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Hi, everyone,

Ever since Mathfanatic was thinking out loud about whether to go to Exeter (by the way, how are you, Mathfanatic?), I've been wondering about the prep school path to learning more mathematics. I recall that there are some other AoPSers in various prep schools. What do you think of the prep school experience? Would you recommend it to your best friend or your younger sibling? What are the advantages of prep school? Are there any disadvantages? How was the application process? What else should people who have never been to prep school know about prep school?

Curious minds would like to know more. Please let us know.

Could you please define "prep school?" I am still quite unsure about the differences in schools. For instance, there are magnet schools, but there are also publicly funded arts schools... are these magnet schools as well? Are there private, non-parochial, non-prep schools, or are all private non-parochials considered prep schools? Or is there a standard for prep schools, such as offering a certain number of AP or IB courses? Does one, perhaps, get college credit for taking certain classes at a prep school (that are not part of the AP/IB systems)? And what about homeschooling? It seems like there might be some blurriness between a large homeschool system and a small private school... is there some sort of home-school registery? Ok, I realize I am now going overboard with questions about schooling, but if you could, perhaps, answer my first question ("What is a prep school?") I would be most grateful.

How about this? If you think your school is a prep school, please tell us about it.

For what it's worth, I don't think the homeschooling I'm doing now is the same thing as what I am asking about.

Yeah, I know it's kind of cheap to double-post to bump up a thread, but I'm really curious about what the math-liking people here who go to prep schools think of them. One kind of school I have in mind is [url=http://www.exeter.edu]Phillips Exeter Academy[/url], which is conspicuous for the AMC scores of some of its students. I'm sure that there are AoPSers who go to other schools, who can perhaps relate why they didn't stay in the local public school, and perhaps why they didn't go to some other prep school they could have gone to.

Thanks to anyone who comes out to represent alma mater and tell us like it is.

My kids (a high school freshman in IB program and one in 3rd grade) are not in prep schools. We visited one of our local prep schools and also looked into Exeter a year ago when my older one was applying for high school. The local prep school is a very good one with mostly rich kids there (very few on need based scholarship). The enrollment is on the competitive base. Good prep schools do ask for PSAT or SSAT in addition to GPA, recommendation, writing assays, and tuition, etc. In our region, we have private schools, two prep schools, two IBs (students in the same enrollment process divided into two IB schools by east or west of a specific road for transportation purpose) and other subject-specific magnet schools, such as music and art. So the choices are plenty. My older one chose IB over the prep school since she did not felt that the prep school would be able to provide better education for her. The IB class size per year (around 110/yr) is about the same as the prep school. Her IB graduating class has12 national merit scholar finalists, which at least double the number of the prep school. Our school district, including private and prep schools (a very large school district, about the 10th largest one in the country) did not have any USAMO qualifier for 2003 and 2004. She is the only qualifier from here for 2005. Math wise, she is probably better than anyone from that prep school. However, that prep school had one finalist in the US bio Olympiad last year (top 20 students in the country are finalists), and one this year too. That is easy to imagine though, considering large number of students in that prep school are children of lawyers and doctors. Teachers in the prep school drive cars in the level of Toyota Corolla. But many students in the upper class drive BMW and Mercedes.

We looked into Exeter since Exeter made half of the US IMO team last year (Tony Zhang, Oleg Golberg, TianKai Liu, all earned gold medals). Those three were not initially from Exeter but were attracted there probably because of Zuming Feng. Exeter is a prep but a boarding school. Going there is more like to start your kids with college life at an age of 14. Maybe it is truly worth of going Exeter if he or she makes the IMO team. Even making the team, in the long run, I don’t know whether your children get more benefits or harm by starting college-like life that early. But we are still thinking about Exeter, probably for junior and senior years.

[quote="qvc122"]In our region, we have private schools, two prep schools, two IBs (students in the same enrollment process divided into two IB schools by east or west of a specific road for transportation purpose) and other subject-specific magnet schools, such as music and art. So the choices are plenty.[/quote]

Thanks for your reply. It sounds like you have a good experience base for making comparisons among schools. Here in Minnesota we have public school open enrollment, which means in principle that students could attend any public school that they can reach by their own transportation. (There are a few exceptions, but that is the general rule.) Alas, none of the Minnesota public schools have consistently strong math teams, although there are always multiple USAMO qualifiers here. Nor do any of privately operated schools here place a high emphasis on math preparation--no matter what they say to the contrary. I have been trying to develop more mathematical culture in our local homeschooling network, but my own limitations as a math coach mean that I am looking EVERYWHERE for other possibilities for my son's education. It seems that most of Minnesota's math stars have parental influence on their education from math-educated parents, and I am not that kind of parent.

[quote="qvc122"]We looked into Exeter since Exeter made half of the US IMO team last year (Tony Zhang, Oleg Golberg, TianKai Liu, all earned gold medals). Those three were not initially from Exeter but were attracted there probably because of Zuming Feng.[/quote]

You aptly described Exeter is being much like going to college early in the part of your post that I didn't quote. Exeter is intriguing because of the strong math team there. My son could not find classmates like that anywhere in our state, or at least not that many. This is a lot to think about during the next year.

Well I'd actually like to comment on the other path, of going to a public high school, the path i did, and why i wouldnt recommend it and wish i had been less lazy in gr8 (wow, i must really have been lazy cuz im still really lazy and can still call the lazy). Anyway, I never thought of private or prep schools, thinking they were just where the rich kids went to stay away from the rest of us. Big Mistake! Here I am, at the end of my junior year and guess what, I've learned absolutely nothing. I've spent the past 3 years of my sleeping through classes and wondering why is wasnt on drugs. I never really get along with the people here becasue we just don't think the same. Sure, I have friends and have had quite a bit of fun, but I still feel that I can't tell people about my interest in advanced math, except for my physics teacher, but he's special. There are several good teachers, but also many who hate intelligent people and are intimidated by us so the best you can do with them is mid 80s even when you deserve well into the 90s (e.g. in my freshman year I got an 87 in science, now I have a 100 in chem, 96 in bio and 97 in physics). The rare intelligent students are not encouraged to pursue advanced learning and all of our schools funds go to the football team.  I was the first student in my school's history to write the COMC and will the first to try for any Olympiad. I have no competition in the classes i bother to wake up for and even in the ones i never wake up in i beat 95% of the people in it. I don't think I would have enjoyed the homeschooling experience because I would miss socializing and gambling (it really is an addiction:)). I wish i had been exposed to this level of knowledge before i was a sophmore and even then i was not prepared to realize or accept the quality of the individuals that were out there, including the vast number of elementary school kids who could teach me a lot about math. If you have the choice, i'd recommend going to a private/prep/gifted school where you will be challenged and exposed to the competition and learning that a young mind needs, don't kill your or your child's love of learning.

Yes, I think it is very important, at least, to do no harm. I had some school experiences that almost turned me off to learning, and I wouldn't want anyone in the current generation to have that kind of school experience.

Some of the public school programs, such as IBs are very similar to local prep school if not stronger, but most of IBs are not as strong as Exeter. In fact, IBs are public fund college prep programs, which have quite vigorous selection standard. Exeter is almost like a private national prep school.

Our state has 13 USAMO qualifiers this year (and 3 MOPpers from our state this year), but our city only has one USAMO qualifier.

If A+ math is determined to prevail in math, maybe he should go to Exeter. In our situation, even my husband can coach my daughter; I still think she should go to Exeter if she is truly serious about any chance of making IMO team in the future, since Zuming Feng is so experienced in making national winners.

Tokenadult, I wish you and A+ math have a great time with Mathpath in Colorado.

i feel like i should post since i have gone to both public school and a prep school (exeter).  for 9th and 10th grade i went to a decent public school in colorado, but i didn't feel challenged academically or even encouraged.  my parents had to pay for me to take the amc, because my school wouldn't.  i felt like my interests and goals were fundamentally different than almost everyone in my school.

i went to canada/usa mathcamp after 9th grade (and after 10th), and that's when i started thinking about changing schools.  several students there were going to boarding schools and seemed to like them, and i saw a prep school experience as being something like a year-round mathcamp.

later that summer i received a letter from exeter that outlined some of the math opportunities there, and i immediately wanted
to apply.  and after one year there, i know it was the right decision.  i have learned so much, about math and otherwise, from going to exeter.  fall term i had mr. feng for calculus, and i have been in math club the whole year.  i've had the opportunity to go to HMMT and ARML (we were third!!), which i never would have been able to do in colorado.

but the prep school experience isn't for everyone, and i don't think i would have been ready to leave home as a freshman.  but i would definitely recommend applying to exeter to anyone who has a serious interest in academics (not just math) and feels capable of living on their own.  if IMO is the goal, exeter's definitely a great place to be.  i know i have improved a great deal.  but exeter academics are very demanding and that can make it difficult to dedicate as much time as one might want to math.  exeter wants its students to be somewhat well-rounded, which is important to remember.  but the opportunities for advancement in math are completely worth it, and there is definitely something to be said for being well-rounded.

i'd be happy to try to answer other questions about the pros/cons of public/prep schools, admissions process, etc.  it was a big decision in my life and there's quite a bit more i could say about it.

[quote="dancer87"]i feel like i should post since i have gone to both public school and a prep school (exeter).  for 9th and 10th grade i went to a decent public school in colorado, but i didn't feel challenged academically or even encouraged.[/quote]

Thanks for replying. I'm especially interested in your experiences because I know Colorado has an unusually large number of schools that purport to be academically advanced for highly gifted students (most private, some charter schools). And I liked hearing about how you learned about Exeter--by meeting classmates at summer programs. After all, the main issue is the overall experience at the school, and most of what I hear about Exeter so far sounds quite positive.

Thanks for the previous responses. Does anyone else have something to say about this issue? Current students?

[quote="chess64"]..... what's a prep school?  :?  :D Is it like a private school?[/quote]

I'd be happy to hear definitions from other people. The word "prep" comes from "preparatory," and comes from the idea that some schools intentionally prepared students for college, while other schools were designed to be last stops before getting a job with only a high school diploma. I don't know if all prep schools are privately operated, or if some are government-operated ("public").

Like Chigr, I've never gone to prep/private school, but I can give a rough description of my experience (which was much better than Chigr's) in public school.  My school has a math culture, but it was definitely not as strong a culture as you would find even at other public schools.  The last year or two, it was a bit frustrating that the math culture here didn't match my desire to improve, but I was nowhere near the situation Chigr mentioned in which he was unable to share his mathematical interest.  People here definitely have an appreciation for math and mathematical talent, it's just that there are not as many people as I would like who really want to develop that talent to its full potential.  My biggest regret is that I didn't step up and take a leadership role early on and push to develop that mathematical culture until the last year or so.  As a result, when I was ready to become more of a leader, there wasn't time to do nearly as much as I would have liked.  Fortunately, it doesn't seem like you'll have any problem with lack of ability/willingness to try to start a math culture.

If you want to take advantage of open enrollment here, some schools you might want to start with are Eden Prairie, St. Paul Academy, Wayzata, and maybe Mounds View.  Those four have had the most consistently strong math teams in the state.  I would also recommend looking into UMTYMP.  I've never actually been in the program (I took the qualifying test once and managed to fail it), but from what I've heard, it seems to be a very good program (if A+MATH doesn't mind a lot of work).  I think there tends to be a good chunk of the ARML team each year that has taken classes there.

about schools in colorado....yes, there are a lot charter schools and other schools that are not quite open enrollment.  however, perhaps because of colorado's location, there doesn't seem to be a huge interest in extracurricular math.  in general though, i think colorado has a very good school system, but my school was not one of the better ones in the state.  going to exeter has definitely opened up so many more opportunities than i would have had at my old school, and i feel like my college options are also more open.

