well some of us were forced to at the beginning :D

my parents are not gifted with maths, and they didn't like it at all during high-school. they both pursued diplomas in areas that did not even intersect with any of the "exact" sciences. they thought that I will have their skills and lacks, so they decided to teach me from a very young age to ALSO like maths (they assumed I like the non-"exact" sciences as they did :D). In fact when I became older (4th grade or so) I did not wanted to go to a math class, but they insisted.

So when I was moving to 5th grade, a special maths class was formed in my school, and a test was given at the end of 4th grade. I could have easily solved all the problems from the test, because it was easier than the olympiad (yes in Romania we have olympiad since 4th grade! :D) but I did not want to, so I didn't present to the test, and told my parents that I didn't make it. At the beginning they only showed the list with the people that entered the class (took a score of 5/10) and so I did foul them. But then, a couple of days after, the teacher called at my home and asked my parents why didn't I go to the test :( and that's how I got discovered .. :)

anyway, the ocasion (is this the right spelling?) was lost and I was happy with my punishment (not to go outside of the house for 1 month!) and that I have eluded the special math class. BUT, (because there's always a but :D) the number of students reaching more than 5 at the test was below 20, and usually a class is supposed to be at least 30 people. So they made another test with 3 days before the beginning of 5th grade. This time my parents walked me there, and forced me to take it. I couldn't think of what the punishment would be not to pass the test, so I had to do my best, and I got an A, entered the class, and then I got used with the idea.

afterwards, after 2 more years of olympiad (the Romanian olympiad stops at the district phase when your below 7th grade), in 7th grade, me and my deskmate qualified to the NMO after weeks of preparations and learning. That's the moment I can truly say that I began to like maths :D

Hmm..I don't really have a story, or, at least, not such an adventurous one as Valentin's (:D). I went to olympiads before 7'th grade but never got any truly good results. Then, in 7'th grade, came a huge dissapointment: I didn't qualify for the NMO. That was hard to cope with, but I got over it pretty quickly (since I was just a kid and all :)). I didn't qualify because I got 2'nd place at the district round, but only 1 student in 7'th grade got to go to the NMO (so it was a really close one).

I managed to qualify for the NMO in 8'th grade and I've loved math ever since, although I didn't exactly know that in 8'th grade. My teacher in 8'th grade helped a lot: she would let me not take exams and not do any homework and just look at problems and try to solve them, so she let me experiment. That's important if you want to discover what youy like. The number of options increases as you get older, of course.

That's about it.

P.S. I think it's "occasion", Vali :D:D

yes. the NMO changes you a lot. not only in maths insight, but as a person too. I recall that the NMO was my first camp :) I was so glad to have finally gotten rid of my parents, that I could stay up all night (which btw I did in the night before the contest :D) and go to the disco and all.

in bucharest it was much easier to qualify to the NMO. In 7th I was 7th (what a coincidence) on Bucharest :) ... but 11 people went. Also in 8th grade I was 1st but 13 (!!) people went to NMO. these days a few students are interested in olympiads :( ...

other persons, besides Romanians, have any insights on this subject? :) if so, feel free (and compelled :D) to write about them ..

well well my story isn't half as interesting as the stuff posted here.but one of my friends (he was a late bloomer in maths but managed to make it to the last 25) had something very interesting to say
(quote) "it's a bit hard on you when you know your girlfriend is better than u in mathematics" yeah and that's when he took mathematics seriously(grade 10 to be precise)
 :D  :D

Well I'm romanian but I think I've got a small story to share.

     My parents are the classic intelectual Romanian couple...my father is a teacher of maths and my mother is a journalist. Both of them forced me since I was young to do a lot of things: play the piano, skate, ski, they made me study maths.

     When i was a small boy everything worked out fine. Qualifying and winning competitions wasn't very difficult because there weren't many competitors and I had good teachers. In the 5-th grade i was the first in the district olympiad (I think that's the way you call 'olimpiada pe municipiu' in English, and btw, Andrei Stefanescu was second then :D ).

     In 7-th grade I didn't qualify for the NMO. My father thought I'd probably suffer some kind of nervous breakdown but it didn't affect me at all.That was when my parents realized that pushing me wasn't going to help me at all so they let me do whatever I wanted to. Luckily i had a little sense of responsability and I managed to qualify in the 8-th grade. I think I started to love maths the first time i encountered some very very difficult problems that seemed both beautiful and impossible to solve. That was in the 9-th grade, during the first selection test.

     I had 4 problems in front of me, all of them being too difficult for me at that time. And I just sat for 2 and a half hours staring at them and then all of a sudden I just felt very happy to be able to ponder upon those problems and to participate in such a competition. In the end I managed to solve one of them and got points for 2 others, but since then my whole perspective on maths changed. I had found something I liked even though it was very probable I'm never going to be very good at it.

My story is probably quite unique compared to the rest of you . . .

Throughout my life, I've never been encouraged in math at all.  In elementary school (grades 1-5), I was pushed to the lowest math class because my parents hadn't forced me into doing math before I entered school (the blame is not on my parents though).  Throughout elementary school, I never learned anything besides addition, subtraction, multiplication, and division.

When I got to junior high school, I got an "A" in the 6th grade math class.  As much as I tried, the school system would not let me move into the accelerated "gifted" math program.  During the next 3 years at junior high school, all I learned was addition, subtraction, multiplication, and division (fundamentals of arithmetic).  It was such a shock knowing that my friend had already passed algebra, geometry, and number theory while I was still learning the basics (he was in the accelerated math program).  During these years, I did quite poorly because I had no motivation to do good because they wouldn't allow me in the upper class no matter how good I performed.

Upon entering into high school, I entered Prealgebra/Algebra, 1 a very NON-simulating math class that taught the students how to do one-variable equations the whole year.  Again, I felt like I was in the wrong class especially since the students in it had no care for math whatsoever.  I begged my teacher to put me into Geometry Accelerated but to no avail I was placed in Geometry "regular."  Again, I felt quite depressed that no matter how hard I tried to move up in the maths and the sciences, I was kept down.  In my Sophomore year, I had a nervous breakdown . . . my grades slipped and my interest in school and life in itself waned.  After doing things I won't mention and learning from the consequences of my depressed actions, I pulled out of my doldrums and in my junior year I took 3 sciences (I forced the school system to allow me) and got my first "A" in math since junior high school.  After receiving all "A's" my junior year, I forced the school system to allow me into 3 maths, 2 sciences, and 2 languages . . . all at the accelerated advanced level, and that is where I am now.  My senior year of high school.

Great pains have I suffered, great pains that I am suffering now, and great pains I will suffer in the future . . . but that's life . . . DEAL WITH IT!

REMEMBER . . . the best lesson in life is failure.

"Far better it is to dare mighty things, to win glorious triumphs, even though checkered by failure, than to take rank with those poor spirits who neither enjoy much nor suffer much, because they live in the gray twilight that knows not victory nor defeat." -Theodore Roosevelt, 26th president of the United States of America.

Mom got me a book for the first three grades of school when I was 5 or 6. I worked it trough (it was one large book -- of Polish origin) by the first year in school. I wanted to advance to the next two classes -- but strangely enough -- mom didn't consider that a good idea because she thought I would lack the social part among older pupils.

   So there I was, doing nothing special up to and including junior high school (we have three years here, 7,8 and 9). Probably the cause was partly because our math teacher was loosing his eyesight slowly, and he was depressed. The classes didn't consist of anything more than us pupils coming there, sitting down and chatting. He didn't have any lectures at all.

   Obviously this laid it's marks on my maths when I entered high school. I didn't know how to solve even an basic linear equation of one degree etc. But I worked that trough somehow and were a normal student in maths.

   Then one day in second high school grade (we have three), something like in November 2002 I watched 'A Beautiful Mind' -- of course it made me rethink about mathematics, and so I also read Simon Singhs book and concurently watched Pi. Those three elements made me realise that maths is what I wanted to do. Now I'm in last grade in high school doing algebraic geometry, elliptic curves theory, in one week algebraic topology and some functional analysis/integration theory. One thing I obviosly lack is the tradition in problem solving as you others -- but I figured a good substitute are IMC/Putnam level problems.

My story:

I was taught arithmetics by my grandfather when i was 4 1/2 - 5 (this is the period when i began to read too,however i knew basic arithmetics before reading) and i enjoyed adding and subtracting numbers very much (freaky  :D ).When i went to school (6 years) i was able to do the arithmetics in my head (with larger numbers).I was the best student in my school in math (as my teacher said) until the 5th grade,when my interest for math vanished.I was never forced to do anything by my parents.So,i dedicated myself to literature ( and won a few prizes at national level).In the 12th grade,i had to prepare for my final highschool exams and the admission exam to the university which was math exam (Algebra,Analysis,Trig+Analytic Geometry).I had the best private math teacher i could imagine and he motivated me in doing math.Another great contribution for determining me to do and love math had the math forums on the web (cheapest math assistance around  :D ).What can i say...i scored max at my exams and since them i am still doing math  :)

P.S.:my highschool math teacher was awful.In the 9th grade,i got some new schoolmates,one was the daughter of my math teacher.She was kinda stupid and i made fun of her in the first school days.This unleashed the wrath of her mother upon me.As a result,my first mark in math during highschool was a 1,25 on a scale from 1 to 10 (where 1 is the worst)[img]http://www.ectona.org/Mysmilies/Smilies/hau.gif[/img]
Math classes started to be really annoying since then.I wasnt allowed to enter any math olympiads during high school.She really enjoyed kicking me in the nuts.My deskmate always copied the solutions from my testpaper and he received,almost always 9-10 and i had always 5-7
[img]http://www.stopstart.freeserve.co.uk/smilie/dark2.gif[/img]

Just curious, what is NMO? Is it like a math camp for 7th and 8th grade students?

--------------------------------------------------------------------------------------
Here is my experience..

I had the lucky opportunity to always go to good schools throughout my life. In the early grades (ages 4 to 10) math was my favorite subject, so I always did my work for math class first (and sometimes only). Then at the age of 11, I was accepted to a special math program for gifted students. It drew students from all over the city to this one school to study and take math. I had a great teacher there and my interest in math began to grow exponentially.

In this special math program we learned basic (single variable) algebra and elementary geometry at the age of 11 (sixth grade), naive set theory, number theory, combinatorics, probability, and rudimentary abstract algebra at age 12 (seventh grade), and basic mathematical logic (propositional and predicate) at age 13 (eigth grade).

It was by far the best mathematics program for middle age students that I have seen. I was lucky to be a part of it. We used the Comprehensive School Mathematics Program's "Elements of Mathematics" (EM) books for the course. I really think that other schools in the US should try to adopt this program. It really works to get bright kids excited about math and to think abstractly.

I was never really involved in competition culture at a young age. I participated in Mathcounts (the biggest math competition in the US for middle age students), but it never really interested me, because I felt it was too based on speed (and I've never been very fast). But still I came one point shy of making it to the National Level. However there was another boy in my class who was a year ahead of me in math, and who was very good (but also very cocky). I always wanted to beat him. In middle school I never did.

In high school (starting at grade 9), I tested out of the subjects of geometry, algebra II, and trigonometry and started taking calculus. I got somewhat more interested in math competitions - mostly local competitions (I finally beat my rival in some of them). But soon after that I was starting to get more absorbed into learning new subjects of math, like calculus, than on some of the questions typically asked in math competitions. So I never practiced for competitions, and I never made it past the AIME to the Olympiad level in my four years of high school.

In my last two years of high school I went to another specialized school for Math and Science. We did not compete in any competitions at all (our math teacher believed they were distracting from learning higher mathematics). So I studied Multivariable Calculus, Linear Algebra, rudimentary Analysis, and Abstract Algebra there.

After high school, I went to study at Princeton University where I had the best math teacher of my life - Peter Sarnak. He taught my freshman class in mathematics and that solidified my interest in the subject (instead of physics). After having some bad teachers in my second year (or at least teachers who did not live up to the expectations set by Prof. Sarnak) I lost interest in math for a while. I took mostly science and humanities classes in my second and third year.

