geometry
algebra
polynomial
MATHCOUNTS
function
analytic geometry
3D geometry
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Hi,

I am a parent of an 11yo who is getting to the point where he no longer has mathematical peers in his small K-8 school.  We are looking into options for him for next year or possibly even this spring, and one  possibility is something like AoPS classes.  He really likes being part of a group process, so I think this is better for him than an independent program like EPGY.

In looking over the "Are You Ready" quizzes for the Introductory courses, I don't think he has all of the prerequisite topics covered yet.  Is there a place on the site that covers an overview of these preliminaries?  We also just got AoPS book 1 so we could review them using that text as well if that is the best approach.

Finally, we're curious as to whether Introductory level students are often on the forums during school time.  Although the classes themselves are all in the evenings, it would be great if he could connect with classmates when it is math time at his school (generally 10:30 - 11:30 in the morning).

Thanks!

The AoPS Volume 1 is a fine way to cover the prelims for the intro classes.  Focus on the first 5 chapters, particularly those on linear equations and word problems.  The classes are quite challenging for such young students - you should encourage him to ask questions.  Bright young students are often not used to having to ask questions in a math class and sometimes need extra encouragement to get them to do so.

We do have many active homeschoolers, so if he posts anything in the Getting Started area, he probably will be able to often get answers during school hours.  Also, plenty of our students access the site from school.

[quote] The classes are quite challenging for such young students - you should encourage him to ask questions. Bright young students are often not used to having to ask questions in a math class and sometimes need extra encouragement to get them to do so.
[/quote]

Thanks for this suggestion/point.  I am worried that he might be intimidated by the level of these courses.   It is true that he is not used to having to ask questions, or indeed to not "getting" everything instantly.  He's had some work on counting/combinatorics before, so that one would be a good place to start.

I'm not sure if we'll push to start in January, or give him more time to polish the preliminaries first.  He has had some introductory algebra, but I don't [b]think[/b] he has done the multiplication of polynomials, or much work with fractions containing algebraic expressions, though these topics seem like things he would master very quickly.

Thanks again!

I have another question about the Introductory level classes -- about how much work is involved in addition to the class time?

Thanks!

Most students seem to spend anywhere from 1 to 4 hours per week working on the material and problems outside of class.  Those who put forth the most effort get the most out of the classes.

Just to mention -- we allow students to enroll in classes to test the waters and then drop the class for a full refund within the first ten days.  This allows a student to attend one or two classes to see what it's like taking a class in the virtual classroom and solving problems on the class Message Board.  I think this option is especially important for 6th and 7th grade students who might be ready for the classes but might also be a few months or a year away from taking the introductory classes on.

[quote]we allow students to enroll in classes to test the waters and then drop the class for a full refund within the first ten days. This allows a student to attend one or two classes to see what it's like taking a class in the virtual classroom and solving problems on the class Message Board. [/quote]

Thanks.  I was thinking about this possibility.  My son is so far lukewarm about this option, and I don't want to push him into it if he isn't interested/ready.

I may try to get him to sign onto the message boards to see if he can get to know some of the members his age who might be taking the classes.  Then he wouldn't feel like he was taking a class with a bunch of random strangers online, but rather with kids he already "knew" in some way.

Thanks again for everyone's thoughts so far.

Hello,

I m a student who has taken a bunch of AoPS classes.  I thought you might like to here my advice on some of the points you brought up and just the classes in general.  This is my opinion.

First of all, it is important to understand AoPS.  AoPS primary purpose is to teach top notch students in math so that they may excel in math competitions.  They do not have a grading system for their classes.   There is no need for one.  These classes don't count for credit (except the Independent Study Programs might).  Students take these classes because they want to excel even further in math.  The courses that AoPS offers are much different from a regular school's math class.  Foremost, they emphasize that the student do NOT just merely memorize formulas.  Instead the student will benefit much more greatly if they understand the formulas.  I totally agree with this philosophy.

You should definitely be prepared to ask questions in these classes.  I almost never ask questions in math classes at school.  On the other hand, I usually ask lots of questions on these forums and in the classes.  If you don't understand something this is a great way to learn.  Also, usually you don't have to ask questions in school because most of the problems given to the student are exercises that have been burned into the student's brain.  The problems that AoPS deals with require problem solving skills.  You have to be able to use the basic methods creatively to solve good problems.  Also, if you do decide to enroll him in a course.  Neither you nor him should expect to solve every single problem immediately.  A lot of the problems require that you play around with them for a little while and then you can solve them.

I hope this helps a little bit!

[quote="joml88"]
AoPS primary purpose is to teach top notch students in math so that they may excel in math competitions.[/quote]

While we do guide students using a curriculum of contest problems, I wouldn't say that our purpose is to teach students math with the goal of excelling in contests.  However, we think it is true that well-written contest problems teach students to think more deeply and therefore excel in all areas of problem solving leading to a better quality of life.

I'm sorry.  I shouldn't have said that.  What I really wanted to say although I really messed up the meaning is that the AoPS courses prepare students for math competitions far better than most other math classes do.

Don't worry about it.  It certainly is consistent with the angle we take at building our curriculum.

Hi robynkoz,

Just to add my two cents here, I think it's best to start as soon as possible given that your child is ready at that time. Like the first day of school, no one in that classroom (except maybe the instructors and those students who have taken 10 or more classes here  :blush: ) will be completely confident and willing to submit questions/responses at first. But gradually, you'll get to know the folks in your class some and giving input will come a lot easier. The classroom software is designed so that any mistakes (such as calculation errors) can be caught by instructors before they go out to the rest of the class, which should greatly reduce the fear of "being laughed at". It's important to remember that everyone, even the students in a class, all have different levels of problem solving ability. The point is to be able to learn from each other by working together. From the looks of your post, I'm assuming that sooner or later you'll be enrolling in one or more of these classes. Personally, I'd recommend you make that sooner instead of later because one day you might find that your child simply no longer has the time to work on problems.

[quote="MithsApprentice"]From the looks of your post, I'm assuming that sooner or later you'll be enrolling in one or more of these classes. Personally, I'd recommend you make that sooner instead of later because one day you might find that your child simply no longer has the time to work on problems.[/quote]

That does make sense, but there is a balance, and if he is not "ready" he may just be intimidated and feel that he would never enjoy these courses, and get unnecessarily "turned off", which I wouldn't like.

He will need to be self-motivated, and at the moment doing "extra" math work in the evenings (he doesn't usually have math homework) is not an appealing idea for him, so I don't think it's worth signing him up yet.  But we'll see...

He is doing MOEMS, division E and will be doing Mathcounts so I'm waiting to see if he gets motivated to do some extra practice/learning on his own.  When that happens, then I think he will be ready for AoPS classes. :)

Trying not to be a pushy mom here...

I agree -- don't be a pushy mom.  The kids who do the best are the ones who choose to take problem solving seriously on their own.  MATHCOUNTS is indeed a great place to find out if they have that passion.  Both Richard and I were personally very energized by the MATHCOUNTS competitions in middle school and from there we each ran with that energy.

[quote="robynkoz"]I am a parent of an 11yo who is getting to the point where he no longer has mathematical peers in his small K-8 school.  We are looking into options for him for next year or possibly even this spring, and one  possibility is something like AoPS classes.[/quote]

I'll jump in here as a parent of a boy just one year older. We've been participating on the site about a year now, so we started out with my oldest son at about the same age yours is now.

The AoPS online community is INVALUABLE just for showing a math-liking boy (or girl) that there are other young people who like math. That's the day-in-day-out benefit of this site that will do incalculable good for your son. I learned in a hurry that there are some very phenomenal math-liking young people in this country, people my son will probably be classmates with at college someday or colleagues with in the business world--if my son can keep up with them. Being part of the community is good for motivation and helps in talent development.

My son is in a very intensive standard curriculum math program

http://www.itcep.umn.edu/umtymp/

and thus is already putting in as much as ten hours per week at learning math. I am happy to have him go up to twenty-four hours a week, but not more. (That specific figure has a research base behind it, which I will explain upon request.) We have been trying a little of this, a little of that for supplementation over the past couple years. I LOVE the AoPS courses, but we are fairly new to them. I thought my son could handle the AoPS Intermediate Independent Study A class this year in addition to his standard curriculum class, but that was a little too busy. We will try the new Introductory Independent Study A class next semester, after conferring with the AoPS staff about what fits my son's level. I enjoy presentations of math content that look at the "same" material from a new angle. I like that so much that my son has had FOUR geometry courses since the beginning of last school year. (Of course he thought the last two of those courses were rather boring, but I think he got a little bit out of them that he didn't get from the first two.) In the next several years, I won't be planning quite that much doubling-up of courses with the same course title, but I will be seeking out a lot of supplemental learning resources, and I expect to have my guy take quite a few more AoPS online courses, both in the subject series and in the problem-solving series. The courses are plenty challenging for a boy who has already aced the SAT I math section, and they bring up new ideas that will never be brought up in the standard curriculum classes available to my son, even though those classes are quite good.

Good luck on hitting the right level. I STRONGLY recommend the Gelfand Correspondence Series books for your son, especially

[url=http://www.amazon.com/exec/obidos/tg/detail/-/0817636773/learninfreed]Algebra[/url]

[url=http://www.amazon.com/exec/obidos/tg/detail/-/0486425649/learninfreed]Functions and Graphs[/url]

and

[url=http://www.amazon.com/exec/obidos/tg/detail/-/0486425657/learninfreed]Method of Coordinates[/url]

in that order. (You can start one book before finishing the previous book.)

[quote="joml88"]Could you explain about that 24 hours a week figure?  Is that for people of all ages or what?[/quote]

I am interested as well.  I can remember times back in middle school when I studied math far more than that (especially during the summer).  If the worry is not being involved in other activities, I never felt deprived.  I played basketball and other sports nearly every day, held a job, had a group of friends, and never felt like my communication skills were suffering as a result of my focus on math.

My personal feeling about what the right amount of study time: Don't study to the point that you burn out or that it becomes a chore.  Because nobody ever forced math on me, the more than 24 hours per week that I worked seemed like play time.  I think that was helped by the facts that (1) I was part of a good local math team circuit, which makes math more fun and (2) I would make games out of it.  It would take a long time to explain the amount of number theory I learned from license plates on cars sitting in my mom's car on the way to and from running errands.

It just seems strangely artificial to put a number on something like that.  Not to mention that I'm always skeptical of scientific studies that assume all humans are congruent round spheres that react the same to every environmental stimulus.  While 4 hours of math a day isn't everybody's cup of tea, there are certainly kids who spend that much time -- at least for some period of their intellectual development -- doing math out of the love of the math itelf.  It's like they get their video game time doing something productive.

That said, assigning 24 hours of math work (including class time or not) per week to a middle school child would be absurd.  If they don't choose to concentrate on it, it's almost certainly harmful.

[quote="EFuzzy"]I don't understand how you can spend more 24 hours a week doing math. That would mean you need to spend 3 hours on the weekdays and 4 on the weekends. But you also need to eat, exercise, do homework, practice instruments, and remain sane. How did you manage your time?[/quote]

For the three years during which I studied math most growing up I would study math in the car on trips, study math while watching television, and study math...in both math and math team classes.  I know teachers would hate to hear me say this, but I would always keep an interesting problem with me during each class where I knew I wouldn't be learning anything new no matter what subject it was.  In some sense, that created a cycle of spare time because learning math makes all the sciences easier meaning that there is even more time during class to think about more interesting tangents to the material being discussed.

I do however recognize that these days more homework is assigned to students.  A good chunk of it is busy work to top students.

I'm not really recommending X amount of study time for anyone.  My point was mostly that I don't think that it's fair to draw conclusions about how X amount of study time works for students who are all individual and unique.

[quote="joml88"]Could you explain about that 24 hours a week figure?  Is that for people of all ages or what?[/quote]

The twenty-four hours per week figure for deliberate practice is based on research on "expert performance" and how that is acquired. The way to become an expert is to practice. The way to practice is deliberately, steadily over a long span of years. Most domains of eminent adult performance have a "ten-year rule," such that people don't reach the peak level (winning an IMO gold medal, writing a masterpiece musical composition, or whatever) until after ten years of practice.