Um... call me stupid, but where [i]is[/i] Exeter? It's a prep school, right? Is it like a math/science oriented school?

Exeter is a private school in New Hampshire. It's one of the most popular for students strong in math and sciences. Its rival is Andover. Exeter (I think) has the largest library of all high schools around the country, but Andover has one of the biggest stuffed animal collections. You pick which is better.

Also, Exeter gets a huge endowment each year, one substantially bigger than almost all its sister schools.

[quote="yif man12"]Exeter is a private school in New Hampshire. It's one of the most popular for students strong in math and sciences. [/quote]

I'll note that this popularity among the math set is a relatively new phenomenon, and I think it almost entirely due to Zuming Feng's fantastic efforts there in attracting and educating top students.  (For those of you who don't know who Zuming is, he's the leader of the US IMO team.  He's also doing much of the work to build the tests for [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]WOOT[/url].)

As a rising junior at [url=http://www.choate.edu]Choate Rosemary Hall[/url] in Connecticut, I'll add my two cents.

[quote="tokenadult"]What do you think of the prep school experience?[/quote]
It's changed my life. I started Choate interested in math and not much else--you could have fairly characterized me as one-dimensional. But I started taking advantage of opportunities around campus and found that I could maintain my interest in math in addition to developing new interests.

Freshman year I was a day student, which means that I commuted to campus each day. I live about 45 minutes away from Choate, and I was lucky enough to be in a carpool which made life easier for my parents. I joined the math team early on and enjoyed it a lot, but competition math was still my sole interest.

Sophomore year I became a boarder and my life began to change. I joined several new clubs, played interscholastic squash, and was elected to the Student Council. The most valuable change that happened to me, however, was that I became a more relaxed and outgoing person. I'd been quiet and shy for much of my life, but living in a dorm helped me learn how important being social is to being happy. Prep school students who spend their time solely pursuing academics are missing the greatest opportunity that their schools offer--the opportunity to grow as a person, not only as an academic machine.

[quote="tokenadult"]Would you recommend it to your best friend or your younger sibling?[/quote]
I would recommend Choate to almost anyone. The only exception I could think of is if a prospective student has only one interest and wants to spend all of his or her time on it and pursue nothing else. This is simply impossible at Choate (or any prep school) because the workload for all classes is too much to entirely focus on only one thing.

[quote="tokenadult"]What are the advantages of prep school?[/quote]

Opportunities! Students at Choate can sign up for over thirty different non-academic clubs, from a cappella singing to the Spanish club to the debate team. In one day, a student can attend five or six interesting classes, practice with their sports team, have dinner, and go to a club meeting--all before 7:00! This is something that's hard to find anywhere else.

[quote="tokenadult"]Are there any disadvantages?[/quote]
At Choate, the workload is so intense that I haven't been able to spend as much time preparing for math competitions as I would have liked to. However, some of this was self-imposed--I chose to take six courses this year instead of the standard five, so I had more work than was necessary--and there have also been opportunities for me to weave my interest in competiton math into my life at Choate. I'm an active member of the math team (which recently took 1st place in the Connecticut championships for the [i]n[/i]th year in a row), and for my final project in BC Calculus I studied Jensen's Inequality. I'm also trying to set up a Directed Study for next year in which I would take the WOOT class on Art of Problem Solving and, under the guidance of the coach of our math team, solve as many problems as possible and develop my proof-writing skills. Overall, it's not easy to continue a thorough study of competition math, but with dedication it is possible.

[quote="tokenadult"]How was the application process?[/quote]
It wasn't particularly grueling. Students who want to apply to Choate take a student-guided tour of campus and have an interview with a member of the Admissions department. The application was pretty standard...if I remember correctly, it asked for teacher recommendations, a grade transcript, awards and extracurricular activities, and a few essays.

I encourage you to visit [url=http://www.choate.edu]Choate's newly-designed website[/url] for more detailed information, and I'd also be happy to answer any other questions you may have. I wish you and your son the best of luck in whatever path you decide to take.

Thank you to JSRosen3 for the detailed description of Choate.

Could you attend Exeter if you lived in, say, Delaware? :? I don't think I want to do prep school, though... it's like boarding school...

If you lived in Deleware, you'd definitely have to board there. Most of the prep schools around here (New England) have programs for day students, who commute each day to and from school. In most of them, boarders make up about 60 to 80 percent of the students.

I don't think admissions are very difficult at all. It's just standard application materials. Teacher reccomendations, grade transcripts, some information about yourself, what you like to do and what your hobbies are, what extracurriculars you do, and one or two essays. I believe that the belief that it's extraordinarily difficult to get in is false. True, you have to have pretty good grades and have a few distinctions, but you don't have to have been getting straight As since 4th grade. I got accepted to the schools I applied to despite being an A-/B+ student for pretty much all of 7th grade and even after getting a B- in English in my first term of 8th grade. My advice would be to apply to a couple schools instead of just one. Get to know the schools. You never know how much you might like one school. I first wanted to go to Exeter like a lot of other people but after visiting, Choate, I found that I liked it a lot. So when I got accepted to both, I ended up choosing Choate.

[quote="qvc122"]Exeter does have need based scholarship, then you have to be outstanding for them to accept you and fund you.[/quote]

The way Exeter puts it is that admissions is more competitive for applicants who request financial aid. That is because Exeter has not yet fully endowed its financial aid fund, so it thinks it can't afford to make financial aid offers to all the needy students who apply. On the other hand, there are definitely quite poor students who attend Exeter on full financial aid.

Off this subject: have any of you ever seen the reference book about independent schools called the "Blue Book"? I saw it today in a public library, and I was astonished to notice that it made no mention of Exeter whatever.

[quote="Chigr"]The rare intelligent students are not encouraged to pursue advanced learning and all of our schools funds go to the football team.[/quote]

:/ that must suck especially if you enjoy math. not all high schools are like that though... tons of schools in California especially the Bay Area are really great about academic activities.

[quote="rrusczyk"][Exeter's] popularity among the math set is a relatively new phenomenon, and I think it almost entirely due to Zuming Feng's fantastic efforts there in attracting and educating top students.[/quote]

It is new that people go there specifically for IMO-level preparation.  But before Z.F., there was certainly no shortage of elite (both in ability level and financial resources) math/science types who went there for at least the last 1-3 years of school as phase 1 of a sequence leading through high school competitions, Harvard/Princeton/MIT/etc, the Putnam, and a PhD at similar university.  There was another teacher there at the time who was an active organizer of math competitions, co-authored some books.  I don't remember the name.

It could also be that Exeter hiring Feng was an effect as well as a cause of this concentration of talent --- the school realizing it had high-level young mathematicians, and keeping an eye out for somebody who could meet their needs.

[quote="qvc122"]For Exeter upper class (11th 12th grades), you need SAT or PSAT as part of the application requirements. For lower class (9th and 10th), you need SSAT. Your parents also need to pay tuition and fees which is somewhere around $35,000 per year. Exeter does have need based scholarship, then you have to be outstanding for them to accept you and fund you.[/quote]

Unless a student is specifically on the IMO track, getting the full value of the 35K(arat) would require [i]not[/i] excelling academically, as it takes too much time from networking.  The future Bush II and Kerry types (scholastically mediocre, socially enthusiastic, ambitious, alcohol-tolerant) are the ones who get their money's worth from the fancy private schools.

[quote="tokenadult"]Off this subject: have any of you ever seen the reference book about independent schools called the "Blue Book"? I saw it today in a public library, and I was astonished to notice that it made no mention of Exeter whatever.[/quote]
We have a book sent by TIP program (I could not find that book at this moment), which listed most of the private schools and boarding schools. Exeter is among them. Exeter was also mentioned in one of the bestseller books (Brining Down the House), written by Ben Mezrich. Exeter caught my attention from last year when I visited AMC site and found many USAMO qualifiers from there.
My understanding is some of the IMO members from Exeter paid full tuition and fees to Exeter (if their parents were capable of affording the cost). However, if a student is at that caliber and wants to attend Exeter and has a financial need, Exeter will support that student.

Regarding Bush and Kerry, I always thought that Kerry was a scholar. However, not long time ago, NPR news said that his first year transcript at Yale showed 4 Ds, worse than Bush's. I was shocked.

[quote="qvc122"]
Unless a student is specifically on the IMO track, getting the full value of the 35K(arat) would require [i]not[/i] excelling academically, as it takes too much time from networking.  The future Bush II and Kerry types (scholastically mediocre, socially enthusiastic, ambitious, alcohol-tolerant) are the ones who get their money's worth from the fancy private schools.[/quote]

i'll be a senior at exeter this coming year.  i've have not met anyone who went for the sake of networking.  (read dancer87's post.)

most people go there for a challenging academic environment.  we learn a lot from each other, especially since we use the harkness method.

from the exeter website:
[i]"Since the arrival of 'Harkness tables' on campus, the principal mode of instruction at Exeter has been discussion around an oval table. The Harkness table is central to both the Exeter classroom and the Exeter curriculum. Though teaching and learning look different in different disciplines and at different levels of study, all Exeter teachers and students are committed to an ideal of active, participatory, student-centered learning which values teaching students not just a given course's content but the skills required to become their own and each others' teachers. As the physical table itself implies, learning at Exeter is a cooperative enterprise in which the students and teacher work together as partners.

The academic life of the school depends upon the quality of each student's preparation for active participation in learning. As one of the schools legendary teachers said, "What happens in class depends upon what the students have done before the class begins." A place at the Harkness table requires students to exercise a high degree of self-discipline, and to engage eagerly and energetically with both peers and instructors. Exeter is a school for students who take pleasure in this distinctive mode of teaching and learning in which each member of each class is in some measure a teacher of all the others."[/i]

i think that going to exeter for math is great, even if you're not at the olympiad level.  if i went to public school, i would have taken AB calc senior year.  but there are accelerated courses and more options.  and the department is pretty nice about letting you skip (not much public school bureaucratic nonsense).  so i got to take BC calc junior year.  we also have multivariable calculus and other courses (analysis, linear algebra, non-euclidean geometry this year... in the past they have offered combinatorics, topology, etc).
for the most part, my teachers have been outstanding.  maybe mr. parris was the teacher fleeing_guest was wondering about?  he is incredibly smart, as well as a very nice guy.  and in my opinion, mr feng cares about more than the success of future IMO teams.  he makes math club great with his elegant solutions and ridiculous jokes.  i hadn't even heard about the AMC competitions or anything before exeter, so mr feng was first introduced to me as "the guy who does math club."  last year, there was a group of people who met every day to do olympiad problems.  any math clubber who was willing to put in the time and effort was welcome... even people who didn't make usamo went.

and exeter also has many other fantastic departments!  we have so many talented musicians, and i have had a great time in the sciences.  i used to hate english; now i love it!

there are downsides, too.  6 ish hours of homework a night (?).  AP season was wiiiiild trying to prep on top of that.  it can be hard to keep up with extracurriculars.  but as long as you stay committed to everything you do, i think it's worth it.  you learn how to think, how to study... you can feel yourself improving.  except for selected high-stress times during junior year when everyone is a sleep-deprived zombie, it is a great environment.

i'd be happy to answer any questions!
email: slcampbell@exeter.edu
IM: saracampbellsoup

Sara, thanks for your comments.

What do you think of Exeter's library? When I land on its online catalog, it doesn't seem to have many books on the subjects I search for. It is said to be, credibly, the world's largest high school library, but we have access to a very large research university library here and I'm still trying to figure out if going to Exeter means the improved classroom environment (e.g., smart classmates) and flexible, rigorous curriculum will make up for much more limited library resources.