Then I took a Junior Seminar in Cryptography and Number Theory from Andrew Wiles, and regained my interest. So I went on to obtain my degree in mathematics, and now I am taking the time to teach high school before going back to graduate school for my degree in mathematics.

My advice to all young students, is to study broadly! If you study for the Olympiad, also try to learn as many areas of mathematics as possible. And also put effort into studing other subjects - such as computer science, physics, chemistry, biology, and economics as well. In the long run they will be helpful to you, even as a mathematician - and will make you more employable.

[quote="gauss202"]Just curious, what is NMO? Is it like a math camp for 7th and 8th grade students?[/quote]

NMO = National Mathematical Olympiad ~ USAMO for other countries (Romania for example).

It the final stage of the olympiad. Usually there is the selection for the large training camp (~ of mop in usa).

the same in romania. we have the same curriculum all school, untill college. even there everyone makes the same thing ... :( ... that's why US Universities are the best in the world. Take a look at http://ed.sjtu.edu.cn/rank/rank-2003.mht

not even 1 (one) Romanian University in the first 500 :( ...

They only considered number of Nobel Prize, not Fields Medals or other
math prize (Nevanlina, Crafoord,...)

http://en.wikipedia.org/wiki/Crafoord_Prize

-------------------------------------------------------------------------------
http://ed.sjtu.edu.cn/rank/FAQ.htm#6

6. Why other international awards are not included?

In addition to Nobel Prizes, there are other well-known international awards in various subjects, such as Fields Medal in mathematics. However, it's extremely difficult for us to select, because it's always going to be controversial about any selection. Nobel Prize is the least controversial one.
--------------------------------------------------------------------------

[quote="belenos"]>> accelerated "gifted" math program

>> she would let me not take exams and not do any homework and just look at problems and try to solve them

For me these are the most surprising part of the stories !!

In my country, everybody is supposed to do exactly the same "program" at the same age - for so-called equality reasons.[/quote]

The same applies to Sweden. They actually make try to make you more retarded if you work ahead, they make you suffer.

Hi :)
I'm a new comer here and I want to make friend to all of you.
I remember that I was really bad at math when I was in the elementary school. I could say that at that time I didn't like math a lot. Then when my parents saw that my school result was too bad they sent me to a extra math class. At the beginning, things was terrible,I couldn't understand what the teacher is talking about. however,all the things have changed, I found out that math was really funny so I decided to choose math as my main subject.  Just one month ago, I failed to enter the team of my city and I was really disapointed. Hope that I will be more lucky next time!

Welcome silversunboy :) I sure you get lots of friends here. Also I am sure that will practice you'll make it to the Nationals. Let's not forget China MO is probably the hardest Math Olympiad in the world, so in another country perhaps things would be different :).

I have the same experience: In my country, everybody had to follow the same program while in primary school, and we were molded not to think outisde the box  :( .
 In primary school, if someone used more advanced methods, the teacher blamed him/ her that he/she copied the answers . I have no idea how on earth she deducted that, because if you had used something advanced, it is at least one level above the methods in the book, so you hadn't copied it...  I've always liked math, but that was so discouraging. I hated the feeling of being laughed at because I wanted to know more, and the other students, even those who were good at math, just didn't care.It was so frustrating... I competed while I was in primary school, but I didn't do  too well.
  However, when I got to high school, things got much better. I went to school for natural sciences and mathematics,and I chose the program with emphasis in math.  We had the freedom to use any method we liked, as long as it was plausible.And we had summer camps  where we were introduced to many topics& we had a fab time. (I loved those camps , for many reasons)
  We got extra credit for elegant solution,so bit by bit, I started searching for elegant solutions, and I went  into higher level math. And that was the moment I fell in love with math. I used to stay up all night, trying to work out a problem, and I loved  the thrill I used to get when I finally got the solution.  I did well on competitions in highschool, even got medals at the JBMO & BMO, and a honorable mention on the IMO. I was never an IMO medalist, but when I  remember that I failed most of the tests while I was in first grade because  I could add 13 and 5 without writing 13+5=10+3+5=10+8=18, I'm more than content.
  I'm pursuing a degree in business economics now, but I still love math, and math knowledge is extremly helpful in othe areas ( I find my knowledge of  linear algebra & analysis very helpul- it helps me solve econ problems in 30 sec, by applying simple methods, instead of long& tedious calculations)
 Sometimes I'm sorry I didn't go for a math major ...I guess I'll always miss math  :(   Everything boils down to math. Too bad some people can't understand :(

ya, don't be frustrated by the fact that you are doing a different major. This is the forum just for you, mathematics is fun. One thing I learned during High School was a saying that my teacher put up in the classroom: Mathematics is not an spectator sport. Cheer up, no matter where you go, mathematics is always with you.

I have a simpler experience.
My arithmetic is bad when I was in the first grade of primary school. Everyone got over 90 points while mostly I got only less than 70, sometimes even less then 60. I don't like arithmetic at all during that time.

My father bought me some popular math books when I was in 5th or 6th grade. The books were good and later on I found myself become interested in math.

And my father is a math prof. (so he knows which kind of math books are suitable for me), there are a lot of math books in my home. I could not understand them, so then I began to study math.

My story is probably the most boringest out of all yours.  I'd bet almost anything on it.

When I was is elemtary school (6 or 7) I was good at math, so naturaly I liked it.

I'm from the UK, don't really ahve a story for olympiads though having never sat one.

I always liked maths through school but was never really pushed, there were no special classes or anything in my school and even in my current college. In my GCSEs (I don't know grades, I took them when I was 15 last year) I did pretty badly in maths, or at least thought I did until I got an A. But I never really understood what it was all about and how it could be fun.

This year I started normally at A-level maths and didn't really do [i]that[/i] well. But this winter something happened... I was always alright at maths but suddenly I could do everything! Something clicked just in time for my mock exam where I got 98% quite unexpectedly. Since then I've been learning maths properly and not like they teach us at college, from books and things although most of the olympiad problems are out of my range.

Not the most interesting story, but looking back it was pretty weird how one day I was just better at maths and this spurred me into learning more.

when i was 14 years old, i bought a contest book with problems from all countries in the world, i like it but i'm not serious with math at that time

but i first become serious in math when i bought contest books that contains all the long list problems of imo 1989 and imo 1990, at that time i was really terrible, i can only solved 3 problems from the imo 1989 long list out of 108 problems, the sequence problem proposed by iceland, the sequence problem proposed by cuba and a quadratic equation problem proposed by portugal

the first imo problem that i solved is imo 1992/1, i remember it because it was a memorable event, it was the first time i solve a problem from the best contest in the world, i remember that i was half asleep half awake and suddenly i got the idea in my mind and i get up to write it down

now i can solve an average of 3 imo problems a year, but there was one time when imo was so difficult to me that it seems impossible

"Anonymous" is the user you are 'logged into' when you are signed out (of your account). In the old AoPS, you could post while logged out under the account/name "Anonymous."

[quote=tigeryong]how does @bove have 335 posts and not exist????[/quote]

why did you bump a post 18 years ago?

Originally I was forced, but later I grew to appreciate it. As an India-born, US-raised person, my family has always been particular in STEM. When I was 5 and a half years old, I had been in the US for a few months, and my father put me in Kumon. He then switched to many other classes when I was 7, as I grew to despise Kumon. My father had a habit of teaching me concepts long before they were taught in school. I eventually grew up with this.

However, one day, when I was 9 years old, my father decided to take me to the nearest Barnes & Noble, in the hope I would take some intellectual books for myself(at this point I was fairly disciplined). I picked up an Algebra I cliff notes textbook, which was not too different. It was a normal textbook. I read it and then realised how easy everything in 4th grade maths became. So then I read more textbooks.

While I was reading many of these, I eventually reached the unreturnable place of Algebra II and Calculus, where I discovered the conceptual part of mathematics. This was propelled by 3Blue1Brown, Numberphile, Mathologer, as well as Maths competitions like the AMC, AIME, Olympiads, etc.

That is when I realised what mathematics truly was: the study of formal(meaning imaginary) systems. The name sounds boring, but what it really means is very intriguing. It means that Maths is the study of everything stripped of the contaminations of reality, where only consistency brings order. It was the patterns and new insights that interested me in it. Since empirical reality is not considered in mathematics, many things that were in one area, showed up in a completely different area, all because of consistency. This pulled me in. It was also the insightfulness brought my mathematics, though that is more of an interest in everything intellectual, which is a different story. From then on, mathematics was one of my favourite disciplines, and it stayed that way all the way up to today.

[quote=Ansh2020]That is when I realised what mathematics truly was: the study of formal(meaning imaginary) systems. The name sounds boring, but what it really means is very intriguing. It means that Maths is the study of everything stripped of the contaminations of reality, where only consistency brings order. It was the patterns and new insights that interested me in it. Since empirical reality is not considered in mathematics, many things that were in one area, showed up in a completely different area, all because of consistency.[/quote]

Your interpretation of mathematics is very in line with that of Plato's, I will say, especially where his theory of forms is concerned. From the technical point of view, yes, mathematics is essentially a set of pre-defined axioms (we haven't defined all of them yet, though), upon which more complex fields are built, a bit like Conway's Game of Life, a game where the initial rules are very simple, but from which very complex designs can be built.

I was forced to "love" math and I never really did. My dad is a software engineer and mother was a banker, so my parents were the "intellectual" type. Only recently when I was watching an old video by a soviet-union math professor (English subs though) and he just somehow made me so interested. The thing is parents force children to learn math from a young age (for me it was 5 years old) and I hated it! My parents would force me to remember, but my memory only lasted for 10 mins on the topic. Genuinely, I feel I am stupid at math despite being barked about math since I could think. I was also forced to do national math competitions, and I just got barely anything. But, I always get straight A for all my subjects.

In primary school, I decided to prove my parents wrong, that I can get solve math. Miraculously I got a high distinction in AMC (Australian Math Comp) in 6th grade. It's quite a big deal in Australia. My teachers were all suspicious that I copied because I wasn't in the Advanced math class. So they made me do a really tough paper (it was 8th grade algebra) and I passed. I felt so good and my parents even gave me a Nintendo ;)

But I again lost interest when I couldn't deliver any good results in high school. I believe the main problem was getting good teachers, really. I then took initiative to self-study and slowly I gained interest again! That's how I liked math for the first time in 9th grade (now in the 10th grade).

How I began to hate math? When I was younger, I did MATHCOUNTS, and MATHCOUNTS is a corrupt organization.

Don't believe me? Let me explain. Probably about 6 or 7 years ago, I participated in MATHCOUNTS, and I am shocked by the lack of integrity that the test had. My sister and I were on the same school team(a Christian Academy), and there was this one kid from a rival school... I don't remember what the circumstances were, but he eventually had one of the proctors pinned against a wall, and he was threatening her. Now... here's the thing. I would go defend her, but I'm tiny compared to him. He was literally 6'3, and at the time, I was 5'9. Even my sister was taller than me at the time. He was able to get the answer key out of her, and eventually, he landed top of the Countdown Round with a perfect score. I ended up in second. My sister ended up in third. I beat my sister in the Countdown round, and I think she still holds a grudge against me to this day. Then I matched him, but he had the answers to the countdown Round, so he answered every question in less than three seconds. I was trying to buzz in for one second just to beat his time and try to get an answer in, but obviously, one second is not enough time for me to solve a problem because I'm not a pro at math. It was an easy 3-0 kill for him, and that was it. My sister and I were deprived of Mathcounts glory because of a cheater who basically rigged the competition.