Most people don't get as good as they could become because they don't practice enough. But it is possible to practice too much. A lot of studies of different domains suggest that four hours a day of practice is a good maximum amount. In physical performance domains (including playing musical instruments), it is necessary to rest sometimes to avoid injury. Many mathematicians have written that one can only "do mathematics" for about four hours a day. So I figure that the kids on my math team could get up to six days of practice a week, times four hours of practice a day, for a maximum (not yet achieved) of twenty-four hours of practice of math problem-solving a week. The rest of the week they should be doing something else (including eating and sleeping  :)  ) for proper well-balanced development. Even summer camps for math-eager young people don't usually go above twenty-four hours per week of actual practice time.

The seminal article on the role of deliberate practice in developing expertise is

http://www.cebiz.org/cds/ericsson_charness.pdf

which has spawned a lot more research on expert performance and deliberate practice in the last decade.

Thank you for sharing that article.  I think there is a lot of interesting and good information in there, but I think the author also has some misunderstandings.  For instance, at one point the author compares reciting music upon hearing it to mentally calculating the day of the week for a date on the calendar.  Certainly we can't expect the author to have a fine understanding of every expert process, but the ability to calculate the day of the week from a calendar date is not at all a feat of intelligence on the level of reciting music upon hearing it once.  In fact, it is something that I could teach virtually any student who has completed our Introduction to Number Theory class.  In fact, I plan to add it to the curriculum as a fun game.  It's nothing more than addition and basic modular arithmetic.

But, getting to the real details, I think I can refute a great deal of the hard-cap study time by personal experience.  Though before I do I want to make a disclaimer: Parents should never EVER [i]force[/i] their child to study any subject for 4 hours a day.  While that is only my opinion, it is a very strong one.

Until I was in the sixth grade, I never studied math in any great detail.  I was always ahead of my peers.  I was moved up a year and was still at the head of the class, but I can't say that I spent any serious time outside of math class thinking about math and even during class I would simply tune out the teacher, get the homework done during class, and think about games or chat with the person next to me (when I could get away with it).  I did read bits of the Saxon Algebra book when I was around 8 years old, but I didn't understand much of it at all.

Yeah, there are going to be a lot of teachers mad at me for all I'm about to say...

In the sixth grade I started Pre-Algebra and math team.  I won only one of about a dozen math team events that year, but I wasn always thirsty for interesting problems.  From that point I estimate that I studied math at least 6 hours a day for the next 4 years.  The summer after 7th grade I probably studied over 10 hours per day.  I was very motivated after getting a perfect score at the national MATHCOUNTS tournament and getting invited to go the Mu Alpha Theta national convention with the high school math team.

[b]The energy of continuous events made it much easier to work hard.[/b]

By the end of the 9th grade I was 12th for the selection of the U.S. IMO team.  If I hadn't written my first proof on the USAMO that year and had been exposed to resources to study for Olympiad level problem solving, I think it is very possible that I could have won a gold medal at the IMO.

After the 9th grade I barely ever studied math at all.  I went to the Math Olympiad Program again and worked a few Putnam problems in college and technically I was a math major, but I was learning almost nothing new except a few definitions here and there.  It was all still the same math with different applications.

I spent my time thinking about economics, debating policy (in high school and college), working various jobs to support myself, and finding other ways to apply the problem solving skills I had developed.  None of my college classes were new material and while I learned a little bit of problem solving, I didn't learn much from the end of the 9th grade until I started at AoPS as I did in the 9th grade.

At AoPS I don't have a lot of time to study math, but I probably learn as much now each month as I did in a week when I was seriously studying math.  I feel like if I had all of the books and resources I have at my disposal now back then that I would have had a very good chance of winning a gold medal at the IMO in the 9th grade.  (Of course, the resources are much better these days and the problems have become accordingly significantly harder over these past 12+ years.)

I think 4 hours a day seems reasonable in some sense.  A child definitely needs to recharge.  It takes down-time for neural pathways to solidify.  In general I probably only rarely studied math for more than a couple of hours continuously.  During the summer I would study in the library for as long as I felt like it.  I would then walk next door and play basketball for as long as I felt like it.  I would then study more math.  When I felt like it I would hang out with friends or ride my bike around town.  And then when I felt like it, I would study math again.

Part of the key to all this was that I had math team teachers who were always willing to copy off stacks of problems for me to take home.  They would give me new material as often as I asked for it.  When it was clear that I needed more advanced material they would copy off stacks of AHSMEs (the old AMC) and other high school contests.  It was always fun and I never failed to enjoy solving more problems.  It was my video arcade.

(Here is where the teachers will want to kill me.)

When it came to class time I learned to completely tune out the teachers.  I would sit inconspicuously in the back of class and get the homework done as quickly as I could and then spend my time thinking about math (or play games or doodle -- you get the idea).  This kind of regimen was very tiring of course, so I slept in some of my classes.  I would pick out the teachers who would ignore me or give me some allowance and take every inch of it that I could.  Then I would be working on math again during the next class.  My grades didn't suffer because in middle school the work wasn't very challenging for anyone who enjoys to think. (Perhaps that's different at some schools.)

Even in the 9th grade I would hide math problems in my English textbook.  I probably worked nearly every AIME problem that had ever been written in about a month during the beginning of that year.

While many might think this would cause me to get behind in my other classes, it wasn't nearly as hard as it sounds.  Studying math up to a high level allows a student to think more broadly in every subject.  Science classes became a breeze.  Sure I'd have to memorize a bit for Biology, but the equations in Chemistry and Physics were silly compared to the math I was studying.  Even in English -- most of analyzing a book is just logically connecting the details such that the story makes sense.  That's [b]exactly[/b] what we do with math problems.  [i]We make analogies[/i].  We say 10 students are 10 objects and selecting a committee of 3 of them is the same as choosing 3 objects out of 10.  And then we apply the combinatorial mechanics to count the number of choices and we're done.

Ignoring school allowed me a continual cycle of studying and downtime.

Sure, there were teachers who hated me.  However, I felt (and still feel) that creating my own program was far better for me than "playing the game."

I didn't sacrifice time with friends or time playing sports and I even played my fair share of computer games.

Having finished with this anecdote, I'm not about to tell students to ignore the entirety of school.  I had a unique situation in that my parents ignored my schooling almost entirely because it was clear that I was learning well out of my own motivation.  I was able to weather minor trouble with teachers because I trained myself to care about my own interests more than the system.  But the system has changed and institutional pressures have never been higher.  These days I might have been sent to an alternative school long before I had the chance to excel (there are other facts behind that assertion, though all of them are essentially related to my insistance in creating my own educational paradigm).

At this point I don't know what to even recommend to students who want to make more time to study problem solving -- something that will help them in virtually any modern field even if it is completely outside of mathematics.  The call of the elite schools looms heavily over their heads and many of the students and parents are too scared not to follow "the golden path."  There is a gun to the back of the heads of over-achievers and there is far less freedom to deviate than I had.

There is this enormous myth about how going to school X guarantees a higher level of success and how taking 4,319 AP exams, being the president of 14 clubs, and running a marathon at the age of 15 makes a student well-rounded.

Well rounded congruent spheres.

It's not so valuable to be the everyman.  Our value is greatest in our uniqueness.  Diversification helps a system retain value.

And it's a shame.  It's a shame because [b]students do far more with their talents when they find their own paths[/b].  Perhaps this is what researchers have never isolated.  Yes, the training is necessary for expert status, but creative thinking is only maintained by those who train for the expertise they find a passion for on their own whether or not it happens to coincide with the desires of their parents.  Perhaps these desires are the "gifts from the gods" that we are yet to scientifically put our finger on.  Certainly some seems environment and some seems inherited, but whatever the source, it's the passion that lends itself to a rigorous academic regimen.

This is exactly why I suggest that parents test the waters with their kids.  Let them give MATHCOUNTS or an AoPS class a try.  If it's their child's passion, the child will continue to ask for more.  We as adults do our best as providers when we open the door to positive pursuits lovingly and then step out of the way...patiently.

Children understand far more than most adults believe.  They aren't mature enough to express it.  They also face a mountain of pressure not to express it honestly when they do find ways to express it.  Frankly, I think chatspeak pigin developed for exactly this reason -- our language is absurd in spelling and grammatical structure and the kids create a hybrid that suits them better.

Children understand the directions they should take with their passions far more than adults tend to understand.  If we let them run with them, they will often break the mold of the congruent round spheres with limitations that test to the norms of the societal institutions that...are the mold.

4 hours a day for 10 years is probably a fantastic estimate for the students of their own passion within this modern society.  But it's certainly not a theorem.

Mathew, that was a fascinating description of how you managed to take ownership of your education, despite the constraints of having to operate within the structured timetable of a typical middle-school/highschool class schedule.

I've seen a number of cases of mathematically talented students doing similar things, though not necessarily focusing in a single area.

One case I have in mind is a young woman currently studying in a top doctoral program in a math-related area who was allowed to do her own thing back when she was in middle school and high school  (e.g.,she was highly accelerated--3 years in overall grade placement and several years even beyond that in her math placement, but even so she was tacitly allowed to  read interesting books in other--unrelated to math--subjects  hidden inside her math book when her honors math class got boring.)  I thought the math teacher was remarkably enlightened to allow her to do this.

I think there is an art to figuring out the right educational path for any given student.    The ideal is that a student's educational enviorment will inspire the student to challenge themselves and allow them to take ownership of their education.  No easy formulas for that...what works for one will not work for another.

I do the same thing in math class. My "effort" grades don't suffer because my teacher now expects that I won't study for quizzes or tests, so I am certainly exceeding expectations by not thinking about it.

[quote="sophia"]Mathew, that was a fascinating description of how you managed to take ownership of your education, despite the constraints of having to operate within the structured timetable of a typical middle-school/highschool class schedule.

I've seen a number of cases of mathematically talented students doing similar things, though not necessarily focusing in a single area.
[/quote]

Thanks.  If I had any one single success with my own education, it was taking control of it myself.

I would not however say that I focused on a single subject except insofar as math is the mechanics behind an enormous number of subjects -- many of which I was interested in.  I never felt like I excluded any other area except perhaps that I never had a foreign language program that motivated me.

I should really probably say that my greatest focus was problem solving and that problem solving is (in most cases) best studied through mathematics until a person decides to specialize.

[quote="Rep123max"]Just a couple of questions. First, out of curiousity, were you good in English in high school? (meaning did you have A's). Actually that was just one question, but I also have a comment. You are definetly correct that the education system isn't the same. Had I chosen to continue public school, I could probably pull off what you were doing, but for private schoolers (which I bet a good percentage on this site are), that is nearly impossible. In math class I sit in the front, and half the time, people count on me to answer the question so we can move forward. And in English class we have one of those round table (Harkness or something), so there is no way I can not pay attention. And one more thing, am I right to conclude that chemistry and physics will be much easier than biology?[/quote]

I got A's in English so long as I turned in enough homework.  My college English/Writing grades were all A's, though I didn't manage to get enough homework in to get an A every time in high school.

It was fortunate for me that there were enough good math students in my school that I wasn't really counted on the keep a class moving forward.

I can't say for sure whether you will find physics and chemistry easier than biology, but at the high school level physics and chemistry depend more on your ability to set up and solve equations while biology involves a far greater degree of memorization.  Biology doesn't become very computational until college and even then it depends on what you study.

Mathematical problem solvers tend not to have a lot of trouble with physics and chemistry relative to other students.  I think the same is true for economics after the very basic level.

[quote]After the 9th grade I barely ever studied math at all. I went to the Math Olympiad Program again and worked a few Putnam problems in college and technically I was a math major, but I was learning almost nothing new except a few definitions here and there. It was all still the same math with different applications. [/quote]

So did you just start from calculus to college? Is that why you didn't learn anything new. Beacause I'm sure advanced math classes teach you something new.

In college I continually found myself required to take prerequisites for classes that I wanted to take down the road, so I only ever took one math class that I enjoyed.  I would have eventually taken some better classes, but I got bored and left for Wall Street.

I am quite amazed by some of the experiences accounted for here, and it makes me wonder about my schedule.

At best, I can fit in 2 hours of math on each weekday, but I usually manage about 3 hours per Saturday and Sunday. With that, I think I am maxing myself out, even though I would rather forget school and stay home thinking about problem solving.

My school is public, so I really don't study for anything. The bad thing is though, teachers really get steamed when they catch you doing non-related things. And you then get detentions for disrespect. So, haven't managed too much time except in math for non-school mathematics. Since my teacher thinks I'm doing the board problems, when I'm not.