BTW, is mathfanatic just too busy these days to post much? I'm especially interested in his reaction to Exeter after his first year there, since he was thinking out loud about it on AoPS a year ago. I also appreciate hearing from AoPSers who toured Exeter but enrolled elsewhere--what was the difference between your school and Big Red that prompted you to enroll where you enrolled?

people don't seem to have a problem with the library... i don't feel limited.  what subjects were you searching for?
mathfanatic had a very impressive first year; perhaps he is still catching up on sleep.
my parents wouldn't let me be a boarding student, so i only applied to exeter.  i was kind of torn between staying with my friends at public school and going to exeter, but only until i went to the preview day and saw everything in action.  i didn't know that there was a place where so many people could genuinely care about so many things.

Just to keep this a little less prep-school biased, there ARE some very, [i]very[/i] good public schools that provide many, if not all, of the resources the aforementioned prep schools provide, for nothing more than the matriculation fees. for those of you out there whose families can't afford you 12 years of college education (4 years of prep, 4 years of undergrad, and 4 years of grad school), you should REALLY look into public magnet schools in your state. some states, like my own, are blessed with superb public magnets (although im not really sure one would call our school a "magnet"), some of which even offer boarding (that, of course, costs extra money, but still nowhere near 35k). if i'm not mistaken, two of our favorite admins are graduates of public schooling, and look what became of them. you make your own opportunities.

[quote="saracampbellsoup"][quote="fleeting_guest"]Unless a student is specifically on the IMO track, getting the full value of the 35K(arat) would require [i]not[/i] excelling academically, as it takes too much time from networking.[/quote]

i've have not met anyone who went for the sake of networking.  (read dancer87's post.)
most people go there for a challenging academic environment.[/quote]

Networking does not imply carrying around business cards, or a specific intention to form potentially exploitable relationships.   Most people don't have the purpose or even the concept of "networking" during their schooling or participation in activities such as summer camps and math competitions.  Nevertheless, years later that is one of the principal outcomes, and in many (economic and non-economic) ways the most valuable one.

For any given academic or intellectual goal of a prep school (including IMO preparation, though not including access to specific IMO coaches such as Mr. Feng), there is usually a surrogate or even a superior means of attaining it for far less than the cost of attendance.  What is actually purchased is rarefaction of one's peer group (which to some extent, and certainly in the hopes of some parents, will set the peer group for the student later in life).  To the extent that one concentrates on the books and doesn't make peer connections, one doesn't reap anywhere near the same advantages.  The intellectual benefits are easier to substitute than the social ones.

 Please note that there is no implied opinion of whether the attendance is "worth it" at 35K or any other price, and no evaluation for or against networking.  I am pointing out only that, whatever the value may be, the bulk of it ultimately lies in the peer relationships, and that people who operate in ways that develop the relationships are the ones getting their money's worth.   It may be counterintuitive, but studying to get top grades may actually be against your interests at (for example) Exeter.

[quote="fleeting_guest"]  It may be counterintuitive, but studying to get top grades may actually be against your interests at (for example) Exeter.[/quote]

Depends on who you are studying with ;)

As for Zuming as a cause or effect, check out the record of Exeter before and after Zuming (not to mention the degree to which they recruit top math students before and after).  Indeed, they had strong students before Zuming, but their track record post-Zuming is much more impressive, and not only at the very top - their depth is leading them to be competitive at Harvard-MIT and ARML, as well.

[quote="rrusczyk"] As for Zuming as a cause or effect, check out the record of Exeter before and after Zuming (not to mention the degree to which they recruit top math students before and after).  Indeed, they had strong students before Zuming, but their track record post-Zuming is much more impressive, and not only at the very top - their depth is leading them to be competitive at Harvard-MIT and ARML, as well.[/quote]

Yes, apparently he had a big effect.  The speculation about the cause (of his hiring) was that Exeter may have found itself in a position similar to that of, say, Harvard (at undergraduate level) some years ago, with a large pool of talented students who were underserved by the existing options, which at that time consisted of advanced courses and participation in the standard competitions.  Other than Stuyvesant, I can't think of any other US schools that would be such a good match for an IMO expert.

As you mention, the talent pool grows through recruitment when people notice the AMC results.  This was also what happened for Harvard with the Putnam competitions, but Exeter is in an even better position in this respect as it has (as far as I know) no comparable national-level rivals analogous to MIT, Princeton, etc who might balance the scales a bit over time.

There are several other schools which get a similar level of talent as Exeter.  To name a few: Thomas Jefferson in VA (2 IMO team members this year, more USAMO participants than any other school for years), AAST in NJ, IMSA in IL, Montgomery-Blair in MD.  Even Saratoga HS, which seems to do very little with the talent it brings in, arguably has the same caliber of students coming into the school (3 students there were top 5 in National MATHCOUNTS - not even Exeter can say that right now).  Several of the other schools in the Bay Area also have outstanding talent, such as Gunn and Lynbrook (the latter benefits from receiving students from one of the top middle school programs in the country).

Those schools draw on a county or state population.  Exeter is, as far as I know, the only with both a national reach and the concentration of mathematical talent.  They can also attract some of the talent once it appears at other schools, either by reputation or by recruitment (offering scholarships).  If you live in Colorado, you can't go to TJHS, but you can attend Exeter; and if you happen to also have an IMO gold medal, they may hook you up with a scholarship.

With that said, I think it's easier to make the IMO from a regular public high school, as one has more free time.

I'm just noting that Stuyvesant (which also does not have a national reach) is not the only other great school for olympiad math students.  Indeed, Exeter has a big advantage in being able to recruit nationally, but these other schools are free, and many parents are not ready to send their kids off to a boarding school.  It's not so clear that Exeter has a natural advantage over TJ.

As for regular public school being your best bet, I would have agreed, but the past few years' teams suggest otherwise.  This year: 4 from magnet schools, 1 from Exeter, 1 from a regular public school.  Though I do agree in general that the highly self-motivated student is probably better off in the long run being at a school that doesn't make so many time demands.  One problem these days is that even the 'regular schools' have absurd AP/IB requirements, so these students have the same time pressure as those at a TJ or Exeter, without the excellent peer group or instruction to help them achieve their goals.

Stuyvesant and maybe a couple of the public magnet schools have some of the peer-group-rarefaction (or the potential for it), as well, though in a more haphazard way than at a place like Exeter, and with different (or more particular) demographics.  And you are right that Stuyvesant is not the only public school with some math culture.

About the number of IMO students from magnet vs other type of schools, it makes sense, as most people do not have definite enough IMO ambitions (or even know what IMO is) at the age of high-school selection to actually opt for an easier regular school; but anyone able to make IMO would at that age be a likely candidate for the local magnet school, if one exists.  I suppose that families of IMO candidates are also more liable to move to communities that have special schools, or good private schools.

Exeter is a prep high school. It does not have middle school portion. So they do not participate in Mathcounts.


If $a,b$ be natural numbers such that $ab|(a^2+b^2+1)$ then
prove that $\frac{a^2+b^2+1}{ab}=3$

http://www.mathlinks.ro/Forum/viewtopic.php?t=71760
http://www.mathlinks.ro/Forum/viewtopic.php?t=40207

Yimin

And one more http://www.mathlinks.ro/Forum/viewtopic.php?t=91590 :D


I think someone has asked this but I cannot seem to find it.
There seems to be no more ... option...

In class, Mr. Vornicu told us that only mods and admins have it now :(

Well I know some people put really really big annoying pictures on...

Or animated pics...

[quote="Valentin Vornicu"]Some people were abusing the privelages[/quote]

So sad but so true

So we probably (regular users) won't be able to have this feature back?

[quote="snee"]Hmm, I hope it's not my fault.  :P
Animated avatars... (Never big though) and text in front of my name. :P
If so, sorry![/quote]
Nah probably not. It was probably a lot of different people.
But don't the avatars have a size limit? Then how can it be *too* big?

Can there at least be an option to have no avatar? I accidentally chose a default avatar and now I cant change it back to no avatar because it won't let me. (I despise those default avatars...)

[quote="AIME15"]They don't have a size limit.  I forget who, but someone had a huge aops logo before...at least it was after class.[/quote]
I think they should. Like 80 by 80 or so.

buzzer: you can. (1) log out and log back in. (2) open the avatar panel again and click on the default avatars again. Then hit OK.

ducky: It's actually 80x100

Well I think it would be helpful if you didn't have to log out completely to change your avatar to a blank one...

[quote="coolts"]we should all have a huge voting rebellion to have the avatars back...[/quote]We would just punish the starter/leader of the rebellion :)

But can we use our AoPS avatar in the classroom? I understand people were abusing the privilege of using URLs but how about our forum avatar?

[quote="ernie"][quote="mathemonster"]admins are evil and must be stopped... except when teach Intermediate NT[/quote]

That might get you warned or even banned. :)[/quote]He was making a subtle joke ..

Anyway we've had enough discussion on this topic, it ends here.


quick, tell me how to solve this problem:
Let a, b, and c be real numbers such that a-7b+8c=4 and 8a+4b-c=7. Then a^2-b^2+c^2 is what?

I know it's a problem involving factoring techinques. However, when I did the problem, I was unable to find a way to combine them nicely into factorable terms. I know the answer, but I don't know how to make such insight in combining equations in such ways in order to solve the problem. Can anyone please enlighten me?

This is how I did it...

You see that there are 3 unknowns and 2 equations so you can plug in anything for the variable that suits you.

So I plugged in 0 = b

a + 8c = 4

8a - c = 7

Here solve for a and c and find the sum of their squares

c = 8a - 7 = (4 - a )/8 ...  64a - 56 = 4 - a so a = 60/65 = 12/13

c = 8*12/13  - 7 = 5/13

So a :^2:  + c :^2: = 169/169 = 1

Thats how I first did it, and thats probably all you need for that type of competition (I think that ones multichoice).

However, heres a nicer solution, and one that would also be enough for a proof.

[hide]

Basically I just tried a few things, adding the given equations, etc. I also wanted to get things with a-b and a+b in it. In any case, I just happened to notice that if you double the first equation and add the second, you get 10a - 10b + 15c = 15, that looks very nice so we simplify it to 2(a-b) = -3(c-1).

Now double the second one and subtract the first. Then 15a + 15b - 10c = 10, so 3(a+b) = 2(c+1). Multiply our two equations together, and 6(a^2 - b^2) = -6(c^2 - 1), so a^2 - b^2 + c^2 = 1.

[/hide]

Thanks for the helps!! Now I understand...


A point $ (x,y)$ is randomly picked from inside the rectangle with vertices
 $ (0,0)$, $ (3,0)$, $ (3,2)$, and $ (0,2)$.  What is the probability that
 $ x < y$?

[quote="GameBot"]A point $ (x,y)$ is randomly picked from inside the rectangle with vertices
 $ (0,0)$, $ (3,0)$, $ (3,2)$, and $ (0,2)$.  What is the probability that
 $ x \& lt; y$?[/quote]

Draw the line $ y = x$.  You want to compute the area less than that line in the rectangle with points $ (0,0) (3,0), (3,2), and (0,2)$.  This area is $ 4$, if you use the formula.  The total area is $ 6$, so I think the probability is $ \boxed{\frac{2}{3}}$.