And... that is [hide=why]I want to thank MATHCOUNTS for ruining math competitions for me. If they hadn't done that, I wouldn't be pursuing an acting job today. :)[/hide]

To be real though, I hate math mainly because after that experience I stopped caring about math. Sure, I got all A's in all my math classes, but that was only due to intense studying. Additionally, I suffer from a disorder called Blonde Syndrome where the color of your hair immediately makes people think that you are stupid so that mental image didn't really help either. On top of all of that, I missed a problem of the Math section of my PSAT and I think it cost me free money. Plus, there is really no use for math. I've taken a couple of Math classes to satisfy my minor, and they are dumb. Linear Algebra? Are you serious? When am I ever going to use the eigenvector of a Euler's Theorem Matrix in my life? Also, how many times do I have to repeat to myself the formula $\frac{1}{b-a}\int_a^b f(x) dx$? Absolutely useless. I don't think any of that is useful. Don't even get me started on Bayes Theorem, Hore Tuples, and Chebyshev Inequalities. Those are literally so useless. Put an end to math and the public school system.

[s]Also... burn your Volume 1 while you still have the chance.

lmao I so much agree with you. The education system is actually a bit stupid.

I'm only 14, so I don't have as much experience with math as many others.
My parents didn't really force me to love math at all.  They are obsessive about my grades, but equally obsessive for all of my classes (even gym).  I just find math relaxing.  If I can't stop thinking about something, doing math always lets me focus on the problems instead.
My parents aren't really good at math or care about it.  They are English teachers.  As someone who aims to become an investment banker or work in science/mathematics fields, mathematics is central to my life goals as well (which my parents very much disapprove of - they keep telling me to become a doctor, lawyer, scientist, or engineer).

[quote=Cygnet]I agree, callie313! I find mathematics relaxing, despite my stubborn tendency to aim for a 100% average score.[/quote]
After getting a 96 in math pre-weight (I think it's around 101-102) after weight, I basically stopped thinking for several days and started googling what averages valedictorians usually have with massive amounts of worrying.

[quote=callie313][quote=Cygnet]What is "pre-weight"? :)[/quote]
The class is weighted, so the average is multiplied by a certain amount.  I'm pretty sure the weight is something like multiplied by 1.05, 1.06, or 1.07 for my class, but my district doesn't actually have a way that you can check the weight for sure.[/quote]

Oh! I should have remembered. Thank you very much. :)

[quote=callie313][quote=Cygnet]What is "pre-weight"? :)[/quote]
The class is weighted, so the average is multiplied by a certain amount.  I'm pretty sure the weight is something like multiplied by 1.05, 1.06, or 1.07 for my class, but my district doesn't actually have a way that you can check the weight for sure.[/quote]

A public school, I assume?

[quote=callie313][quote=pezzeepenguin69][quote=callie313]Now I'm obsessively checking the averages usually needed for different ivy league/top colleges.[/quote]

You really shouldn't do that. Firstly, college admissions are very unpredictable. Secondly, college doesn't even matter once you get your job.[/quote]

College does matter to get the job in the first place, and the field I am interested in, investment banking, recruits from top tech and mathematics colleges very selectively.[/quote]

You are a communist, but you are interested in investment banking? How?


Let $\Gamma_1, \Gamma_2, \Gamma_3$ be three circles in the plane, each with radius $R$ and having the point $O$ in common. Let $D$ be the set of the interior points to at least two of the circles.
How do we have to choose the positions of the circles so as to maximize the area of $D$?

Pierre.

It’s not hard to find, in any case, a circle of the same radius as the three given circles which covers the set we’re trying to maximize (sometimes one of the three circles, sometimes the circumcircle of the triangle formed by the three intersection pts of the three circles, other than the point which is common to all of them), so the area is less than or equal to the area of such a circle, and we reach equality when the three circles coincide.

The area can also be minimized, and this happens when the centers of the three circles are the vertices of an equilateral triangle, if I remember correctly (it is an old Romanian TST problem).


Can you just clarify your question abit...it's not clear!!

Oh sorry, I accidentally forgot the brackets for the exponents.

Regarding the question, find what the units digit (ones digit) of the operation would be.

I'm using congruences..

$ 19 \equiv -1 \bmod{10} $
Thus $ 19^{82} \equiv 1 \bmod{10} $

also $ 7^2 \equiv -1 \bmod{10} $
thus, $ 7^{42} = (7^{2})^{21} \equiv -1 \bmod{10} $

Thus $ 19^{82} - 7^{42} \equiv 2 \bmod{10} $

Thus the required remainder is 2

Well done. Would anyone like to post a different solution?

Mods are usually the easiest way to solve problems, but heres an alternate approach.

[hide]  Notice that the units digit of powers of 9 repeat in the following fashion:

$9^1=9$

$9^2=1$

$9^3=9$

$9^4=1$

  And that the units digit of powers of 7 repeat in the following fashion:

$7^1=7$

$7^2=9$

$7^3=3$

$7^4=1$

  $\frac{82}{2}$ goes in perfectly while,  $\frac{42}{4}$ leaves a remainder of two.

  Thus we will end up with $1-9$ as the units digit, but after we carry a ten over to the 1 (known as borrowing), we have $11-9=\boxed{2}$ as our units digit. [/hide]

[quote="4everwise"]Find the units digit of $19^{82}-7^{42}$[/quote]
[hide]
9x9=1
9x9x9=9

7x7=9
7x7x7=3
7x7x7x7=1
7x7x7x7x7=7

82/2=o remainder => units digit is 1
42/4=remainder 2 => units digit is 9

1-9=2

[/hide]


analytic geometry
algebra
linear equation

The two equations x + y + z = 10 is referred as the blue plane and the equation x = 3z = 50 is referred to as the red plane.

Find a t least 5 points which lie on both planes (please explain).

Two planes intersect in a line, so if you can determine the equation of the line where they intersect, then you can easily find five points that lie on that line.

Except I don't know how to find the equation of the line of intersection...

[quote="guile"]x = 3z = 50 is referred to as the red plane.[/quote]

x is constant, z is constant, y isn't, it's a line.

x=50, z=50/3, then y=-170/3

I think he meant x[b]+[/b]3z=50.  But that's still a line, isn't it?

Wait... is x+3z=50 in the plane of x+y+z=10?

[quote="cf249"]I think he meant x[b]+[/b]3z=50.  But that's still a line, isn't it?[/quote]

Probably and no: a single linear equation in three variables gives you a plane.  Two equations gives you the intersection of two planes, i.e. a line (barring the special case of parallel planes).

OK, then my original post was correct.  Let me see if I can find the equation of the line by using systems of equations...

x+y+z=10
x+3z=50
Multiplying both sides of the 2nd equation by -1 gives -x-3z=-50
Adding the equations, gives: y-2z=-40

x+y+z=10
x+3z=50
Multiplying both sides of the 1st equation by -3 gives -3x-3y-3z=-30
Adding the equations, gives: -2x-3y=20

y-2z=-40
-2x-3y=20
Multiplying both sides of the 1st equation by 3 gives 3y-6z=-120
Adding the equations gives -2x-6z=-120

x+3z=50
-2x-6z=-120
Multiplying both sides of the 1st equation by 2 gives 2x+6z=100
Adding the equations gives 0=-20, a contradiction.

...what did I do wrong?

You added $ 20$ to $ \minus{} 120$ and got $ \minus{} 120$.  More importantly, you're just running around in circles: there is no "the equation" of a line in 3-space that you're going to find by manipulating these equations; you need two equations to describe a line, and you already have them.  If you want to find points, you need to start picking values for the coordinates, or do something equivalent.  (This just adds a third equation, or equivalently a third plane; in general position three planes give you one point of intersection, or equivalently three equations in three unknowns give you a unique solution.)

guile, do your own homework.

[quote]You added 20 to -120 and got -120.[/quote]

OOPS!!!  :blush:


limit
ratio
geometry
calculus
calculus computations

I read this in a notebook of a friend and I was confused; where did $ x$ go in the solution below?

Q: Determine the complete interval of convergence of $ \sum_{k \equal{} 1}^\infty \frac {k!x^k}{k^k}$.

Solution:

$ \rho \equal{} \lim_{k \to \infty} \displaystyle|\frac {k!}{k^k} \cdot \frac {(k \plus{} 1)(k \plus{} 1)^k}{(k \plus{} 1)k!}| \equal{} ...$

$ \equal{} e$

Interval of convergence: $ \minus{} e < x < e$

Is the solution correct?

It's incomplete. The ratio test tells you that it converges for $ |x|<e$ and diverges for $ |x|>e$. It doesn't say anything about convergence on the boundary $ |x|\equal{}e$.

[quote="jmerry"]It's incomplete. The ratio test tells you that it converges for $ |x| < e$ and diverges for $ |x| > e$. It doesn't say anything about convergence on the boundary $ |x| \equal{} e$.[/quote]

yes. there is supposed to be an $ x$ at the numerator.

That's merely a clumsiness in the presentation; it's not what jmerry was talking about. You have found the radius of convergence but you have not identified how the series behaves at the boundary points.

The ratio test (or root test) is normally incapable of distinguishing the boundary behavior. The ratio test and root test weigh and balance everything that grows or shrinks geometrically fast or faster. But things that grow or shrink at a less-than-geometry rate, say $ k^p$ for some $ p,$ are invisible to the root and ratio tests. Slower growing factors make no difference if you are strictly inside or strictly outside the radius of convergence where they are overwhelmed by things that grow or shrink geometrically. But right on the boundary, all the bigger factors balance out, and these slower factors become what determines convergence or divergence.

This radius of convergence result can be restated as this: $ k! \equal{} k^ke^{ \minus{} k}g(k),$ where $ g(k)$ grows or shrinks as a slower than geometric rate. To actually decide what goes on at the boundary, we need more detail about $ g(k).$

[i]Stirling's approximation[/i] says that $ g(k)$ behaves asymptotically like $ \sqrt {2\pi k}.$ And that suffices to tell us that this power series diverges at both $ e$ and $ \minus{} e$ because its terms don't tend to zero.


[b]Rigorously prove[/b] that if a rectangular prism is resized into another rectangular prism that its volume remains the same.

Resized implies that volume remains the same... is there anything to prove?

Yes. But my science teacher says the volume increases. Can anyone prove using theorems and postulates that it stays the same. Non-rigorous proofs are also welcome for now...

[quote="paladin8"]The question is not clear. Define "resize." It could mean: preserving volume, preserving surface area, preserving diagonal length, preserving total side length, etc.[/quote]

I'm pretty sure none of those would lead to preserve volume (other than preserve volume itself, which makes this silly and trivial).

Then your teacher should get fired and look for a job in fashion or politics  :lol:

I think I know what your talking about. It has something to do with "shearing" (is that a word?) a rectangular prism, and then you know that the volume of the initial figure and sheared figure are the same by looking at infinitely many cross sections.

[quote="anirudh"]I mean to keep the same amount of volume used. My science teacher thinks that by "stretching out" a certain amount of clay, teh volume increases.[/quote]

Well clay does not occupy a fixed amount of volume... stretching out clay CAN increase the volume. The only thing that is fixed with an amount of clay is the mass.

EDIT: Consequently, with a fixed mass and increasing volume, the clay will become less dense.

[quote="anirudh"]How? How can stretching clay increase its mass? :maybe:[/quote]

... umm, you could read my post before asking a question like that. I said "The only thing that is [b]fixed[/b] with an amount of clay is the [b]mass[/b]."