So, I get home, do homework for an hour. Play saxophone for an hour. Read, because I like to read almost as much as I like math. Then, I work on perparing my cases for LD debate. With this, and other extra-curriculars, I'm finding it more and more difficult to spare time.

Should I drop some extra-curriculars, even if I have a passion for those too?

[quote="white_horse_king88"]
Should I drop some extra-curriculars, even if I have a passion for those too?[/quote]

Personally, I would say no, you shouldn't.  But I suppose that the answer really depends on what your goals are.  If you *really* want to go to IMO, for example, perhaps you would need to drop other passions and concentrate exclusively on math.  I don't know folks here well enough yet to know whether or not that is a reasonable goal for you or not.  I know it is for many participants here!

Overall, though, I think staying rounded will make you happier in the long run than participating in IMO, especially since you had to ask the quesiton.  If math was your one overriding passion, I would guess you wouldn't have had to ask the question.

Personally, I love math and problem solving, and did some contests in high school (Canadian contests through Waterloo, though I am back in the US now).  I was at a school that didn't take these things too seriously, so I didn't get much preparation, but I did well enogh to get a decent scholarship to Waterloo and was invited to participate in the Canadian Math Olympiad once (but I don't even think I knew I was taking that until the day of!  so clearly I had no preparation for that and I don't think I got a full solution to any of the problems!)  But I also love computer programming and ened up pursuing a career in computers (artificial intelligence).  Now, to be honest, my math is very rusty.  I still have excellent problem solving skills, which are useful everywhere, but I don't remember much of my trig, my calculus, etc...  I just don't use those in my life these days.  In high school I'm sure I would have thought this impossible!

Anyhow, I digress...

I think that unless you can imagine yourself doing notihing but math for the rest of your life, you shouldn't drop your other interests to pursue math to that pinnacle of competitiveness, but rather enjoy the time you do spend on it, and enjoy time spent on other things too.  24 hours or more per week seems extreme to me, though  I'm sure it is common among the best in the world.  How badly do you want counted among them?

Good luck sorting through everything.

[quote="white_horse_king88"]At best, I can fit in 2 hours of math on each weekday, but I usually manage about 3 hours per Saturday and Sunday. With that, I think I am maxing myself out, even though I would rather forget school and stay home thinking about problem solving.

So, I get home, do homework for an hour. Play saxophone for an hour. Read, because I like to read almost as much as I like math. Then, I work on perparing my cases for LD debate. With this, and other extra-curriculars, I'm finding it more and more difficult to spare time.

Should I drop some extra-curriculars, even if I have a passion for those too?[/quote]In my case, I studied at most 2 hrs / day while in high-school in off-olympiad season, and during olympiad season (march-july) I would increase this up to 6 hrs / day (all numbers are in average of course, there were days in which I was extremely lazy, and others in which I spent 12 hours :D).

On topic, I think that you shouldn't give up your extras.

I strongly agree with both mathmom and Valentin.  The fact that you spend the time on the activities you do suggests a broad range of interests.

Personally, I too spent less time on math as my interests broadened.  Part of the reason I didn't go the route of becoming a math professor was that I enjoyed the pursuit of careers that had both problem solving and other elements.

Part of the point I tried to make in my earlier essay was that it's great that we have such a diverse group of problem solvers.  The fact that each is unique with different interests means that we move forward in more ways as a society -- and individuals are happier having pursued their own interests.

[quote="white_horse_king88"]My school is public, so I really don't study for anything. The bad thing is though, teachers really get steamed when they catch you doing non-related things. And you then get detentions for disrespect. So, haven't managed too much time except in math for non-school mathematics. Since my teacher thinks I'm doing the board problems, when I'm not.[/quote]
Can you do math in detention?

If you can remember an interesting problem...how can anybody stop you from doing math in detention?

Beyond that -- why would anybody want to?

[quote="MCrawford"]If you can remember an interesting problem...how can anybody stop you from doing math in detention?[/quote]There were ways to do that in the Dark Ages (the Chinese water torture for example) :D
However I don't think that they can be applied in our times anymore.

[quote="Valentin Vornicu"]There were ways to do that in the Dark Ages (the Chinese water torture for example) :D [/quote]

Well as long as you're not doing any scratch work to speak of (some people are good at mentally doing things) you're good to go, even if any water torture it applied. I'm sure there were plenty of um...creative solutions back then. In fact, we even had a poll on our favorite tortures back in the days of old AoPS (with the Etc. forum and all)


trigonometry AMC AIME calculus inequalities

On the interval $ 0 < y < \frac{\pi}{2},$ let $ \frac{16 y^2 \cos^2 y \plus{} 9}{y \cos y} \geq m$ where $ m$ is the minimum value that can be attained. Compute $ \binom{m}{m\minus{}2}$.

Level: Early AIME

[quote="Silverfalcon"]On the interval $ 0 < y < \frac {\pi}{2},$ let $ \frac {16 y^2 \cos^2 y \plus{} 9}{y \cos y} \geq m$ where $ m$ is the minimum value that can be attained. Compute $ \binom{m}{m \minus{} 2}$.

Level: Early AIME[/quote]
[hide="Intended solution"]Let $ x\equal{}y \cos y$. Then $ \frac{16x^2\plus{}9}{x}\geq m$. Using either calculus, or the fact that $ (4x\minus{}3)^2\equal{}0$, we get $ m\equal{}24$ (achieved when $ x\equal{}\frac{3}{4}$), and our final answer is $ \frac{24\cdot23}{2}\equal{}\boxed{276}$.[/hide]
[hide="However..."]In the look-alike AIME problem, it has something like $ y \sin y$ instead of $ y\cos y$. In the AIME problem, you could use the fact that $ y \sin y$ is equal to $ 0$ when $ y\equal{}0$, and is equal to $ \frac{\pi}{2}$ when $ y\equal{}\frac{\pi}{2}$, guaranteeing the existence of a value of $ y$ such that $ y \sin y\equal{}\frac{3}{4}$ by the Intermediate Value Theorem. However, I noticed that $ y \sin y$ is equal to $ 0$ at both endpoints of the given interval, and became suspicious and launched my own calculator investigation...
The maximum of $ f(x)\equal{}x\sin x$ in the interval $ \left(0,\frac{pi}{2}\right)$ is approximately $ 0.56109633819104$, meaning that the value of $ \frac{3}{4}$ is unattainable. Plugging in this value into the equation gives $ m\approx 25.01756767288487$, which would be hard to put in $ \binom{m}{m\minus{}2}$ :P
(Note that changing the problem's inequality to $ \frac{16y^2 \sin^2 y \plus{}9}{y \sin y}\geq m$ would make it valid.)[/hide]

This is strangely familiar to an old AIME problem...

[url]http://www.artofproblemsolving.com/Wiki/index.php/1983_AIME_Problems/Problem_9[/url]

[hide]Letting $ y \cos y\equal{}a$, we have $ 16a \plus{} \frac{9}{a}$.  Conveniently, $ \cos y$ must be positive by restraints.  Using AM-GM minimizes the expression at $ 24$.  $ \binom{24}{2}\equal{}276$.[/hide]

Curses...someone beat me to it.

Thanks, Brut3Forc3. I didn't really bother to find if the value WAS really attainable other than existing after applying AM-GM. Thanks again for pointing that out.

azn6021023, I already said that this problem is modification from an AIME problem, which indeed is the one you linked. It is pretty famous problem and a good application of AM-GM.


calculus integration algebra function domain probability logarithms

I am reading a paper about moving averages, the author writes the following without any citation:

......
Let $ (X_t)_{t\geq0}$ be the Riemann-Liouville fractional integral given by

$ X_t \equal{} \int_{0}^{t}(t \minus{} s)^{\gamma} d L_s$,  $ t \geq 0$,
where $ (L_t)_{t\geq0}$ is a Levy Process with characteristic triplet $ (0, b, \nu)$ (i.e. no Brownian component). For
$ (X_t)_{t\geq0}$ to be well-defined it is necessary that one of the following (a)-(c) are satisfied:
(a) $ \gamma > \minus{}\frac{1}{2}$,

(b) $ \gamma \equal{} \minus{}\frac{1}{2}$ and $ \int_{[\minus{}1,1]}x^2 |\log|x\parallel{}\nu(dx) < \infty$,

(c) $ \gamma < \minus{}\frac{1}{2}$ and $ \int_{[\minus{}1,1]}|x|^{\minus{}1/\gamma}\nu(dx) < \infty$
......

I don't know how it comes? Someone help me , thanks!

I do not understand the formulation completely.
(1) Are you sure that it is said 'necessary'?
(2) If yes, than the author should understand the intgral in some standard sense. In what sense? If its the Ito integral, then the standard domain of integration for $ \int_0^tf_sdL_s$ is $ \{f$ predictable $ : \int_0^tf_s^2d\langle L\rangle_s<\plus{}\infty$ a.s. $ \}$, which gives case (a) because $ d\langle L\rangle_t\equal{}t\int x^2\mu(dx)$. And its unclear where (b) and (c) come from.
Maybe, the point is that one can invent some procedure that allows one to extend the definition of the integral in the cases (b) and (c)? Then it should be written 'sufficient' ... Maybe, you can find some further tips in the paper?
Anyway, this looks interesting: let us invent some 'integral' that would cover the cases (b) and (c)   :-)

Ok, let's try.. Note that   $ \gamma$ should be greater that $ -1$ unless the trivial case $ \nu\equiv 0, b=1$ gives a nonsense. $ Z$ can be decomposed into a sum $ Z^{\leq 1}+Z^{>1}$, where $ Z^{\leq 1}_t=\int_0^t\int_{|u|\leq 1}u\tilde \mu(ds,du)$ ($ \mu$ is the Poisson point measure associated to $ Z$, $ \tilde \mu(ds,du)=\mu(ds,du)-ds\nu(du)$ is respective compensated measure)), and $ Z^{>1}_t=ct+$'finite number of jumps'. Integral w.r.t. $ Z^{>1}$ under assumption $ \gamma>-1$ is defined trivially. Hence in the sequel I assume
$ Z=Z^{\leq 1}$.
For every $ \epsilon>0$ the integral $ \int_0^{t-\epsilon}$ is well defined. Let us define $ \int_0^t$ as a limit in probability of $ \int_0^{t-\epsilon}, \epsilon \to 0$. We need to show that this limit exists; it is sufficient to show fundamenrtality; that is, that for every sequences $ \epsilon_n,\delta_n\to 0$
$ \int_{t-\delta_n}^{t-\epsilon_n}$ tends to 0 in probability. To simplify notation, I prove further on that $ \xi_n=\int_{\delta_n}^{\epsilon_n}s^{-\alpha}dL_s\to 0$ in probability when $ \alpha\in [1/2,1)$ (this is an equivalent formulation).
One has the estimate for the deviations of a r.v. $ \xi$: $ P(|\xi|\geq a)\leq{7\over a}\int_0^a[1-\mathrm{Re}\phi_\xi(z)]\, dz$, where $ \phi_\xi(z)=Ee^{i\xi z}$. Therefore, in order to prove $ \xi_n\to 0$ in probability it's enough to show that $ \phi_{\xi_n}\to \phi_\xi$ point-wise (by the dominated convergence theorem).
We have $ \phi_{\xi_n}(z)=\exp\left[\int_{\delta_n}^{\epsilon_n}
\int_{|u|\leq 1}(e^{izus^{-\alpha}}-1-izus^{-\alpha})\nu(du)ds\right]$, hence it's enough to prove that
\[ \int_{\delta_n}^{\epsilon_n}
\int_{|u|\leq 1}(e^{izus^{-\alpha}}-1-izus^{-\alpha})\nu(du)ds\to 0,\quad n\to \infty\] for every $ z\not=0$ (the case $ z=0$ is trivial). To simplify notation, I assume further $ z=1$. Rewrite, after Fubini and change of variables $ r=us^{-\alpha}$, lefthand side of (*) to \[ \int_{|u|\leq 1}u^{1\over \alpha}
[K(u\epsilon_n^{-\alpha})-K(u\delta_n^{-\alpha})]\nu(du)\quad \quad (**)\] with ${ K(y)=(1/\alpha)\mathrm{sign}(y)\int_1^|y|(e^{ir}-1-ir})r^{-1/\alpha-1}\, dr$. Note that ${ k(r)=(e^{ir}-1-ir})r^{-1/\alpha-1}$ is absolutely integrable on $ [1,+\infty)$ hence $ K(y)\to C\in \Re, y\to +\infty$. Then the term under the integral in (**) converges to 0 point-wise. To get (*) we need to dominate this term by something integrable. If $ \alpha>1/2$, then $ k(r)$ is absolutely integrable on $ \Re$, hence $ K$ is bounded, and condition $ \int_{|u|\leq 1}|u|^{1\over \alpha}\nu(du)<+\infty$ gives (*). This is exactly the case (c) from the 1st post.
In the case $ \alpha=1/2$, $ K(y)\sim C\mathrm{sign}(y)\ln (1/|y|), y\to 0$ and hence the ratio $ [K(u\epsilon_n^{-\alpha})-K(u\delta_n^{-\alpha})]\over \ln (1/|u|)$ is bounded (one should take into account that $ \epsilon_n,\delta_n\to 0$ and $ |K(y)|\to C, y\to \infty$) . Then assumption $ \int_{|u|\leq 1}|u|^{2}\ln(1/|u|)\nu(du)<+\infty$ gives (*). This is exactly the case (b) from the 1st post.