Someone check my work please  :wink:

Geometric probability.
[asy]size(250);
import graph;
xaxis("$x$",-0.5,6,Arrows);
yaxis("$y$",-0.5,5,Arrows);
dot("$(0,0)$",(0,0),SE,black);
dot("$(3,0)$",(3,0),S,black);
dot("$(0,2)$",(0,2),W,black);
dot("$(3,2)$",(3,2),NE,black);
draw((0,0)--(3,0)--(3,2)--(0,2)--cycle);
draw("$x<y$",(-0.5,-0.5)--(2.5,2.5),NW,dashed,Arrows);[/asy]
The small triangle is the set of points that satisfy. Its area is $ 2\times2\equal{}4$ and the area of the whole rectangle is $ 2\times3\equal{}6,$ so the probability is $ \frac{4}{6}\equal{}\boxed{\frac{2}{3}}.$

Your work is correct :)

Actually, the answer should be $ \boxed{\frac{1}{3}}$ since the area of the triangle is 1/2bh, or 2.


geometry
geometric transformation
reflection
inequalities
number theory
perpendicular bisector
Diophantine equation

Is it the right section or not? ;)

The convex quadrilateral $ABCD$ is inscribed in the circle with diameter $[AD]$.
Assume that $AB=BC=a,CD=b$ and $AD=d$ where $a,b,d$ are integers and $a \neq b$.
Determine the least possible value for $d$.

Pierre.

[color=red][Author edit: Corrected a calculation mistake.][/color]

Thanks, Pierre, I will remember this problem as one of the first nontrivial number theory problems I solved...

The least possible value for d is d = 8. Hereby, a = 2 and b = 7.

Indeed we begin with some geometry. Since the points A, B, C, D are supposed to lie on the circle with diameter AD, we have OA = OB = OC = OD, where O is the midpoint of the segment AD.

Now, from OA = OC and AB = BC, it follows that the two points O and B lie on the perpendicular bisector of the segment AC. In other words, the line OB is perpendicular to the line AC and passes through the midpoint M of the segment AC.

Now, since the point O is the midpoint of the segment AD, we have $\displaystyle OD = \frac{AD}{2}$. Since OA = OB = OC = OD, this entails $\displaystyle OA = OB = OC = OD = \frac{AD}{2}$.

On the other hand, the point M is the midpoint of the segment AC, and thus $\displaystyle AM = \frac{AC}{2}$.

Finally, $\displaystyle MO = \frac{CD}{2}$, since the points M and O are the midpoints of the sides AC and AD of the triangle ACD. Hence, $\displaystyle BM = OB - MO = \frac{AD}{2} - \frac{CD}{2} = \frac{AD-CD}{2}$.

The Pythagoras theorem, applied to the right-angled triangle AMB, yields

$\displaystyle AB^2 = BM^2 + AM^2 = \left( \frac{AD-CD}{2} \right)^2 + \left( \frac{AC}{2} \right)^2 = \frac{\left( AD-CD \right)^2 + AC^2}{4}$
$\displaystyle = \frac{AD^2 + CD^2 - 2 \cdot AD \cdot CD + AC^2}{4} = \frac{AD^2 + \left( CD^2 + AC^2 \right) - 2 \cdot AD \cdot CD}{4}$.

Now, by the Pythagoras theorem in the right-angled triangle ACD (this triangle is right-angled since < ACD = 90, what is because the point C lies on the circle with diameter AD), we have $CD^2 + AC^2 = AD^2$, and thus

$\displaystyle AB^2 = \frac{AD^2 + AD^2 - 2 \cdot AD \cdot CD}{4} = \frac{AD^2 - AD \cdot CD}{2} = \frac{AD \cdot \left( AD-CD \right) }{2}$.

Thus, $2 \cdot AB^2 = AD \cdot \left( AD-CD\right)$. In other words, $2a^2 = d \left( d-b \right)$. Hence we have proven:

[b]Lemma 1.[/b] In the configuration of your problem, the equality $2a^2 = d \left( d-b \right)$ holds.

Now we will show:

[b]Lemma 2.[/b] Conversely, if a, b, d are any three positive real numbers satisfying the equation $2a^2 = d \left( d-b \right)$, then there exists a convex quadrilateral ABCD such that this quadrilateral is inscribed in the circle with diameter AD, and the equations AB = BC = a, CD = b and AD = d hold.

[i]Proof.[/i] Let AD be an arbitrary segment in the plane such that the length of this segment is AD = d. Construct the circle with diameter AD. Now, since $2a^2 = d \left( d-b \right) = d^2 - db < d^2$, we have $\displaystyle a < \frac{\sqrt2}{2} d$, and thus a < d. Hence, the circle with diameter AD (of length d) must have two common points with the circle with center A and radius a. Let B be one of these two common points. Then this point B lies on the circle with diameter AD, and AB = a. Let O be the midpoint of the segment AD, i. e. the center of the circle with diameter AD. If we reflect the point A in the line OB, we get a new point C which also lies on the circle with diameter AD (just because the circle with diameter AD is symmetric with respect to the line OB, since this line passes through the center of that circle). Moreover, from the reflection, we have AB = BC. Combined with AB = a, this gives AB = BC = a.

Hence we have obtained a quadrilateral ABCD inscribed in the circle with diameter AD, and satisfying the equations AB = BC = a and AD = d.

Now, we can use our inequality $\displaystyle a < \frac{\sqrt2}{2} d$ to show that the quadrilateral ABCD is convex: In fact, if we consider the arc ABD on the circle with diameter AD, and if K is the midpoint of this arc, then the triangle AOK is right-angled and isosceles, so that $\displaystyle AK = \sqrt2 \cdot OA = \sqrt2 \cdot \frac{AD}{2} = \sqrt2 \cdot \frac{d}{2} = \frac{\sqrt2}{2} d$, so that $\displaystyle a < \frac{\sqrt2}{2} d$ implies a < AK, so that AB < AK. Thus, arc AB < arc AK. But since the point C is the reflection of the point A in the line OB, we have arc AC = 2 arc AB, so that arc AC = 2 arc AB < 2 arc AK = 2 $\cdot$ 90 = 180 = arc AD. Hence, the quadrilateral ABCD is convex.

Finally, we have to show that CD = b. In fact, from Theorem 1, applied to the quadrilateral ABCD, we obtain $2a^2 = d \left( d-CD \right)$. On the other hand, we know that $2a^2 = d \left( d-b \right)$. This immediately yields CD = b. Hence Lemma 2 is proven.

Now, Lemma 1 and Lemma 2 show that for four positive numbers a, b, c, d, the existence of a convex quadrilateral ABCD such that this quadrilateral is inscribed in the circle with diameter AD, and the equations AB = BC = a, CD = b and AD = d hold, is equivalent to the equation $2a^2 = d \left( d-b \right)$. Therefore, your problem reduces to the question of finding the least positive integer d such that there exist two positive integers a and b with $a \neq b$ and satisfying this equation.

Well, the equation $2a^2 = d \left( d-b \right)$ is a diophantine equation, and there are lots of methods for solving such ones. Actually, you should note that always $d \geq b$ (otherwise, d (d - b) would be negative, but the equation $2a^2 = d \left( d-b \right)$ implies that d (d - b) is twice a perfect square, and perfect squares can't be negative!). Hence, if d is a fixed natural number, then the only positive integers b coming in question for the equation $2a^2 = d \left( d-b \right)$ are the numbers b = 1, b = 2, ..., b = d. These are finitely many numbers and we can check them through.

By this kind of checking, we see that for every of the numbers d = 1, d = 2, d = 3, d = 4, d = 5, d = 6 and d = 7, there do not exist two positive integers a and b with $a \neq b$ and satisfying the diophantine equation $2a^2 = d \left( d-b \right)$. In the case d = 8, we find the solution a = 2, b = 7, d = 8. Hence, d = 8 is the required least positive value.

  Darij

Nice job Darij but, unfortunately, it seems that you made some error in your checking, because for $d=8$, we may choose $a=2$ and $b=7$...
Thus, the minimal value is in fact $d=8$.

Moreover, i think some computations can be done with less efforts. Recall that $AB=BC=a, CD=b$ and $DA=d$.
Thus, from Pythagore's theorem : $BD= \sqrt {d^2 - a^2}$ and $AC = \sqrt {d^2-b^2}$.
Since $ABCD$ is inscriptible, we may use Ptolemee's theorem, which leads to :
$(ab+ad)^2 = (d^2-b^2)(d^2-a^2)$ that is $2a^2 = d(d-b)$ as you noticed.

And if you want to avoid some boring cases to be checked, and to prove an extra result, we may prove that in that case $d$ cannot be prime. Indeed, first note that (1) implies that $d^2 = 2a^2 + bd > 2a^2$ so that $d> a \cdot \sqrt {2}$.
Now assumethat $d$ is a prime :
Clearly $d=2$ leads to $a=b=$ which contradicts that $a \neq b$. Thus, $d \geq 3$ and since $d$ divides $2a^2$, it follows that $d$ divides $a$.
But, in that case : $a \geq d > a \cdot \sqrt {2}$ which is absurd.

Pierre.

[quote="grobber"]I did almost the same thing, getting $(a,b,d)=(2,7,8)$ (I missed this case too at first :)).[/quote]

Oh, thanks Grobber, I have now corrected the bug.

Other solutions are a = 3, b = 7 and c = 9, or also a = 6, b = 1 and c = 9. It could be interesting to parametrize all solutions, but I am too tired now...

[quote="pbornsztein"]Nice job Darij but, unfortunately, it seems that you made some error in your checking, because for $d=8$, we may choose $a=2$ and $b=7$...[/quote]

Thanks to you too... anyway, I have corrected it now.

[quote="pbornsztein"]Moreover, i think some computations can be done with less efforts. Recall that $AB=BC=a, CD=b$ and $DA=d$.
Thus, from Pythagore's theorem : $BD= \sqrt {d^2 - a^2}$ and $AC = \sqrt {d^2-b^2}$.
Since $ABCD$ is inscriptible, we may use Ptolemee's theorem, which leads to :
$(ab+ad)^2 = (d^2-b^2)(d^2-a^2)$ that is $2a^2 = d(d-b)$ as you noticed.[/quote]

Indeed, this is shorter, but my proof doesn't use Ptolemy. I actually see from your post and from Grobber's post on another thread that I should learn solving problems with Ptolemy; I am currently not so good at that...

[quote="pbornsztein"]And if you want to avoid some boring cases to be checked, and to prove an extra result, we may prove that in that case $d$ cannot be prime.[/quote]

This is what I actually did! But then I didn't want to write it up...

  Darij


Given that both the roots of the quadratic equation $x^2+bx+c=0$ are greater than 1 . Show that $b+c+1>0$   :)

Since $a=1 >0$, so $f(x) >0 \Longleftrightarrow x \in (-\infty, x_1) \cup (x_2, \infty)$. Since the root of the function's roots are all greater than $1$, so $f(1)>0$. So $1+b+c>0$.

[quote="shyong"]Well done shobber , you got it right  :D . By the way , what program you use to plot the graph ?[/quote]
It is called the geometer's sketchpad. There is a topic in the Other Problem Solving Topic discussing it. I've just bumped it up.


Suppose that
$ f(x)\equal{}c_{0}\plus{}c_{1}x\plus{}c_{2}x^{2}...$
is an even function. Show that
$ f(x)\equal{}c_{1}x\plus{}c_{3}x^{3}\plus{}c_{5}x^{5}...$

[quote="AstroPhys"]Show what?  :maybe:[/quote]

Show that the first function is congruent to the second if f(x) is even.

Really? That second function sure looks odd to me.

... I guess I meant that in two ways.

If you meant even function and $ f(x) \equal{} c_{0}\plus{}c_{2}x^{2}\plus{}c_{4}x^{4}$, then this is easy by noticing that $ f$ is even iff $ c_{1}\equal{} 0$, so then just let $ c_{4}\equal{} 0$ and you're done.

If you meant odd function and what you have above, then
$ f(x) \equal{}\minus{}f(\minus{}x)$
$ c_{0}\plus{}c_{1}x\plus{}c_{2}x^{2}\equal{}\minus{}c_{0}\plus{}c_{1}x\minus{}c_{2}x^{2}$
$ c_{0}\plus{}c_{2}x^{2}\equal{} 0$
Thus $ f(x) \equal{} c_{1}x$, so let $ c_{3}\equal{} 0$, $ c_{5}\equal{} 0$, and you're done.