I meant increasing the volume :blush:  :oops:

The volume would only increase if there are airholes in the clay when you're stretching it out, which there usually is unless you pounded it.


linear algebra
matrix
vector
inequalities
linear algebra unsolved

$A \in M_n(R)$ a symetric matrix with eigenvalues $a_1,..,a_n$.
$f(x)=(\frac{{a_1}^x+{a_2}^x+..+{a_n}^x}{n})^{1/x}$

$D$ is the matrix that has the same diagonal as $A$ and $0$ everywhere else.
The eigenvalues of $D$ are : $b_1,..,b_n$. We define similarly
$g(x)=(\frac{{b_1}^x+{b_2}^x+..+{b_n}^x}{n})^{1/x}$

Prove that $f(x) \geq g(x)$ for $x \geq 1$ and $f(x) \leq g(x)$ for $x \leq 1$

Doesn't this follow from the fact the vector formed by the diagonal of the matrix majorizes the vector of eigenvalues and Karamata inequality?


function
number theory
least common multiple
induction
algebra proposed
algebra

Find all functions $ f$, defined on the set of ordered pairs of positive integers, satisfying the following properties:
$ f(x,x) \equal{} x$
$ f(x,y) \equal{} f(y,x)$
$ (x\plus{}y)f(x,y) \equal{} yf(x, x\plus{}y)$

$ LCM\equal{}$ Last Commun Multiple .
Let's define the total ordered relation ship : $ (x,y) \leq (u,v) \Longleftrightarrow (x\plus{}y < u\plus{}v)or((x\plus{}y\equal{}u\plus{}v)\&(x\leq u))$
By induction you can prove that $ f(x,y)\equal{}LCM(x,y)$ where $ LCM\equal{}$ Last Commun Multiple . :cool:

Another solution:

As $ \gcd(x,y) = \gcd(x,x+y)$, we have that
1) $ \mathrm{lcm}(x,x) = x$,
2) $ \mathrm{lcm}(x,y) = \mathrm{lcm}(y,x)$, and
3) $ (x+y)\cdot\mathrm{lcm}(x,y) = \frac{xy(x+y)}{\gcd(x,y)} =  \frac{xy(x+y)}{\gcd(x,x+y)} = y\cdot\mathrm{lcm}(x,x+y)$.
Therefore, $ f\text{lcm}(x,y)$ is a solution to these conditions.

Now, let $ h(x,y): =\frac{f(x,y)}{\mathrm{lcm}(x,y)}$, then, for all $ x,y\in\mathbb{N}$,
a) $ h(x,x)=1$,
b) $ h(x,y)=h(y,x)$, and
c) $ h(x,y) = h(x,x+y)$.
The items above will lead to $ h(x,y)=1$.  Thus, $ f(x,y) = \text{lcm}(x,y)$ is the only solution.


Find all $m,n\in\mathbb{N}$ such that $gcd(m,n)=1$ and $m^4+n^4$ be a perfect square.

[quote="ZetaX"]That means that $m^4+n^4=k^2$. This can and has already been solved by infinte descent.[/quote]

 :huh:  :huh:

[b]Main problem:[/b]

Suppse that $a,b \in \mathbb{Z}$ and $gcd(a,b)=1$ and $a+b=x^2$ .Prove that there is not exist $y\in\mathbb{Z}$ such that $ab=y^4$

Well, the main problem is equivalent to the one given in your first post, so what's the question¿
One thread where it's shown is http://www.mathlinks.ro/Forum/viewtopic.php?highlight=descent+Fermat&t=3542


Prove that there does not exist $A\in M_{2}(R)$ such that $A^{2}\not=0$ but $A^{3}=0$.
(equivalently, show that $A\in M_{2}(R)$ and $A^{3}=0$ implies $A^{2}=0$.)

Suppose $A^{3}=0.$ Then all eigenvalues of $A$ (even all complex eigenvalues) must satisfy $x^{3}=0.$ Thus, the only eigenvalues are zero, which means that the characteristic polynomial for $A$ is $x^{2}.$ Then, by Cayley-Hamilton, $A^{2}=0.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128281]A more general result[/url] has been discussed.


geometry
trapezoid
circumcircle
power of a point
radical axis
geometry unsolved

give trapezoid ABCD such AB is parellal to CD the perpendicular bisecter of AB intersect DC at P the circumscribed of triangles APD,PCB intersect at Q. prove that AC,BD,PQ concur at one point.

AC，BD insert at E，AC insert circle_BPC at F，BD insert circle_APD at G，then we have ABFG is on a circle，so CDFG is on a circle
......

how did you prove that ABFG is on circle  :!: . i think you have amistake

where are you lovers of geometry? this is too hard for you? :mad:

[b]lym is right![/b]
and:
BD-radical axis of circles CDFG and ADP;
AC-radical axis of circles CDFG and BCP;
PQ-radical axis of circles ADP and BCP;
$ \Rightarrow$ DB, AC and PQ concur in point E.

Call  the circumcircles of $ APD$ and $ BPC$ respectively by $ C_1$ and $ C_2$. Let the radical axis $ PO$ of these two circles meet $ AB$ at $ Q$.
Let $ X$ and $ Y$ be the second points of intersections (possibly coincident) respectively of the line $ AB$ with $ C_1$ and $ C_2$. Observe that $ XYCD$ is an isosceles trapezoid. (Since $ \angle PDX \equal{} \angle PAX \equal{} \angle PBY \equal{} \angle PCY$)

Since $ POXD$, $ POYC$ and $ DXYC$ are cyclic, $ DX,PQ,CY$ concur and the point of concurrency is the center of similarity that carries $ XQY$ into $ DPC$ . Hence, $ \frac {XQ}{QY} \equal{} \frac {DP}{PC}$
We know that $ QX.QA \equal{} QY.QB$ . Using these, we have $ \frac {AQ}{QB} \equal{} \frac {CP}{PD}$.  Hence, $ AQB$ is carried into $ CPD$ by a central similarity and hence, $ AC,QP,BD$ concur

[img]http://img6.imageshack.us/img6/4945/imagerr.png[/img]

the problem has been solved but you didn't thank.


geometry
3D geometry
sphere
algebra solved
algebra

Given a cube in which you can put two massive spheres of radius 1.
What's the smallest possible value of the side - length of the cube?
Prove that your answer is the best possible.

No try for this one ? It's not difficult but amusing !

I understand.
We may assume that each ball tangent to three planes of cube ==> there are three different positions to analyze ==> possible values for length of cube's side are 4, 2+\sqrt 2, (2+2 \sqrt 3 )/\sqrt 3  ==> minimum is  (2+2 \sqrt 3 )/\sqrt 3.

Yeah, that's pretty intuitive. The min side is reached when each ball is tg to 3 planes and the balls are alligned along a main diagonal of the cube, but that's not rigourous...

grobber, my solution is rigorous, because any configuration of two balls we may transform by series of movements in x,y,z directions (each movement saves non-overlapping state of the balls) to described above configuration, i.e each ball tangent to 3 planes of cube. This solution is easy to write in russian rather than in english, but I hope you understand me.

To put a marble with radius $1$ cm in a cube it is obvious that the cube must have an edge with at least a length of $2$ cm. What is the minimum length of the edge of a cube which can contain two marbles of radius $1$ cm? Prove your answer.


AMC
AMC 8
inequalities
Support
blogs
FTW
Gauss

The old Corrupt a Wish Game hasn't been posted on in a month, so here is a new one. :)

Same Rules: you corrupt the person above you's wish and then write your own wish. I'll start:

i wish  :rotfl: would stop laughing.

He stops laughing for a week, but cannot hold it in, and laughs 7 times as hard as ever, driving you totally insane!:rotfl: :rotfl: :rotfl:

I wish I had insane amounts of power!

Wish granted. However, the government unveils a super top-secret technology that allows mind control, and they use you for evil purposes.

I wish that I would have no homework.

Yeah, that's and easy one.

You don't have [i]home[/i]work, but they force you to do tons of extra work in school until 12:00 AM >:)

you are not granted a wish, but then a magical telepathic genie comes and grants all you ever wanted in life..

i wish my boarding school application essays were written perfectly so i could get admitted into the best schools...

wish granted. but a week after you enter the school, you get kicked out because you cheated on a test

i wish i could fly WITHOUT repenting it

[quote="kimmystar94"]you are not granted a wish, but then a magical telepathic genie comes and grants all you ever wanted in life.[/quote]

WOOHOO!!!!!

Wish granted. You don't forget flying, because you have a heart attack and die instantly.

Wish granted. You scream WOHOOO and run out of breath, and die of heart failure.

I wish I could go back in time and meet Isaac Newton.

WG. he kills you

i wish i got a 25/25 on the AMC8

WG. You die of shock.

I wish I controlled time.

with granted. you went into the future and lost control. byebye!

I wish  :)  and  :(  and  :rotfl:  would get a grip and a life.

WG. but afterwards they realize life sucks and they all murder you and then commit suicide.

i wish, i wish i was a fish.

wish granted. but you are become a pet of a family that forgets to feed you. You die of starvation.

WG.  There is, however, a building known as "school".

I wish  :mad: was angrier.

WG. He turns into :furious:
I wish :wow: wasn't so hyperactive (and neither was ernie)

WG. Instead they are just hyper.

I wish that  :(  and :)  and :| weren't so ordinary-looking.

WG. They become  :noo: ,  :yup: , and  :no:.

I wish I hadn't run out of smilies.

WG.  Instead, you run [i]into[/i] smilies and get a concussion.

I wish all smilies were holographic.

WG. When they turned holographic, they got holes in them.

I wish i didn't have school tomorrow.

WG. Today is now Friday.

I wish my homework wasn't so annoying.

WG. It's now college-level difficulty, but you aren't annoyed about it and just happily flunk all your classes.

I wish :!:

WG. You get nothing.

I wish ak909090 would not make such a big deal out of wishing for nothing

WG. He makes a big deal out of wishing for something, and a war is started.

I wish the war would end.

WG The war has just started so it cannot end.

i wish i was still in Los Alamos.

WG. You get run over by a car.

I wish I was like my avatar.

WG. Note that your avatar is a huge head and has a tiny green tail. A shark eats you up.

I wish I for an A++ in all my grades by the end of the quarter.

WG. you get an A++ in P.E. and fail english as well as all your other classes. (your grammar wasn't very fabulous so i just assumed you meant you wanted one A++ among the grades you got, that's also why i picked english)

I wish I had $ \$$100


calculus
function
limit
real analysis
real analysis unsolved

In spivaks calculus book he gives a problem:

consider a function f whose only discontinuities are removable discontinuities, that is, the limit of f exists everywhere. let  g(x) = the limit of f(y) as y approaches x. Prove that g is continuous everywhere.

I didn't feel too comfortable with his explaination, maybe you can help.

Since f is continuous at all points but the set of removable discontinuities, we see g is continuous at all those points as well since f=g at continuous points by definition of continuity.  Let a be a removable discontinuity.  Then, by definition of removable there is a value of $\lim_{{y}\rightarrow a}f(y)= g(a)$ that would make f continuous by putting g(a) in for f(a). Thus, a is an isolated discontinuous point and as before continuity implies f=g, so $\lim_{{y}\to a}g(y)= g(a)$ so g is continuous at a.  Thus, g is continuous at removable discontinuities of f.


inequalities proposed
inequalities
algebra

Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{4abc}\plus{}\frac{4abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge \frac{a\plus{}b}{b\plus{}c}\plus{}\frac{b\plus{}c}{a\plus{}b}\plus{}\frac{1}{2}\]
:)

Let $ a,b,c$ be positive real numbers. Prove that
$$\frac{(a+b)(b+c)(c+a)}{abc}+\frac{abc}{(a+b)(b+c)(c+a)} \ge \frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{49}{8}$$
$$\frac{(a+b)(b+c)(c+a)}{2abc}+\frac{2abc}{(a+b)(b+c)(c+a)} \ge \frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{9}{4}$$


calculus
derivative
function
analytic geometry
calculus computations

Find the equation of the line tangent to $ y\equal{}f(x)$ at $ x\equal{}1$, where $ f(x)\equal{}6e^{5x}\plus{}e^{\minus{}x^2}$.

First you need to find the derivative of the function at the point $ x\equal{}1$: we have $ f'(x)\equal{}30e^{5x}\minus{}2xe^{\minus{}x^{2}}$.  Now just plug in $ x\equal{}1$, yielding $ f'(1)\equal{}30e^{5}\minus{}2e^{\minus{}1}$.  Now you just need to determine the line.  We have $ x\equal{}1$ so let's get the $ y$-coordinate, which is just $ 6e^{5}\plus{}e^{\minus{}1}$.  Now the equation of a line is $ y\minus{}y_{0}\equal{}m(x\minus{}x_{0})$, so we plug in what we have:
$ y\minus{}6e^{5}\plus{}e^{\minus{}1}\equal{}(30e^{5}\minus{}2e^{\minus{}1})(x\minus{}1)$.
You can rearrange to make it look nicer.