Alex, thanks. i don't know if your approach is exactly what the author mean. Maybe i upload this paper, so you can read and check it.  :) My citation is on page 7.

Aha, the definition of the integral is given explicitly (p. 3). It is formulated in terms of convergence in probability of appropriate approximations, so I believe it's essentially the same what was used in my post.
I think, the paper is not written completely rigorously: there are claims which are stated withot proofs and, for sure,  are non-clear. I believe that the text from my previous post is enough to state sufficiency -- and in the paper they state sufficiency -- without any discussion -- under additional assumption that $ \nu$ is symmetric. It is non-elementary to prove necessity -- and again this is not discussed at all ...  Maybe it is supposed that (2.1) -- (2.3) should be used?  Then where is a reference?
Anyway, these  details are not important; the main answer is that conditions on the Levy measure in cases (a) -- (c) are determined by the behaviour of the integrals
\[ \int_{\delta_{n}}^{\epsilon_{n}}(e^{izus^{ \minus{} \alpha}} \minus{} 1 \minus{} izus^{ \minus{} \alpha})ds \equal{} u^{1\over\alpha}[K(u\epsilon_{n}^{ \minus{} \alpha}) \minus{} K(u\delta_{n}^{ \minus{} \alpha})]\]


geometry
rectangle
combinatorics proposed
combinatorics

a rectangle may be split up in disjoint rectangles all with an integer dimension. prove that the  big rectangle has an integer dimension.

well-known problem.

place the rectangle on a checkerboard whose squares are $ 1/2\times 1/2$ so that one corner of the rectangle is on a lattice point and the edges meeting at that corner are along lattice lines. since each tile has an integer dimension, it covers equal areas of black and white. thus the large rectangle covers equal areas of black and white, and it's easy to see that this can happen only if at least one of its dimensions is an integer.


search
inequalities unsolved
inequalities

Let $ x,y,z>1$ satisfy $ \frac{1}x\plus{}\frac{1}y\plus{}\frac{1}z\equal{}2$. Prove that:
$ \sqrt{x\plus{}y\plus{}z}\geq\sqrt{x\minus{}1}\plus{}\sqrt{y\minus{}1}\plus{}\sqrt{z\minus{}1}$

It has already been posted. just searching for your title I found this for example  :wink:
http://www.mathlinks.ro/viewtopic.php?search_id=1411496829&t=144


a band wants to distribute its music on CDs. the equipment to produce the CDs costs 250 dollars, and blank CDs cost 5.90 dollars for a package of 10. give an expression for the total cost, in dollars, to produce $n$ CDs, where $n$ is a multiple of 10.

i thought the answer was 250+5.90n

but the book says 250+0.59n

i think its because it asks for a expression for cost per cd, not package, so you have to divide 5.9 by 10 because there are 10 cds per package

yes, you could only plug in multiples of 10 for $n$ because cds can only be bought in packages of 10.


Good catch. It is $x=\frac{19}6$ in my solution file but I typed $\frac{17}6$ here. :oops:  Unfortunately, I no longer can edit my old post.

[quote=10000th User]I'm amazed to find this one unsolved until now  :)  . Here is my solution straight from my files  :lol:

[hide]Clearly, $[x]\ne0$ so this means there are no solutions in
the interval $[0,1)$. Let $x=[x]+r$, for some positive real number
$r\in[0,1)$. We have 2 cases:

[b]Case 1[/b]: $r\in[0,\frac12)$. Then, $2r\in[0,1)$ and since
$2x=2[x]+2r$, $[2x]=2[x]$. So, \[\begin{gathered}\frac1{[x]}+\frac1{2[x]}=r+\frac13\\ r=\frac32\frac1{[x]}-\frac13 \end{gathered}\] Since $r\in[0,\frac12)$, we find that $\frac95<2\le [x]\le4<\frac92$.
For $[x]=2$, $r=\frac5{12}$, and $x=[x]+r=\frac{29}{12}$. For
$[x]=3$, $r=\frac16$, and $x=\frac{19}6$. For
$[x]=4$, $r=\frac1{24}$, and $x=\frac{97}{24}$.

[b]Case 2[/b]: $r\in[\frac12,1)$. Then, $2r\in[1,2)\Rightarrow 2r-1\in[0,1)$ and since $2x=2[x]+2r=(2[x]+1)+(2r-1)$,
$[2x]=2[x]+1$. So, \[\begin{gathered}\frac1{[x]}+\frac1{2[x]+1}=r+\frac13\\ r=\frac1{[x]}+\frac1{2[x]+1}-\frac13 \end{gathered}\] Since $r\in[\frac12,1)$, we find that \[\frac12\le\frac1{[x]}+\frac1{2[x]+1}-\frac13<1\] Notice that since $[x]$ is an integer and \[\frac12+\frac13\le\frac1{[x]}+\frac1{2[x]+1}\] then $[x]<2$ since $\frac12+\frac13>\frac12+\frac15$, and
$[x]$ cannot be negative as well. So, $[x]=1$. But, from the other
end of the inequality, we have \[\frac1{[x]}+\frac1{2[x]+1}<1+\frac13\] so that $[x]<1$. Remember that $[x]\ne0$. Hence, there are no
solutions when $r\in[\frac12,1)$.

Therefore, the real solutions to the original equation are
$x=\frac{29}{12},\frac{19}6,\frac{97}{24}$.[/hide][/quote]

[quote=10000th User]:lol: No one seemed to have caught that I had a missing solution. Third solution updated and the solution is correct :)[/quote]

Which third solution??

[quote=10000th User]I'm amazed to find this one unsolved until now  :)  . Here is my solution straight from my files  :lol:

[hide]Clearly, $[x]\ne0$ so this means there are no solutions in
the interval $[0,1)$. Let $x=[x]+r$, for some positive real number
$r\in[0,1)$. We have 2 cases:

[b]Case 1[/b]: $r\in[0,\frac12)$. Then, $2r\in[0,1)$ and since
$2x=2[x]+2r$, $[2x]=2[x]$. So, \[\begin{gathered}\frac1{[x]}+\frac1{2[x]}=r+\frac13\\ r=\frac32\frac1{[x]}-\frac13 \end{gathered}\] Since $r\in[0,\frac12)$, we find that $\frac95<2\le [x]\le4<\frac92$.
For $[x]=2$, $r=\frac5{12}$, and $x=[x]+r=\frac{29}{12}$. For
$[x]=3$, $r=\frac16$, and $x=\frac{17}6$. For
$[x]=4$, $r=\frac1{24}$, and $x=\frac{97}{24}$.

[b]Case 2[/b]: $r\in[\frac12,1)$. Then, $2r\in[1,2)\Rightarrow 2r-1\in[0,1)$ and since $2x=2[x]+2r=(2[x]+1)+(2r-1)$,
$[2x]=2[x]+1$. So, \[\begin{gathered}\frac1{[x]}+\frac1{2[x]+1}=r+\frac13\\ r=\frac1{[x]}+\frac1{2[x]+1}-\frac13 \end{gathered}\] Since $r\in[\frac12,1)$, we find that \[\frac12\le\frac1{[x]}+\frac1{2[x]+1}-\frac13<1\] Notice that since $[x]$ is an integer and \[\frac12+\frac13\le\frac1{[x]}+\frac1{2[x]+1}\] then $[x]<2$ since $\frac12+\frac13>\frac12+\frac15$, and
$[x]$ cannot be negative as well. So, $[x]=1$. But, from the other
end of the inequality, we have \[\frac1{[x]}+\frac1{2[x]+1}<1+\frac13\] so that $[x]<1$. Remember that $[x]\ne0$. Hence, there are no
solutions when $r\in[\frac12,1)$.

Therefore, the real solutions to the original equation are
$x=\frac{29}{12},\frac{17}6,\frac{97}{24}$.[/hide][/quote]

Here in case 2, \[\frac1{[x]}+\frac1{2[x]+1}<1+\frac13\]
 Therefore I think it should be $[x]>1$

I don't think that answer will vary as still
${[x]=1}$ union ${[x]>1}$ will be a null set


Slove equation:  $ 31(xyzt\plus{}xy\plus{}yz\plus{}zt\plus{}1)\equal{}40(yzt\plus{}y\plus{}t)$, with $ x,y,z,t \in Z$.

[quote="nsato"]Looks like this problem:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201349[/url][/quote]

Thanks nsato!! :lol:


geometry
poles and polars
symmetry
projective geometry
geometric transformation
incenter

[quote="grobber"]Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the side $BC$ at $A^{\prime}$, and the line $AA^{\prime}$ meets the incircle again at a point $P$. Let the lines $CP$ and $BP$ meet the incircle of triangle $ABC$ again at $N$ and $M$, respectively. Prove that the lines $AA^{\prime}$, $BN$ and $CM$ are concurrent.[/quote]

Pretty tough to start. For clarity purposes, let A' = Z. Let E be the intersection of the incircle with AB and F the intersection of the incircle with AC. Let O be the intersection of AZ and the incircle. We want to prove that OM / MB * BZ / ZC * CN / NO = 1 (@). (By Ceva's, that implies BN, CM, and AZ are concurrent.)

angle BOZ = angle BZM (incircle is tangent to BC at Z) and
angle COZ = angle CZN.
Thus, triangle BOZ ~ triangle BZM and triangle COZ ~ triangle CZN.

We then have BO / BZ = BZ / BM and CO / CZ = CZ / CN.

Substituting in @, we see that we must prove

BO / BZ * OM / ON * CZ / CO = 1.

Again, from similar triangles, we have BO / BZ = OZ / ZM and CO / CZ = OZ / ZN.

Substituting again, we see that we must prove

OM / ZM * NZ / NO = 1, or OM * ZN = ON * ZM.

For this, we use inversion.

Invert about point O with arbitrary radius.

We have the following figure:

The incircle I is inverted into a line that contains E', M', Z', N', and F' (in that order) and line BC is inverted into a circle that is tangent to I' at Z'. Line AB is inverted into a circle that is tangent to I' at E' and meets circle (BC)' at B' and O. Line AC is inverted into a circle that is tangent to I' at F' and meets circle (BC)' at C' and O. Finally, the ray Z'O goes through the other intersection of circles (BC)' and (AC)' at A'.

Now, we claim that M'Z' = Z'N'.

Z'E' = Z'O * Z'A' = Z'F'.

M'E'^2 = M'B' * M'O = M'Z'^2 implies M'E' = M'Z'. Likewise, N'Z' = N'F'.

Thus, Z'E' = Z'F' implies 2M'Z' = 2N'Z' implies M'Z' = Z'N'.

By inversive distances, OM / OZ' = MZ / M'Z' and ON / OZ' = NZ / N'Z'.
Because M'Z' = Z'N', we have MZ / MO = NZ / NO.

Then OM * ZN = ON * ZM, as desired.

