

Which one's right?[/quote]
Your generating function solution is fine -- your ability to compute double sums, however, seems questionable  :P  :wink:

Edit: I take it back: your ability to interpret ambiguous notation is questionable, not your ability to add.  It's $ \sum_{i \equal{} 2}^{9}\sum_{j \equal{} 3}^8 \left((25 \minus{} i \minus{} j) \plus{} 1\right)$.  (You should, of course, think some about [i]why[/i] this sum gives the correct answer.)

ooo I see now. That $ \plus{}1$ applies to all combinations of the restricted kids, not just one for the entire set  :oops:


How can I set an image on the left and other one on the right, but both of them on the same line?

Use minipages or a table to contain them

Okay.

Anyway, would you mind to show an example please?

Adjust the widths (3cm) to suit your images and page:
[code]\begin{minipage}{3cm}
\includegraphics{first image}
\end{minipage}
text goes here \hfill
\begin{minipage}{3cm}
\includegraphics{second image}
\end{minipage}[/code]or
[code]\begin{tabular*}{\textwidth}{@{\extracolsep{\fill}}lcr}
\includegraphics{first image} &
text goes here &
\includegraphics{second image}
\end{tabular*}[/code]


trigonometry
limit
logarithms
function
real analysis
real analysis unsolved

Let $ f(x)\equal{}\lim_{n\to\infty}f_n(x)\equal{}\sum_{k\equal{}1}^{\infty}\frac{\cos kx}k.$

As it happens, we can explicitly compute that $ f(x)\equal{}\minus{}\ln\left|2\sin\left(\frac{x}{2}\right)\right|.$

This is a decreasing function on $ \left(0,\frac{\pi}2\right),$ and hence its minimum value on that interval is $ \minus{}\frac12\ln2.$

That doesn't necessarily mean that that's the infinum of $ a_n.$ I did that just to give some structure to how to think about the problem. I see two avenues to explore to possibly decide what happens.

(1) Is $ f_n(x)$ a decreasing function on that interval? Or not?

(2) Explore the size and behavior of $ \sum_{k\equal{}n\plus{}1}^{\infty}\frac{\cos kx}k.$

Concerning (1): $ f_4$ attains its minimum at $ 1.2566\dots$, where its value is $ \minus{} 0.2879\dots < f_4(\pi/2) \equal{} \minus{} 1/4$.

It seems that $ f_2(\pi/2) \equal{} \minus{} \frac {1}{2}$ is the correct answer.

Yes,I think that $ \minus{}\frac {1}{2}$ be the correct answer.
For the prove:
Let $ f(x) \equal{} \frac {1}2 \plus{} \cos x \plus{} \frac {\cos 2x}2 \plus{} \cdots \plus{} \frac {\cos nx}n$
We should consider the extremal point in $ [0,\frac {\pi}2]$ and the boundary.
For the extremal point we have $ f'(x) \equal{} 0$ which is equal to $ \sin \frac {(n \plus{} 1)}2 \sin \frac {n} {2}x\equal{}0$
we can prove that:$ \cos nx \ge 0$ and combined the induction.
For the boundary,I think it is obvious.


integration
calculus
calculus computations

If $ g'\left(x\right)\equal{}\sqrt{x^3\plus{}2}$ and $ g\left(3\right)\equal{}5$, find $ g\left(1\right)$.

Never mind; the answer would be $ 5 \minus{} \int_1^3 \sqrt {x^3 \plus{} 2}\,dx$.


What is the remainder of $ \frac {x^{10}}{x^2 \minus{} a}$ ?

hello, do you mean $ \binom{x^{10}}{x^2\minus{}a}$?
Sonnhard.

No, I mean $ \frac {x^{10}}{x^2\minus{}a}$ this is the Remainder Theorem.

Anyway, the answer is $ a^5 \minus{} 1$
I don't know why.

Hmm... If we let $ x^2 \equal{} y,$ we have $ \frac {y^5}{y \minus{} a}.$ So, if $ f(y) \equal{} y^5,$ dividing $ f(y)$ by $ y \minus{} a$ must have a remainder of $ f(a)$ by the remainder theorem. But $ f(a) \equal{} a^5 \ne a^5 \minus{} 1.$

I also tried it with synthetic division, and it is also yielding $ a^5.$

lol cliu you copied the question down wrong. Its x^10-1, not x^10.


[quote="gracexcor3"]can someone explain to me how to factor x(squared)+2x-20=28?[/quote]

$x^{2}+2x-20=28$

First, set the whole equation to $0$, so to do that, subtract 28 from both sides, leaving:

$x^{2}+2x-48=0$. Now, this equation is in the form of $ax^{2}+bx+c$. To factor and equation, find two numbers that multiply to $c$ and add to $b$. In this case, the numbers would be $8$ and $-6$. The factored version is

$(x+8)(x-6)=0$

Based on the problem, it seems like you would also want to solve.  You usually don't simply factor problems which are equations.

To solve, set each resulting factor equal to zero and solve those problems.  You will almost always get two answers, the exception being trinomial squares in which there is only one solution.

$(x+8)(x-6)=0$

Solve:
$(x+8)=0$ and $(x-6)=0$

Solving for the first factor yields $x=-8$ and solving for the second factor yields $x=6$.

The reason this works is due to the multiplication property of zero.  You are setting each factor to zero individually, and you don't care what the value of the other factor is because anything times zero is zero.