You've posted problems nearly identical to this one, and the technique is basically the same each time.  Please actually try to understand what other people have posted instead of just posting basically the same problems over and over again.

I know I have posted similar problems and I did try this one and had a similar answer however I was told that both the way I did it and the way you did it is incorrect. Additionally I have to report the answer in the form $ y\equal{}$blah$ x\minus{}$blah.

I'm pretty sure that the method I used is correct-maybe there's an arithmetic error but the method is certainly correct.  With regards to your last sentence,
[quote="JRav"]You can rearrange to make it look nicer. [/quote]

Hmm... could have been a miscalculation on my part the first time also.

[hide]$ y\equal{}e^{\minus{}1}(2(15e^6\minus{}1)x\minus{}24e^6\plus{}1)$

$ y\equal{}4451.66x\minus{}3561.55$.[/hide]


My mom doesn't work and my dad's a doctor (almost, right now he's an admin cause he doesn't have a license to practice in the USA).

[quote="nick42"]my mom & dad both are computer specialists.

 :gleam:[/quote]
same here.

Dad: Nuclear engineer.  Mom: First grade teacher.

My dad's a computer technician for Bank of America and my mom's an accountant.

Dad: Consultant Petroleum Engineer
Mom: Middle School Math Teacher

Dad: not really sure but something to do with computer programing
Mom: chinese teacher

My dad is some sort of homeowner helper guy. My mom is a Consumer Specialist.

My mom is a computer programmer (data warehouse) at MIT and my dad is a professor of mathematics at Brandeis University.

[quote="Aylsero"]My dad is some sort of homeowner helper guy.[/quote]

That must be a depressing job these days...

Dad: Professor

Mom: Kindergarten teacher

Dad: Systems Analyst

Mom: Vice President at a hospital.

(Wow,  everyone's parents seem to be very succesful...I wonder if that does not bode well for aspiring mathematicians in destitute areas  :o )

My dad is a computer engineer/programmer.

My dad is a journalist, my mom is a psychologist.

My dad is an Operations Research specialist, my mom was a Drama teacher.

My dad is an electrical engineer.  My mom is an accountant.

My dad works in finance, previously worked as an experimental physicist

Well, in my family (Indian), we have this whole thing about doctors.  My dad is one.  His brother is one.  His brother's two children are in medical school.  His sister's two children are in medical school.  Many of his predecessors were doctors.  Same for my mom's side.  I have this huge pressure to become a doctor.  But anyways, my mom stays at home (she could work as an engineer though).

My Mom works at sorting greeting cards for HallMark, and my Dad works for Procter and Gamble pushing P&G crud around.

My Dad used to work for RGIS.

Math talent comes from the weirdest places.

Dad:surgeon
Mom:Lawyer
[quote]Well, in my family (Indian), we have this whole thing about doctors. My dad is one. His brother is one. His brother's two children are in medical school. His sister's two children are in medical school. Many of his predecessors were doctors. Same for my mom's side. I have this huge pressure to become a doctor. But anyways, my mom stays at home (she could work as an engineer though)[/quote]I'm Egyptian, but same thing here

Dad: Computer Engineer
Mom: Nutritionist...is that doctor?

My dad's a software engineer.

I'm not sure what my mom works as, in some sort of minor customer service thingy for computers i guess...

[quote="gauss1181"][quote="Poincare"]both parents hate math.[/quote]
How could they hate math if they are involved in scientific careers?[/quote]
they're not involved in scientific careers. I mean my dad is a doctor by he doesn't need math at all (except maybe arithmetic) so he hates it and my mom is in showbiz but can't work in the USA cause she has an Indian accent and all the machines here are a lot more advanced than in India.

See splatyango's post (she's my sister)
My mom works as... I think helpdesk, and it's in a paper company.

[quote="Poincare"]my mom is in showbiz but can't work in the USA cause she has an Indian accent and all the machines here are a lot more advanced than in India.[/quote]
 :huh: ? What do you mean?

Dad: software engineer
Mom: technical writer

[quote="evilhamster"]See splatyango's post (she's my sister)
My mom works as... I think helpdesk, and it's in a paper company.

[quote="Poincare"]my mom is in showbiz but can't work in the USA cause she has an Indian accent and all the machines here are a lot more advanced than in India. (I used to live in India)[/quote]
 :huh: ? What do you mean?[/quote]
1. The people in the USA have trouble understanding her accent
2. So say there was a new camera released in India, then it would have been in the USA for over a year.

My dad is a professor teaching theology, religion, ethics, philosophy...

He always confuses me.

Dad: Did you know that most Americans get to ride a limo twice?
me: Really? When?
Dad: Once when they're married, another when they're dead.
me: ... :(

Nothing against him, he's a great dad, but i think he reads too much Plato or Socrates or whatever.

MY dad is a dentist who specializes in surgical implnts and my mom helps take care of my sister at home.  No math at all, so neither of them can help me (except my Dad found AOPS.   :D ).

My dad joined the Army when he was 16 and switched to the Air Force later and is now retired. My mom is in the Air Force. My mom knows only basic math and my dad never knew math beyond basic algebra.

[quote="1=2"]
Math talent comes from the weirdest places.[/quote]
Yep.

Mom=accountant (lots of math)
Dad=Computer programmer (Lots of math..idk?)

but yea my sister got the creativity genes and i got the very limited number of math genes.

Dad= Actuary
Mom= back at uni getting PHD in Philosophy


Prove that for any $n\in\mathbb{Z}$ number $(\sqrt{2}-1)^n$ can be presented in that way:
$\sqrt{m+1}-\sqrt{m}$, $m\in\mathbb{N}$

Expanding with binomial we have that
$(\sqrt{2}-1)^n + (\sqrt{2}+1)^n$ is an [b]even number[/b]

Since $\sqrt{2}-1 = 1/ (\sqrt{2}+1)$ we have a binomial equation which root is  $(\sqrt{2}-1)^n$

conclusion follows from $(b \pm \sqrt{\Delta})/2a$  ;)

Hmm... $(\sqrt{2}-1)^n + (\sqrt{2}+1)^n$ is even when $n$ is odd?

I think this is what you meant:

Let $x = (\sqrt{2}-1)^n$, so say $x+\frac{1}{x} = (\sqrt{2}-1)^n + (\sqrt{2}+1)^n = k$.  Thus, $x^2-kx+1 = 0$, and

$x = \frac{k}{2} - \sqrt{\left(\frac{k}{2}\right)^2-1}$

If $n$ is even, then $k$ is an even integer like you said.  If $n$ is odd, then $k = l\sqrt{2}$ where $l$ is even, and the result follows.


I can trade 3 fish for 2 loaves of bread.

I can also trade a loaf of bread for 4 bags of rice.

I can trade 2.5 bags of rice for 4 blocks of salt.

a) How many bags of rice can I get for 1 fish?

b) How many blocks of salt can I get for 1 fish?

Answers are in spoiler format:

F=Fish,L=Loaves,R=Rice,S=Salt

a) [hide] Since 3F=2L and L=4R, then 3F=8R or F=8R/3
So you can get 8/3 bags of rice for 1 fish[/hide]

2) [hide] Start with the 3rd equation: 2.5R=4S, so R=8S/5
We also know that 8R/3=F Since we want to know what S is in terms of F we substitute for R
(8/3)*(8S/5)=F 64S/15=F
So you can get 64/15 blocks of salt for 1 fish[/hide]


geometry
perimeter
trigonometry
Support
rectangle
analytic geometry
function

A convex 2n-gon has its vertices at lattice points. Prove that its
area is not less than (n^3)/100.

Lemma 1. For any convex $n$-gon $F$ with perimetr $P$ some three vertices of $F$ form a triangle of area at most $\pi\cdot P^{2}/n^{3}$.

Proof. Let $a_{1},\,a_{2}\dots,a_{n}$ be sides lengths of $F$, and $\alpha_{i}$ be external angles. Then $2S_{i}= a_{i}a_{i+1}\sin \alpha_{i}$ equals to an area of triangle formed by three consecutive vertices of $F$. So,  by using $\sin x<x$, AM-GM and $\sum \alpha_{i}=2\pi$ we get
\[\prod (2S_{i})\le \prod(a_{i})^{2}\prod \alpha_{i}\le \left((P/n)^{2}(2\pi/n)\right)^{n},\]
hence there exist $j$ such that $2S_{j}<2\pi P^{2}/n^{3}$.

Lemma 2. For any convex $n$-gon $F$ of area $S$ there exist an affine transfrom $T\in SL(2,\mathbb{R})$ (i.e. affine transform which preserve the area) such that  $T(F)$ has perimetr at most $4\sqrt{2S}$.

Proof. Consider support lines to $F$, parallel and orthogonal to its diameter (diameter is the segment between two most distant points of $F$). They form a rectangle of area at most $2S$. Choose $T$ such that $T$ maps this rectangle onto the square of the same area. Such $T$ is desired map.

Combining lemmas we see that for any convex $n$-gon of area $S$ we may find three of its vertices which form a triangle of area at most $32\pi S/n^{3}$. For an $n$-gon with integer coordiantes, this area may not be less then $1/2$, so $1\le 64\pi S/n^{3}$.

Sorry for reviving this thread, but recently I stumbled on the official solution and it was too nice not to post...

[hide="Solution"]
[hide="I. Reduction to the centrally symmetric case"]
Pick a diameter (=longest diagonal) of the convex $2n$-gon, splitting it into a $k+1$-gon and a $l+1$-gon ($k+l=2n$). Now we consider the two polygons formed by rotating each half about the center of the diameter $180^{\circ}$ and then merging it with the original, which gives us a $2k$-gon and a $2l$-gon, both centrally symmetric.
Assuming our bound to be of the form $cn^3$ (constant $c$), since the function is convex, we have $cn^3\ge ck^3+cl^3$. Thus we may assume that the polygon in question is centrally symmetric.
[/hide]
[hide="II. Decomposition of the Polygon into Parallelograms"]
Assuming that the $2n$-gon is centrally symmetric, let the side vectors be $\vec{a_1},\vec{a_2},...,\vec{a_n},-\vec{a_1},...,-\vec{a_n}$ in that order clockwise.
Now we break the $2n$-gon up with the following algorithm:
1. Pick a pair of opposite sides. They will split the perimeter into two halves upon omitting this pair of sides, a left arc and a right arc.
2. Translate a copy of the left arc so that the endpoints meet. This will cut the $2n$-gon into a $(2n-2)$-gon and a parallel section, which can be decomposed into $(n-1)$ parallelograms.
3. Repeat steps 1 and 2 on the new centrally symmetric polygon formed, ignoring the parallelograms.
From this, it should be clear that the area of the $2n$-gon is in fact \[S=\sum_{i<j}|\vec{a_i}\times\vec{a_j}|\] since $|\vec{a_i}\times\vec{a_j}|$ is the area formed by the parallelogram defined with the vectors $\vec{a_i}$ and $\vec{a_j}$.
We realise however that this sum is symmetric, so it will be true under any reordering of the indices.
[/hide]
[hide="III. A lower bound for the cross products"]
We firstly reorder the indices such that $|\vec{a_1}|\le|\vec{a_2}|\le...\le|\vec{a_n}|$.
Suppose some $\vec(a_j)$ has $k+1$ lattice points on it (including endpoints). Then our main means of estimation is the following lemma: there are at most $4k$ indices $i$ such that $i<j$ and $|\vec{a_i}\times\vec{a_j}|$ are all the same value.
[hide="Proof"]
So suppose there are $4k+1$ such indices. We translate all these vectors to start from the origin. The endpoints must lie on two parallel lines (parallel to $\vec{a_j}$), so one of them will have $2k+1$ points. This means that there exists vectors $\vec{a_p}$ and $\vec{a_q}$ such that $|\vec{a_p}-\vec{a_q}|>2|\vec{a_j}|$. But this violates our initial increasing ordering, thus contradiction.
[/hide]
We also have the fact that $k$ must divide $|\vec{a_i}\times\vec{a_j}|$ (by slicing it up into more parallelograms).
Now we have that for a fixed $j=4kr+s+1$, where $0\le s<4k$, $r,s$ integers:
\[S_j=\sum_{i<j}|\vec{a_i}\times\vec{a_j}|\ge (4k)(k)(1+2+...+r)+(s)(k)(r+1)\]
[/hide]
[hide="Calculations"]
$\sum_{i<j}|\vec{a_i}\times\vec{a_j}\ge (4k)(k)(1+2+...+r)+(s)(k)(r+1)$
$=k(r+1)(2kr+s)=\frac{1}{8}(4kr+4k)(4kr+2s)$
$=\frac{1}{8}(j+4k-s-1)(j+s-1)=\frac{1}{8}\left((j+2k-1)^2-(2k-s)^2\right)$
$\ge \frac{1}{8}\left((j+2k-1)^2-(2k)^2\right)=\frac{1}{8}(j+4k-1)(j-1)\ge \frac{1}{8}(j+3)(j-1)$ (since $k\ge 1$.)
Now we have
\[S=S_1+S_2+...+S_n\ge \frac{1}{8}\sum_{j=1}^n(j-1)(j+3)=\frac{1}{8}\left(\frac{n(n+1)(2n+1)}{6}+n(n+1)-3n\right)\ge \frac{n^3}{24}\] for $n>1$.
[/hide]
[/hide]