I'm very sure I meant even function, and yes, that is correct.

I'm afraid I don't follow.  That second function you have written is very clearly an odd function.


Is anyone familiar with a program called Scilab (which I'm told is really similiar to Matlab), to write a program to compute the exponential function e^x using only the +,-,*,/ operations (ie. no logs or exponents)?

 The program is supposed to have the same accuracy and limitations as the built-in routine in Scilab or equivalent, so i am told.

How does one in general write a program to solve a math question?

What about the function e^x? Can we approximate it using its Taylor series expansion? Does anyone know the accuracy or limitations that the built-in program of Scilab has?

I am not familiar with Scilab, but this link might help

http://en.wikipedia.org/wiki/Exponential_function#Computing_exp.28x.29_for_real_x

There should be something in the command list: http://www.scilab.org/product/man-eng/index.html


Prove the ineq: $ \sqrt[n]{n+ \sqrt[n]{n}}+ \sqrt[n]{n- \sqrt[n]{n}} \leq 2 \sqrt[n]{n}$ , $ \forall n \in N$, $ n \geq2$.

cheers! :D :D :cool:

The problem is trivial:
  Just use the fact that x^1/n+y^1/n<=2*((x+y)/2)^1/n (Jensen).

You are too hard with Moubi, Harazi... :D

Pierre.

Not at all (or at least I really didn't have any intention), but I apologise if this is the impression. Anyway, you must admit that the problem is very simple. Moubi has found a beautiful solution, so why should I be hard with him?


geometry
trapezoid
trigonometry
geometry unsolved

M is the midpoint of side AB of triangle ABC.The inscribed circle touches AB and AC in P and Q. The excircle that lies in different halfplane from A with respect to BC touches AB and AC in R and S.Prove that the perpendicular distance from M to PQ equals the perpendicular distance from M to RS

Are you sure about this problem??? I think this statement doesn't hold

In my opinion, $M$ is the midpoint of the side $BC\ !$

[hide="Here is the corrected enunciation and its solution."]

[color=blue]Let $ABC$ be a triangle. [u]Denote:[/u] the distance $\delta_{d}(X)$ of the point $X$ to the line $d$; the midpoint $M$ of the side $[BC]$; the points $P\in AB$, $Q\in AC$ where the incircle $C(I,r)$ touches the respective sides; the points $R\in AB$, $S\in AC$ where the exincircle $C(I_a,r_a)$ touches the respective sides. Prove that $\delta _{PQ} (M)=\delta_{RS} (M).$[/color]

[b]A solution.[/b] The quadrilateral $PQSR$ is isosceles trapezoid ($PR=QS=a$,

$PQ\parallel RS$, i.e. $\frac{PQ}{p-a}=\frac{RS}{p}=2\sin \frac A2$) and $PB=SC=p-b$.

So that, the point $M$ is the middlepoint of the segment $[PS]$, i.e.

 $\delta_{PQ}(M)=\delta_{RS}(M)\ \left(\ =\frac 12\cdot BC\cdot \cos \frac A2\ \right).$

[u]Remark.[/u] You can use and the following property:
$D\in BC\cap C(I,r)$, $E\in BC\cap C(I_a,r_a)$, $\{D,D'\}=ID\cap C(I,r)$, $\{E,E'\}=I_aE\cap C(I_a,r_a)$, $\Longrightarrow$ $A\in ED'\cap E'D.$ [/hide]

Let $D\equiv AI \cap \odot(ABC)$; it is the midpoint of $II_a$, hence it it is projected onto $AB$ and $AC$ at the midpoints $K$ and $L$ of $PR$ and $QS$ respectively. $PRSQ$ is an isosceles trapezoid, so $KL \parallel PQ \parallel RS$. See that $M$ is the projection of $D$ onto $BC$, therefore $K, M$ and $L$ are collinear (Simson line), done.

Best regards,
sunken rock


function
trigonometry
algebra unsolved
algebra

[quote="antinoise007"](tan2x)^2 + 2(tan2x)(tan3x) = 1

after arranging, I get

7T^3 - 35T^2 + 21T - 1 = 0  (T = (tanx)^2)

Now..... how?

In my opinion, using cardano method & arctan function is not elegant.

Help me, plz~[/quote]

the equation gives

$\tan 2x \frac{\tan 2x+\tan 3x}{1-\tan2x \tan 3x}= 1$

provided ,$1-\tan2x \tan 3x\neq 0$
which gives two cases

$\tan 2x = \cot 5x$
or
$\tan 2x = \cot 3x$

then i suppose u can proceed .....


[quote="Tiks"]Find a group of $4$ positive integer numbers $(a;b;c;d)$ that satisfies the following equation $a^{4}+b^{7}+c^{9}=d^{11}$.[/quote]
If $(a,b,c,d)$ is solution, $(at^{m/4},bt^{m/7},ct^{m/9},dt^{m/11})$ is solution too, were $m=4*7*9*11$.
We can find solution $a=3^{63},b=3^{36},c=3^{28},d=3^{23}$.
We can find solutions (3 typs), when all numbers are powers of 2. For example $a=2^{646},b=2^{369},c=2^{287},d=2^{235}$.


geometry
perimeter
modular arithmetic
combinatorics unsolved
combinatorics

Please, don't call problems like "scary problem", "hard problem" or something like that. Problem's title should describe its essence.

Read http://www.mathlinks.ro/viewtopic.php?p=140098

It depends on what your are calling 'triangles'.
Are $(a,a,b)$ and $(a,b,a)$ supposed to be distincts or not?

Pierre.

hey what's the matter? nobody can solve it or nobody tried?

helllllllllllllllllllllllp , u r wrong myth , scary is the best name we can give to this problem , come on guys , grobber ,orl, myth , bornzstein , einstein????

[quote="wisemaniac"]what is the number of triangles whose side lenghts are integers and whose perimeter is n???!!!!!!!![/quote]

so you seek the number of solutions to $a+b+c = n$ with $a \leq b \leq c$ and $a+b > c$.  these conditions are together equivalent to $ [(n+2)/2] \leq a+b \leq [2n/3] $ and $ a \leq b $.  in general, the number of solutions to $a+b = k$ with $ a \leq b $ is $[k/2]$; and the problem is basically solved from there...

Ok. Let's count the number of all triples $(a,b,c)$ with $a\leq b\leq c$, $a+b+c=n$, $c<a+b$.

Let's count the number of triple with $a+b+c=n$ and $a\leq b\leq c$.
There are $C_{n-1}^2$ solutions to the $a+b+c=n$. Each such solution we can rearrange to $a\leq b\leq c$. If $a\neq b\neq c$, then there are 6 rearrangements, but if two or three of these numbers are equal then the number of rearrangemts is less than 6.
The number of triples with two equal numbers is $\left[\frac{n-1}{2}\right]-[3\mid n]$. They were counted in $C_{n-1}^2$ thrice.
Therefore the number of solutions $a+b+c=n$ with $a\leq b\leq c$ is \[\frac{C_{n-1}^2-3\cdot \left[\frac{n-1}{2}\right]+3\cdot[3\mid n]}{6}+\frac{\left[\frac{n-1}{2}\right]-[3\mid n]}{3}+[3\mid n]=\frac{C_{n-1}^2-\left[\frac{n-1}{2}\right]+7\cdot [3\mid n]}{6}.\]
Now let's count the number of solutions of $a+b+c=n$ with $a\leq b\leq c$ and $c\geq a+b$.
Actually we need to find the number of triple with $n-2>c\geq \left[\frac{n+1}{2}\right]$. For fixed $c$ we have $\left[\frac{n-c}{2}\right]$ choices for pair $(a,b)$.

Thus final answer is $\frac{C_{n-1}^2-\left[\frac{n-1}{2}\right]+7\cdot [3\mid n]}{6}-\sum_{c=[\frac{n+1}{2}]}^{n-2} \left[\frac{n-c}{2}\right]$.

Of course, you can find closed form for it, but I am not sure it is nice looking (something depening on $n\pmod{4}$ :? )

a friend have just told told me that it's for sure exactlythe integer part of:
1/48(n²+3n+21+((-1)^n)*3n). what do u think?

the number of triples $(x,y,z)$ with all the entries positive is simply
$t_n=\binom{n-1}{2}$.
now among these we will have to discard the triples that don't form triangles;this simply has to throw away the triples with (exactly) one of $x,y,z$ being $\geq n/2$; suppose all $x,y,z<n/2$ but say $x+y\leq z$; then $x+y+z\leq 2z \Rightarrow 2z\geq n$ and that is ruled out.
counting the number of triples with say $x\geq n/2$ is the same as having the number of triples with $y+z\leq n/2$ and that is the same as the number of triples satisfying $x+y+z'=[n/2],x,y>0$ and this number is $\binom{[n/2]}{2}$.

i see that many others have already written.....

[quote="Myth"]You count some solutions twice. Right? How do you distinguish $b\leq c$ or $c\leq b$?
I am writing my own solution right now. But your approach is nice, though it needs some corrections.[/quote]

no... the conditions $[(n+2)/2] \leq a+b \leq [2n/3]$ and $a \leq b$ guarantee that $a \leq b \leq c$


A King gathers 100 men from his country in a room and says, "I will give 1,000,000 dollars to the Laziest Person here". He then asks them to raise their hands if they're the laziest person here. Everybody except for one person raise their hands. When the King asks him why he didnt raise his hand, the persons response makes the King give him the 1,000,000 dollars. What was the mans response?

I think this is a pretty cool problem, I think its well known though.

[quote="chess64"][quote="Hamster1800"]uhh... is it[hide="?"]"I'm too lazy to raise my hand."[/hide][/quote]

That's exactly what I thought.[/quote]
That's exactly what I was going to post.

hehe guess it was well known, i never heard it untill my dad told me  :blush:

[quote="Hamster1800"]actually i'd never heard it until you told it to us but the answer was really obvious. Unfortunately, the answer actually WOULDN"T work as raising your hand takes less energy than to speak that response. Therefore, he also must be too lazy to answer so another answer might be he says nothing?[/quote]

Hahaha...so I guess the King will get angry at him for not answering, chop his head off.  Then he'll keep the money.

Thank you Davendra for so graciously posting the solution, after it has already been posted, and confirmed for the (something like...) 5th time.  :)

I think the guy said "I'm too lazy to raise my hand!"


Given that $x^3+3x^2+3x+1=(x+3)^3$, find all real $x$ such that $x^3+3x^2+3x=1$

is there a way to factorize x^3+3x^2+3x=14? i can't bring it down to a quadratic equation.......

Looking at the $/LaTeX$ code, it says the thrid degree equation is equal to $(x+3)$ to the power of #

I don't know what that means...  :?

is # a variable or something that didn't appear?

i';m pretty sure he meant that \[\text{given  } x^3+3x^2+ 3x +1 = {(x+1)}^3\]

so, then its find all real x if (x+1)^3-1=14?

Well, that's a pretty nasty radical answer.

Sorry everybody, I fixed the typo. Pressing down the shift key too often is not a good thing.  :oops:

Oh, and one more thing. X isn't a nasty radical, it's a clean radical.  ;)

I don't believe this problem is on the level of MOEMS.

If you are not certain of the MOEMS curriculum level, please do not post problems in the MOEMS forum.


function
limit
real analysis
real analysis unsolved

Everyone likes to study the asymptotics of the Riemann zeta function near the critical strip, because that's what's actually got deep meaning in analytic number theory. But what about in the "boring" parts of the plane?