I know that in graph theory, you can easily tell if in a network you can travel over each path exactly once. But I often see puzzles that require you to only visit each node exactly once, not having to use all paths. How would you find out if this was possible and how to do it?

For example, the grid below: you can start at any x, move up, down, left, or right to another x and touch each x exactly once.

x x x x x x x
x x x _ _ _ x
x x x _ _ _ x
x x x _ _ _ x
x x x x x x x
x x x x x x x
x x x x x x x

Well the example you gave is obvious... just start from the outside, and kinda "outline" the whole x's towards inside... you will see it easily

as generally though, I cannot answer your question.

You are asking a question about finding a [url=http://mathworld.wolfram.com/HamiltonianCircuit.html]Hamiltonian circuit[/url] on a general graph.  This question is what is known as [url=http://mathworld.wolfram.com/NP-CompleteProblem.html]NP-Complete[/url], meaning that there is no known algorithm that can solve the question efficiently, although it has not been proven that none exists.  (A proof that none exists or the demonstration of an algorithm that does solve the question efficiently would be equivalent to a solution to the Millenium P vs. NP problem.)  What this means as a practical matter is that there isn't any way to do better than simply checking all possible paths to see if one works.


geometry
3D geometry
tetrahedron
algebra
polynomial
function
angle bisector

[quote="nr1337"][quote="MCrawford"]I will be attending the Mu Alpha Theta national convention next week and giving two talks.  If you will be in attendance, feel free to introduce yourself to me.[/quote]

Sounds interesting.  Where is it and when?[/quote]

This year it's in Huntsville and starts on July 27th and lasts about 4 1/2 days.

Darn.. I don't suppose that's anywhere near cleveland?

Yeh, its alabama, and btw, it costs close to 500$ to attend, which is paid before the school year ends.

[quote="codeblue87"]i had a good time at the convention. thanks for coming and talking.
-ben[/quote]

It was a lot of fun meeting many of the AoPSers at Mu Alpha Theta.  Hopefully I will be able to attend again sometime in the future.

I also think the first Mu Alpha Theta Olympiad was a very good idea and went very well.

Yes, I concur. For me, the convention last year was more enjoyable (but that could be chalked up to the fact that last year my team had about 20, and this yuear we had 6.), but meeting Mr. Crawford and certain other people was worth it all. I thank Mr. Crawford and Mr. Ruszcyk very much for creating the olympiad there, but I must say that I condemn you, Sir Ruszyck, for creating number 1 on there. I couldn't for the life of me leave that problem, and due to that, I didnt even look at number 2 (which I got in 10 mins during the awards) and wasnt able to see the solution for 5, which I have seen about 300 million times before. Bah to number 1, bah to it. But as Mr. Newton says, "Danke Schoen, Thank you for all the joy and pain".

Is there a site somewhere where they would have the problems posted?  I m just curious to what it was like.

These were the problems:

1. Out of the first 36 pos ints, pick any 7. Prove that for any such subset, there are 4 elements a,b,c,d such that ab-cd = 5k, for some non-zero int k.

2. P is any point in tetrahedron ABCD. If the altitude to side ABC from D is b_1 and the altitude to side ABC from P is b_2, then prove that b_2/b_1 + a_2/a_1 + c_2/c_1 + d_2/d_1 = 1.

3. z = 3+4i, and |w| = 10. Find the imag number w such that |w-z| is maximal.

4. A of triangle ABC has an angle bisector that meets BC at D. The line L that is tangent to the circle that circumscribes ABC at A is parallel  to another line R that intersects D. Prove that R is tangent to the incircle of ABC.

5. The polynomial function f(x) is such that f(1)=f(2)=f(3)=f(4)=2004. Prove that there is no int such that f(m) = 1.

my opinion of the problem difficulty...
3 is by far the easiest.
2 comes next. Its intimidating a little bit, but brutally easy also when attempted
1 is next with the only tricky part coming at the very end
5 after that with a solution that is very common.
4 i couldnt figure out what to do with the angle bisector. The person who won it (he got a perf) came up with a solution that just involved the angles (i was trying to find one with symmetry).

Thank you Sir Ruszyck for including number one on the mathematics Olympiad. Although it did take me one and a half pages, I did finally write a proof for that question.

Also, I have to thank Mr. Crawford for the Algebra class that I am taking; without it, I would not have gotten correct the two sigma questions on the individual written test and, therefore, would not have gotten first place for that exam. Art of Problem Solving has definately been a great resource for students interested in mathematics.

I'm glad that your guys all enjoyed Mu Alpha Theta; it seems that our city was a pretty good host.

[quote="xxreddevilzxx"]I enjoyed this year's MAO at alabama. this was my first year and i was happy that i qualified for the olympiad. I thought I had partials on every problem, but I think i only got one for problem 1, which i figured out in like 2 minutes. anyhow, it was a good convention.[/quote]

I'm glad you enjoyed it.  Olympiads tend to be graded very harshly in that work presented must actually pertain to a possible solution in an organized way.  For this reason there tend to be a lot of 1/7's, very few 2/7's and 3/7's.  4/7 is typically a full solution minus a minor step.  5/7's and 6/7's are full solutions with some minor error or inconsequential lack of rigor.

I am glad they held the Olympiad and I hope they continue to do so in the future.