analytic geometry
graphing lines
slope
calculus
derivative
function
limit

use the "four step process" to find a slope predictor formula for the graph
y=f(x) of

f(x)= X-X^3

"Slope-predictor formula" and "four-step process" must be the peculiar language of some particular course or some particular textbook, one that I'm not familiar with. They're not universal nomenclature. Ask us if we can use the definition of the derivative to find the derivitative of this function (which is my best guess as to what you're talking about) and we can do that.

And that's a problem. If you use the commonly accepted names for things (and there are worldwide commonly accepted names for the concepts of calculus) then you can have a conversation with anyone. Use special language that differs from the commonly accepted names, and you've greatly reduced the number of people you can talk to.

sorry ....i meant any more limit laws to find the slope- predictor formula


Why do so many different phenomena match the normal distribution so well? I assume it has something to do with the central limit theorem, but still- what makes this particular curve so special, that so many different distributions in nearly every science match it so well?

If this is intended to be a discussion, I think it's better off in a higher level forum.


function
integration
logarithms
calculus
complex numbers
complex analysis
complex analysis unsolved

1. Show that any function $ F$ is analytic in a larger disc than $ \mathbb{D}$ and has no zeros in $ \mathbb{D}$ is an outer function

2. Show that the statement is no longer true if we just assume that $ F$ extends continuously to $ \overline{\mathbb{D}}$.

Well 1 is only true modulo a unimodular constan. For 1, we need to show that
$ G(z) \equal{} \exp\left( \int_0^{2\pi} \frac{e^{it}\plus{}z}{e^{it}\minus{}z}\log |F(e^{it})| \frac{dt}{2\pi}\right)$
assumes the right values. I have huge problems with this when $ F(z_0) \equal{} 0$ for some $ z_0 \in \partial \mathbb{D}$. By studying $ H(z) \equal{} \int_0^{2\pi} \frac{e^{it}\plus{}z}{e^{it}\minus{}z}\log |F(e^{it})| \frac{dt}{2\pi}$ when $ z$ tends to a boundary point I can show that $ G$ extends continuously to the boundary with the right values in modulus, but this is still not enough to finish it off by applying the maximum principle since it doesn't provide us with enough info to say that $ F/G$ extends continuously to the boundary. I know some stronger versions of the maximum principle, but I don't see how they can be applied either.

Furthermore, even in showing that if $ F(z_0) \equal{} 0$ then $ \Re H(z_n) \to \minus{}\infty$ for $ z_n \to z_0$  I have to use that $ F$ extends analytically so I know well how $ F$ tends to zero.

This leads me to questions about 2, what is it that can go wrong? One thing that might go wrong is that we might not have $ \log |F(e^{it})|$  integrable, although I don't know how to construct such a function. If it is integrable, can the result still fail?

Figured out some improvement...

Let $ G_{\epsilon}$ be the outer function with $ |G_{\epsilon}(e^{it})| \equal{} \max \{|F(e^{it})|, \epsilon \}$ a.e.. $ G_{\epsilon}$ is now bounded from below, so $ F/G_{\epsilon}$ is analytic and bounded. By the maximum principle, the supremum in the disc of such a function is the essential supremum of the nontangential limits, and this implies that $ |F(z)| \leq |G_{\epsilon}(z)|$ in the disc. Passing $ \epsilon \to 0$ we get $ |F(z)| \leq |G(z)|$.

Still don't know why $ G/F$ is bounded. Also am fairly convinced that $ \log |F(e^{it})|$ can not fail to be integrable in the extending continuously case, so indeed this part is the crucial step.

For 1): If you have a few zeroes on the boundary, just divide by them! All you need to show to get them back is that $ z\minus{}1$ is an outer function. That can be done in a variety of ways.

For 2): Do you know an inner function without zeroes? What is the simplest one? What can you say about its continuity on the boundary?

For 1, yes I did try to use that approach, but I found it more or less as hard as the original problem.  I don't see what particular properties I can use about this function.

For 2, I shall wait until I've read about the singular-inner factorization.

You have trouble showing that $ z\minus{}1$ is outer?  :o All you need to do is to show that the family $ \log|rz\minus{}1|$, $ 0<r<1$ has an integrable majorant to pass to the limit under the integral sign! (as usual, considering $ f(rz)$ helps :))

Huh, yeah, I had trouble. What can I say? I've only just seen the definition of an outer function, I think they're less intuitive than you. I was so obsessed with trying to determine the boundary values and using the maximum principle, and that's harder than just seeing directly that they have the right value in the interior  :blush:

Anyway, I understand how to do 2 now, by choosing a simple analytic function whose real part may not be represented as a Poisson integral, exponentiating it and killing singularities with outer functions.


limit
floor function
calculus
derivative

Let $ \mathop {\lim }\limits_{n \to \infty } (1 \plus{} \frac{1}{n})^n  \equal{} e$(n in N)
Prove that
$ \mathop {\lim }\limits_{x \to \infty } (1 \plus{} \frac{1}{x})^x  \equal{} e$ :)
I need many diferent solutions.

[hide]$ (1\plus{}\frac{1}{\lfloor x \rfloor \plus{}1})^{\lfloor x \rfloor}\le(1\plus{}\frac{1}{x})^{x}\le (1\plus{}\frac{1}{\lfloor x \rfloor})^{\lfloor x \rfloor \plus{}1}$

Then take $ lim$, thereby sandwiching the middle term to $ e$[/hide]

[quote="chrischris"][hide]$ (1 \plus{} \frac {1}{\lfloor x \rfloor \plus{} 1})^{\lfloor x \rfloor}\le(1 \plus{} \frac {1}{x})^{x}\le (1 \plus{} \frac {1}{\lfloor x \rfloor})^{\lfloor x \rfloor \plus{} 1}$

Then take $ lim$, thereby sandwiching the middle term to $ e$[/hide][/quote]
Yes, but you have to prove that f '(x) is positive. :lol:

:?:   But I was not using $ f(\lfloor x \rfloor)$ or $ f(\lfloor x \rfloor \plus{}1)$. The powers are different from the denominators.

[quote="chrischris"]:?:   But I was not using $ f(\lfloor x \rfloor)$ or $ f(\lfloor x \rfloor \plus{} 1)$. The powers are different from the denominators.[/quote]
Can you be more detailed? :lol:

[quote="TRAN THAI HUNG"]Let $ \mathop {\lim }\limits_{n \to \infty } (1 \plus{} \frac {1}{n})^n \equal{} e$(n in N)
Prove that
$ \mathop {\lim }\limits_{x \to \infty } (1 \plus{} \frac {1}{x})^x \equal{} e$ :)
I need many diferent solutions.[/quote]

This can be shown by using first principles of differentiation:

Prove lim(n->inf) (1+x/n)^n=e^x

Solution: d/dx(Inx)=1/x.

d/dx(Inx)=lim(n->inf) [In(x+h)-In(x)]/h

1/x= lim(n->inf) In((x+h)/x)/h

1/x=lim(n->inf) In(1+h/x)^1/h

Ley y=1/x and let n=1/h

y=lim(n->inf) In(1+y/n)^n

rearranging

lim(n->inf) (1+x/n)^n=e^x

[quote="chrischris"]:?:   But I was not using $ f(\lfloor x \rfloor)$ or $ f(\lfloor x \rfloor \plus{} 1)$. The powers are different from the denominators.[/quote]
Yes I have got it thanks. :lol:


There must be at least a lot of chess players on AoPS! Appealing to these people, I am starting a chess game...with a minor twist.

ANYONE can play. Whoever's turn it is, (black or white), move for them.
Don't post twice in a row.
No edits. That's cheating.
Post the chessboard after you move.
Post your move in bold.

I will start.

[b]e4[/b]

Black to move.
________________________
|r _|n_|b_|q_|k_|b_|n_|r _| 8
|p_|p_|p_|p_|p_|p_|p_|p_| 7
|__|__|__|__|__|__|__|__| 6
|__|__|__|__|__|__|__|__| 5
|__|__|__|__|[color=red]p[/color]_|__|__|__| 4
|__|__|__|__|__|__|__|__| 3
|[color=red]p[/color]_|[color=red]p[/color]_|[color=red]p[/color]_|[color=red]p[/color]_|__|[color=red]p[/color]_|[color=red]p[/color]_|[color=red]p[/color]_| 2
|[color=red]r [/color]_|[color=red]n[/color]_|[color=red]b[/color]_|[color=red]q[/color]_|[color=red]k[/color]_|[color=red]b[/color]_|[color=red]n[/color]_|[color=red]r [/color]_| 1

|r _|n_|b_|q_|k_|b_|__|r _| 8
|p_|p_|__|p_|p_|p_|p_|p_| 7
|__|__|__|__|__|n_|__|__| 6
|__|__|p_|__|__|__|__|__| 5
|__|__|[color=red]p[/color]_|__|[color=red]p[/color]_|__|__|__| 4
|__|__|[color=red]n[/color]_|__|__|__|__|__| 3
|[color=red]p[/color]_|[color=red]p[/color]_|__|[color=red]p[/color]_|__|[color=red]p[/color]_|[color=red]p[/color]_|[color=red]p[/color]_| 2
|[color=red]r [/color]_|__|[color=red]b[/color]_|[color=red]q[/color]_|[color=red]k[/color]_|[color=red]b[/color]_|[color=red]n[/color]_|[color=red]r [/color]_| 1

[b]c3[/b]
Can we use capital and small letters for black and white? It's too complicated else.

1. d4 d5
2. Nc3 Nf6
3. Bg5 c5
4. Rb1 b6
5. e3 h6
6. Bxf6 exf6
7. dxc5 bxc5
8. Qxd5 Qxd5
9. Nxd5 a6

rofl

1. d4 d5
2. Nc3 Nf6
3. Bg5 c5
4. Rb1 b6
5. e3 h6
6. Bxf6 exf6
7. dxc5 bxc5
8. Qxd5 Qxd5
9. Nxd5 a6
10. Nb6

 :P  :rotfl:

1. d4 d5
2. Nc3 Nf6
3. Bg5 c5
4. Rb1 b6
5. e3 h6
6. Bxf6 exf6
7. dxc5 bxc5
8. Qxd5 Qxd5
9. Nxd5 a6
10. Nb6 Ke7

:rotfl:

1. d4 d5
2. Nc3 Nf6
3. Bg5 c5
4. Rb1 b6
5. e3 h6
6. Bxf6 exf6
7. dxc5 bxc5
8. Qxd5 Qxd5
9. Nxd5 a6
10. Nb6 Ke7
11. Nxa8

:rotfl:  :P  :rotfl:


Two circles are tangent to each other in the xy-plane. One of the circles has center (1,1) and radius  1. The other circle has center (0,-1). What are the possible radii of the other circle?

how do you do this?

its sqrt 5 plus or minus 1

Just find the distance between the centers, and then subtract the radius to find the other radius.