My solution:

I did use some things I won't prove here. For example:

[b]Lemma

Let ABCD be a cyclic quadrilateral. Assume the tangents at A and C to its circumcircle and BD are concurrent. Then the tangents at B and D to the circumcircle and AC are also concurrent.[/b]

Let A', B', C' be the points where the incircle touches BC, CA, AB respectively. In order to prove that AA', BN, CM are concurrent me must prove (by Desargue's thm) that the intersection pts of the pairs of lines (MN, BC); (A'N, BP); (A'N, CP) are collinear. Let A'P \cap MN=X. It's well known that the line formed by the intersection pts of (A'N, BP) and (A'N, CP) is the polar of X with respect to the circle, so what we want is for MN \cap BC to be on the polar of X, which is equivalent to X being on the polar of BC \cap MN, so the tangents to the circle at A' and P and MN must be concurrent. What we've proved here is that this condition is equivalent to what we initially wanted to prove.

I'll come back later. As you can see, it's going to take a while to explain everything.

[quote="grobber"]I did use some things I won't prove here. For example:

[b]Lemma

Let ABCD be a cyclic quadrilateral. Assume the tangents at A and C to its circumcircle and BD are concurrent. Then the tangents at B and D to the circumcircle and AC are also concurrent.[/b]

Let A', B', C' be the points where the incircle touches BC, CA, AB respectively. In order to prove that AA', BN, CM are concurrent me must prove (by Desargue's thm) that the intersection pts of the pairs of lines (MN, BC); (A'N, BP); (A'N, CP) are collinear. Let A'P \cap MN=X. It's well known that the line formed by the intersection pts of (A'N, BP) and (A'N, CP) is the polar of X with respect to the circle, so what we want is for MN \cap BC to be on the polar of X, which is equivalent to X being on the polar of BC \cap MN, so the tangents to the circle at A' and P and MN must be concurrent. What we've proved here is that this condition is equivalent to what we initially wanted to prove.

I'll come back later. As you can see, it's going to take a while to explain everything.[/quote]

OK. Before, that, did you mean (A'N, BP) and [b](A'M, CP)[/b], or (A'N, BP) and (A'N, CP) as you stated? Because (A'N, CP) = N, in which case the line formed by the intersection pts of those two points is A'N.

I am familiar with that lemma. Perhaps it can be posted as a problem on its own.

Also, what is the "polar" of a point? =D Thanks!

I have solved this one before ( it's really a nice one):

We name the intersection of MN and BC, D.
Instead of showing that AA', BN and CM are concurrent, we show that
(DBA'C) = -1.
First we will prove that KP is tangent to the circle: K is on the polar of A, so its polar passes through A, it passes also through A' so AA' is the polar of K. So KP is the second tangent from K to the incircle.

Now to finish the proof I will prove that P (DBA'C) = -1. We intersect the rays of this system with center P with the line MN, (let X = (AA', MN) so we have to prove that P(DMXN) = -1 so we have to prove that A'(DMXN)= -1.We intersect the rays with the incircle, so we have to prove that (A'MPN)= -1.So we have to prove that
P(A'MPN) = -1 .We intersect the rays with BC ,so we have to prove that
(KBA'C) = -1,and that is obvious because AA', BB' and CC' are concurrent.
 :D

[quote="hossein11652"]We name the intersection of MN and BC, D.
Instead of showing that AA, BN and CM are concurrent, we show that
(DBAC) = -1.
First we will prove that KP is tangent to the circle: K is on the polar of A, so its polar passes through A, it passes also through A so AA is the polar of K. So KP is the second tangent from K to the incircle.[/quote]

I think you have forgotten to say: Define K as the point of intersection of B'C' and BC. Hereby, B' and C' are the points where the incircle of ABC touches CA and AB.

[quote="hossein11652"]Now to finish the proof I will prove that P (DBAC) = -1. We intersect the rays of this system with center P with the line MN, (let X = (AA, MN) so we have to prove that P(DMXN) = -1 so we have to prove that A(DMXN)= -1.We intersect the rays with the incircle, so we have to prove that (AMPN)= -1.So we have to prove that
P(AMPN) = -1 .We intersect the rays with BC ,so we have to prove that
(KBAC) = -1,and that is obvious because AA, BB and CC are concurrent.[/quote]

Thanks, a very nice proof!

  Darij

well done! I think we should have more problems which can be solved by this method

The term (ABCD) denotes the double ratio of four points A, B, C, D, i. e.

  (ABCD) = (AB / BC) / (AD / DC).

The notion P (ABCD) means the same as (ABCD), but with the collateral signification that we project the points from the point P.

  Darij

I have found a merely philosophical solution to the problem. It uses nothing but the following lemma from projective geometry:

Given a circle and a triangle inscribed in it, there always exists a projective transformation letting the circle fixed and mapping the triangle to an equilateral triangle.

Now, consider the projective transformation letting the incircle of triangle ABC fixed and mapping the triangle A'B'C' (formed by the points of tangency of the incircle with the sides BC, CA, AB) to an equilateral triangle. Then, the image of triangle ABC will be an equilateral triangle, too (since the tangents at the vertices of an equilateral triangle to its circumcircle form an equilateral triangle themselves), and everything follows from simple symmetry observations.

This strategy helps killing several problems. Are we allowed to use it at the IMO?

  Darij

Are we allowed to use affine transformations, for example, to reduce a triangle geometry problem to the isosceles case?
And when exactly is it "correct" to use this technique?

Yes one of of the contestantants solved this with affine transformations,and reduced the problem to isoceles triangle. He got the perfect mark

[quote="DusT"]I think in Iran a lot of things are considered well - known. I saw some official solutions that used some hard to prove facts.
Anyway, can somebody tell how much of projective tranformations can be used at IMO?[/quote]Any solution that uses well-known mathematical facts, even mathematicals facts that are way further the imo level (advanced calculus, etc.) are taken into consideration, but obivously you will never get partial credit for stating something like "this problem CAN be solved using advanced calculus easy, we are leaving the details for the reader ..." :). You are more likely to receive partial credit if you state something as above, but instead of "advanced calculus" put a smart idea upon which a solution can be developed.

In case of solved problems you will receive full-score if the mathematics you use are correct.

Hi, all. I have a solution to this one (hope it is different from Art of problem solving) :
        Let the incircle touch AB and AC at B', C'.
    [b][u]Lemma[/u][/b] :
          If $A_1, B_1, C_1$ be points lying on the incircle, then
  $AA_1, BB_1, CC_1$ are concurent if and only if $A'A_1, B'B_1, C'C_1$ are concurent.
    This one is easy to prove, just use Ceva trig.
         Therefore, $MC', NB', AP$ are concurent at $K$ since $BM, CN, AP$  are concurent at P. On the other hand, according to Pascal 's theroem for 6 points $M, N, B', C', B', C'$ we obtain $T$ lies on $AK$, where $T$ is the intersection of $C'N$ and $B'M$. Thus, T is on AA'.
         Applying our lemma again we finally get $AA', BN, CM$ are concurent since $A'P, C'N, B'M$ are concurent at $T$.

May or may not have taken me a few months

Let $B'$ and $C'$ be the points at which the incircle is tangent to $AC$ and $AB$, respectively. Notice that by Gergonne Point, $AA'$, $BB'$, and $CC'$ concur, so by Ceva/Menelaus configuration, if we let $B'C'\cap{}BC=X$, we have that $(XA';BC)=-1$. \\

Since $AB'$ and $AC'$ are tangent to the incircle, and $A$, $P$, and $A'$ are collinear, we have that $A'PC'B'$ is a harmonic quadrilateral! Projecting this quadrilateral to the line $BC$ gives us that $A'PC'B'$ must project to $(A'X;BC)$, since $(XA';BC)=-1$, meaning that the tangent to the incircle of $\triangle{}ABC$ at $P$ intersects $BC$ at $X$. \\

If we then project $XA'BC$ back through $P$ to the incircle, we get that $PA'MN$ is also harmonic!! Which means that $MN$ passes through $X$(by Ceva/Menelaus configuration), and by Ceva/Menelaus configuration again, we have that $AA'$, $BN$, and $CM$ are concurrent, and we are done!

Why is no one literally using Desargues' Theorem?  :maybe:

[b]Remark:[/b] E to dekhchi puro horipodo problem bhai (google translate messes up the actual meaning of this, so trying to translate this is of no use [hide=.](yes I myself translated to check that if you try to translate this, would you actually be able to get what I meant [hide=.](yeah I am actually this jobless [hide=.](bro are you still reading this? Hats off to you :omighty: )[/hide])[/hide])[/hide]!

[img]https://i.imgur.com/nJoiai1.png[/img]

Define the other two contact points $B'$ and $C'$ symmetrically. Now it is well known that $T=PP\cap B'C'\cap A'A'$. Also let $K=PA'\cap MN$, $X=MP\cap A'N$ and $Y=NP\cap A'M$.

Now Pascal on $PPNA'A'M$ gives that $\overline{T-Y-X}$ are collinear. So we have that the triangles $\left\{\triangle PBC,\triangle A'NM\right\}$ are perspective from the line $\overline{T-X-Y}$ and now from Desargues' Theorem, we now get that $\left\{\triangle PBC,\triangle A'NM\right\}$ are perspective from a point which means that the lines $\left\{A'P,BN,CM\right\}$ concur at a point and we are done! :stretcher:

Let $B'$ and $C'$ be the $B$- and $C$-intouch points respectively, and let $\overline{BC} \cap \overline{B'C'}=X$. Since $X$ lies on the polar of $A$, $A$ lies on the polar of $X$, which should include $A'$ as well, hence this polar is just $\overline{A'P}$ and $\overline{XP}$ is tangent to the incircle. Then, since $\overline{AA'},\overline{BB'},\overline{CC'}$ concur,
$$-1=(X,A';B,C)\stackrel{P}{=}(P,A';M,N),$$
hence $\overline{MN}$ passes through the intersection of the tangents to the incircle at $A'$ and $P$, which is just $X$. Therefore, $\overline{PA'},\overline{BN},\overline{CM}$ concur, since $(X,A';B,C)=-1$ (here we view $X$ as $\overline{BC} \cap \overline{MN}$), as desired. $\blacksquare$

[quote=yofro][quote=Megus][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28189]I'm asking someone for the proof of the following lemma[/url]:

[quote="darij grinberg"]
Given a circle and a triangle inscribed in it, there always exists a projective transformation letting the circle fixed and mapping the triangle to an equilateral triangle.
[/quote]
[/quote]
I have found a very nice proof of this lemma.
[hide=Solution]
[b]Lemma:[/b] Let $ABC$ be a triangle inscribed in circle $\omega$. Then there exists a projective transformation sending $ABC$ to an equilateral triangle fixing $\omega$.
[i]Proof:[/i] Let $\ell$ be the line containing $AA\cap BC$, $BB\cap AC$, $CC\cap AB$. Take a projective transformation sending $\ell$ to $\ell_{\infty}$. Then take an affine transformation sending the image of $\omega$ (an ellipse) back into a circle. Hence $\omega$ is preserved and $AA\cap BC, BB\cap AC, CC\cap AB$ are sent to infinity, i.e. $AA\parallel BC, BB\parallel AC, CC\parallel AB$ which implies $\triangle ABC$ is equilateral, as desired.

Define $B',C'$ as $A'$ was defined. Apply the lemma to $A'B'C'$ and the incircle. The problem is now obvious.
[/hide][/quote]

How can we be sure that $\omega$ is an ellipse and not some other conic? I am not too familiar with projective transformations but this does not seem immediately clear to me

@above Projective images of the circle under transformations in which a line not intersecting the circle is mapped into the line at infinity are ellipses. This is a fundamental fact of homographies. If you want a proof of even this fact you may consider the central projection definition of projective transformation. See [url=https://math.stackexchange.com/a/2126301/815074]this[/url]. Thanks for pointing this out though, I've updated [url=https://artofproblemsolving.com/community/c3381235h3120795_projective_transformations_part_ii]my blog[/url].