Let's just work on the real line. We know that $ \lim_{x \to \infty} \zeta(x) \equal{} 1$. Following up on this, here's the problem I propose:

Find (explicitly) an elementary function $ g(x)$ so that $ \lim_{x \to \infty} \frac{\zeta(x) \minus{} 1}{g(x)} \equal{} 1$.

Now what is $ \sum_{n\equal{}2}^\infty (2^{n} (\zeta(n) \minus{} 1) \minus{} 1)$ ? ;)

$ \frac{2^n}{3^n}$, of course. We could go on like this forever.

Dang. I should put less scary words in my topics. Probably the only people who look at it are the ones for whom it's trivial...  :(


Let $ i$, $ n \in\mathbb{N}^{*}$. And let $ a_{i}= \min\{k+\frac{i}{k}| k \in\mathbb{N}^{*}\}$. Find the value of $ \sum_{i=1}^{n^{2}}[a_{i}]$, where $ [x]$ means the largest integer no larger than $ x.$


Juanita paid $ \$18$ for a book after a $ 10\%$ discount. What was the number of dollars in the original price of the book?


floor function
geometry
3D geometry
pigeonhole principle
combinatorics proposed
combinatorics

Let $S$ be the set of $n$-uples $\left( x_{1}, x_{2}, \ldots, x_{n}\right)$ such that $x_{i}\in \{ 0, 1 \}$ for all $i \in \overline{1,n}$, where $n \geq 3$. Let $M(n)$ be the smallest integer with the property that any subset of $S$ with at least $M(n)$ elements contains at least three $n$-uples \[\left( x_{1}, \ldots, x_{n}\right), \, \left( y_{1}, \ldots, y_{n}\right), \, \left( z_{1}, \ldots, z_{n}\right) \] such that
\[\sum_{i=1}^{n}\left( x_{i}-y_{i}\right)^{2}= \sum_{i=1}^{n}\left( y_{i}-z_{i}\right)^{2}= \sum_{i=1}^{n}\left( z_{i}-x_{i}\right)^{2}. \]
(a) Prove that $M(n) \leq \left\lfloor \frac{2^{n+1}}{n}\right\rfloor+1$.
(b) Compute $M(3)$ and $M(4)$.

Call two n-tuples neighbors if they differ by exactly one element. If $M(n) > \lfloor \frac{2^{n+1}}{n}\rfloor+1$, then for the corresponding S, each element of S has n neighbors. Now $nM(n) > n \lfloor \frac{2^{n+1}}{n}\rfloor+1 \ge 2^{n+1}$, so there is some n-tuple x such that three elements a, b, and c of S are neighbors of x. Then these three elements eacher differ from each other in two places, a contradiction.

Now to find M(3) and M(4). For n = 3, we can have 000, 100, 010, 110. If we could have 5 elements, we can't have three elements with the same number of 1's. Also, either some two have exactly one 1, or some two have exactly one 0, and WLOG we'll take the latter. Then we can not have 000, so then we must have some pair having exactly one 1. But that means we can't have 111, so either some three have exactly one 1 or some three have exactly one 0, contradiction.

For n = 4, we can have 0000, 1000, 0100, 1010, 0101, 0111, 1011, 1111, and by the arguments in proving the general bound, we can show that this is optimal.

[quote="probability1.01"]If $M(n) > \lfloor \frac{2^{n+1}}{n}\rfloor+1$, then for the corresponding S, each element of S has n neighbors. Now $nM(n) > n \lfloor \frac{2^{n+1}}{n}\rfloor+1 \ge 2^{n+1}$, so there is some n-tuple x such that three elements a, b, and c of S are neighbors of x. Then these three elements eacher differ from each other in two places, a contradiction.[/quote]

I don't get a single thing from it, could you elaborate on it?  maybe I'm just blind at the moment :maybe:

I think I proved M(3) = 5. Except I thought M(n) was the largest integer such that you don't have three elements of equal distance to each other, so it looks like I have M(3) = 4. Unless you really meant M(3) = 6? To Megus, I just used the pigeonhole principle...  :maybe:

[quote="probability1.01"]$ n \lfloor \frac {2^{n \plus{} 1}}{n}\rfloor \plus{} 1 \ge 2^{n \plus{} 1}$, [/quote]

This is not true! Take $ n\equal{}5$ as an example.

i think he wanted to write $ n(\lfloor \dfrac{2^{n \plus{} 1}}{n} \rfloor \plus{} 1)$
then there is no error in the solution :lol:

and maybe $ M(3)\equal{}5$ $ M(4)\equal{}9$

[quote=pohoatza][quote="probability1.01"]
Now to find M(3) and M(4). For n = 3, we can have 000, 100, 010, 110. If we could have 5 elements, we can't have three elements with the same number of 1's. Also, either some two have exactly one 1, or some two have exactly one 0, and WLOG we'll take the latter. Then we can not have 000, so then we must have some pair having exactly one 1. But that means we can't have 111, so either some three have exactly one 1 or some three have exactly one 0, contradiction.
[/quote]

actually $M(3)=5$ is the sollution.  :wink: and a hint for finding this is by considering a cube and translating the problem geometrically.[/quote]
What are you talking about?How can you get 5?


How do you take a screen shot of the page you are on and then put it in a message?
I think this was asked before if so im sorry :(

Press Print Scr on your keep board to take a picture and ctrl-v to paste.

[url=http://www.google.com/search?q=how+to+screenshot&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]Are you new to the internet, by chance? Google is your friend[/url]

i mean putting a picture in an AoPS message.

You can upload it to another host (such as tinypic) and embed it using the Img tags and then place it in within your image wherever you like, or you can send it as an attachment and click "Add an Attachment" next to "Message Body" when composing a message.

so u can save it as a picture? how would u do that?

Did you even click that link? *sigh*

1) Press PrintScreen
2) Open Paint (or another iamge editing program)
3) Right-click -> Paste or Press Ctrl-V
4) Click Save As and save where you want

Don't forget to crop out the stuff you don't want. No point in uploading a massive PNG file onto AoPS.  :wink:

thanks!! (by the way that link didnt work Anaxerzia.)

..dude the link works :huh:

FYI, on a Mac, it's Cmd+Shift+3 for a fullscreen capture, and Cmd+Shift+4 for a window. Pressing spacebar after Cmd+Shift+4 should enable you to screenshot a selection.

on Linux with GNOME, you can go to Accessories > Take Screenshot. dunno about KDE.

I've also managed to tweak settings so that I can drag my cursor across the screen while holding the Command key to capture a portion.. forgot how. google it.

is cropping a pic like erasing the stuff u dont want?

Yes; when you go onto a graphics program, for example the GIMP, you use the crop function and select an area you want to keep. Everything else will be deleted.

Or, on some computers, like mine, when you take a screen shot, you can just choose a option that only takes a picture of an open window.


The ring is a two dimensional figure. Given that the line charge density "U", what can we say about the Eletric field everywhere inside such a ring of charge?

Yes i did. The goal of  shell theory is to describe the energy levels of nuclei and transitions between these levels. Ideally the theory is based on known nuclear forces.

I believe the shell theorem states that inside a uniform circular shell, the gravitational field is 0.
And electric fields work a lot like gravity...

Shell Theory and Shell Theorem are two different things, I believe... :)

Anyway, daveed is right. You should try to prove it, it's not that hard.

Ooops, actually - daveed is not so right. The shell theorem works in 3D with all $F=\frac{k}{x^2}$ forces but not in 2D! In 2D, it could probably work if the force was $F=\frac{k}{x}$.

oof.  My apologies.  If that's so, maybe you could get the result from some integration?

The electric field inside any conductor, when a constant current is travelling through it, is zero. Now, that and any charge distribution usualy creates a field which is found by solving Poisson or d'Alembert equations with proper boundary conditions.

Ooops, i think the problem mentions no conductor.But it's true that the potential at every point on the ring is the same.Take it to be 0. Inside, the potential satisfies Laplace's equation.


What is the square root of the cube root of 64?

This is the sixth root of 64=2^6, which is 2.

64's cube root is 4, because 4*4*4=64.  4's square root is 2, because 2*2=4, hence the answer is 2. :surf:


trigonometry
trig identities
Law of Cosines
geometry
congruent triangles

Ten equally spaced points are plotted on a circle's circumference.  Connecting each point to each other point 3 points away, a regular convex decagon is formed in the center of the circle. Show the length of a side of the decagon is equal to the length of the radius.

i dont think its possible...because that would mean each sector containing 2 radii and one side of the decagon must be equilateral, and its not.

Let the length of the radius and decagon's side be $ r$ and $ s$ respectively. Consider the triangle made by two radii and a side. Because the sides span 3 points, the central angle measures $ 3/10 \cdot 360 \equal{} 108$ degrees. By the law of cosines, $ 2r^2 \minus{} r^2 \cos 108 \equal{} s^2$, so $ \frac{s}{r} \equal{} \sqrt{2 \minus{} \cos 108}$. It's easy to see this is not $ 1$.

Obviously there was a transcription error somewhere. I tried hard to figure out what the right wording should be. This version works.
[hide]Ten equally spaced points are plotted on a circle's circumference.  Connecting each point to the next, a regular convex decagon is formed. Connecting each point to each other point 3 points away, a smaller regular convex decagon is formed in the center of the circle. Show the length of a side of the larger decagon is equal to the length of the radius of the circle circumscribed about the smaller decagon.[/hide]
There may be a simpler, more straightforward wording that works too. Comments?

sorry it should read as follows:

Ten equally spaced points are plotted on a circle's circumference These points connect to form a regular convex decagon. Connecting each point to each other point 3 points away, a regular convex decagon is formed in the center of the circle. Show the difference between the lengths of the sides of the decagons is equal to the length of the radius.

By the side lenght of the larger decagon, i mean the length of one point to another point 3 points away.

The larger decagon is a regular 10-pointed star

Maybe something is lost in translation. The way I understand the drawing it still does not add up.
If you draw it, and add segments going from one point of the 10-point star to the opposite one (diameters), you will find isosceles triangles all over.
The angles can be figured as 36, 72 and 108 degrees (pi/5, 2pi/5. and 3pi/5).
There are many congruent triangles and many congruent sides.
Consider one side of the inner decagon. I find two segments of the same length along each of the segments drawn from one point in the circumference to another 3 points away ("sides of the larger decagon").
Consider a radius drawn from one of the points of the star. There is a segment of equal length on the adjacent "side of the larger decagon".
The problem is that the two segments I find along the "side of the larger decagon" (one as long as the radius, the other as long as the side of the small decagon) do not add up to the "side of the larger decagon".


Prove: If Q is an orthogonal matrix, then each entry of Q is the same as its cofactor if det(Q)=1 and is the negative of its cofactor if det(Q)=-1.

Prove: Let $ u$ be a vecter in an inner product space V, and let $ \{v_1,v_2,...,v_n\}$ be an orthonormal basis for V, and if $ \alpha_i$ is the angle between $ u$ and $ v_i$, then

$ \sum_{i\equal{}1}^n\cos^2 \alpha_i\equal{}1$.

[b]a)[/b]Let $ A$ be an invertible matrix. Then $ A\text{cof}(A)^{t}\equal{}\det(A)I$. Now, compare it with the definition of an orthogonal matrix.

[b]b)[/b]Let $ 0\neq u\in V$. $ u\equal{}\sum_{i\equal{}1}^{n}c_{i}v_{i}$ for some scalars $ c_{i}$ in the ground field. Use $ \cos(\alpha_{i})\equal{}u.v_{i}/|u|$ to conclude.


All are questions from the path to combinatorics for undergraduates:

1)Determine the no. of functions f: {1,2,3,...,1999) -> {2000,2001,2002,2003} satisfying the condition that f(1) + f(2) + f(3) ....+ f(1999) is odd.