...bearing in mind that they could be internally or externally tangent (giving two answers).


Let $ a,b,c$ be nonnegative real numbers such that $ a+b+c=3$. prove that
\[{ \frac{bc}{\sqrt{4a^2+3}}+\frac{ca}{\sqrt{4b^2+3}}+\frac{ab}{\sqrt{4c^2+3}} \le \frac{3\sqrt{3}}{4}}\]
:)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers such that $ a + b + c = 3$. prove that
\[{ \frac {bc}{\sqrt {4a^2 + 3}} + \frac {ca}{\sqrt {4b^2 + 3}} + \frac {ab}{\sqrt {4c^2 + 3}} \le \frac {3\sqrt {3}}{4}}
\]
:)[/quote]

Can Hang , can you please tell me, if there is a general way to solve inequalities like this?
What is the method is preferred for this kind of "root" inequalities?

Thanks!

The following inequality is also true？
[quote=can_hang2007]Let $ a,b,c$ be nonnegative real numbers such that $ a+b+c=3$. prove that
\[{ \frac{bc}{\sqrt{a^2+1}}+\frac{ca}{\sqrt{b^2+1}}+\frac{ab}{\sqrt{c^2+1}} \le \frac{9}{4}}\]
:)[/quote]


it hasn't come up yet and is going to come in like another 20 or so days

but that's okay

so who's going

i mean come on everyone has to go

or else no arml for you

unless you made mop or usamo or something

sheesh our sponsor still hasnt told us if we've been invited

My teacher is hosting a play-off thing to see which four members of our math team will make up our GCTM team, and even though I've beaten everyone on our team at every tourament that any of us has been to the entire year, I still have to do well in the play-off to go to GCTM. I'm afraid that I'll get uber-nervous and fail. :(

[quote="Twiz"]I'm afraid that I'll get uber-nervous and fail. :([/quote]

Don't worry. If you fail, just come as an individual, own everyone on ur team, then laugh. That's what I'd do.

[quote="gtf"]Some of the invitations just went out today. From Cobb County:

Lassiter
Walton
Wheeler
Pope
Kell

have been invited. This is not necessarily the final list, just who's been invited as of today.

Fulton[/quote]

W00t! Our Varsity team did well enough to qualify, instead of our JV team qualifing. Looks like we might have more than 1 student to score over 100 this time.   :)  :)

[quote="Xantos C. Guin"][quote="Twiz"]I'm afraid that I'll get uber-nervous and fail. :([/quote]

Don't worry. If you fail, just come as an individual, own everyone on ur team, then laugh. That's what I'd do.

W00t! Our Varsity team did well enough to qualify, instead of our JV team qualifing. Looks like we might have more than 1 student to score over 100 this time.   :)  :)[/quote]

Two things, you can't just come as an individual if you want to. Only invited team can come and only 4 students are allowed per team. As ARML coaches, we are allowed to invite certain individuals that are "of interest" to us if either their team didn't qualify or the team wouldn't have that person on the four person team. (Miles and Santosh are/might have been great examples of this. They could easily been left off of their varsities because they were only in Algebra II.)

Santosh, Wheeler's varsity qualified last year, NOT the JV. You were invited because of your quality. Somehow that all got translated into Wheeler's JV was invited. That was NOT the case. But Wheeler was entitled to send any 4-person team they chose. And they did! :-)

Norcross/I am going!!!!
This will be my fourth and final shot at ARML... I'm SO making it this time. ARRR.

Hmm... when will the individuals of interest be emailed/given to schools?

I don't have an answer for that, but assume that it will be reasonably soon.

Fulton


Find all positive integers n such that $ 2^n|(3^n\minus{}1)$

[url]http://www.mathlinks.ro/viewtopic.php?t=150455[/url]
Trivialized by "Lifting the Exponent" lemma (see the link)

Ya bambaman, great move

Another solution
n=1,2 are solutions. Assume n>2 and we see that for the number to be divisible by 4, n has got to be even. Let $ n = {2^{u}v}$ with v odd.
$ 3^n - 1 = (3^v - 1) \prod_{i = 0}^{u - 1}(3^{v2^i} + 1)$
We know that $ 3^{2k} + 1$ is fully divisible by 2. Now, considering the term $ 3^v + 1$, we see that it is congruent to either 2 or 4 mod 8.
Summarizing, we get that the RHS of the equation has an exponent of 2 at most $ u + 2$ whereas the problem requires it to have an exponent $ 2^{u}v$. So, we've got to have $ u + 2 \geq2^{u}v$.
We see that this is possible only for (u,v)= (1,1) , (2,1)
So we have the solutions n=1,2,4


$\exists S=\{1,...,3000\}$. Does there exist a set $A\subset S$ with $|A|=2000$ such that $\forall x\in A\implies 3x\notin A$?

Masoud Zargar

See also : http://www.mathlinks.ro/Forum/viewtopic.php?t=547 .


How would you find the LCM of a fraction? I came up across this in one of my practices.

Find the Least common multiple of 74/3 and 24.

I am completely confused------ish. :?:   :(

I'm not 100% sure on this but in my opinion,
multiply both numbers by three, find the LCM of those numbers, then divide that by three(just my opinion)

[hide]You only need to multiply 74/3 by 3, so then find the LCM of 74 and 24.  ;) [/hide]

[hide]
This is true because you can't get a fraction from 24
[/hide]

multiply 74/3 by 3 and 24 by 3. then you get the numbers 72 and 74 find a comman factor which in this case would be two. 72/2= 36, and 74/2= 37. Now multiply 2 by 36 which is 72 and 72 by 37which is 2664. That is the lcm

Note: the method that i used was the L method to find the lcm

:first:  :ninja:

However, 888 is divisible by both of the numbers

I suppose it would be the lcm of 74 and 24 since it obviously has to be a whole number. The least whole number multiple of 74/3 is obviously 74. The lcm is 888.

just turn them both into a fractions with the same denominator, and find the lcm of the numerators


A car, a van, a truck and a bike are all travelling in the same direction on the same road, each at its own constant speed. At 10am, the car overtakes the van; at noon it overtakes the bike. At 4pm the truck overtakes the bike and at 6pm the van overtakes the truck.

The speed of the car is 120km/h and the speed of the truck is 80km/h

a) Find the speeds of the van and the bike?
I think the van's speed is 90 km/h and the bike's speed is 40 km/h please could someone verify?

b) Find the time at which the van overtakes the bike.

Let c and T represent the speeds in km/h of the car and the truck respectively.

c)Find the speeds of the van and the bike in terms of c and T

d) Show that the time when the van overtakes the bike is the same regardless of the speeds of the car and the truck.

I really need some help with this problem, help much appreciated!!!

Infinity

please ban this topic. answers shouldnt be given. ;)

ruins the integrity of the maths challenge event runnign in Australia.


I just got an acoustic guitar today! Its my first time playing it!  :D

Are there any good guitar sites where I can learn some basic stuff?

here is a site:
[url]http://www.acguitar.com/leveltwo/ForBeginners-428.asp[/url]

try something like ultimateguitar.com or any other random tab sites, you can find some easy songs which you can learn easy

Yea ultimate-guitar.com is definitely the best out there, I'd also suggest getting a tab program too. Best out there now is Guitar Pro, but that costs money. There's an open source one called Power Tab which isn't as good, but it does the job. Search it on google for a download link.

Wow, a topic nearly a year old.  :)

Thanks.  I'm much better now, and I take private lessons.  :D

guitarplayertv.com has some good free stuff

for technique you can check out ibreathmusic.com ... there's tons of sites on the web


Hello,

I am 45, I have taught math from fifth grade (secondary schools) up to Master's level. In Romania and also, two years, as a teaching assistant in the USA. I got a Ph. D. in math (geometry) in Romania. I have also trained students for math contests, in Romania.

Is it possible to come to work in a high school or at college level in the USA? If yes, how?

Thanks,

Jean

It is certainly possible, and this country needs more good math teachers.  The first issue you'll have to deal with, though, is being legally able to work in the United States at all.  This is a visa issue.  Do you have legal status to work in the US?

It is hard, though there are some programs that facilitate such enterings on the market.

[hide="The text that follows is in Romanian"]Eu personal as participa la loteria vizelor printre alte chestii. [url]http://www.loteriavizelor.com[/url] este un site care explica ce si cum. Altfel sansele sunt minuscule de a primi o viza semi-permanenta. Mai mult cred ca un prof de mate bun castiga mai mult in Romania din meditatii decat ar putea sa castige in SUA :)[/hide]

Se poate. Eu de ani de zile nu mai dau meditatii si nu am tangenta cu preuniversitarul decat ca parinte. Castiga in general cei care au posturi la licee bune. Oricum consider industria meditatiilor o chestie nedreapta pentru ca cei saraci nu mai au nici o sansa. La scoala profesorii predau putin si cer mult, fortand copiii sa vina la meditatii. La toate materiile de bac. Ca parinte gust din plin tehnica asta. Si in universitati, politehnici, etc fac la fel. Doar ca acolo dau mai rar meditatii, se prefera spaga direct. Nu mai zic nimic.

Ma gandeam sa mai schimb tara. ..

Ce se stie atunci despre Canada? Acolo au nevoie de profesori? La liceu sau colegiu? Si cum se poate ajunge si lucra?

Multumesc,

Jean

[quote="jeanvaljean"]Thanks for your answer.

No, I guess I don't have any visa and stuff. I am a Romanian citizen. I've been to USA on a graduate student visa, F1 (?) I don't remember, between 1995-1997, when I also worked as a teaching-assistant in New Mexico (UNM).

So, what possibilities could be for me?

Jean[/quote]

Sorry for not answering sooner - I didn't see that you'd answered.

A couple avenues to consider are:

1) Talk to the folks at Math for America (http://www.mathforamerica.org) - I don't think they take applicants from people who don't have visas, but they may have suggestions for people in this situation.

2) Look at the department of education webpages for big states (CA, NY, NJ, FL).  There are occasionally incentives for people who want to teach math/science.  Help with a visa may be one.

3) It's a longshot, but Teach for America might be another place to do some research.

If you want to become a high school or middle school math teacher in the USA, most states will require you to take either the Praxis II ® Mathematics: Content Knowledge (0061) exam for high school, or the Praxis II Middle School Mathematics (0069) exam for middle school. Some states name this exam differently, but it is basically the same exam.

This exam tests high school mathematics up to a tiny bit of first year calculus, first year statistics, and first year matrix algebra.  Most members of this board would consider it an easy exam, but how many of us have touched high school geometry for example in many years?

How do you prepare for the Praxis exam?  Like any math exam, besides knowing the material, you should practice with many problems.  One inexpensive source of such problems can be found at http://www.praxisiimath.com


Two n-digit integers (leading zero allowed) are said to be equivalent, if one is a permutation of the other. For instance, 10075, 01075 and 00751 are equivalent 5-digit integers.


What would the units digit be for 3 raised to the 9999th power?

[hide]pretty easy one.
3 powers go in a pattern of 3, 9, 7, 1, 3, 9, ... for the units digit.
9999/4 is something r 3, so the units digit would be 7.[/hide]

Or you can use Euler's Totient Function, $x^{40} \equiv 1 \mod {100}$  :lol: but you're only finding the units digits so factorial's method is easier

lol what is euler's totient function... answer 7 btw (easy pattern)


Does anyone have a way to plot statistical distributions in Mathematica.

For instance, how would I plot the area under the curve to the right of a z-score of 0.97 on a normal distribution?

Does Mathematica have a command to shade the integral of part of a curve?


I would like to ask a series of questions about the taking of the AMC 12 B.  Please feel free to answer all, none or some of them as the case may be.