If you want to see why the line is outside the circle, it's sort of obvious: it's the polar of the symmedian point. The symmedian point is clearly inside the circle so the result follows.

forgot that Ceva-Menelaus works in the other direction  :coolspeak:

[asy]
unitsize(4cm);

pair A = dir(120); pair B = dir(205); pair C = dir(335); pair I=incenter(A, B, C); pair AA=foot(I, B, C); pair BB=foot(I, C, A); pair CC=foot(I, A, B); pair X=extension(B, C, BB, CC); pair P = IP(Line(A, AA, 10), circumcircle(AA, BB, CC), 0); pair M = IP(Line(P, B, 10), circumcircle(AA, BB, CC), 1); pair N = IP(Line(P, C, 10), circumcircle(AA, BB, CC), 0);

draw(A--B--C--cycle, heavygreen); draw(circumcircle(AA, BB, CC), heavygreen); draw(P--X--C, red); draw(X--N, red); draw(X--BB, red); draw(A--AA, red); draw(B--P--C, heavygreen); draw(B--N, red); draw(C--M, red);

dot("$B$", B, SSW); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$I$", I, NE); dot("$A'$", AA, dir(AA)); dot("$B'$", BB, NE); dot("$C'$", CC, SW); dot("$X$", X, dir(X)); dot("$P$", P, dir(P)); dot("$N$", N,dir(N)); dot("$M$", M,NE);
[/asy]

Let $B'$ and $C'$ be the points at which the incircle meets $\overline{AC}$ and $\overline{AB}$, and $X$ be the intersection of the tangents to the circle at $P$ and $A'$. Then $A'P$ is the polar of $X$, so by La Hire's we find that $X$ lies on line $B'C'$ (the polar of $A$). Therefore, since $AA'$, $BB'$, and $CC'$ concur at the Gergonne point, by Ceva-Menelaus we get
$$-1 = (X, A'; B, C) \overset{P}{=} (P, A'; M, N),$$
so $PMA'N$ is harmonic and $X$ lies on line $MN$. Now, from Descargue's on $\triangle PBC$ and $\triangle A'NM$, it suffices to show that $X$, $\overline{PB} \cap \overline{A'N}$, and $\overline{PC} \cap \overline{A'M}$ are collinear; but this is true by Pascal's on $A'A'NPPM$. So, we are done.

Let $\ell$ be the tangent to the incircle at $P$ and $D=\ell\cap BC$. The incircle touches sides $CA$, $AB$ in points $E$, $F$, respectively. Note that line $A'P$ is the pole of $D$ in the incircle. Since $A\in A'P$ we have from La-Hire's theorem that $D$ lies on pole of $A$, which is line $EF$. So, $D,F,E$ are collinear. Since $AA',BE,CF$ are concurrent, we conclude from Ceva-Menelaus' configuration that $(D,A';B,C)=-1$. From perspectivity at $P$, we have that $PMA'N$ is a harmonic quadrilateral. So, lines $\ell$, $MN$ and $BC$ concur (At $D$), which means $D,M,N$ are collinear. Again from Ceva-Menelaus' configuration, we have that $BN,CM,AA'$ concur at some point. $\square$

Denote $B'$ and $C'$ as the touch points of the incircle with $AC$ and $AB$. Also, let $K=B'C'\cap BC$, $I$ be the incenter, and $I'$ be the foot of $I$ onto $AA'$.\\

AS $AA'$, $BB'$, $CC'$ are concurrent, we can say:
\[(KA';BC)=-1.\]

Using the radical center on the incircle, $(II'A)$, $(B'II'C)$, we get that $II'$, $B'C'$, and $BC$ concur at $K$.

Also, as $PB'A'C'$ is harmonic, $KP$ tangent to the incircle. Therefore:
\[-1=(KA';BC)\overset{P}{=}(PA';MN),\]
implying that $M,N,K$ are collinear.

As a result, by Ceva-Menelaus, and $(KA';BC)=-1$, the desired line concur.

Define $B'$, $C'$ analogously.
Let $D$ be the polar of $APA'$ with respect to the incircle, which lies on the tangents to the incircle at $P$ and $A'$.
Then, since $A$ lies on the polar of $D$ it follows that $D$ lies on the polar of $A$, which is $B'C'$.
Since $AA'$, $BB'$, $CC'$ concur at the Gergonne point, it follows that by Ceva-Menelaus that $(DA';BC) = -1$.
As such, it follows that \[ (DA';BC) \overset{P}= (PA';MN) = -1 \] and thus $D$ also lies on $MN$. As such, by the converse of Ceva-Menelaus this implies that $AA'$, $BN$ and $CM$ concur.

Let $\omega$ be the incircle of $\triangle ABC$, and let $B'$ and $C'$ be the points of tangency of $\omega$ to $CA$ and $AB$, respectively. Also, define $X=CM\cap AA'$, $Y=BN\cap AA'$, and $G=AP\cap BB'\cap CC'$. Since $A$, $P$, and $A'$ are collinear, it follows that $G$ must exist, and is exactly the Gergonne point of $\triangle ABC$.

Since $BA'$ and $BC'$ are both tangents to $\omega$ and $B$, $M$, and $P$ are collinear, it follows that $A'PC'M$ is a harmonic quadrilateral. Projecting from $C$ onto $AA'$, we find that
\begin{align*}
    (PX;GA')&\overset{C}=(PM;C'A')\\
    &=-1.
\end{align*}

By a symmetric argument, we also have that $A'NB'P$ is harmonic, so $(PY;GA')=-1$. Thus $(PX;GA')=(PY;GA')=1$, implying that $X=Y=AA'\cap BN\cap CM$. Hence $AA'$, $BN$, and $CM$ are concurrent, as desired. $\blacksquare$

- Jörg

Let $B'$ and $C'$ be defined analogously to $A'$. Let $T$ be the point on $BC$ such that $\angle TPI = 90$. It is well known that $TB'C'$ are collinear. Projecting through $P$ we find,
\begin{align*}
-1 = (TA', BC) \overset{P}{=} (PA', MN)
\end{align*}
so $PMA'N$ is harmonic. Then we claim that $T$, $M$ and $N$ are collinear. Let $G$ be the point such that $MG$ and $NG$ are tangents to the incircle. Clearly then $G$ lies on $AA'$ from harmonic $MA'NP$. Thus $G$ lies on the polar of $T$ . Then by La Hires we know that $T$ lies on the polar of $G$, or $MN$.

Now we can conclude from $T$, $M$, $N$ collinear and $(TA', BC) = -1$ that in triangle $PBC$ the cevians $PA'$, $NB$ and $MC$ are concurrent.

Let $B'$ and $C'$ be the other intouch points. Since $PB'A'C'$ is harmonic, we have $B'C'$, $\overline{PP}$, and $\overline{A'A'}$ concur at $X$. By Ceva, $AA'$, $BB'$ and $CC'$ concur so by Ceva-Menelaus we have $(\overline{B'C'} \cap \overline{BC}, A'; C, B) = -1 \implies \overline{B'C'} \cap \overline{BC} = X$. Then notice that $(X, A'; C, B) \overset{P}= (P, A'; N, M) = -1$, so $PA'NM$ is a harmonic quadrilateral. This implies that $\overline{A'A'}$ and $\overline{MN}$ intersect at $X$, so by Ceva-Menelaus we have $BN$ and $CM$ and $AA'$ concurrent.

Very simple to be honest if you know basic projective geometry.

Let $B'$, $C'$ be the tangency points of the incircle with $AC$ and $AB$ respectively.

Now define $T=\overline{PP} \cap \overline{B'C'} \cap \overline{BC}$ (this exists since $(P,A';C',B')=-1)$.

It is equivalent to show that $T$, $M$ and $N$ are collinear; and this is true since \[-1=(T,A';B,C) \overset{P}{=} (P,A';M,N)\] which implies our desired.

Let $B'$ and $C'$ be the $B$-intouch point and the $C$-intouch point respectively.
Note that $(A',P;B',C')=-1$. Thus define $T$ as the intersection between $BC$, $B'C'$, and the tangent to the incircle at $P$.
It is well-known that cevians $AA', BB', CC'$ of $\Delta ABC$ concur at a point, so $(A',T;B,C)=-1$.
Since $(A',P,M,N) \stackrel{P}{=} (A',T;B,C)=-1$, $T$ lies on $MN$. So the cevians $PA',BN,CM$ of $\Delta PBC$ are concurrent.


Let $a$ be a positive integer such that $24|a+1$.
Prove that the sum of all divisors of $a$ divisible by 24.

$24|a+1\Longrightarrow$ a is not square. Therefore for all divisor k, divisors $k,\frac ak$ are distinckt. And $k+\frac ak=\frac{a+k^2}{k}=0(mod \ 24)$, because $(k,24)|(a,24)=1, a+k^2=0(mod 24)$.


[quote="asethi01"]These are AHSME problems that I don't understand how to do. (hard AMC 10/middle AMC12 level)

1. The sequence a1,a2,a3,..., statisfies a1=19 and a9=99. Also, a(n) is the arithmetic mean of the first n-1 terms wherever n_>_3. What is a2? (By the way in these:a1,a2,a9, the numbers are the subscripts same with a(n), where n is the subscript). I didn't know how to put in Latex   :(

2. A list of integers has mode 32 and mean 22. The smallest number on the list is 10. The median m of the list is a member of the list. If the list members m were replaced by m+10, the mean and median of the new list would be 24 and m+10 respetively. If m were instead replaced by m-8, the median of the new list would be m-4. What is m?[/quote]1. [url=http://www.mathlinks.ro/viewtopic.php?p=1372978#1372978]Click me[/url]

2. [url=http://www.mathlinks.ro/viewtopic.php?t=252075]Click me again[/url]

Please use the SEARCH function and help posting less CLUTTER.

You can learn $ \text{\LaTeX}$ [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]here[/url]

Please ask questions in those topics. Topic can be locked.

Thanks, I didn't know there would be other topics about this.


I, I have some problems when I to draw graphs that extend "too far" in the positive of negative y-direction.

Say that I would want to graph y=x^3 from x=0 to x=4. I would then write

[code]
size(500,500);
import graph;
real f(real x) {return x^3;}
draw(graph(f,0,4,operator ..),linewidth(1)+red);
xaxis("$x$",0,5,linewidth(1)+fontsize(20),EndArrow);
yaxis("$y$",0,80,linewidth(1)+fontsize(20),EndArrow);
[/code]

to get the below picture. My problem is that I want the x-axis and y-axis to be of roughly the same length, but how do I accomplish this?

[asy]size(500,500);
import graph;
real f(real x) {return x^3;}
draw(graph(f,0,4,operator ..),linewidth(1)+red);
xaxis("$x$",0,5,linewidth(1)+fontsize(20),EndArrow);
yaxis("$y$",0,80,linewidth(1)+fontsize(20),EndArrow);[/asy]

Thanks in advance!


I don't think it's worth it to do geometric series here, so just brute force: $ \frac{27}{81}\plus{}\frac9{81}\plus{}\frac3{81}\plus{}\frac1{81}\equal{}\boxed{\frac{40}{81}}$.

[quote="GameBot"]What is the sum of $ \left(\dfrac{1}{3}\right) \plus{} \left(\dfrac{1}{3}\right)^2 \plus{} \left(\dfrac{1}{3}\right)^3 \plus{} \left(\dfrac{1}{3}\right)^4$?[/quote]

This is $ \frac{1}{3} \plus{} \frac{1}{9} \plus{} \frac{1}{27} \plus{} \frac{1}{81} \equal{} \frac{40}{81}$.

EDIT: Beaten


Fie x>0 un numar real dat.Studiati convergenta sirului $a_n=\{ \frac{{nx + 1}}{{x + n}}\} ,n \ge 1$

Fie $x > 0$ numar real dat. Să se studieze convergenţa şirului $a_n  = \left\{ {\frac{{nx + 1}}{{x + n}}} \right\},n \ge 1,n \in \mathbb{N}$
($\left\{ \omega  \right\}$ am notat partea fracţionară a lui $\omega  \in \mathbb{R}$ iar cu $\left[\omega  \right]$ am notat partea fracţionară a lui $\omega  \in \mathbb{R}$)

[u]Soluţie:[/u]
Avem succesiv:
$\left\{ {\frac{{nx + 1}}{{x + n}}} \right\} = \frac{{nx + 1}}{{n + x}} - \left[ {\frac{{nx + 1}}{{n + x}}} \right] = $
$ = \frac{{nx + 1}}{{n + x}} - \left[ {\frac{{nx + x^2  + \left( {1 - x^2 } \right)}}{{n + x}}} \right] = $
$ = \frac{{nx + 1}}{{n + x}} - \left[ {x + \frac{{1 - x^2 }}{{n + x}}} \right] = $
$ = \frac{{nx + 1}}{{n + x}} - \left[ {\left[ x \right] + \left\{ x \right\} + \frac{{1 - x^2 }}{{n + x}}} \right] = $
$ = \frac{{nx + 1}}{{n + x}} - \left[ x \right] - \left[ {\left\{ x \right\} + \frac{{1 - x^2 }}{{n + x}}} \right]$

Deoarece pentru $n \in \mathbb{N}$ suficient de mare avem $0 \le \left\{ x \right\} + \frac{{1 - x^2 }}{{n + x}} < 1$ şi deci avem:
$\mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{{nx + 1}}{{x + n}}} \right\} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{nx + 1}}{{n + x}} - \left[ x \right] - \left[ {\left\{ x \right\} + \frac{{1 - x^2 }}{{n + x}}} \right]} \right) = x - \left[ x \right] = \left\{ x \right\}$

[u]Intrebare:[/u]
Este condiţia $x > 0$ esenţială?


ratio
geometry
rectangle
parallelogram
rhombus
analytic geometry
Pythagorean Theorem

Let $AE$ be the crease, where $E$ is the point on $BC$. Let $F$ be the midpoint of $CD$.