[hide]My answer :$ 1000. 2^{1999}$[/hide]

2) In how many ways can COMPUTER be spelled by moving either down or diagonally to the right in fig:

C
O O
M M M
P  P  P  P
U  U  U  U  U
T  T  T  T  T  T
E  E  E   E  E  E  E
R  R  R   R  R  R  R  R

[hide]I tried generating function[/hide]

3) A basketball team consists of 12 pairs of twin brothers. On the first day of training, all 24 players stand in a circle in such a way that all pairs of twin brothers are neighbors. In how many ways can this be done?

[hide]My answer :$ 11! . 2^{12}$[/hide]

[hide="Solution to 1"]Instead consider functions $ g: \{1,2,3,\ldots ,1999\}\rightarrow\{0,1\}$.  We want the number of such functions with $ g(1) \plus{} g(2) \plus{} g(3) \plus{} \cdots \plus{} g(1999) \equal{} n$ for $ n$ odd.  For any $ n$ the number of functions is $ \binom{1999}{n}$, so we want $ \binom{1999}{1} \plus{} \binom{1999}{3} \plus{} \binom{1999}{5} \plus{} \cdots \plus{} \binom{1999}{1999}$, which is $ 2^{1998}$ (Try to prove this  :) ).  Now we want to looks at functions $ f: \{1,2,3,\ldots ,1999\}\rightarrow\{2000,2001,2002,2003\}$ such that the sum of function values is still odd.  There is a $ 2^{1999}$ to 1 correspondence between the number of functions $ f$ and the number of functions $ g$, since if $ f(i) \equal{} 0$ then we make $ g(i) \equal{} 2000,2002$ and if $ f(i) \equal{} 1$ then we make $ g(i) \equal{} 2001,2003$ (this is also clearly reversible).  Therefore, the answer is $ 2^{3997}$.[/hide]
EDIT:
[hide="Much nicer solution to 1"]There are $ 4^{1999}$ such functions ignoring the sum condition.  Now given a function $ f$ consider the function $ g$ such that $ g(n) \equal{} 4003 \minus{} f(n)$ for each $ n$.  Then $ g(n)$ has different parity than $ f(n)$ for each $ n$, so $ g(n)\equiv f(n) \plus{} 1999\pmod{2}$, so exactly one of $ f(1) \plus{} \cdots \plus{} f(1999)$ and $ g(1) \plus{} \cdots \plus{} g(1999)$ is odd.  We can pair up all the functions in this way, and exactly one of each pair of functions will satisfy the desired property.  Therefore, the answer is $ 4^{1999}/2 \equal{} 2^{3997}$.[/hide]
[hide="Hint to 2"]Consider the number of choices at each step in spelling the word.  :wink: [/hide]
Your answer to 3 is correct.


linear algebra
matrix
combinatorics unsolved
combinatorics

Two players fill an $ n\times n$ square board with $ 0$ and $ 1$. As soon as the board is filled, they calculate the sum $ r_i,c_i$ of numbers in $ i$th row and $ i$th column, respectively. Let

$ R\equal{}\sum_{i\equal{}1}^{n}ir_i, C\equal{}\sum_{i\equal{}1}^{n}ic_i$. The first player wins if $ R<C$, and the second wins otherwise.

Who has a winning strategy?

[b]Solution:[/b] Nobody has the winning strategy. Consider the gain matrix of the 1st player on a $ nXn$ board when every entry in the board is made 1. Let $ g(a,b)$ be the gain of the first player if 1 is entered in the square $ (a,b)$ (considering leftdown square to be $ (1,1)$).
\[ g(a,b) \equal{} a \minus{} b
\]
Hence the game is equally balanced.

I don't really get it... what is the gain matrix? Why the game is equally balanced?

Please explain a bit more  :)

The picture shows the $ 3X3$ gain matrix for first person. For the first person (who tries to maximize the column sum) 1 in (3,1) gives him the score of 3 while it gives the second person score 1. It is clear from the gain matrix that game is equally balanced.

Assuming both players make intelligent moves (like in all game problems), [b]a[/b] strategy is as follows:
If the P1 puts 1 in (3,1) or a 0 in (1,3)then P2 has to put a 1 in (1,3) or 0 in (3,1)

If the P1 puts 1 in (2,1) or a 0 in (1,2)then P2 has to put a 1 in (1,2) or 0 in (2,1)

If the P1 puts 1 in (3,2) or a 0 in (2,3)then P2 has to put a 1 in (2,3) or 0 in (3,2)

I hope now it is clear ?

Umm... it's not really clear. It doesn't seem rigorous at all.

How do you know those are the most intelligent moves?

[quote="easternlatincup"]Two players fill an $ n\times n$ square board with $ 0$ and $ 1$.[/quote]
{Both players can use both $ 0$ and $ 1$} or {The first player puts $ 0$ and the 2nd player puts $ 1$}?

Both players can use both $ 0$ and $ 1$  :)

I think Srikanth is right.

To make his solution more understandable and more rigorous we need to show that each player has a not-loosing(winning or drawing) strategy.

Suppose a player decides that he wont let the other player win and he is intelligent enough. In each move he might copy the other player's move in the centrally symmetric cell. It might be that there is no such centrally symmetric cell (For example, He is the 1st player making his 1st move, or the previous move had been in the center-cell etc.). In that case he may put a 1 in his most-gain cell. His most-gain cell will have a non-negative value because in such a situation the board would be completely centrally symmetric. Thus the game must end in a draw or a win in favor of the discussed player. (I am not sure if it can be a win but that doesn't matter.)

(a,b) and (c,d) are centrally symmetric iff a+c=n+1 and b+d=n+1.

Well, I wrote my solution short and sweet in my first post (username: srikanth) as we typically write solutions to most game problems ( especially Russian olympiads).
Here are more details which have could have been got from the gain table.

The gain table for the 1st and 2nd player look like

-2 -1  0
-1  0  1
 0  1   2

2  1  0
1  0 -1
0 -1 -2

By intelligent move I mean operations like {p1 put 1 in (1,3) that gives him a gain of -2} are not done by p1.

We call (1,3) and (3,1) to be corresponding squares. If P1 puts 1 in (3,1) , p2 retaliates by putting 1 in (1,3). Like wise {(1,2) and (2,1)} {(2,3) and (3,2)} can be corresponding squares. The other possibility is {(1,2) and (3,2)} and {(2,1) and (2,3)}. This new possibility merely comes as the number in the squares are same(here 1).

I have been as explicit as I could have been. Hope the solution is clear now. generalization to n is trivial.


Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate
six more bananas than on the previous day. How many bananas did Big Al eat
on May 5?

[quote="$LaTeX$"]Let x equal the number of bananas he ate on May 3. Thus we have $ x\minus{}12\plus{}x\minus{}6\plus{}x\plus{}x\plus{}6\plus{}x\plus{}12\equal{}100$, or $ x\equal{}20$. Thus Big Al ate $ 20\plus{}12\equal{}\boxed{32}$ bananas on May 5.[/quote]
[hide="A Solution That May Be A Tiny Bit Quicker"]We know that on the day in the middle (which is May 3), Big Al must have ate the mean of the number of banana's he ate on each day.  Since we are already given the total, we can just divide by the number of days and get $\frac{100}{5}=20$ bananas.  Then we see that May 5 is only 2 days after May 3 so we just add $6\times2=12$ to the number of bananas Big Al ate on May 3.  So Big Al ate $20+12=\boxed{32}$ bananas on May 5.[/hide]


calculus
integration
function
calculus computations

Given $ f(x)$ is an odd function defined everywhere,periodic with period $ 2$ and integrable on every interval.

Let $ g(x)\equal{}\int_{0}^x f(t) dt$.

Then prove that $ g(2n)\equal{}0$ for every integer $ n$.

Hi :) !!

$ f$ is odd and $ 2$-periodic function in everywhere than $ \int_{2(k \minus{} 1)}^{2k} f(t)dt \equal{} 0$  for every integr $ k$

than

$ g(2n) \equal{} \int_0^{2n}f(t)dt \equal{} \sum_{k \equal{} 1}^{n} \int_{2(k \minus{} 1)}^{2k}f(t)dt \equal{} 0$

...
i taked that $ n \geq 1$ because $ g$ is no-odd  function .

PS: we can prove that by a lof of methode .

and thanks

$ \int_0^{2n} f(x) \ dx \equal{} \int_{\minus{}n}^{n} f(x\minus{}2n) \ dx \equal{}  \int_{\minus{}n}^{n} f(x) \ dx \equal{} 0 \ (\ \because $ f(x) has period 2 and integration of an odd function over a symmetric interval about 0 gives 0)

[quote="pankajsinha"]Thank you mathema.
Can you show any of the other methods[/quote]
OK :) !!

after the methode proposed by [b]hsbhatt[/b] i give one other nice :) !!

if $ h$ is a odd and $ T$-periodic function than we have $ \int_0^T h(t)dt \equal{} 0$

than our exemple ($ T \equal{} 2$) :

$ g(x) \equal{} \int_0^x f(t)dt \equal{} x \int_0^1 f(tx)dt$

than $ g(2n) \equal{} 2n \int_0^1 f(2nt)dt$

we know that the fuction: $ h\quad : \quad t \longrightarrow f(2nt)$ is odd and $ 1$-periodic fuction than $ g(2n) \equal{} 0$ ...

and thanks :)

For $ n\in N$ and $ f(t)$ periodic with period $ 2$
$ g(2n) \equal{}\int_{0}^{2n}f(t)dt \equal{}n\int_{0}^{2}f(t)\equal{}n\int_{\minus{}1}^{1}f(t)\equal{}0  .$ (as $ f(x)$ is an odd function)
Note for periodic function with period $ T$, we have $ \int_{a}^{a\plus{}T}f(t)dt \equal{}\int_{0}^{T}f(t)$
I chosen $ a\equal{}\minus{}1$ in last integral


function
symmetry
complex analysis
complex analysis unsolved

Prove or give a counterexample:

If an analytic function mapping the unit disk into itself has two fixed points in the unit disc, then it must be the identity function.

From Schwarz lemma it's clear that it's an automorphism of the unit disc (and in particular a Möbius transform). Such an automorphism maps the unit circle onto itself as well, and such a map cannot have two fixed points in the unit disc without being the identity function. One argument is that möbius transforms preserve symmetry w.r.t. circles


If $ a,b,c\in {\rm Z}$ and
\[ \begin{array}{l} {x\equiv a\, \, \, \pmod{14}} \\
{x\equiv b\, \, \, \pmod {15}} \\
{x\equiv c\, \, \, \pmod {16}} \end{array}
\]
, the number of integral solutions of the congruence system on the interval $ 0\le x < 2000$ cannot be

$\textbf{(A)}\ 0  \qquad\textbf{(B)}\ 1  \qquad\textbf{(C)}\ 2  \qquad\textbf{(D)}\ 3  \qquad\textbf{(E)}\ \text{None}$

[hide="Solution"]if $ a$ is odd abd $ c$ is even, there is no solution. Solution set can be empty.