1.  If I take the B at another school, how does MAA associate me with my true school?

2.  If my score on the B is higher than that on the A, must I take the AIME II?

3.  Is it contrary to the spirit of the contest to take both the A and B exams, as my school's math department suggested to me?

I would be grateful for your thoughtful, elegantly-articulated, and perhaps comically-undertoned responses.

I don't know the answer to number one, but:

2.) No. You can take the AIME I.
3.) Some people would say so. But it's up to you to make that call.

3) What if someone is sick or they just can't focus on the A date? Is it their fault they have a bad day?

Well in my case, I have no excuse.  I'm just not satisfied with my 121.5, because I practiced kind of hard.

> 1. If I take the B at another school, how does MAA associate me with my true school?

We don't, unless you pay the Visitor fee for special handling.

> 2. If my score on the B is higher than that on the A, must I take the AIME II?

No.

> 3. 3. Is it contrary to the spirit of the contest to take both the A and B exams, as my school's math department suggested to me?

The original, and continuing, intent of having two official adminstration dates is to allow shools some degree of choice and flexibility in adminstering the AMC contests, while at the same time confining each adminstration to a reasonably small time to frame because of the speed and widespread availability of communication, such as this forum.  The intent was never, and is not now, to provide opportunities for additional test-taking by individuals.  We allow it because we do not have the ability to control it.

Whether it is "contrary to the spirit of the contest" is a value judgement.

Steve Dunbar
AMC Director


Asymptote
analytic geometry
Support
geometry
rectangle
function
real analysis

Does Asymptote support Cartesian coordinate stuffs??  Like can it draw the x and y axis with lattice points (although I realize lattice points can be done with a simple for loop)??

Or even more, can I draw a graph of $ x^{2}$ and then rectangles under it for Riemann Sum pic??

Thank you.

Directly from the asymptote gallery: ([url]http://asymptote.sourceforge.net/gallery/[/url])

[asy]size(300,150,IgnoreAspect); real f(real x) {return 1/x^(1.1);} pair F(real x) {return (x,f(x));} dotfactor=7; void subinterval(real a, real b) { guide g=box((a,0),(b,f(b))); fill(g,lightgray); draw(g); draw(box((a,f(a)),(b,0))); } int a=1, b=9; xaxis("$x$",0,b); yaxis("$y$",0); draw(graph(f,a,b,operator ..),red); int n=2; for(int i=a; i <= b; ++i) { if(i < b) subinterval(i,i+1); if(i <= n) labelx(i); dot(F(i)); } int i=n; labelx("$\ldots$",++i); labelx("$k$",++i); labelx("$k+1$",++i); labelx("$\ldots$",++i); arrow("$f(x)$",F(i-1.5),NE,1.5cm,red,Margin(0,0.5));[/asy]

or, for $ x^{2}$:

[asy]size(300,150,IgnoreAspect); real f(real x) {return x^2;} pair F(real x) {return (x,f(x));} dotfactor=7; void subinterval(real a, real b) { draw(box((a,0),(b,f(b)))); guide g = box((a,f(a)),(b,0)); fill(g,lightgray); draw(g); } int a=1, b=9; xaxis("$x$",0,b); yaxis("$y$",0); draw(graph(f,a,b,operator ..),red); int n=2; for(int i=a; i <= b; ++i) { if(i < b) subinterval(i,i+1); if(i <= n) labelx(i); dot(F(i)); } int i=n; labelx("$\ldots$",++i); labelx("$k$",++i); labelx("$k+1$",++i); labelx("$\ldots$",++i); arrow("$f(x)$",F(i-1.5),NE,1.5cm,red,Margin(0,0.5));[/asy]

Doesn't quite have everything you want, but lattice points shouldn't be hard. I say use a for loop. (actually 2 ;))

Cool.

Anyone knows if I have to include any packages for this to work?

You may want to read the description of the graph package in the standard help file accompanying Asymptote to see what is needed for what here but the good thing about Asymptote is that you can program all such things fairly easily by yourself and if you want to do the same thing again and again, you can write your own small package containing all the functions you need. The graph of $ x^{2}$ with Riemann sum can be done in about 10-15 minutes from scratch (I used cse5 because I like it but it is not really necessary here):
[asy]import cse5; size(256); pen b=black, t=blue+linewidth(1.5), r=red+linewidth(1); path xaxis=(-2,0)--(2,0),yaxis=(0,-0.3)--(0,4); real u(real x) { return x^2; } path GF (real a, real b,real f(real), int n=200) { path P; real h=(b-a)/n; for(int k=0;k<n+1;++k) P=P--(a+h*k,f(a+h*k)); return P; } D(xaxis,b); D(yaxis,b); void PLRS(real a, real b, real f(real), int m) { real h=(b-a)/m; for(int k=0;k<m;++k) { real x=a+k*h; path P=(x,0)--(x,f(x))--(x+h,f(x))--(x+h,0)--cycle; fill(P, yellow); D(P,r); MP("t_"+string(k),(x,0),7); } MP("t_"+string(m),(b,0),7); } PLRS(0.3,1.7,u,6); D(GF(-1.8,1.8,u),t);[/asy]
and after you did it once in a reasonable way, you can just modify your program very easily to make new pictures. The gallery mentioned by Hamster1800 is a good source of code pieces for such things but what you can produce by yourself in half an hour or so will be hardly any worse    :)

As to packages, it never hurts to start your code with import graph; and import math;. Both are the standard ones included in any Asymptote distribution. But, as I said, you can do everything from scratch as well.


An integer n is good if |n| is not the square of an integer.  Determine all integers m with the property:
m can be represented in infinitely many different ways as a sum of three distinct good integers whose product is the square of an odd integer.

I saw this problem two times here on forum before. But now I can't find appropriate topics (one of which contains my solution, though I don't remember whether my solution correct).

[quote="darij grinberg"]I have found one of the threads: http://www.mathlinks.ro/Forum/viewtopic.php?t=5682

I think your last solution is correct, Myth, since I have checked it (in fact, I had learnt it by heart since I expected the problem to come in one of our TSTs  :D )...

  Darij[/quote]

//speechless ;)

I was doing some digging in the forum archives, because I also remember having seen the proof of this some while ago, and the irony of it is that I checked out all the possible threads except that one because for some reason, there was some voice in my brain telling me it couldn't possibly be that thread :( (Must be the effect of silly school projects which take away all my free time :( ) That thread is one of my favourites because we've been playing tricks on Pierre again :D .

I am very glad that I correctly solved this problem. I often forget my own solution in last time. Maybe I need a rest after summer?  :lol:

...as a sum of three distinct good integers whose product is the square of an odd integer....

(1) n=2(n+2k2)-(n+2k2)-2k2
(2) n=4(n+3k2)-3(n+3k2)-3k2

these are myth constructions, and their product is not odd...

Maybe i am locking to the wrong solution, however, the idea seems nice.

It was solution for http://www.mathlinks.ro/Forum/viewtopic.php?t=5682, where we no need product is odd.

For current problem.

It is clear n is odd, moreover n=1 mod 3, otherwise only case n=1+3+3 mod 4 --> product = 3 mod 4 -- is not perfect square.

So we assume n=2 mod 4.
We will represent n as kx+ky+xy. It implies n+k2 =(k+x)(k+y).
Let x=-(k-2) and y=(n+k2-2k)/2.
Suppose (n,k)=1, (n,k-2)=1 and k and k-2 are not perfect squares, k is odd. It is easy to see that k,x,y are odd.
It is clear kx is a good number. Numbers ky and xy are good numbers too, because (k,y)=1, (k-2,y)=1 and k and k-2 are not perfect squares.
So it is sufficient to find k. For example, k=t2n(n-2)+5, where t is an arbitrary large even number.

I hope that is correct.


geometry perimeter rectangle MATHCOUNTS calculus integration

Find all possible sizes of rectangles with integer sides such that the numerical value of the area is the same as the numerical value of the perimeter.

(Please - don't copy everything you see from that other thread. No one should use the symbol "$\infty$".)

6 by 3 is the only none square one :D  :D  :D

Arts -
[hide]Correct  but A square is a rectangle too..and need not be ignored.. [/hide]

[hide]So, there is 6*3 (18) and 4*4 (16)
All I have.
[/hide]

Please provide (MATHCOUNTS-level) solutions, not just answers.  It will help everybody else, and you might get some feedback on your method.

If you used "Guess and check", there is no shame.  If you used Dr. Merryfield's neat [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=160004#p160004]factoring trick[/url], that's great, too.  But please help the rest of the community see how it was done.

[hide]We have sides x and y. The perimeter is 2(x+y) and the area is xy, so we have 2(x+y)=xy. If x=y, the easy solution, 4x=x^2, so x=y=4. Else if x=2y, (6,3). If x=3y, 8=3x, which has no integral solutions. If x=4y, 10=4x, ditto. I think that is all[/hide]

Solafide -- there are lots and lots of rectangles that don't fit into those categories you named, such as a 2x5 or 3x15.  How do you know they won't work?

I think the solutions that you all proved are the only soluitons, although it's based on interpretation whether you will list three or two combinations for your answer:
If you consider the 6*3 and 3*6 the same, there will be two, otherwise, your answer will be 6*3, 3*6, and 4*4...  :)


function
calculus
integration
real analysis
real analysis unsolved

Suppose $ f$ is a continuous function on the circle and that the Fourier series of $ f$ is absolutely convergent, $ \sum_{n\equal{}\minus{}\infty}^{\infty}|\hat{f}(n)|<\infty$. Prove that the Fourier series converges uniformly to $ f$, i.e. $ S_N(f)(\theta) \rightarrow f(\theta)$ uniformly in $ \theta$ as $ N\rightarrow \infty$; where $ S_N(f)(\theta)\equal{}\sum_{n\equal{}\minus{}N}^N \hat{f}(n)e^{in\theta}$.

There are multiple ways of proving that if $ f$ is integrable on the circle and $ \hat{f}(n)\equal{}0\ \forall n\in\mathbb{Z}$, then $ f\equal{}0$ at all points of continuity.

If $ \sum_{n\in\mathbb{Z}}|\hat{f}(n)|<\infty$, then $ g(\theta)\equal{}\sum_{n\in\mathbb{Z}}\hat{f}(n)e^{in\theta}$ is absolutely and uniformly convergent, and so you may interchange integrals with the sum.  As a result of integration, you obtain $ \hat{g}(n)\equal{}\hat{f}(n)\ \forall n\in\mathbb{Z}$.  So $ f\equal{}g$, as required. $ \Box$

why does interchanging integrals with sum gives us $ \hat{g}\equal{}\hat{f}$?

Since $ \left|\widehat{f}(n)e^{inx}\right|\le\left|\widehat{f}(n)\right|$ and $ \sum_{n\equal{}\minus{}\infty}^{\infty}\left|\widehat{f}(n)\right|<\infty,$ we have that the series $ \sum_{n\equal{}\minus{}\infty}^{\infty}\widehat{f}(n)e^{inx}$ converges absolutely and uniformly by the Weierstrass M-test.

This gives us that $ g(x)\equal{}\sum_{n\equal{}\minus{}\infty}^{\infty}\widehat{f}(n)e^{inx}$ is a continuous function on $ \mathbb{T}.$

$ \widehat{g}(n)\equal{}\frac1{2\pi}\int_0^{2\pi}\sum_{k\equal{}\minus{}\infty}^{\infty}\widehat{f}(k)e^{ikx}e^{\minus{}inx}\,dx.$

Since the series converges uniformly on a finite interval, we can justify interchanging the sum and the integral. (This is the standard Riemann integral theorem.) Computation after the interchange then gets us $ \widehat{g}(n)\equal{}\widehat{f}(n)$ for all $ n$ since
\[ \frac1{2\pi}\int_0^{2\pi}e^{ikx}e^{\minus{}inx}\,dx\equal{}\begin{cases}1&\text{if }n\equal{}k\\0&\text{if }n\ne k\end{cases}.\]

From there, the rest of boxedexe's comments should make sense.