Let $x=BE(=EF)$ and $y=BC$. Then we have $EC=y-x$.
Also, $BC=\frac{1}{2}CD=\frac{1}{2}BC=\frac{y}{2}$

Using Pythagorean Theorem to $\triangle ECF$,

$(y-x)^2+\left(\frac{y}{2}\right)^2=x^2$
$\Longleftrightarrow y^2-2xy+x^2+\frac{y^2}{4}=x^2$
$\Longleftrightarrow 2xy=\frac{5y^2}{4}$
$\Longleftrightarrow x=\frac{5y}{8}$

$\therefore EC=y-x=\frac{3y}{8}$

Thus, $BE: EC=5: 3$

Very nice problems in this fresh section ! Congratulations, Vornicu, for the your initiative.

[u]The "my" solution.[/u]

I suppose w.l.o.g. that $AB=2$. I note the middlepoint $M$
of the side $CD$ and the point $N\in (BC)$ which belongs to the crease. Thus,
$BN=MN=x,\ CN=2-x,\ CM=1,\ MN^2=CN^2+CM^2$
$\Longrightarrow x^2=(2-x)^2+1\Longrightarrow x=\frac 54$, i.e. $\frac{NB}{NC}=\frac{x}{2-x}=\frac{\frac 54}{2-\frac 54}=\frac 53.$

[u]Remark 1.[/u] If the crease cuts the side $AD$ in the point $P$, then $\frac{PA}{PD}=\frac 18$. If I note the intersections $S\in AB\cap NP$, $R\in AD\cap MS$, then $\frac{SA}{AB}=\frac 16$, $\frac{RA}{RD}=\frac 13.$

[u]Remark 2.[/u] This problem can be generalized to other quadrilaterals. I wish to see a extension of this problem to the rectangle, parallelogram deltoid, rhombus (and $\frac{MC}{MD}=r$) a.s.o.

[quote="Rushil"]A square sheet of paper $ABCD$ is so folded that $B$ falls on the mid point of $M$ of $CD$. Prove that the crease will divide $BC$ in the ration $5 : 3$.[/quote]

This problem can be solved easily with the help of a diagram.(Although I am not able to draw it for you but try it yourself.)

[geogebra]54d49e6e43eead3a0ad7f05e07fe272ccef2bcf5[/geogebra]
A trigonometric short process is there......
taking tan of angle CBM in triangle BMC and then cos of angle CYM gives the result easily. :)

I have found a solution to this problem long ago.
[hide]
Let $ ABCD$ be the square. Let its sides be = $ a$ units.
$ M$ is the midpoint of $ CD$. $ EF$ is the crease formed.
Join $ BM$ and $ EM$.
We may take that $ BE: CE=m: n$.
Hence $ BE=ma, CE=na$
See that $ ma+na=a\Longrightarrow m+n=1$.....................(*)
Also, $ BE=EM$.
In $ \Delta EMC$, $ \angle MCE=90$
$ %Error. "hence" is a bad command.
EM^2=MC^2+EC^2$
$ %Error. "hence" is a bad command.
m^2a^2=\frac{1}{4}a^2+n^2a^2$
$ %Error. "hence" is a bad command.
m^2-n^2=\frac{1}{4}$
$ %Error. "hence" is a bad command.
m-n=\frac{1}{4}$...............(**)

Add and then substract (**) from (*) to get :
$ m=\frac{5}{8}, n=\frac{3}{8}$
$ %Error. "hence" is a bad command.
m: n=BE: CE=5: 3$
[b]QED[/b]
*My solution is just like
[/hide]
[hide="PS"]
I think there is a very easy solution with coordinate geometry but (needless to say), I cannot find it.[/hide]

Let A,B,C,D be A$(0,1)$, B$(0,0)$, C$(1,0)$, D$(1,1)$. Then, CD's middle point is M$\left(1,\frac{1}{2} \right)$,
and line BM is $y=\frac{x}{2}$. BM's middle point is N$\left( \frac{1}{2},\frac{1}{4}\right)$.
The crease is a perpendicular bisector of BM, thus the crease is $y=-2\left(x-\frac{1}{2}\right)+\frac{1}{4}=-2x+\frac{5}{4}$.
Thus, the crease cross the x-axis at $\left( \frac{5}{8},0\right)$, which divides BC in the ratio $5:3$.

barycentrics, cartesians, trig bash all works.. try anything.
(HINT:- for using cartesians prefer choosing C as the origin)

Dear Mathlinkers,
another way is to use the pytagorician triple 3-4-5...

Sincerely
Jean-Louis

It's clear that $\Delta BMH$ is isosceles $\implies \boxed{BH=HM}$
Let, $$\angle HMC =x \implies \frac{1+\tan \tfrac{x}{2}}{1-\tan \tfrac{x}{2}}=2 \implies \boxed{\tan \frac{x}{2}=\frac{1}{3}}$$
Therefore, $$\tan x =\frac{3}{4} \implies \text{ clearly } \sin x =\frac{3}{5}=\frac{HC}{HM}=\frac{HC}{BH} \implies \boxed{BH:HC=5:3}$$

A similar kind of problem appreared in V.A. Yasinsky Geometry Olympiad 2017 VIII-IX p1 [Ukranie], here,
https://artofproblemsolving.com/community/u378952h1711824p11269422

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20III.pdf  p. 31

Sincerely
Jean-Louis

Has no one used coordinate bash in this? I’m surprised


Find all $x,y\in\mathbb{C}$ such that $x^2+y^3=17xy+5\sqrt{x}$.

Why on earth would you want so solve such an ugly thing?  For each $x$, we have a cubic equation in $y$ which we can solve by any method we happen to like, so for each $x$ other than 0 there are 6 (not necessarily distinct) possible values of $y$ (6, because we can take two different square roots).

[quote="JBL"]Why on earth would you want so solve such an ugly thing?quote]

Why not?  :P


first smallest n such that n! begins with 1993

Hm ...
I saw a proof somehere of the fact that for a certain n, n! begins with 1993. But I don't know how you can find the smallest such n.

i bet that was a problem posed in say... 1993??

you saw a proof on the IMO mailing list which is a proof that nobody I showed to that could understand

I'm also having difficulties to understand it.

Can you PLEASE give a clearer proof ?

You're a bit severe with my post on imo-problems I think (but probably fair :( ). Remember, English is not my mother language so I have often great difficulties to make myself understood.
Furthermore my proof may well be wrong !

Proofs given on imo-problem tend to be very short. And rather rare, people with talent to write clear proofs should not refrain ...

Referring to that post :

The key idea is to start with A!
with A = 10^5 + 10^9 = 1000100000

It doesn't matter what A! is, watch what's going on when you multiply A! with A+1, then with A+2, ... A+d, with d "small" : the first four digits of the result increase by 0 or 1, not more.

But these first four digits should increase enough otherwise we will reach d > 10^5 without having all the succesive numbers we need to be sure 1993 (or what ever) is reached (long english sentence  :( , hope it makes sense).

The second part of the post is to prove that those first 4 digits increase enough : at least 1 every 10 numbers.

I.E. if A! starts with 1000.... for instance, then (A+10)! must be at least 1001..., (A+20)! must be at least 1002...

Conclusion : if when we multiply by A+d, the first four digits of (A+d-1)! increases by 0 or 1 and we are sure that if we multiply by A+d, A+d+1, .., A+d+10 the first four digits will have increased by 1.

Then the first four digits of (A+d)! we will go through every 4 digits numbers for d = 0 .. 10^5-1.

Hope it's clearer.

I think I've found the answer:
http://www.wolframalpha.com/input/?i=11402!


Let $ a,b,c$ and $ p$ be nonnegative real numbers. Find the largest range of $ p$ such that

$ \frac {a}{pa^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {b}{pb^2 \plus{} c^2 \plus{} a^2} \plus{} \frac {c}{pc^2 \plus{} a^2 \plus{} b^2} \le \frac 9{(p \plus{} 2)(a\plus{}b\plus{}c)}$


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

i got a 5(but made a careless error and should've had 6) on aime last year, with a 118.5 on AMC 12, how much do I need to improve this year to get into USAMO?

A ten should probably be the maximum floor if the difficulty is right, maybe an 8 or 9 will get you in, but don't count on anything lower than 8

[quote="Karth"]Take your AMC score and add it to 10 times your AIME score. If this 'index' is around 227 - 233, you'll probably make it. Note how I say [b]probably[/b] because the floor index changes every year.[/quote]Last year the index was somewhere between 210 and 220, I forgot exactly.

i think i may get a 9 or 10 considering i never prepped for aime and got a 5 and im prepping this yr

A $15$ will most likely get you into USAMO, but failing that, aim for a $9$ on the easier years and less for the harder years.

AoPS has raised the level of competition considerably in the last few years. I would advise anyone who is seriously interested in qualifying for USAMO to aim this school year for consistent scores of 10 or higher on practice AIME tests. That's because when you take the test for real, you need to have a margin for careless mistakes or unfamiliarity with new question types.

Good luck to everyone preparing this year.

[quote="Karth"][quote="rrusczyk"]We've also lowered the bar by funding a doubling of the size of the USAMO ;)

I agree with tokenadult about the training target of aiming for consistent 10+ in practice...[/quote]

How much will the bar difference be between the floor score (just a guess) if you guys didn't have the extra graders (and thus didn't sponsor the expansion) and now when you do have the expansion?[/quote]
I think it's been said to be about 1 point for floor and about 6-8 for index. Not sure though...

[quote="Phelpedo"][quote="bpms"][quote="13375P34K43V312"]Do they still have the thing where 7th-10th graders getting a score of X on the AIME, no matter what their score on the AMC 10/12, they make USAMO?[/quote]
Do you mean the floor? :huh:
Yes, they still have it.[/quote]

How do you know?[/quote]
If the floor wasn't there it would be pretty likely to be announced. Also, the floor has been there since 00, yes?

The AMC 10/12 teachers manual says the final USAMO rules will be announced in the AIME teachers manual (which appears to be not yet posted). But there are still old references up on the AMC Web site to the floor value rule, so that is likely to continue.

[quote="rrusczyk"]We've also lowered the bar by funding a doubling of the size of the USAMO ;)  [/quote]

Yes, that was quite sporting of you to respond to wrecking the curve with your alumni by opening up more spots for taking USAMO.

 :thumbup:

 :clap:

[quote="1234567890"][quote="rrusczyk"]We've also lowered the bar by funding a doubling of the size of the USAMO ;)

I agree with tokenadult about the training target of aiming for consistent 10+ in practice...[/quote]

So this year, the number of qualifiers will go from 350 to 700ish?

That would be awesome!  :lol:  :D  I barely missed the cutoff last year (216.5).[/quote]

I think he meant that for last year, not this year...but I don't know...

If Mr. Rusczyk is referring to a doubling, I'm sure that's with reference to the ORIGINAL, pre-AoPS number of USAMO qualifiers, which was about 250 or so.

[quote="bpms"][/quote][quote="Phelpedo"][quote="bpms"][quote="13375P34K43V312"]Do they still have the thing where 7th-10th graders getting a score of X on the AIME, no matter what their score on the AMC 10/12, they make USAMO?[/quote]
Do you mean the floor? :huh:
Yes, they still have it.[/quote]

How do you know?[/quote][quote="bpms"]
If the floor wasn't there it would be pretty likely to be announced. Also, the floor has been there since 00, yes?[/quote]

It was almost not there last year.

[quote="Calculus"]when did the AMC announce they were considering removing the floor?[/quote]
They didn't.  I think Phelpedo is merely referring to the fact that at first the CAMC was going to use index as a cutoff for USAMO qualification via floor score.  I certainly wouldn't go so far as to put it that way.