Let $ a \equal{} 2a'$ and $ c \equal{} 2c'$, then the congruence
\[ \begin{array}{l}{x\equiv 2a'\,\,\,\left(mod\, 7\right)} \\
{x\equiv b\,\,\,\left(mod\, 15\right)} \\
{x\equiv 2c'\,\,\,\left(mod\, 16\right)}\end{array}
\]
has a solution in the form $ x \equal{} 7 \cdot 15 \cdot 16 \cdot k \plus{} m \equal{} 1680k \plus{} m$ (Chinese Remainder Theorem). The number of solution on the interval $ 0\le x < 2000$ is $ 1$ or $ 2$.
We will show that the solution of the second congruence is also the solution of the first congruence.
From the second congruence, $ x \equal{} 1680k \plus{} 7p \plus{} 2a' \equal{} 1680k \plus{} 16q \plus{} 2c' \equal{} 1680k \plus{} 15r \plus{} b$.  $ RHS$ is even, so $ p$ must be even. $ x \equal{} 1680k \plus{} 14p' \plus{} 2a' \equal{} 1680k \plus{} 16q \plus{} 2c' \equal{} 1680k \plus{} 15r \plus{} b$. If we rewrite the equation as congruence, we get
\[ \begin{array}{l}{x\equiv 2a'\,\,\,\left(mod\, 14\right)} \\
{x\equiv b\,\,\,\left(mod\, 15\right)} \\
{x\equiv 2c'\,\,\,\left(mod\, 16\right)}\end{array}
\]
Similar approach for $ a$ and $ c$ are odd.
So the answer is $ C$ which is $ 3$.[/hide]


geometry
trapezoid
symmetry
rectangle
AMC
AMC 10

The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $ 9$ trapezoids, let $ x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $ x$?
[asy]unitsize(4mm);
defaultpen(linewidth(.8pt));
int i;
real r=5, R=6;

path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0);
for(i=0; i<9; ++i)
{
draw(rotate(20*i)*t);
}
draw((-r,0)--(R+1,0));
draw((-R,0)--(-R-1,0));[/asy]$ \textbf{(A)}\ 100 \qquad
\textbf{(B)}\ 102 \qquad
\textbf{(C)}\ 104 \qquad
\textbf{(D)}\ 106 \qquad
\textbf{(E)}\ 108$

[hide]Extend the legs of the trapezoids: by symmetry, they all meet.  There are 9 trapezoids, so there are 9 congruent angles in 180 degrees: hence the small angles are each 20 degrees.  By symmetry, the smaller angles of the trapezoids are equal and the sum of two of them and 20 is 180, so one of them is 80 degrees.  Thus the larger angle is this amount from 180, or 100 degrees.  $ \boxed{A}$[/hide]

[hide]The arch is half of a regular $ 18$-gon. Thus the measure of each interior angle is $ 180 \minus{} \frac {360}{18} \equal{} 160$ degrees. Because corresponding base angles of an isosceles trapezoid are congruent, each of the base angles on the outside of the arch is $ 160/2 \equal{} 80$ degrees. Because opposite angles of an isosceles trapezoid are supplementary, the other two angles are $ 180 \minus{} 80 \equal{} 100$ degrees, or $ A$.

Was it just me or was this problem ridiculously easy for #24. The answer choices were hardly even good distractors as far as I can tell. [/hide]

I was pretty annoyed with this problem. For some reason, I was forgetting to subtract 80 from 180. If they had put 80 as an answer, I bet less people would have gotton this right. How would someone get 106?!?!?!?!

@henry To stop clueless people from protractoring-it, of course :]. I didnt even realize I had a protractor until too late xD.
Here's the stupid way that I did it
[hide]
Since the trapezoids were isosceles, drop their altitudes to cut them into rectangles and right triangles. Now, if we remove all the rectangles, the triangles will form a 180 degree 9-sided "loop" made up of 18 triangles. So, the smaller angle of each triangle is 10 degrees, and thus the larger angle of the trapezoid is 90+10=100 degrees.

Didn't think to use my protractor.
Then I looked back and realized that it was impossible to get 102 or 104 xD.
Also, I didn't even notice that the figure was half of a regular 18-gon.

[/hide]

If you look closely at the diagram on the actual page it is slightly distorted - they tried to prevent people from protractoring it.

[quote="l Bob l"]I was surprised how easy this problem was, as a #24 question. It's just isoceles triangles.[/quote]

I agree. This was the easiest #24 on an AMC10 I'd seen in all my years. I literally solved this in 1 minute.

Same, there was nothing to this problem, :P #7,10 and others, while they were easy deffinitely much harder than #24, which any moronic sophomore can do...

[hide=Solution]Let the answer be $x.$ This means that $$\frac{8(180)-2(180-x)}{8}=360-2x\Leftrightarrow180-\frac{180-x}{4}=360-2x\Leftrightarrow135+\frac{x}{4}=360-2x\Leftrightarrow225=\frac{7x}{4}\Leftrightarrow x=\frac{4\cdot225}{7}=100\Longrightarrow\boxed{\mathbf(A)}.$$[/hide]

[quote=asdf334]$$180-\frac{180-\tfrac{180}{9}}{2}$$[/quote]

and this is a problem 24

The shape made by the outer side lengths is half a 18 sided regular polygon so it has angle sum $\frac{16 \cdot 180}{2} = 1440$. The small interior angle is $180-x$, so $2(180-x) + 8(2(180-x))=18(180-x) = 1440 \implies x = \boxed{100}$.

[hide=Solution]Reflect the shape over the bottom line and notice that it is a regular $18$-gon. Each interior angle has measure $180-360/18=160,$ so $$2(180-x)=160\implies 180-x=80\implies x=\boxed{100\quad\text{(A)}}.$$[/hide]

Because the interior polygon is a decagon, we know $$180(10 - 2) = 2(180 - x) + 8(360 - 2x)$$ which yields $x = \boxed{100.}$


Let's make a thread about good books that would be ideal Christmas presents.  Don't limit yourself to just mathematics books!

-America (The Book)
-Anything Dilbert-related
-Dave Barry stuff is good.
-The Onion (Ad Nauseam, Our Dumb Century, etc.)
-various math textbooks, although they are pretty expensive
-Art of Problem Solving!

Probably depends on people's interests (don't give anything Michael Savage-related to a liberal kid)

personally, I absolutely loved "The Time Traveler's Wife", by Audrewy Niffenegger

What a great book.  I picked it up while walking around at the bookstore... read a paragraph and decided I had to have it.

It's really good  :)

I've been reading books by Steinbeck lately, my favorite so far is East of Eden.  I don't know if these are really great christmas ideas since the books aren't jolly and joyful.

[quote="SnowStorm"]I've been reading books by Steinbeck lately, my favorite so far is East of Eden.  I don't know if these are really great christmas ideas since the books are jolly and joyful.[/quote] They are or they arent?

Whoops.. meant for it to say "aren't" sorry

It's about time this thread was made :lol:

I'm one of those people that reads about 2 pages of a particular book, and then immediately decides whether it's boring or not.  Generally, I like books with an interesting concept, otherwise, I stick with Mathematical texts.

One of my favourite books is "Battle Royale", an English translated Japanese depiction of a modern fascist society (well, a particular event of that society).  There is a movie based on this, and I'd have to say that both movie and novel are well done.

While I won't spoil the plot, I will say that YOU MUST READ THIS BOOK.  While not Christmas "jolly", the novel plot, ending, and concept are absolutely gratifying.

Anyone ever read any of Tom Clancy's books? I like Patriot Games and the Hunt for Red October

Also:
LotR
Bartimaeus Trilogy
Harry Potter

etc. etc. anything big that'll last.

The Revelation Space triolgy is great! Chasm city isn't in the triolgy, but it still takes place in the same universe. It's writing stlye is way different, an assasin's diary instead of a space opera.

Pink Floyd and White Stripes cd's are good. So are compilations of Queen, Led zepplin, and other bands.

Chess, math, programming, and electronics books are always good.

Music method books and song books for assorted instuments are great. I plan on getting a guitar  :) Some jazz lead sheets for pianists are great :lol:

Rubiks cubes, juggling balls, and other skills toys are great.

I love giant puzzle compilations, especially computer programming contest compilations.

If you wanted to delve deeper into nerdery, MtG and D&D are options, I guess.]

Microchip programming is fun. So are burnable gameboy carts. Programming gaming consoles= fun. If you wanted to avoid specs or installing linux on to some things, the Xgamestation is fun.

Nerf and water guns are fun, a staining agent like kool-aid makes it even more fun.

Oh, an external hard drive, more RAM, and a new GPU are also high on my list.

Les Miserables is a really good book, its a depiction of French society of back then and some stuff, but really really long, although quite exciting....

AOPS and AACOPS (art & craft of problem solving) are good books also, ith addition of USSR Math Olympiad Book. which has some pretty cool problems...


algebra
polynomial
system of equations
real analysis
real analysis unsolved

\[ \sum\limits_{k \equal{} 0}^n {k^p }
\]

bernoulli, but I can't get it...
thx

better state it as $ \sum_{k \equal{} 1}^n \; k^p,$ where $ p\in\mathbb Z$

see
http://www.trans4mind.com/personal_development/mathematics/series/sumsBernoulliNumbers.htm

also see equation $ (13)$
http://mathworld.wolfram.com/PowerSum.html

The name for the general formula is [url=http://mathworld.wolfram.com/FaulhabersFormula.html]Faulhaber's formula[/url].

An outline of the proof:

Write down the polynomial formula $ \sum_{k\equal{}1}^n k^p\equal{}\frac1{p\plus{}1}n^{p\plus{}1}\plus{}\sum_{j\equal{}0}^{p\minus{}1}a_jn^{p\minus{}j}$. Take its difference, and expand the $ (n\plus{}1)^r\minus{}n^r$ terms on one side with the binomial theorem. You get $ n^p$ from the $ \frac1{p\plus{}1}n^{p\plus{}1}$ term, and the other terms should balance to zero. That leaves a system of equations in the $ a_j$; insert the appropriate factors to turn it into the defining equation of the Bernoulli numbers $ (e^x\minus{}1)\sum_{j\equal{}0}^{\infty}B_nx^n\equal{}x$.


[color=red][quote]Vanessa has N pieces of wedding cake. If she arranges them in groups of 5, she winds up with 4 left over. If she arranges the pieces in groups of 8, she ends up with 7 left over. If Vanessa decides to arranges the pieces of cake in groups of 20, how many pieces will be left over?[/quote]

ANSWER: 19

well, since all multiples of 5 end with 0 or 5 in the units digit N must end with a 4 or 9. all multiples of 8 end in 8, 6,4,2, and 0. N can end in 5,3,1,9. or 7. one can see that  9 is shared between the two. so we can conclude that N must end in a 9. if N gives a remainder of 7 when divided by 8 one can tell that N is one less than a multiple of 8. as long as it ends in a 9. so 9+1=?, anyone, anyone, okay i can see that you dont know, the world these days, okay, 9+1=10. so the multiple of 8 must end in a 0. i just randomly picked 80 as my number. so 80-1=79. N can be many different numbers like 39. but 80 came to my head before 40. were not done with the question yet. we still have to find how many pieces will be left over if we Vanessa arranges the pieces of cake in groups of 20. its pretty straight forward from here. but ill tell anyhow. 79/20=3r19. the 19 is what we are looking for here, not the 3.  wow. that took some typing to explain. my way may not be the fastest, but it didnt take me more than 30 seconds to solve. it just took some time to type. lool.  i remember in this year's state mathcounts there was a question like this except it was like n^2 divided by a number gave a remainder of something and n^3 divided by a number gave a remainder of something else, find n. anyways. ummmm. hope that helped. i am wanting to know if there is a faster way. oh yeah, that question that i mentioned that came from the state competition was also a countdown round question in nationals.  ummmm. and if i dont see you again good morning good evening and good night. nuff said. wait. the answer is 19. just so you wont have to look through my passage again. now anwho. nuff said

[hide]

n=5m + 4 = 5m - 1
n also = 8x + 7 = 8x-1

thus n+1=5m=8x.  the smallest integer n satisfying this is 39 (n+1)=40.

upon division by 20, the remainder is 19.
[/hide]


modular arithmetic
number theory
number theory unsolved

p is a prime number
$4^{p-1}-1$ is $0$ (mod $p$) is easy to prove by Fermat's little theorem
but I don't know how about in mod$p^{2}$