[quote="5alive"]my scores:
9th grade : AIME 2
10th grade: AIME 10, AMC10 144
11th grade: AIME 10,  AMC12 115.5 or something (off day)
yet i can't get into the usamo
i cant figure out why i didn't get in especially if the so called floor was in place for 10th graders and under
neway to check ur scores? college apps ask for them and im not sure of mine

brayden ware[/quote]

You have to check with the teacher or whoever registered for your results. You may be able to contact the AMC people about it, but I don't know how that works.

This year I would guess that you should get at least a 9 if you want a solid chance of getting into USAMO, an 8 will probably get you in if you have high enough AMC score, and there's a slight chance that 7 AIME + 150 USAMO might work.

The reason people are talking of removing the floor system is that last year it didn't work out because too many people made the floor, thus requiring them to set an index for 9th and 10th graders qualifying by floor score. However, this unfairly favored those who took the AMC 10, and thus (correct me if this isn't the reason) they decided to let in all 10th graders and below who got above the floor score, giving them too many students (around 430 instead of 375).

I think what they should do this year is that they should simply set an AIME qualifying score for 10th graders and below that produces an adequate number of qualifiers, regardless of what the "floor" or the lowest AIME score among the index qualifiers is. I think this would preserve the essence of the old system, by allowing 10th graders and below to qualify solely by their AIME score, without paying any attention to the current definition of the "floor score," which I believe is unnecessary and problematic in the current situation. As well, this would give 10th graders or below who take the AMC 12 a slight advantage over those taking the 10, as they would be able to qualify even if their AIME score wasn't quite high enough, i.e. if their index made the index score for 10th and 12th graders.


I have found quite simmilar (I guess) equations which need to be proven.

a) $\displaystyle\sum_{x=1}^n\frac{C^{x-1}_{n-1}}{C^x_{2n-1}}=\frac{2}{n+1}$

b) $\displaystyle\sum_{x=1}^n\frac{C_{n-1}^{x-1}}{C_{n+q}^{x}}=\frac{n+q+1}{(q+1)(q+2)}, n\geq 1$

c) $\displaystyle\sum_{x=0}^n\frac{C_n^xC_n^r}{C_{2n}^{x+r}}=\frac{2n+1}{n+1}$

d) $\displaystyle\sum_{k=0}^{n}k^2*C^k_n=n(n+1)2^{n-1}$

I try to solve them, but have no idea where to start. The most difficult for me is to get rid of that sigma and to reduce (2n)! from combinations.

Great. Really good. Just 3 more to go and there are only an avarage 33 minutes left :)

:idea: Awsome.

I have one question according to previous one (in the second, you have used that too):

${\sum_{x=1}^n{xC^{n-1}_{2n-x-1}}}=\sum_{y=1}^n{(n+1-y)C^{n-1}_{n+y-2}}=\sum_{y=1}^n\sum_{k=1}^yC^{n-1}_{n+k-2}=\sum_{y=1}^nC^{n}_{n+y-1}$

I often see that way of changing sum with x to sum with y (here as: (n+1-y)) and then to sum of sums with y and k. What are the formulas you have used here (especially the first one)? I know these are like a basis for those equations, but I can't figure them out and those are points where I stuck - my book doesn't containt information about it. Could you explaing that in the simplest possible math's way? :).

c)It's a little bit difficult,but I finally solved it.

[quote="Bourne"]Let y=n+1-x[/quote]

Well, I see that, but shouldn't we change the range of sum? (I guess the answer is obvious - "No", but anyway..)

//OK. Now I understand it.


factorial
number theory
Diophantine Equations
pen

We shall prove that $x$ is odd. Indeed, suppose that $x$ is even. Then $y$ is also even, so it's divisible by $16$. It means that $x^2 \equiv 12 \mod{16}$ but that's impossible (it can be checked by hand very easily).
Now, we see that the equation is equivalent to the following:
$(x+2i)(x-2i) = y^5$
Now let $d=(x+2i, x-2i)$
It follows that $d|4i$
But because $x$ is odd we must have $d|i$. It implies that $x+2i$ and $x-2i$ are coprime, so they have to be the fifth powers.
$x+2i = (a+bi)^5$ and
$x-2i = (a-bi)^5$
(this is because $x+2i$ and $x-2i$ are conjugates)
Now we substract the second equation from the first and, after doing some boring, bot fortunately not too long, computations, we obtain a equation which has no solutions (after comparing the real and imaginary parts or so...).

[quote="Hawk Tiger"]I remmeber (mod 11) can solved the problem.
But Which book introduced the knowledge of $Z[i]$?I want to learn it by myself.it seems very useful. :???:[/quote]

I strongly recommend the book [b]Theory of Numbers: A Text and Source Book of Problems[/b]  written by Andrew Adler and John E. Cloury.

my solution is more simple.
first, is easy to see that x is odd.

now, we see that $x^2 + 36 = (y+2)(y^4-2y^3+4y^2-8y+16)$

and that, if p is a prime of form 4k+3, and p|a+b, we have that p|a and p|b! (the two factors that compose the rhs, say A and B, have that 3 don't divide A and B at the same time. so, because A and B are of form 4k+3, one of this have one factor p of this form, different of 3, that divide x+36)

I think it's more usefull to directly learn superior algebra, at least the standard stuff on rings like to be factorial and on fields like Galois stuff.
Then you can take a look at J. Neukirch: "Algebraic Number Theory" (best book ever  :grin: ) and Ram Murty: "Problems in algebraic number theory" (being on a lower level and having lots of good problems, a good amount of them is here on the site).

[quote="Peter"]Prove that there are no integers $x$ and $y$ satisfying $x^{2}=y^{5}-4$.[/quote]

   Here is a standard proof :

Actually ; the initial equation is equivalent to :

     $ x^2 + 4 = y^5   \ \ (*) \implies y^5 + 2^5 = x^2 + 36$

$\implies  x^2 + 6^2 = (y+2) \cdot (y^4 - 2y^3 + 4y^2 - 8y + 16 ) $

  Consider $ b = y^4 - 2y^3 + 4y^2 - 8y + 16 $ ; we have :

$ b \equiv y^4 - 2y^3 = y^4 - 2y^3 + y^2 - y^2 = y^2 ( y-1)^2 - y^2 \ \ \mod 4 \ \ (**)$

   From $(*)$ ; we can deduce that $2 \not  | y$ since if $y$ is an even number ;

we can have a contradiction as given in Tomicio's post ( $x$ modulo $8$).

   So we only need to condider the case : $y = 4k \pm 1$

  Combine this with $(**)$ ; we always have : $ b \equiv -1  \ \ \mod 4  \ \ (1)$

   So ; $b$ have at least one prime divisor $p$ of the form $4m +3$

$ \implies p | x^2 + 6^2 \implies p | x ; 6 \implies p = 3 ; x = 3t $ ( $2 \not | t$)

  If $ 3 | y+2  \implies y \equiv -2  \ \ \mod 3 \implies b \equiv 5 \cdot 2^4 ( \mod 3)$

   The last congruence cannot hold since $ 3 | b$

  Thus ; $3 \not | y+2  \implies 3^2 | b \implies b = 9k ; k \in \mathbb{N}$

$ \implies x^2 + 6^2 = 9(t^2 + 4) = (n+2) n = 9(n+2)k \implies t^2 +2^2= (n+2)k$

   So ; It's very easy to see that $k$ doesn't have any prime divisor of the form
$4k+3$

 ( $2$ is not divisible by any odd prime :D)

 $ \implies k \equiv 1  \ \ \mod 4 \implies b = 9k \equiv 1  \ \ \mod 4$

   This congruence also cannot hold due to $(1)$

    So ; we have done .

extending on the comment of mod 11. We are motived to use mod 11 because by fermats little theorem we can see that y^5 = $\pm$ 1 mod 11 which is very constrained which is good in this situation. y^5-4 = 6,8 mod 11 however a simple check shows that the squares mod 11 are 1,4,9,5,3. Thus there are no solutions.

(modulo11)
it's have x^5 congruence -1,1 (mod 11)    <Fermat's little theorem>
and x^2 congruence 8,6 (mod 11)
but, x^2 congruence 0,1,3,4,5,9 (mod 11)
thus,no integers x and y satisfying x^2=y^5-4

It seems [b]---Fermat--- [/b]specializes in providing solutions which were already given in the thread (see the post just above his), but not notice the obvious flaw. The residues modulo $11$ for $y^5$ can be $-1, 0,1$, [b]not just [/b]$-1,1$, as stated. Luckily, that means $y^5 - 4 \in \{6,7,8\}$ modulo $11$, none a square residue, so the method still works.

Let us show that there are even no rational solutions. My solution uses the theory of hyperelliptic curves and the 2-Selmer group from Michael Stoll's lecture notes (google for Skript-ArithHypCurves-pub-screen.pdf - all references refer to that PDF).

Sorry, it is in German, I will translate it later.

\begin{theorem}
	Sei $C/\mathbf{Q}$ die glatte projektive hyperelliptische Kurve vom Geschlecht $2$ zur affinen Gleichung $y^2 = x^5 - 4 =: f(x) \in \mathbf{Q}[x]$ mit dem Grad $5 = 2g+1$ ungerade.  Wir zeigen, dass $C(\mathbf{Q}) = \{[0:1:0]\}$ nur aus dem Punkt im Unendlichen besteht, also $C_\mathrm{aff}(\mathbf{Q}) = \emptyset$, via $C(\mathbf{Q}) \hookrightarrow J(\mathbf{Q}) = 0$ mit $J/\mathbf{Q}$ die Jacobische von $C$, eine Abelsche Varietät der Dimension $g = 2$.
\end{theorem}
\begin{proof}
Die Stellen schlechter Reduktion sind genau $2$ und $5$, die Diskriminante von $f$ ist $2^{16}\cdot5^5$.

Weil $f$ über $\mathbf{Q}_2$ irreduzibel ist (Newton-Polygon hat genau ein Segment mit Steigung $\frac{v_2(-4)}{5} \not\in \mathbf{Z})$, ist $f$ auch über $\mathbf{Q}$ irreduzibel.  Weil der Grad $5 = 2g+1$ von $f$ ungerade ist, ist $J(\mathbf{Q})[2] = (\mathbf{Z}/2)^{1-1} = 0$ (Lemma 5.2).

Es ist sogar $J(\mathbf{Q})_{\mathrm{tors}} = 0$: $J$ hat wie $C$ gute Reduktion bei $3$ und $11$ und $J(\mathbf{F}_3) \cong \mathbf{Z}/10$ und $J(\mathbf{F}_{11}) \cong \mathbf{Z}/41$ haben keine gemeinsame nichttriviale Untergruppe (Theorem 4.21).

Es ist $J(\mathbf{Q}_p)[2] = 0$ für $p = 2,3,5,7,11$, da $f$ über diesen $\mathbf{Q}_p$ irreduzibel bleibt.  Daher ist $\mathrm{Sel}^2(J/\mathbf{Q}) = J(\mathbf{Q})[2] = 0$ (theory from Chapter 5).  Aus der exakten Sequenz
\[
	0 \to J(\mathbf{Q})/2 \to \mathrm{Sel}^2(J/\mathbf{Q}) \to \mathrm{Sha}(J/\mathbf{Q})[2] \to 0
\]
folgt $J(\mathbf{Q}) = 0$ wegen $J(\mathbf{Q})_{\mathrm{tors}} = 0$ (und es folgt auch $\mathrm{Sha}(J/\mathbf{Q})[2^\infty] = 0$).

Also ist der einzige Punkt von $C/\mathbf{Q}$ der Punkt $[0:1:0]$ im Unendlichen, also hat die ursprüngliche Gleichung keine rationale Lösung.
\end{proof}

\begin{remark}
	\begin{enumerate}
		\item Ganzzahlige Lösungen gehen leichter mit imaginär-quadratischen Zahlkörpern (see the posts above).
		\item Chabauty mit $r = 0 < 2 = g$ und $3 > 2r$ eine Stelle guter Reduktion gibt nur $|C(\mathbf{Q})| \leq |C(\mathbf{F}_3)| + 2r = 4 + 0 = 4$ (M. Stoll: Independence of rational points on twists of a given curve, Compositio Math. 142, 1201–1214 (2006)).
	\end{enumerate}
\end{remark}